The tangent line to \(y=f(x)\) at a point should go through the point, and be “in the same direction” as \(f\) at that point. The secant line through \(P\) and \(Q\) is simply the straight line passing through \(P\) and \(Q\text{.}\)
True: since \(y=2x+3\) is the tangent line to \(y=f(x)\) at the point \(x=2\text{,}\) this means the function and the tangent line have the same value at \(x=2\text{.}\) So \(f(2)=2(2)+3=7\text{.}\)
In general, this is false. We are only guaranteed that the curve \(y=f(x)\) and its tangent line \(y=2x+3\) agree at \(x=2\text{.}\) The functions \(f(x)\) and \(2x+3\) may or may not take the same values when \(x\ne 2\text{.}\) For example, if \(f(x)=2x+3\text{,}\) then of course \(f(x)\) and \(2x+3\) agree for all values of \(x\text{.}\) But if \(f(x) = 2x+3 +(x2)^2\text{,}\) then \(f(x)\) and \(2x+3\) agree only for \(x=2\text{.}\)
Since the tangent line to the curve at point \(P\) passes through point \(P\text{,}\) the curve and the tangent line touch at point \(P\text{.}\) So, they must intersect at least once. By drawing various examples, we can see that different curves may touch their tangent lines exactly once, exactly twice, exactly three times, etc.
Since you started and ended in the same place, your difference in position was 0, and your difference in time was 24 hours. So, your average velocity was \(\dfrac{0}{24}=0\) kph.
Objects accelerate as they falltheir speed gets bigger and bigger. So in the entire first second of falling, the object is at its fastest at the onesecond mark. The average speed, which includes the slower beginning of the fall, will be a smaller number then the speed at the onesecond mark.
The slope of a curve is given by \(\dfrac{\mbox{change in vertical component}}{\mbox{change in horizontal component}}\text{.}\) The change in the vertical component is exactly \(s(b)s(a)\text{,}\) and the change in the horizontal component is exactly \(ba\text{.}\)
The slope of a curve is given by \(\dfrac{\mbox{change in vertical component}}{\mbox{change in horizontal component}}\text{.}\) The change in the vertical component is exactly \(s(b)s(a)\text{,}\) and the change in the horizontal component is exactly \(ba\text{.}\)
The velocity is positive when the object is going in the increasing direction; it is going “up” on the graph when \(t\) is between 0 and 2, and when \(t\) is between \(6\) and \(7\text{.}\) So, the velocity is positive when \(t\) is in \((0,2) \cup (6,7)\text{.}\)
Average velocity: \(\dfrac{\mbox{change in position}}{\mbox{change in time}} = \dfrac{s(5)s(3)}{53} = \dfrac{(3\cdot 5^2+5)(3\cdot 3^2+5)}{53}=24\) units per second.
From the notes, we know the velocity of an object at time \(a\) is \[v(a)=\lim_{h \rightarrow 0}\frac{s(a+h)s(a)}{h}\] So, in our case: \[v(1)=\lim_{h \rightarrow 0}\frac{s(1+h)s(1)}{h} = \lim_{h \rightarrow 0}\frac{[3(1+h)^2+5][3(1)^2+5]}{h} = \lim_{h \rightarrow 0}\frac{6h+3h^2}{h}= \lim_{h \rightarrow 0}6+3h = 6\] So the velocity when \(t=1\) is 6 units per second.
Average velocity: \(\dfrac{\mbox{change in position}}{\mbox{change in time}} = \dfrac{s(9)s(1)}{91} = \dfrac{31}{91}=\dfrac{1}{4}\) units per second.
From the notes, we know the velocity of an object at time \(a\) is \[v(a)=\lim_{h \rightarrow 0}\frac{s(a+h)s(a)}{h}\] So, in our case:
\begin{align*}
v(1)=\lim_{h \rightarrow 0}\frac{s(1+h)s(1)}{h}
&=\lim_{h \rightarrow 0}\frac{\sqrt{1+h}1}{h}\\
&=\lim_{h \rightarrow 0}\frac{\sqrt{1+h}1}{h}\cdot \left( \dfrac{\sqrt{1+h}+1}{\sqrt{1+h}+1}\right)\\
&=\lim_{h \rightarrow 0}\frac{(1+h)1}{h(\sqrt{1+h}+1)}\\
&=\lim_{h \rightarrow 0}\frac{1}{\sqrt{1+h}+1}=\frac{1}{2}\mbox{ units per second}
\end{align*}
\(\displaystyle \lim_{x \rightarrow 2} f(x)=1\text{:}\) as \(x\) gets very close to \(2\text{,}\) \(y\) gets very close to \(1\text{.}\)
\(\displaystyle \lim_{x \rightarrow 0}f(x)=0\text{:}\) as \(x\) gets very close to \(0\text{,}\) \(y\) also gets very close to \(0\text{.}\)
\(\displaystyle \lim_{x \rightarrow 2}f(x)=2\text{:}\) as \(x\) gets very close to \(2\text{,}\) \(y\) gets very close to \(2\text{.}\) We ignore the value of the function where \(x\) is exactly \(2\text{.}\)
The limit does not exist. As \(x\) approaches 0 from the left, \(y\) approaches 1; as \(x\) approaches 0 from the right, \(y\) approaches 1. This tells us \(\displaystyle\lim_{x \rightarrow 0^} f(x)=1\) and \(\displaystyle\lim_{x \rightarrow 0^+} f(x)=1\text{,}\) but neither of these are what the question asked. Since the limits from left and right do not agree, the limit does not exist. Put another way, there is no single number \(y\) approaches as \(x\) approaches 0, so the limit \(\displaystyle\lim_{x \rightarrow 0} f(x)\) does not exist.
\(\displaystyle \lim_{x \rightarrow 1^{}} f(x)=2\text{:}\) as \(x\) approaches \(1\) from the left, \(y\) approaches 2. It doesn't matter that the function isn't defined at \(x=1\text{,}\) and it doesn't matter what happens to the right of \(x=1\text{.}\)
\(\displaystyle \lim_{x \rightarrow 1^{+}} f(x)=2\text{:}\) as \(x\) approaches \(1\) from the right, \(y\) approaches 2. It doesn't matter that the function isn't defined at \(1\text{,}\) and it doesn't matter what happens to the left of \(1\text{.}\)
\(\displaystyle \lim_{x \rightarrow 1} f(x) =\) DNE: since the limits from the left and right don't agree, the limit does not exist.
\(\displaystyle \lim_{x \rightarrow 2^{+}} f(x) =0\text{:}\) as \(x\) approaches \(2\) from the right, \(y\) approaches 0. It doesn't matter that the function isn't defined at 2, or to the left of 2.
\(\displaystyle \lim_{x \rightarrow 2^{}} f(x)=0\text{:}\) as \(x\) approaches \(2\) from the left, \(y\) approaches 0. It doesn't matter that the function isn't defined at 2, or to the right of 2.
As \(x\) gets closer and closer to 3, \(y\) gets closer and closer to 10: this shows \(\displaystyle\lim_{x \rightarrow 3} f(x)=10\text{.}\) Also, at 3 itself, the function takes the value 10; this shows \(f(3)=10\text{.}\)
Note that, as \(x\) gets closer and closer to 3 except at 3 itself, \(y\) gets closer and closer to 10: this shows \(\displaystyle\lim_{x \rightarrow 3} f(x)=10\text{.}\) Then, when \(x=3\text{,}\) the function has value 0: this shows \(f(3)=0\text{.}\)
False. The limit as \(x\) goes to 3 does not take into account the value of the function at 3: \(f(3)\) tells us nothing about \(\displaystyle\lim_{x \rightarrow 3} f(x)\text{.}\)
\(\displaystyle\lim_{x \rightarrow 2^} f(x)=16\text{:}\) in order for the limit \(\displaystyle\lim_{x \rightarrow 2} f(x)\) to exist and be equal to 16, both one sided limits must exist and be equal to 16.
Not enough information to say. If \(\displaystyle\lim_{x \rightarrow 2^+} f(x)=16\text{,}\) then \(\displaystyle\lim_{x \rightarrow 2} f(x)=16\text{.}\) If \(\displaystyle\lim_{x \rightarrow 2^+} f(x)\neq 16\text{,}\) then \(\displaystyle\lim_{x \rightarrow 2} f(x)\) does not exist.
\(\displaystyle\lim_{x \rightarrow 0^+} \ln x = \infty\text{:}\) as \(x\) approaches 0 from the right, \(\ln x\) is negative and increasingly large, growing without bound.
\(\displaystyle\lim_{x \rightarrow 0^} \dfrac{1}{x}=\infty\text{:}\) as \(x\) gets closer and closer to 0 from the left, \(\dfrac{1}{x}\) becomes a larger and larger negative number.
\(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x}=\) DNE: as \(x\) gets closer and closer to 0 from the left, \(\dfrac{1}{x}\) becomes a larger and larger negative number; but as \(x\) gets closer and closer to 0 from the right, \(\dfrac{1}{x}\) becomes a larger and larger positive number. So the limit from the left is not the same as the limit from the right, and so \(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x}=\) DNE. Contrast this with Question 1.3.2.15.
\(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty\text{:}\) as \(x\) gets closer and closer to 0 from the either side, \(\dfrac{1}{x^2}\) becomes a larger and larger positive number, growing without bound. Contrast this with Question 1.3.2.14.
\(\displaystyle\lim_{x \rightarrow 3} \dfrac{1}{10}=\dfrac{1}{10}\text{:}\) no matter what \(x\) is, \(\dfrac{1}{10}\) is always \(\dfrac{1}{10}\text{.}\) In particular, as \(x\) approaches 3, \(\dfrac{1}{10}\) stays put at \(\dfrac{1}{10}\text{.}\)
When \(x\) is very close to 3, \(f(x)\) looks like the function \(x^2\text{.}\) So: \(\displaystyle\lim_{x \rightarrow 3} f(x) =
\displaystyle\lim_{x \rightarrow 3} x^2=9\)
Zeroes cause a problem when they show up in the denominator, so we can only compute (1.4.2.1.a) and (1.4.2.1.d). (Both these limits are zero.) Be careful: there is no such rule as “zero divided by zero is one,” or “zero divided by zero is zero.”
The statement \(\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10\) tells us that, as \(x\) gets very close to 3, \(f(x)\) is 10 times as large as \(g(x)\text{.}\) We notice that if \(f(x)=10g(x)\text{,}\) then \(\dfrac{f(x)}{g(x)}=10\text{,}\) so \(\displaystyle\lim_{x \rightarrow} \dfrac{f(x)}{g(x)}=10\) wherever \(f\) and \(g\) exist. So it's enough to find a function \(g(x)\) that has limit 0 at 3. Such a function is (for example) \(g(x)=x3\text{.}\) So, we take \(f(x)=10(x3)\) and \(g(x)=x3\text{.}\) It is easy now to check that \(\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0\) and \(\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\ds\lim_{x \to 3}\frac{10(x3)}{x3}=\ds\lim_{x \to 3}10=10\text{.}\)
As we saw in Question 1.4.2.2, \(x3\) is a function with limit 0 at \(x=3\text{.}\) So one way of thinking about this question is to try choosing \(f(x)\) so that \(\frac{f(x)}{g(x)}=g(x)=x3\) too, which leads us to the solution \(f(x)=(x3)^2\) and \(g(x)=x3\text{.}\) This is one of many, many possible answers.
Another way of thinking about this problem is that \(f(x)\) should go to 0 “more strongly” than \(g(x)\) when \(x\) approaches \(3\text{.}\) One way of a function going to 0 really strongly is to make that function identically zero. So we can set \(f(x)=0\) and \(g(x)=x3\text{.}\) Now \(\dfrac{f(x)}{g(x)}\) is equal to 0 whenever \(x \neq 3\text{,}\) and is undefined at \(x=3\text{.}\) Since the limit as \(x\) goes to three does not take into account the value of the function at 3, we have \(\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=0\text{.}\)
One way to start this problem is to remember \(\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty\text{.}\) (Using \(\dfrac{1}{x^2}\) as opposed to \(\dfrac{1}{x}\) is important, since \(\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}\) does not exist.) Then by “shifting” by three, we find \(\displaystyle \lim_{x \rightarrow 3} \dfrac{1}{(x3)^2}=\infty\text{.}\) So it is enough to arrange that \(\dfrac{f(x)}{g(x)}=\dfrac{1}{(x3)^2}\text{.}\) We can achieve this with \(f(x)=x3\) and \(g(x)=(x3)^3\text{,}\) and maintain \(\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0\text{.}\) Again, this is one of many possible solutions.
Any real number; positive infinity; negative infinity; does not exist.
This is an important thing to remember: often, people see limits that look like \(\dfrac{0}{0}\) and think that the limit must be 1, or 0, or infinite. In fact, this limit could be anythingit depends on the relationship between \(f\) and \(g\text{.}\)
Questions 1.4.2.2 and 1.4.2.3 show us examples where the limit is 10 and 0; they can easily be modified to make the limit any real number.
Question 1.4.2.4 show us an example where the limit is \(\infty\text{;}\) it can easily be modified to make the limit \(\infty\) or DNE.
Since we're not doing anything dodgy like putting 0 in the denominator, \(\displaystyle\lim_{y \rightarrow 0} \dfrac{(y+1)(y+2)(y+3)}{\cos y}
=\dfrac{(0+1)(0+2)(0+3)}{\cos 0}=\dfrac{6}{1}=6\text{.}\)
Since the limits of the numerator and denominator exist, and since the limit of the denominator is nonzero: \(\displaystyle\lim_{x \rightarrow 3} \left(\dfrac{4x2}{x+2}\right)^4 =
\left(\dfrac{4(3)2}{3+2}\right)^4=16\)
If we try to plug in \(x=4\text{,}\) we find the denominator is zero. So to get a better idea of what's happening, we factor the numerator and denominator:
To calculate the limit of a polynomial, we simply evaluate the polynomial: \(\displaystyle\lim_{t \rightarrow 2} \frac{1}{2}t^43t^3+t
=
\frac{1}{2}\cdot2^43\cdot 2^3+2 = 14\)
This is a classic example of the Squeeze Theorem. It is tempting to try to use arithmetic of limits: \(\displaystyle\lim_{x \rightarrow 0} x^2=0\text{,}\) and \(\displaystyle\lim_{x \rightarrow 0}\cos\left(\frac{3}{x}\right)=something\text{,}\) and zero times something is 0. However, this is invalid reasoning, because we can only use arithmetic of limits when those limits exist, and \(\displaystyle\lim_{x \rightarrow 0}\cos\left(\frac{3}{x}\right)\) does not exist. So, we need the Squeeze Theorem.
Since \(1 \leq \cos\left(\frac{3}{x}\right) \leq 1\text{,}\) we can bound our function of interest from above and below (being careful of the sign!):
So our function of interest is between \(x^2\) and \(x^2\text{.}\) Since \(\displaystyle\lim_{x \rightarrow 0} x^2=\displaystyle\lim_{x \rightarrow 0} x^2=0\text{,}\) by the Squeeze Theorem, also \(\displaystyle\lim_{x \rightarrow 0}x^2\cos\left(\frac{3}{x}\right)=0\text{.}\)
Advice about writing these up: whenever we use the Squeeze Theorem, we need to explicitly write that two things are true: that the function we're interested is bounded above and below by two other functions, and that both of those functions have the same limit. Then we can conclude (and we need to write this down as well!) that our original function also shares that limit.
Recall that sine and cosine, no matter what (realnumber) input we feed them, spit out numbers between \(1\) and \(1\text{.}\) So we can bound our horrible numerator, rather than trying to deal with it directly.
As in the previous questions, we want to use the Squeeze Theorem. If \(x \lt 0\text{,}\) then \(x\) is positive, so \(x \lt x\text{.}\) Use this fact when you bound your expressions.
\(\ds\lim_{x\to0}\sin^2\left(\dfrac{1}{x}\right)=DNE\text{,}\) so we think about using the Squeeze Theorem. We'll need to bound the expression \(x\sin^2\left(\dfrac{1}{x}\right)\text{,}\) but the bounding is a little delicate. For any nonzero value we plug in for \(x\text{,}\) \(\sin^2\left(\dfrac{1}{x}\right)\) is a number in the interval \([0,1]\text{.}\) If \(a\) is a number in the interval \([0,1]\text{,}\) then:
\begin{align*}
0 \lt &xa \lt x & &\mbox{ when $x$ is positive}\\
x \lt &xa \lt 0 & &\mbox{ when $x$ is negative}
\end{align*}
We'll show you two ways to use this information to create bound that will allow you to apply the Squeeze Theorem.
Solution 1: We will evaluate separately the limit from the right and from the left.
When \(x \gt 0\text{,}\)
\begin{equation*}
0 \leq x\sin^2\left(\dfrac{1}{x}\right) \leq x
\end{equation*}
because \(0 \leq \sin^2\left(\dfrac{1}{x}\right) \leq 1\text{.}\) Since
Remark: this is a perfectly fine proof, but it seems to repeat itself. Since the cases \(x \lt 0\) and \(x \gt 0\) are so similar, we would like to take care of them together. This can be done as shown below.
Solution 2: If \(x\neq 0\text{,}\) then \(0\leq\sin^2\left(\dfrac{1}{x}\right)\leq1\text{,}\) so
When we plug \(w=5\) in to the numerator and denominator, we find that each becomes zero. Since we can't divide by zero, we have to dig a little deeper. When a polynomial has a root, that also means it has a factor: we can factor \((w5)\) out of the top. That lets us cancel:
Note that the function \(\dfrac{2w^250}{(w5)(w1)} \) is NOT defined at \(w=5\text{,}\) while the function \(\dfrac{2(w+5)}{(w1)}\) IS defined at \(w=5\text{;}\) so strictly speaking, these two functions are not equal. However, for every value of \(w\) that is not 5, the functions are the same, so their limits are equal. Furthermore, the limit of the second function is quite easy to calculate, since we've eliminated the zero in the denominator: \(\displaystyle\lim_{w \rightarrow 5} \dfrac{2(w+5)}{(w1)}
=\dfrac{2(5+5)}{51}=5.\)
So \(\displaystyle\lim_{w \rightarrow 5} \dfrac{2w^250}{(w5)(w1)}=\displaystyle\lim_{w \rightarrow 5} \dfrac{2(w+5)}{(w1)}=5\text{.}\)
When we plug in \(r=5\) to the denominator, we find that it becomes 0, so we need to dig deeper. The numerator is not zero, so cancelling is out. Notice that the denominator is factorable: \(r^2+10r+25 = (r+5)^2\text{.}\) As \(r\) approaches \(5\) from either side, the denominator gets very close to zero, but stays positive. The numerator gets very close to \(5\text{.}\) So, as \(r\) gets closer to \(5\text{,}\) we have something close to \(5\) divided by a very small, positive number. Since the denominator is small, the fraction will have a large magnitude; since the numerator is negative and the denominator is positive, the fraction will be negative. So, \(\displaystyle\lim_{r \rightarrow 5} \dfrac{r}{r^2+10r+25}=\infty\)
First, we find \(\displaystyle\lim_{x \rightarrow 1}\dfrac{x^3+x^2+x+1}{3x+3}\text{.}\) When we plug in \(x=1\) to the top and the bottom, both become zero. In a polynomial, where there is a root, there is a factor, so this tells us we can factor out \((x+1)\) from both the top and the bottom. It's pretty easy to see how to do this in the bottom. For the top, if you're having a hard time, one factoring method (of many) to try is long division of polynomials; another is to factor out \((x+1)\) from the first two terms and the last two terms. (Long division is covered in Appendix A.14 of CLPII Integral Calculus.)
One thing to note here is that the function \(\dfrac{x^3+x^2+x+1}{3x+3}\) is not defined at \(x=1\) (because we can't divide by zero). So we replaced it with the function \(\dfrac{x^2+1}{3}\text{,}\) which IS defined at \(x=1\text{.}\) These functions only differ at \(x=1\text{;}\) they are the same at every other point. That is why we can use the second function to find the limit of the first function.
Now we're ready to find the actual limit asked in the problem:
When we plug \(x=0\) into the denominator, we get 0, which means we need to look harder. The numerator is not zero, so we won't be able to cancel our problems away. Let's factor to make things clearer.
As \(x\) gets close to 0, the numerator is close to 1; the term \((3x^25)\) is negative; and the sign of \(x^3\) depends on the direction we're approaching 0 from. Since we're dividing a numerator that is very close to 1 by something that's getting very close to 0, the magnitude of the fraction is getting bigger and bigger without bound. Since the sign of the fraction flips depending on whether we are using numbers slightly bigger than 0, or slightly smaller than 0, that means the onesided limits are \(\infty\) and \(\infty\text{,}\) respectively. (In particular, \(\displaystyle\lim_{x \rightarrow 0^}\dfrac{x^2+2x+1}{3x^55x^3}=\infty\) and \(\displaystyle\lim_{x \rightarrow 0^+}\dfrac{x^2+2x+1}{3x^55x^3}=\infty
\text{.}\)) Since the onesided limits don't agree, the limit does not exist.
As usual, we first try plugging in \(t=7\text{,}\) but the denominator is 0, so we need to think harder. The top and bottom are both squares, so let's go ahead and factor: \(\dfrac{t^2x^2+2tx+1}{t^214t+49}=\dfrac{(tx+1)^2}{(t7)^2}\text{.}\) Since \(x\) is positive, the numerator is nonzero. Also, the numerator is positive near \(t=7\text{.}\) So, we have something positive and nonzero on the top, and we divide it by the bottom, which is positive and getting closer and closer to zero. The quotient is always positive near \(t=7\text{,}\) and it is growing in magnitude without bound, so \(\displaystyle\lim_{t \rightarrow 7} \dfrac{t^2x^2+2tx+1}{t^214t+49}=\infty\text{.}\)
Remark: there is an important reason we specified that \(x\) must be a positive constant. Suppose \(x\) were \(\frac{1}{7}\) (which is negative and so was not allowed in the question posed). In this case, we would have
The function whose limit we are taking does not depend on \(d\text{.}\) Since \(x\) is a constant, \(x^532x+15\) is also a constantit's just some number, that doesn't change, regardless of what \(d\) does. So \(\displaystyle\lim_{d \rightarrow 0} x^532x+15=x^532x+15\text{.}\)
There's a lot going on inside that sine function... and we don't have to care about any of it. No matter what horrible thing we put inside a sine function, the sine function will spit out a number between \(1\) and \(1\text{.}\) So that means we can bound our horrible function like this:
Since \(\displaystyle\lim_{x \rightarrow 1}(x1)^2\cdot(1) = \displaystyle\lim_{x\rightarrow1}(x1)^2\cdot(1)=0\text{,}\) the Squeeze Theorem tells us that
Remark: there is a technical point here. When \(x\) is a positive number, \(x\) is negative, so \((1)x \lt x\text{.}\) But, when \(x\) is negative, \(x\) is positive, so \((1)x \gt x\text{.}\) This is why we take the onesided limits of our function, and apply the Squeeze Theorem to them separately. It is not true to say that, for instance, \((1)x^{1/101} \leq x^{1/101} \sin\big(x^{100}\big)\) when \(x\) is near zero, because this does not hold when \(x\) is less than zero.
When we plug in \(x=5\) to the top and the bottom, both limits exist, and the bottom is nonzero. So \(\displaystyle\lim_{x \rightarrow 5} \dfrac{(x5)^2}{x+5}=
\dfrac{0}{10}=0\text{.}\)
As \(x\rightarrow2\text{,}\) the denominator goes to 0, and the numerator goes to \(2a+7\text{.}\) For the ratio to have a limit, the numerator must also converge to \(0\text{,}\) so we need \(a=\dfrac{7}{2}\text{.}\) Then,
\(\displaystyle\lim_{x \rightarrow 0} f(x)=0\text{:}\) as \(x\) approaches 0, so does \(2x\text{.}\)
\(\displaystyle\lim_{x \rightarrow 0} g(x)=\) DNE: the left and right limits do not agree, so the limit does not exist. In particular: \(\displaystyle\lim_{x \rightarrow 0^} g(x)=\infty\) and \(\displaystyle\lim_{x \rightarrow 0^} g(x)=\infty\text{.}\)
\(\displaystyle\lim_{x \rightarrow 0} f(x)g(x)=\displaystyle\lim_{x \rightarrow 0} 2x\cdot\dfrac{1}{x}=\displaystyle\lim_{x \rightarrow 0} 2=2\text{.}\) Remark: although the limit of \(g(x)\) does not exist here, the limit of \(f(x)g(x)\) does.
Note that \(\frac{1}{f(x)}\) is undefined when \(f(x) = 0\text{.}\) So \(\frac{1}{f(x)}\) is undefined at \(x=2\) and \(x=2\text{.}\) We shall look more closely at the behaviour of \(\frac{1}{f(x)}\) for \(x\) near \(\pm 2\) shortly.
Plotting the above points, we get the following picture:
Since \(f(x)\) is constant when \(x\) is between 1 and 0, then also \(\frac{1}{f(x)}\) is constant between 1 and 0, so we update our picture:
The big question that remains is the behaviour of \(\frac{1}{f(x)}\) when \(x\) is near 2 and 2. We can answer this question with limits. As \(x\) approaches \(2\) from the left, \(f(x)\) gets closer to zero, and is negative. So \(\frac{1}{f(x)}\) will be negative, and will increase in magnitude without bound; that is, \(\displaystyle\lim_{x \rightarrow 2^}\dfrac{1}{f(x)}=\infty\text{.}\) Similarly, as \(x\) approaches \(2\) from the right, \(f(x)\) gets closer to zero, and is positive. So \(\frac{1}{f(x)}\) will be positive, and will increase in magnitude without bound; that is, \(\displaystyle\lim_{x \rightarrow 2^}\dfrac{1}{f(x)}=\infty\text{.}\) We add this behaviour to our graph:
Now, we consider the behaviour at \(x=2\text{.}\) Since \(f(x)\) gets closer and closer to 0 AND is positive as \(x\) approaches 2, we conclude \(\displaystyle\lim_{x \rightarrow 2} \frac{1}{f(x)}=\infty\text{.}\) Adding to our picture:
Now the only remaining blank space is between \(x=0\) and \(x=1\text{.}\) Since \(f(x)\) is a smooth curve that stays away from 0, we can draw some kind of smooth curve here, and call it good enough. (Later on we'll go into more details about drawing graphs. The purpose of this exercise was to utilize what we've learned about limits.)
We cannot divide by zero, so \(\dfrac{f(x)}{g(x)}\) is not defined when \(x=\pm2\text{.}\) But for every other value of \(x\) that we plotted, \(f(x)\) is twice as large as \(g(x)\text{,}\) \(\dfrac{f(x)}{g(x)}=2\text{.}\) With this in mind, we see that the graph of \(f(x)\) is exactly the graph of \(2g(x)\text{.}\)
This gives us the graph below.
Remark: \(f(2)=g(2)=0\text{,}\) so \(\dfrac{f(2)}{g(2)}\) does not exist, but \(\displaystyle\lim_{x \rightarrow 2}\dfrac{f(x)}{g(x)}=2\text{.}\) Although we are trying to “divide by zero” at \(x=\pm 2\text{,}\) it would be a mistake here to interpret this as a vertical asymptote.
Velocity of white ball when \(t=1\) is \(\displaystyle\lim_{h \rightarrow 0}\dfrac{s(1+h)s(1)}{h}\text{,}\) so the given information tells us \(\displaystyle\lim_{h \rightarrow 0}\dfrac{s(1+h)s(1)}{h}=5\text{.}\) Then the velocity of the red ball when \(t=1\) is \(\displaystyle\lim_{h \rightarrow 0}\dfrac{2s(1+h)2s(1)}{h}=
\displaystyle\lim_{h \rightarrow 0}2\cdot\dfrac{s(1+h)s(1)}{h}=2\cdot 5 = 10\text{.}\)
1.4.2.44.a DNE , DNE\qquad 1.4.2.44.b 0 \qquad 1.4.2.44.c No: it is only true when both \(\displaystyle\lim_{x \rightarrow a} f(x)\) and \(\displaystyle\lim_{x \rightarrow a} g(x)\) exist.
1.4.2.44.c No: this is an example of a time when the two individual functions have limits that don't exist, but the limit of their sum does exist. This “sum rule” is only true when both \(\displaystyle\lim_{x \rightarrow a} f(x)\) and \(\displaystyle\lim_{x \rightarrow a} g(x)\) exist.
Any polynomial of degree one or higher will go to \(\infty\) or \(\infty\) as \(x\) goes to \(\infty\text{.}\) So, we need a polynomial of degree 0that is, \(f(x)\) is a constant. One possible answer is \(f(x)=1\text{.}\)
This will be the case for any polynomial of odd degree. For instance, \(f(x)=x\text{.}\)
Many answers are possible: also \(f(x)=x^{15}32x^2+9\) satisfies \(\displaystyle\lim_{x \rightarrow \infty} f(x) = \infty\) and \(\displaystyle\lim_{x \rightarrow \infty} f(x)= \infty\text{.}\)
As \(x\) gets larger and larger, \(2^x\) grows without bound. (For integer values of \(x\text{,}\) you can imagine multiplying \(2\) by itself more and more times.) So, \(\displaystyle\lim_{x \rightarrow \infty}2^x=\infty\text{.}\)
Write \(X=x\text{.}\) As \(x\) becomes more and more negative, \(X\) becomes more and more positive. From Question 1.5.2.4, we know that \(2^X\) grows without bound as \(X\) gets larger and larger. Since \(2^x= 2^{(x)}= 2^{X} =\frac{1}{2^X}\text{,}\) as we let \(x\) become a huge negative number, we are in effect dividing by a huge positive number; hence \(\lim\limits_{x\rightarrow\infty} 2^x = 0\text{.}\)
A more formulaic way to describe the above is this: \(\lim\limits_{x\rightarrow\infty} 2^x
= \lim\limits_{X\rightarrow\infty} 2^{X}
= \lim\limits_{X\rightarrow\infty} \frac{1}{2^X}
=0\text{.}\)
There is no single number that \(\cos x\) approaches as \(x\) becomes more and more strongly negative: as \(x\) grows in the negative direction, the function oscillates between \(1\) and \(+1\text{,}\) never settling close to one particular number. So, this limit does not exist.
Our standard trick is to factor out the highest power of \(x\) in the denominator: \(x^4\text{.}\) We just have to be a little careful with the square root. Since we are taking the limit as \(x\) goes to positive infinity, we have positive \(x\)values, so \(\sqrt{x^2}=x\) and \(\sqrt{x^8}=x^4\text{.}\)
We have two terms, each getting extremely large. It's unclear at first what happens when we subtract them. To get this equation into another form, we multiply and divide by the conjugate, \(\sqrt{x^2+5x}+\sqrt{x^2x}\text{.}\)
\begin{align*}
\lim_{x\rightarrow \infty}\left[\sqrt{x^2+5x}\sqrt{x^2x}\right]
&=\lim_{x\rightarrow \infty}\left[\dfrac{(\sqrt{x^2+5x}\sqrt{x^2x})(\sqrt{x^2+5x}+\sqrt{x^2x})}{\sqrt{x^2+5x}+\sqrt{x^2x}}\right]\\
&= \lim_{x\rightarrow \infty}\dfrac{(x^2+5x)(x^2x)}
{\sqrt{x^2+5x}+\sqrt{x^2x}}\\
& = \lim_{x\rightarrow \infty}\dfrac{6x}{\sqrt{x^2+5x}+\sqrt{x^2x}}\\
\end{align*}
Now we divide the numerator and denominator by \(x\text{.}\) In the case of the denominator, since \(x \gt 0\text{,}\) \(x=\sqrt{x^2}\text{.}\)
\begin{align*}
&= \lim_{x\rightarrow \infty}\dfrac{6(x)}{\sqrt{x^2}\sqrt{1+\frac{5}{x}}+\sqrt{x^2}\sqrt{1\frac{1}{x}}}\\
&= \lim_{x\rightarrow \infty}\dfrac{6(x)}{(x)\sqrt{1+\frac{5}{x}}+(x)\sqrt{1\frac{1}{x}}}\\
& = \lim_{x\rightarrow \infty}\dfrac{6}{\sqrt{1+\frac{5}{x}}+\sqrt{1\frac{1}{x}}} \\
&=\frac{6}{\sqrt{1+0}+\sqrt{10}}= 3
\end{align*}
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator.
Remember that \(\sqrt{\ }\) is defined to be the positive square root. Consequently, if \(x \lt 0\text{,}\) then \(\sqrt{x^2}\text{,}\) which is positive, is not the same as \(x\text{,}\) which is negative.
The two terms in the denominator grow at order \(x\)that is, their largest power is \(x\text{.}\) For \(x \lt 0\text{,}\) we must remember \(\sqrt{x^2}=x=x\text{,}\) so \(\)\sqrt{4x^2+x} = \sqrt{x^2}\sqrt{4+\frac{1}{x}} = x \sqrt{4+\frac{1}{x}}.\(\) We conclude that \(\)\frac{3x}{\sqrt{4x^2+x}2x} = \frac{3}{\sqrt{4+1/x}2}.\(\) Since \(1/x\to 0\) as \(x\to \infty\text{,}\) we conclude that \(\)\lim_{x\to \infty} \frac{3x}{\sqrt{4x^2+x}2x} = \frac{3}{\sqrt{4+0}2} = \frac{3}{4}\(\)
We have, after dividing both numerator and denominator by \(x^2\) (which is the highest power of the denominator) that \(\)\frac{5x^23x+1}{3x^2+x+7}=\frac{5\frac{3}{x}+\frac{1}{x^2}}{3+\frac{1}{x}+\frac{7}{x^2}}.\(\) Since \(1/x\to 0\) and also \(1/x^2\to 0\) as \(x\to +\infty\text{,}\) we conclude that \(\)\lim_{x\to +\infty} \frac{5x^23x+1}{3x^2 +x+7}=\frac{5}{3}.\(\)
We have, after dividing both numerator and denominator by \(x\) (which is the highest power of the denominator) that \(\)\frac{ \sqrt{4\,x + 2}}{3x+4}=\frac{\sqrt{\frac 4 x + \frac 2 {x^2}}}{3 + \frac 4 x}.\(\) Since \(1/x\to 0\) and also \(1/x^2\to 0\) as \(x\to +\infty\text{,}\) we conclude that \(\)\lim_{x\to +\infty} \frac{ \sqrt{4\,x + 2}} {3\,x+4}=\frac {0}{3} = 0.\(\)
The dominant terms in the numerator and denominator have order \(x^3\text{.}\) Taking out that common factor we get \(\)\frac{4x^3+x}{7x^3 + x^2  2} = \frac{4 + \frac{1}{x^2}}{7 + \frac{1}{x}  \frac{2}{x^3}}.\(\) Since \(1/x^a\to 0\) as \(x\to +\infty\) (for \(a \gt 0\)), we conclude that \(\)\lim_{x\to +\infty} \frac{4x^3+x}{7x^3 +x^22}=\frac{4}{7}.\(\)
Divide both the numerator and the denominator by \(x\) (which is the largest power of \(x\) in the denominator). In the numerator, move the resulting factor of \(1/x\) inside the two roots. Be careful about the signs when you do so. Even and odd roots behave differently see Question 1.5.2.10.
We want to factor out \(x\text{,}\) the highest power in the denominator. Since our limit only sees negative values of \(x\text{,}\) we must remember that \(\sqrt[4]{x^4}=x=x\text{,}\) although \(\sqrt[3]{x^3}=x\text{.}\)
Alternately, we can use the transformation \(\displaystyle\lim_{x \rightarrow \infty} f(x)=\displaystyle\lim_{x \rightarrow \infty} f(x)\text{.}\) Then we only look at positive values of \(x\text{,}\) so roots behave nicely: \(\sqrt[4]{x^4}=x=x\text{.}\)
We have, after dividing both numerator and denominator by \(x^3\) (which is the highest power of the denominator) that: \(\)\lim_{x \to \infty} \frac{5x^2+10}{3x^3 +2x^2+x}=\lim_{x \to\infty}\frac{\frac{5}{x}+\frac{10}{x^3}}{3+\frac{2}{x}+\frac{1}{x^2}}=\frac{0}{3}=0.\(\)
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. It is not always true that \(\sqrt{x^2}=x\text{.}\)
We divide both the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x\text{.}\) Since \(x \lt 0\text{,}\) we have \(\sqrt{x^2}=x=x\text{,}\) so that \(\)\frac{\sqrt{x^2+5}}{x}=\sqrt{\frac{x^2+5}{x^2}}=\sqrt{1+\frac{5}{x^2}}.\(\) Since \(1/x\to 0\) and also \(1/x^2\to 0\) as \(x\to \infty\text{,}\) we conclude that \(\)\lim_{x\to \infty} \frac{3x+5}{\sqrt{x^2+5}x}=\lim_{x\to \infty}\frac{3+\frac{5}{x}}{\sqrt{1+\frac{5}{x^2}}1}=\frac{3}{11}=\frac{3}{2}.\(\)
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. When is \(\sqrt{x}=x\text{,}\) and when is \(\sqrt{x}=x\text{?}\)
We divide both the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x\text{.}\) Since \(x \lt 0\text{,}\) we have \(\sqrt{x^2}=x=x\text{,}\) so that \(\)\frac{\sqrt{4x^2+15}}{x}= \frac{\sqrt{4x^2+15}}{\sqrt{x^2}} =\sqrt{\frac{4x^2+15}{x^2}}=\sqrt{4+\frac{15}{x^2}}.\(\) Since \(1/x\to 0\) and also \(1/x^2\to 0\) as \(x\to \infty\text{,}\) we conclude that \(\)\lim_{x\to \infty} \frac{5x+7}{\sqrt{4x^2+15}x}=\lim_{x\to \infty}\frac{5+\frac{7}{x}}{\sqrt{4+\frac{15}{x^2}}1}=\frac{5}{21}=\frac{5}{3}.\(\)
First, we need a rational function whose limit at infinity is a real number. This means that the degree of the bottom is greater than or equal to the degree of the top. There are two cases: the denominator has higher degree than the numerator, or the denominator has the same degree as the numerator.
If the denominator has higher degree than the numerator, then \(\displaystyle\lim_{x \rightarrow \infty} f(x)=\displaystyle\lim_{x \rightarrow \infty} f(x)=0\text{,}\) so the limits are equalnot what we're looking for.
If the denominator has the same degree as the numerator, then the limit as \(x\) goes to \(\pm \infty\) is the ratio of the leading terms: again, the limits are equal. So no rational function exists as described.
This is the amount of the substance that will linger longterm. Since it's nonzero, the substance would be something that would stay in your body. Something like “tattoo ink” is a reasonable answer, while “penicillin” is not.
The amount of the substance that will linger longterm is some positive numberthe substance will stick around. One example of a substance that does this is the ink in a tattoo. (If the injection was of medicine, probably it will be metabolized, and \(\displaystyle\lim_{t \rightarrow \infty} c(t)=0\text{.}\))
Remark: it actually doesn't make much sense to let \(t\) go to infinity: after a few million hours, you won't even have a body, so what is \(c(t)\) measuring? Often when we use formulas in the real world, there is an understanding that the are only good for some fixed range. We often use the limit as \(t\) goes to infinity as a standin for the function's longterm behaviour.
If we let \(f\) be my height, the time of my birth is \(a\text{,}\) and now is \(b\text{,}\) then we know that \(f(a) \leq 1 \leq f(b)\text{.}\) It is reasonable to assume that my height is a continuous function. So by the IVT, there is some value \(c\) between \(a\) and \(b\) where \(f(c)=1\text{.}\) That is, there is some time (we called it \(c\)) between my birth and today when I was exactly one meter tall.
Notice the IVT does not say precisely what day I was one meter tall; it only guarantees that such a day occurred between my birth and today.
One example is \(f(x) = \left\{ \begin{array}{ll}
0&\mbox{when }0 \leq x \leq 1\\
2&\mbox{when }1 \lt x \leq 2
\end{array}\right.\text{.}\) The IVT only guarantees \(f(c)=1\) for some \(c\) in \([0,2]\) when \(f\) is continuous over \([0,2]\text{.}\) If \(f\) is not continuous, the IVT says nothing.
One example is \(f(x) = \left\{ \begin{array}{ll}
0&\mbox{when }0 \leq x \leq 1\\
2&\mbox{when }1 \lt x \leq 2
\end{array}\right.\text{.}\) The IVT only guarantees \(f(c)=1\) for some \(c\) in \([0,2]\) when \(f\) is continuous over \([0,2]\text{.}\) If \(f\) is not continuous, the IVT says nothing.
No. IVT says nothing about functions that are not guaranteed to be continuous at the outset. It's quite easy to construct a function that is as described, but not continuous. For example, the function pictured below, whose equation format is somewhat less enlightening than its graph: \(f(x)=\left\{\begin{array}{ll}
\frac{26}{5}x+65,&10 \leq x \lt 15\\
\frac{26}{5}x+91,&15 \leq x \leq 20
\end{array}\right.\text{.}\)
Using the definition of continuity, we need \(k=\displaystyle\lim_{x \rightarrow 0} f(x)\text{.}\) Since the limit is blind to what actually happens to \(f(x)\) at \(x=0\text{,}\) this is equivalent to \(k=\displaystyle\lim_{x \rightarrow 0} x\sin\left(\frac{1}{x}\right)\text{.}\) So if we find the limit, we solve the problem.
As \(x\) gets small, \(\sin\left(\frac{1}{x}\right)\) goes a little crazy (see example 1.3.5), so let's get rid of it by using the Squeeze Theorem. We can bound the function above and below, but we should be a little careful about whether we're going from the left or the right. The reason we need to worry about direction is illustrated with the following observation:
If \(a\le b\) and \(x \gt 0\text{,}\) then \(xa\le xb\text{.}\) (For example, plug in \(x=1\text{,}\) \(a=2\text{,}\) \(b=3\text{.}\)) But if \(a\le b\) and \(x \lt 0\text{,}\) then \(xa \ge xb\text{.}\) (For example, plug in \(x=1\text{,}\) \(a=2\text{,}\) \(b=3\text{.}\)) So first, let's find \(\lim\limits_{x\rightarrow 0^}\sin\left(\frac{1}{x}\right)\text{.}\) When \(x \lt 0\text{,}\)
and \(\displaystyle\lim_{x \rightarrow 0^} x = \displaystyle\lim_{x \rightarrow 0^} x = 0\text{,}\) so by the Squeeze Theorem, also \(\displaystyle\lim_{x \rightarrow 0^} x\sin\left(\frac{1}{x}\right)=0\text{.}\)
and \(\displaystyle\lim_{x \rightarrow 0^+} x = \displaystyle\lim_{x \rightarrow 0^+} x = 0\text{,}\) so by the Squeeze Theorem, also \(\displaystyle\lim_{x \rightarrow 0^+} x\sin\left(\frac{1}{x}\right)=0\text{.}\)
Since the limits from the left and right agree, we conclude \(\displaystyle\lim_{x \rightarrow 0} x\sin\left(\frac{1}{x}\right)=0\text{,}\) so when \(k=0\text{,}\) the function is continuous at \(x=0\text{.}\)
Since \(f\) is a polynomial, it is continuous over all real numbers. \(f(0)=0 \lt 12345\) and \(f(12345)=12345^3+12345^2+12345+1 \gt 12345\) (since all terms are positive). So by the IVT, \(f(c)=12345\) for some \(c\) between \(0\) and \(12345\text{.}\)
Since \(f\) is a polynomial, it is continuous over all real numbers. \(f(0)=0 \lt 12345\) and \(f(12345)=12345^3+12345^2+12345+1 \gt 12345\) (since all terms are positive). So by the IVT, \(f(c)=12345\) for some \(c\) between \(0\) and \(12345\text{.}\)
The function is continuous when \(x^21 \gt 0\text{,}\) i.e. \((x1)(x+1) \gt 0\text{,}\) which yields the intervals \((\infty, 1)\cup (1,+\infty)\text{.}\)
The function \(1/\sqrt{x}\) is continuous on \((0,+\infty)\) and the function \(\cos(x) +
1\) is continuous everywhere.
So \(1/\sqrt{\cos(x) + 1 }\) is continuous except when \(\cos x=1\text{.}\) This happens when \(x\) is an odd multiple of \(\pi\text{.}\) Hence the function is continuous except at \(x=\pm \pi, \pm 3\pi, \pm 5\pi, \cdots\text{.}\)
The function is continuous for \(x\ne c\) since each of those two branches are polynomials. So, the only question is whether the function is continuous at \(x=c\text{;}\) for this we need
The function is continuous for \(x \ne 0\) since \(x^2+c\) and \(\cos cx\) are continuous everywhere. It remains to check continuity at \(x=0\text{.}\) To do this we must check that the following three are equal.
The function is continuous for \(x\ne c\) since each of those two branches are defined by polynomials. Thus, the only question is whether the function is continuous at \(x=c\text{.}\) Furthermore,
The function is continuous for \(x\ne c\) since each of those two branches are polynomials. So, the only question is whether the function is continuous at \(x=c\text{;}\) for this we need
This isn't the kind of equality that we can just solve; we'll need a trick, and that trick is the IVT. The general idea is to show that \(\sin x\) is somewhere bigger, and somewhere smaller, than \(x1\text{.}\) However, since the IVT can only show us that a function is equal to a constant, we need to slightly adjust our language. Showing \(\sin x = x1\) is equivalent to showing \(\sin x  x + 1 = 0\text{,}\) so let \(f(x)=\sin x  x +1\text{,}\) and let's show that it has a real root.
First, we need to note that \(f(x)\) is continuous (otherwise we can't use the IVT). Now, we need to find a value of \(x\) for which it is positive, and for which it's negative. By checking a few values, we find \(f(0)\) is positive, and \(f(100)\) is negative. So, by the IVT, there exists a value of \(x\) (between \(0\) and \(100\)) for which \(f(x)=0\text{.}\) Therefore, there exists a value of \(x\) for which \(\sin x = x1\text{.}\)
This isn't the kind of equality that we can just solve; we'll need a trick, and that trick is the IVT. The general idea is to show that \(\sin x\) is somewhere bigger, and somewhere smaller, than \(x1\text{.}\) However, since the IVT can only show us that a function is equal to a constant, we need to slightly adjust our language. Showing \(\sin x = x1\) is equivalent to showing \(\sin x  x + 1 = 0\text{,}\) so let \(f(x)=\sin x  x +1\text{,}\) and let's show that it has a real root.
First, we need to note that \(f(x)\) is continuous (otherwise we can't use the IVT). Now, we need to find a value of \(x\) for which it is positive, and for which it's negative. By checking a few values, we find \(f(0)\) is positive, and \(f(100)\) is negative. So, by the IVT, there exists a value of \(x\) (between \(0\) and \(100\)) for which \(f(x)=0\text{.}\) Therefore, there exists a value of \(x\) for which \(\sin x = x1\text{.}\)
So, because \(f(x)\) is continuous on \((\infty, \infty)\) and \(f(0) \gt 0\) while \(f(1) \lt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (1,0)\) such that \(f(c)=0\text{.}\)
So, because \(f(x)\) is continuous on \((\infty, \infty)\) and \(f(0) \gt 0\) while \(f(1) \lt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\) in the interval \((1,0)\) such that \(f(c)=0\text{.}\)
We let \(f(x)=2\tan(x)x1\text{.}\) Then \(f(x)\) is a continuous function on the interval \((\pi/2, \pi/2)\) since \(\tan(x)=\sin(x)/\cos(x)\) is continuous on this interval, while \(x+1\) is a polynomial and therefore continuous for all real numbers.
We find a value \(a\in (\pi/2,\pi/2)\) such that \(f(a) \lt 0\text{.}\) We observe immediately that \(a=0\) works since
So, because \(f(x)\) is continuous on \([0,\pi/4]\) and \(f(0) \lt 0\) while \(f(\pi/4) \gt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (0,\pi/4)\) such that \(f(c)=0\text{.}\)
We let \(f(x)=2\tan(x)x1\text{.}\) Then \(f(x)\) is a continuous function on the interval \((\pi/2, \pi/2)\) since \(\tan(x)=\sin(x)/\cos(x)\) is continuous on this interval, while \(x+1\) is a polynomial and therefore continuous for all real numbers.
We find a value \(a\in (\pi/2,\pi/2)\) such that \(f(a) \lt 0\text{.}\) We observe immediately that \(a=0\) works since
So, because \(f(x)\) is continuous on \([0,\pi/4]\) and \(f(0) \lt 0\) while \(f(\pi/4) \gt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (0,\pi/4)\) such that \(f(c)=0\text{.}\)
Let \(f(x) = \sqrt{\cos(\pi x)}  \sin(2\pi x) 1/2\text{.}\) This function is continuous provided \(\cos(\pi x)\geq 0\text{.}\) This is true for \(0 \leq x
\leq \frac{1}{2}\text{.}\)
Now \(f\) takes positive values on \([0,1/2]\text{:}\)
(Notice that \(f(1/3)=(\sqrt{2}\sqrt{3})/21/2\) also works)
So, because \(f(x)\) is continuous on \([0,1/2)\) and \(f(0) \gt 0\) while \(f(1/2) \lt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (0,1/2)\) such that \(f(c)=0\text{.}\)
Let \(f(x) = \sqrt{\cos(\pi x)}  \sin(2\pi x) 1/2\text{.}\) This function is continuous provided \(\cos(\pi x)\geq 0\text{.}\) This is true for \(0 \leq x
\leq \frac{1}{2}\text{.}\)
Now \(f\) takes positive values on \([0,1/2]\text{:}\)
(Notice that \(f(1/3)=(\sqrt{2}\sqrt{3})/21/2\) also works)
So, because \(f(x)\) is continuous on \([0,1/2)\) and \(f(0) \gt 0\) while \(f(1/2) \lt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (0,1/2)\) such that \(f(c)=0\text{.}\)
We let \(f(x)=\dfrac{1}{\cos^2(\pi x)}c\dfrac{3}{2}\text{.}\) Then \(f(x)\) is a continuous function on the interval \((1/2, 1/2)\) since \(\cos x\) is continuous everywhere and nonzero on that interval.
The function \(f\) takes negative values. For example, when \(x=0\text{:}\)
We let \(f(x)=\dfrac{1}{\cos^2(\pi x)}c\dfrac{3}{2}\text{.}\) Then \(f(x)\) is a continuous function on the interval \((1/2, 1/2)\) since \(\cos x\) is continuous everywhere and nonzero on that interval.
The function \(f\) takes negative values. For example, when \(x=0\text{:}\)
\(f(x)\) is a polynomial, so it's continuous everywhere. If we can find values \(a\) and \(b\) so that \(f(a) \gt 0\) and \(f(b) \lt 0\text{,}\) then by the IVT, there will exist some \(c\) in \((a,b)\) where \(f(c)=0\text{;}\) that is, there is a root in the interval \([a,b]\text{.}\) Let's start plugging in easy values of \(x\text{.}\)
\(f(0)=15\text{,}\) and \(f(1)=115+918+15=8\text{.}\) Since 0 is between \(f(0)\) and \(f(1)\text{,}\) and since \(f\) is continuous, by IVT there must be some \(x\) in \([0,1]\) for which \(f(x)=0\text{:}\) that is, there is some root in \([0,1]\text{.}\)
That was the easiest interval to find. If you keep playing around, you find two more. \(f(1)=26\) (positive) and \(f(2)=1001\) (negative), so there is a root in \([2,1]\text{.}\)
The arithmetic is nasty, but there is also a root in \([14,15]\text{.}\)
This is an important trick. For highdegree polynomials, it is often impossible to get the exact values of the roots. Using the IVT, we can at least give a range where a root must be.
Let \(f(x)=x^3\text{.}\) Since \(f\) is a polynomial, it is continuous everywhere. If \(f(a) \lt 7 \lt f(b)\text{,}\) then \(\sqrt[3]{7}\) is somewhere between \(a\) and \(b\text{.}\) If we can find \(a\) and \(b\) that satisfy these inequalities, and are very close together, that will give us a good approximation for \(\sqrt[3]{7}\text{.}\)
Let's start with integers. \(1^3 \lt 7 \lt 2^3\text{,}\) so \(\sqrt[3]{7}\) is in the interval \((1,2)\text{.}\)
Let's narrow this down, say by testing \(f(1.5)\text{.}\) \((1.5)^3 = 3.375 \lt 7\text{,}\) so \(\sqrt[3]{7}\) is in the interval \((1.5,2)\text{.}\)
Let's narrow further, say by testing \(f(1.75)\text{.}\) \((1.75)^3\approx 5.34 \lt 7\text{,}\) so \(\sqrt[3]{7}\) is in the interval \((1.75,2)\text{.}\)
Testing various points, we find \(f(1.9) \lt 7 \lt f(2)\text{,}\) so \(\sqrt[3]{7}\) is between \(1.9\) and \(2\text{.}\)
By testing more, we find \(f(1.91) \lt 7 \lt f(1.92)\text{,}\) so \(\sqrt[3]{7}\) is in \((1.91,1.92)\text{.}\)
In order to get an approximation for \(\sqrt[3]{7}\) that is rounded to two decimal places, we have to know whether \(\sqrt[3]{7}\) is greater or less than \(1.915\text{;}\) indeed \(f(1.915) \approx 7.02 \gt 7\text{,}\) so \(\sqrt[3]{7} \lt 1.915\text{;}\) then rounded to two decimal places, \(\sqrt[3]{7}\approx 1.91\text{.}\)
If this seems like the obvious way to approximate \(\sqrt[3]{7}\text{,}\) that's good. The IVT is a formal statement of a very intuitive principle.
If \(f(a)=g(a)\text{,}\) or \(f(b)=g(b)\text{,}\) then we simply take \(c=a\) or \(c=b\text{.}\)
Suppose \(f(a) \neq g(a)\) and \(f(b) \neq g(b)\text{.}\) Then \(f(a) \lt g(a)\) and \(g(b) \lt f(b)\text{,}\) so if we define \(h(x)=f(x)g(x)\text{,}\) then \(h(a) \lt 0\) and \(h(b) \gt 0\text{.}\) Since \(h\) is the difference of two functions that are continuous over \([a,b]\text{,}\) also \(h\) is continuous over \([a,b]\text{.}\) So, by the Intermediate Value Theorem, there exists some \(c \in (a,b)\) with \(h(c)=0\text{;}\) that is, \(f(c)=g(c)\text{.}\)
If \(f(a)=g(a)\text{,}\) or \(f(b)=g(b)\text{,}\) then we simply take \(c=a\) or \(c=b\text{.}\)
Suppose \(f(a) \neq g(a)\) and \(f(b) \neq g(b)\text{.}\) Then \(f(a) \lt g(a)\) and \(g(b) \lt f(b)\text{,}\) so if we define \(h(x)=f(x)g(x)\text{,}\) then \(h(a) \lt 0\) and \(h(b) \gt 0\text{.}\) Since \(h\) is the difference of two functions that are continuous over \([a,b]\text{,}\) also \(h\) is continuous over \([a,b]\text{.}\) So, by the Intermediate Value Theorem, there exists some \(c \in (a,b)\) with \(h(c)=0\text{;}\) that is, \(f(c)=g(c)\text{.}\)
If \(Q\) is to the left of the \(y\) axis, the secant line has positive slope; if \(Q\) is to the right of the \(y\) axis, the secant line has negative slope.
If \(Q\) is to the left of the \(y\) axis, the line through \(Q\) and \(P\) is increasing, so the secant line has positive slope. If \(Q\) is to the right of the \(y\) axis, the line through \(Q\) and \(P\) is decreasing, so the secant line has negative slope.
2.1.2.2.a By drawing a few pictures, it's easy to see that sliding \(Q\) closer to \(P\text{,}\) the slope of the secant line increases.
2.1.2.2.b Since the slope of the secant line increases the closer \(Q\) gets to \(P\text{,}\) that means the tangent line (which is the limit as \(Q\) approaches \(P\)) has a larger slope than the secant line between \(Q\) and \(P\) (using the location where \(Q\) is right now).
Alternately, by simply sketching the tangent line at \(P\text{,}\) we can see that has a steeper slope than the secant line between \(P\) and \(Q\text{.}\)
The slope of the secant line will be \(\dfrac{f(2)f(2)}{2(2)} = \dfrac{f(2)f(2)}{4}\text{,}\) in every part. So, if two lines have the same slope, that means their differences \(f(2)f(2)\) will be the same.
The graphs in (a),(c), and (e) all have \(f(2)f(2)=1\text{,}\) so they all have the same secant line slope. The graphs in (b) and (f) both have \(f(2)f(2)=1\text{,}\) so they both have the same secant line slope. The graph in (d) has \(f(2)f(2)=0\text{,}\) and it is the only graph with this property, so it does not share its secant line slope with any of the other graphs.
A good approximation from the graph is \(f(5)=0.5\text{.}\) We want to find a secant line whose endpoints are both very close to \(x=5\text{,}\) but that also give us clear \(y\)values. It looks like \(f(5.25) \approx 1\text{,}\) and \(f(4.75)\approx \frac{1}{8}\text{.}\) The secant line from \(x=5\) to \(x=5.25\) has approximate slope \(\dfrac{f(5.25)f(5)}{5.255}\approx \dfrac{1.5}{.25}=2\text{.}\) The secant line from \(x=5\) to \(x=4.75\) has approximate slope \(\dfrac{0.5\frac{1}{8}}{54.75}=\dfrac{3}{2}\text{.}\)
The graph increases more and more quickly (gets steeper and steeper), so it makes sense that the secant line to the left of \(x=5\) has a smaller slope than the secant line to the right of \(x=5\text{.}\) Also, if you're taking secant lines that have endpoints farther out from \(x=5\text{,}\) you'll notice that the slopes of the secant lines change quite dramatically. You have to be very, very close to \(x=5\) to get any kind of accuracy.
If we split the difference, we might approximate the slope of the secant line to be the average of \(\frac{3}{2}\) and \(2\text{,}\) which is \(\frac{7}{4}\text{.}\)
Another way to try to figure out the tangent line is by carefully drawing it in with a ruler. This is shown here in blue:
It's much easier to take the slope of a line than a curve, and this one looks like it has slope about 1.5. However, we drew this with a computer: by hand it's much harder to draw an accurate tangent line. (That's why we need calculus!)
The actual slope of the tangent line to the function at \(x=5\) is about \(1.484\text{.}\) This is extremely hard to figure out just from the graphby hand, a guess between \(1.25\) and \(1.75\) would be very accurate.
There is only one tangent line to \(f(x)\) at \(P\) (shown in blue), but there are infinitely many choices of \(Q\) and \(R\) (one possibility shown in red).
There is only one tangent line to \(f(x)\) at \(P\) (shown in blue), but there are infinitely many choices of \(Q\) and \(R\) (one possibility shown in red). One easy way to sketch the secant line on paper is to draw any line parallel to the tangent line, and choose two intercepts with \(y=f(x)\text{.}\)
Any place the graph looks flat (if you imagine zooming in) is where the tangent line has slope 0. This occurs three times.
Notice that two of the indicated points are at a low point and a high point, respectively. Later, we'll use these places where the tangent line has slope zero to find where a graph achieves its biggest and smallest values.
The function shown is a line, so it has a constant slope(a) . Since the function is always increasing, \(f'\) is always positive, so also (d) holds. Remark: it does not matter that the function itself is sometimes negative; the slope is always positive because the function is always increasing. Also, since the slope is constant, \(f'\) is neither increasing nor decreasing: it is the function that is increasing, not its derivative.
The function is always decreasing, so \(f'\) is always negative, option (e). However, the function alternates between being more and less steep, so \(f'\) alternates between increasing and decreasing several times, and no other options hold.
Remark: \(f\) is always positive, but (d) does not hold!
At the left end of the graph, \(f\) is decreasing rapidly, so \(f'\) is a strongly negative number. Then as we move towards \(x=0\text{,}\) \(f\) decreases less rapidly, so \(f'\) is a less strongly negative number. As we pass 0, \(f\) increases, so \(f'\) is a positive number. As we move to the right, \(f\) increases more and more rapidly, so \(f'\) is an increasing positive number. This description tells us that \(f'\) increases for the entire range shown. So (b) holds, but not (a) or (c). Since \(f'\) is negative to the left of the \(y\) axis, and positive to the right of it, also (d) and (e) do not hold.
\(f'(1)\) does not exist, because to the left of \(x=1\) the slope is a pretty big positive number (looks like around \(+1\)) and to the right the slope is \(1/4\text{.}\) Since the derivative involves a limit, that limit needs to match the limit from the left and the limit from the right. The sharp angle made by the graph at \(x=1\) indicates that the left and right limits do not match, so the derivative does not exist.
\(f'(3)\) also does not exist. One way to see this is to notice that the function is discontinuous here. More viscerally, note that \(f(3)=1\text{,}\) so as we take secant lines with one endpoint \((3,1)\text{,}\) and the other endpoint just to the right of \(x=3\text{,}\) we get slopes that are more and more strongly negative, as shown in the picture below. If we take the limit of the slopes of these secant lines as \(x\) goes to \(3\) from the right, we get \(\infty\text{.}\) (This certainly doesn't match the slope from the left, which is \(\frac{1}{4}\text{.}\))
At \(x=3\text{,}\) there is some kind of “change” in the graph; however, it is a smooth curve, so the derivative exists here.
if it exists. We know from our work with limits that if both onesided limits
\(\ds\lim_{h \to 0^}\frac{f(a+h)f(a)}{h}\) and \(\ds\lim_{h \to 0^+}\frac{f(a+h)f(a)}{h}\) exist and are equal to each other, then \(\ds \lim_{h \to 0}\dfrac{f(a+h)f(a)}{h}\) exists and has the same value as the onesided limits. So, since the onesided limits exist and are equal to one, we conclude \(f'(a)\) exists and is equal to one.
where \(f'(x)=1\) whenever \(x \neq 0\) (so in particular, \(\ds\lim_{x \to 0^}f'(x)=\ds\lim_{x \to 0} f'(x)=1\)) but \(f'(0)\) does not exist.
There are two ways to see that \(f'(0)\) does not exist. One is to notice that \(f\) is not continuous at \(x=0\text{.}\)
Another way to see that \(f'(0)\) does not exist is to use the definition of the derivative. Remember, in order for a limit to exist, both onesided limits must exist. Let's consider the limit from the left. If \(h \to 0^\text{,}\) then \(h \lt 0\text{,}\) so \(f(h)\) is equal to \(h\) (not \(h1\)).
\begin{align*}
\lim_{h \to 0^}\frac{f(0+h)f(0)}{h}&=\lim_{h \to 0^}\frac{(h)(1)}{h}\\
&=\lim_{h \to 0^}\frac{h+1}{h}\\
&=\lim_{h \to 0^}1+\frac{1}{h}\\
&=\infty\\
\end{align*}
In particular, this limit does not exist. Since the onesided limit does not exist,
\begin{align*}
\lim_{h \to 0}\frac{f(0+h)f(0)}{h}&=DNE
\end{align*}
The units of the numerator are meters, and the units of the denominator are seconds (since the denominator comes from the change in the input of the function). So, the units of \(s'(t)\) are metres per second.
Remark: we learned that the derivative of a position function gives velocity. In this example, the position is given in metres, and the velocity is measured in metres per second.
We can use pointslope form to get the equation of the line, if we have a point and its slope. The point is given: \((1,6)\text{.}\) The slope is the derivative:
The slope of the tangent line is the derivative. We set this up using the same definition of the derivative that we always do. This limit is hard to take for general \(x\text{,}\) but easy when \(x=0\text{.}\)
For \(f\) to be differentiable at \(x=2\text{,}\) two things must be true: it must be continuous at \(x=2\text{,}\) and the derivative from the right must equal the derivative from the left.
When \(x\) is not equal to 2, then the function is differentiable the only place we have to worry about is when \(x\) is exactly 2.
In order for \(f\) to be differentiable at \(x=2\text{,}\) it must also be continuous at \(x=2\text{.}\) This forces \(x^2\big_{x=2}=\big[ax+b\big]_{x=2}\) or
\begin{equation*}
2a+b=4.
\end{equation*}
In order for a limit to exist, the left and righthand limits must exist and be equal to each other. Since a derivative is a limit, in order for \(f\) to be differentiable at \(x=2\text{,}\) the left hand derivative of \(ax+b\) at \(x=2\) must be the same as the right hand derivative of \(x^2\) at \(x=2\text{.}\) Since \(ax+b\) is a line, its derivative is \(a\) everywhere. We've already seen the derivative of \(x^2\) is \(2x\text{,}\) so we need
The domain of the function is \([1,\infty)\text{.}\) In particular, \(f(x)\) is defined when \(x=1\text{.}\) However, \(f'(x)\) is not defined when \(x=1\text{,}\) so \(f'(x)\) only exists over \((1,\infty)\text{.}\)
Remark: \(\ds\lim_{x \to 1^+}f'(x)=\infty\text{,}\) so the tangent line to \(f(x)\) at the point \(x=1\) has a vertical slope.
From Section 1.2, compare the definition of velocity to the definition of a derivative. When you're finding the derivative, you'll need to cancel a lot on the numerator, which you can do by expanding the polynomials.
You'll need to look at limits from the left and right. The fact that \(f(0)=0\) is useful for your computation. Recall that if \(x \lt 0\) then \(\sqrt{x^2}=x=x\text{.}\)
exists (note that we used the fact that \(f(0)=0\) as per the definition of the first branch which includes the point \(x=0\)). We start by computing the left limit. For this computation, recall that if \(x \lt 0\) then \(\sqrt{x^2}=x=x\text{.}\)
exists (note that we used the fact that \(f(0)=0\) as per the definition of the first branch which includes the point \(x=0\)). We compute left and right limits; so
where in these inequalities we used the fact that \(x^2\ge 0\text{.}\) Finally, since \(\lim_{x\to 0^+}x^2=\lim_{x\to 0^+}x^2=0\text{,}\) the Squeeze Theorem yields that also \(\lim_{x\to 0^+}x^2\cos\left(\frac{1}{x}\right) =0\text{,}\) as claimed.
Since the left and right limits match (they're both equal to \(0\)), we conclude that indeed \(f(x)\) is differentiable at \(x=0\) (and its derivative at \(x=0\) is actually equal to \(0\)).
exists (note that we used the fact that \(f(1)=0\) as per the definition of the first branch which includes the point \(x=0\)). We compute left and right limits; so
where in these inequalities we used the fact that \(x\to 1^+\) yields positive values for \(x1\text{.}\) Finally, since \(\lim_{x\to 1^+}x+1=\lim_{x\to 1^+}x1=0\text{,}\) the Squeeze Theorem yields that also \(\lim_{x\to 1^+}(x1)\sin\left(\frac{1}{x1}\right) =0\text{,}\) as claimed.
Since the left and right limits match (they're both equal to \(0\)), we conclude that indeed \(f(x)\) is differentiable at \(x=1\) (and its derivative at \(x=1\) is actually equal to \(0\)).
The key is to realize that the few points you're given suggest a pattern, but don't guarantee it. You only know nine points; anything can happen in between.
At step (\(*\)), we use the limit law that \(\displaystyle\lim_{x \rightarrow a} F(x)+G(x) = \displaystyle\lim_{x \rightarrow a} F(x)+\displaystyle\lim_{x \rightarrow a}G(x)\text{,}\) as long as \(\displaystyle\lim_{x \rightarrow a} F(x)\) and \(\displaystyle\lim_{x \rightarrow a}G(x)\) exist. Because the problem states that \(f'(x)\) and \(g'(x)\) exist, we know that \(\displaystyle\lim_{x \rightarrow 0} \frac{f(x+h)f(x)}{h}\) and \(\displaystyle\lim_{x \rightarrow 0}
\frac{g(x+h)g(x)}{h}\) exist, so our work is valid.
At step (\(*\)), we use the limit law that \(\displaystyle\lim_{x \rightarrow a} F(x)+G(x) = \displaystyle\lim_{x \rightarrow a} F(x)+\displaystyle\lim_{x \rightarrow a}G(x)\text{,}\) as long as \(\displaystyle\lim_{x \rightarrow a} F(x)\) and \(\displaystyle\lim_{x \rightarrow a}G(x)\) exist. Because the problem states that \(f'(x)\) and \(g'(x)\) exist, we know that \(\displaystyle\lim_{x \rightarrow 0} \frac{f(x+h)f(x)}{h}\) and \(\displaystyle\lim_{x \rightarrow 0}
\frac{g(x+h)g(x)}{h}\) exist, so our work is valid.
2.2.4.25.a Since \(y=f(x)=2x\) and \(y=g(x)=x\) are straight lines, we don't need the definition of the derivative (although you can use it if you like). \(f'(x)=2\) and \(g'(x)=1\text{.}\)
2.2.4.25.b \(p(x)=2x^2\text{,}\) so \(p(x)\) is not a line: we use the definition of a derivative to find \(p'(x)\text{.}\)
2.2.4.25.c No, \(p'(x) = 4x \ne 2\cdot 1 = f'(x)\cdot g'(x)\text{.}\) In general, the derivative of a product is not the same as the derivative of the functions being multiplied.
A generic point on the curve has coordinates \((\alpha, \alpha^2)\text{.}\) In terms of \(\alpha\text{,}\) what is the equation of the tangent line to the curve at the point \((\alpha, \alpha^2)\text{?}\) What does it mean for \((1,3)\) to be on that line?
We know that \(y'=2x\text{.}\) So, if we choose a point \((\alpha,\alpha^2)\) on the curve \(y=x^2\text{,}\) then the tangent line to the curve at that point has slope \(2\alpha\text{.}\) That is, the tangent line has equation
\begin{align*}
(y\alpha^2)&=2\alpha(x\alpha)\\
\mbox{simplified, } \qquad y&=(2\alpha)x\alpha^2\\
\end{align*}
So, if \((1,3)\) is on the tangent line, then
\begin{align*}
3&=(2\alpha)(1)\alpha^2\\
\iff\qquad 0&=\alpha^22\alpha3\\
\iff\qquad 0&=(\alpha3)(\alpha+1)\\
\iff\qquad \alpha&=3, \quad\mbox{or}\qquad \alpha=1.\\
\end{align*}
So, the tangent lines \(y=(2\alpha)x\alpha^2\) are
\begin{align*}
y&=6x9 \quad\mbox{and}\quad y=2x1.
\end{align*}
Using the definition of the derivative, \(f\) is differentiable at \(0\) if and only if
\begin{align*}
\lim_{h \to 0}\frac{f(h)f(0)}{h}&\\
\end{align*}
exists. In particular, this means \(f\) is differentiable at \(0\) if and only if both onesided limits exist and are equal to each other.
\begin{align*}
\\
\end{align*}
When \(h \lt 0\text{,}\) \(f(h)=0\text{,}\) so
\begin{align*}
\lim_{h \to 0^}\frac{f(h)f(0)}{h}&=\lim_{h \to 0^}\frac{00}{h}=0\\
\end{align*}
So, \(f\) is differentiable at \(x=0\) if and only if
\begin{align*}
\lim_{h \to 0^+}\frac{f(h)f(0)}{h}&=0.\\
\end{align*}
To evaluate the limit above, we note \(f(0)=0\) and, when \(h \gt 0\text{,}\) \(f(h)=h^a\sin\left(\frac{1}{h}\right)\text{,}\) so
\begin{align*}
\lim_{h \to 0^+}\frac{f(h)f(0)}{h}&=\lim_{h \to 0^+}\frac{h^a\sin\left(\frac{1}{h}\right)}{h}\\
&=\lim_{h \to 0^+}h^{a1}\sin\left(\frac{1}{h}\right)\\
\end{align*}
We will spend the rest of this solution evaluating the limit above for different values of \(a\text{,}\) to find when it is equal to zero and when it is not. Let's consider the different values that could be taken by \(h^{a1.}\)
\begin{align*}
\end{align*}
If \(a=1\text{,}\) then \(a1=0\text{,}\) so \(h^{a1}=h^0=1\) for all values of \(h\text{.}\) Then
(Recall that the function \(\sin\left(\frac{1}{x}\right)\) oscillates faster and faster as \(x\) goes to 0. We first saw this behaviour in Example 1.3.5.)
If \(a \lt 1\text{,}\) then \(a1 \lt 0\text{,}\) so \(\ds\lim_{h \to 0^+}h^{a1}=\infty\text{.}\) (Since we have a negative exponent, we are in effect dividing by a smaller and smaller positive number. For example, if \(a=\frac{1}{2}\text{,}\) then \(\ds\lim_{h \to 0^+}h^{a1}=\ds\lim_{h \to 0^+}h^{\frac{1}{2}}=\ds\lim_{h \to 0^+}\frac{1}{\sqrt{h}}=\infty\text{.}\)) Since \(\sin\left(\frac{1}{x}\right)\) goes back and forth between one and negative one,
since as \(h\) goes to 0, the function oscillates between positive and negative numbers of everincreasing magnitude.
If \(a \gt 1\text{,}\) then \(a1 \gt 0\text{,}\) so \(\ds\lim_{h \to 0^+}h^{a1}=0\text{.}\) Although \(\sin\left(\frac{1}{x}\right)\) oscillates wildly near \(x=0\text{,}\) it is bounded by \(1\) and \(1\text{.}\) So,
In the above cases, we learned \(\ds\lim_{h \to 0^+}\frac{f(h)f(0)}{h}=\ds\lim_{h \to 0^+} h^{a1}\sin\left(\frac{1}{x}\right)=0\) when \(a \gt 1\text{,}\) and
2.3.3.1.a The average rate of change of the height of the water over the single day starting at \(t=0\text{,}\) measured in \(\frac{\mathrm{m}}{\mathrm{hr}}\text{.}\)
2.3.3.1.b The instantaneous rate of change of the height of the water at the time \(t=0\text{.}\)
2.3.3.1.a The slope of the secant line is \(\dfrac{h(24)h(0)}{240} \quad \dfrac{\mathrm{m}}{\mathrm{hr}}\text{;}\) this is the change in height over the first day divided by the number of hours in the first day. So, it is the average rate of change of the height over the first day, measured in meters per hour.
2.3.3.1.b Consider 2.3.3.1.a. The secant line gives the average rate of change of the height of the dam; as we let the second point of the secant line get closer and closer to \((0,h(0))\text{,}\) its slope approximates the instantaneous rate of change of the height of the water. So the slope of the tangent line is the instantaneous rate of change of the height of the water at the time \(t=0\text{,}\) measured in \(\frac{\mathrm{m}}{\mathrm{hr}}\text{.}\)
\(p'(t) = \displaystyle\lim_{h \rightarrow 0}\frac{p(t+h)p(t)}{h} \approx \frac{p(t+1)p(t)}{1} = p(t+1)p(t)\text{,}\) or the difference in profit caused by the sale of the \((t+1)^{\mathrm{st}}\) widget. So, \(p'(t)\) is the profit from the \((t+1)^{\mathrm{st}}\) widget, so \(p'(t)\) is the profit per widget.
\(T'(d)\) measures how quickly the temperature is changing per unit change of depth, measured in degrees per metre. \(T'(d)\) will probably be largest when \(d\) is near zero, unless there are hot springs or other underwater heat sources.
How quickly the temperature is changing per unit change of depth, measured in degrees per metre. In an ordinary body of water, the temperature near the surface (\(d=0\)) is pretty variable, depending on the sun, but deep down it is more stable (unless there are heat sources). So, one might reasonably expect that \(T'(d)\) is larger when \(d\) is near 0.
\(C'(w)=\displaystyle\lim_{h \rightarrow 0} \dfrac{C(w+h)C(w)}{h} \approx \dfrac{C(w+1)C(w)}{1}=C(w+1)C(w)\text{,}\) which is the number of calories in \(C(w+1)\) grams minus the number of calories in \(C(w)\) grams. This is the number of calories per gram.
The rate of change of velocity is acceleration. (If your velocity is increasing, you're accelerating; if your velocity is decreasing, you have negative acceleration.)
Degrees Celsius temperature change per joule of heat added. (This is closely related to heat capacity and to specific heat — there's a nice explanation of this on Wikipedia.)
The rate of change in this case will be the relationship between the heat added and the temperature change. \(\displaystyle\lim_{h \rightarrow 0} \dfrac{T(j+h)T(j)}{h} \approx \dfrac{T(j+1)T(j)}{1}=T(j+1)T(j)\text{,}\) or the change in temperature after the application of one joule. (This is closely related to heat capacity and to specific heat — there's a nice explanation of this on Wikipedia.)
Number of bacteria added per degree. That is: the number of extra bacteria (possibly negative) that will exist in the population by raising the temperature by one degree.
This is the difference in population between two hypothetical populations, raised one degree in temperature apart. So, it is the number of extra individuals that exist in the hotter experiment (with the understanding that this number could be negative, as one would expect in conditions that are hotter than the bacteria prefer). So \(P'(T)\) is the number of bacteria added to the colony per degree.
\(R'(t)\) is the rate at which the wheel is rotating measured in rotations per second. To convert to degrees, we multiply by 360: \(\boxed{360R'(t)}\text{.}\)
If \(P'(t)\) is positive, your sample is below the ideal temperature, and if \(P'(t)\) is negative, your sample is above the ideal temperature. If \(P'(t) = 0\text{,}\) you don't know whether the sample is exactly at the ideal temperature, or way above or below it with no living bacteria.
If \(P'(t)\) is positive, your sample is below the ideal temperature, because adding heat increases the population. If \(P'(t)\) is negative, your sample is above the ideal temperature, because adding heat decreases the population. If \(P'(t)=0\text{,}\) then adding a little bit of heat doesn't change the population, but it's unclear why this is. Perhaps your sample is deeply frozen, and adding heat doesn't change the fact that your population is 0. Perhaps your sample is boiling, and again, changing the heat a little will keep the population constant at “none.” But also, at the ideal temperature, you would expect \(P'(t)=0\text{.}\) This is best seen by noting in the curve below, the tangent line is horizontal at the peak.
False, in general. The product rule tells us \(\diff{}{x}\{f(x)g(x)\}=f'(x)g(x)+f(x)g'(x)\text{.}\) An easy example of why we can't do it the other way is to take \(f(x)=g(x)=x\text{.}\) Then the equation becomes \(\diff{}{x}\{x^2\}=(1)(1)\text{,}\) which is false.
False: the power rule tells us that, for a constant \(n\text{,}\) \(\ds\diff{}{x}\{x^n\}=nx^{n1}\text{.}\) In the equation shown, the base is a constant, and the exponent is the variable: this is the opposite of the situation where the power rule applies.
We do not yet know how to differentiate this function. We'll learn about it in Section 2.7
We have already seen \(\diff{}{x}\{\sqrt{x}\}=\frac{1}{2\sqrt{x}}\text{,}\) but if you forget the formula, it's easy to find using the power rule: \(\diff{}{x}\{\sqrt{x}\}=\diff{}{x}\left\{x^{1/2}\right\}=\frac{1}{2}x^{1/2}=\frac{1}{2\sqrt{x}}\text{.}\)
We compute the derivative of \(x^3\) as being \(3x^2\text{,}\) which evaluated at \(x=\frac{1}{2}\) yields \(\frac{3}{4}\text{.}\) Since we also compute \(\left( \frac{1}{2}\right)^3=\frac{1}{8}\text{,}\) then the equation of the tangent line is
\begin{gather*}
y  \frac{1}{8} = \frac{3}{4}\cdot \left(x\frac{1}{2}\right).
\end{gather*}
Hence at \(t=2\text{,}\) 2.4.2.9.a the particle has speed of magnitude {4}, and 2.4.2.9.b is moving {towards the left}. At \(t=2\text{,}\) \(f''(2) \gt 0\text{,}\) so \(f'\) is increasing, i.e. becoming less negative. Since \(f'\) is getting closer to zero, 2.4.2.9.c the magnitude of the speed is {decreasing}.
We need \(f'(1)\text{,}\) so first we must find \(f'(x)\text{.}\) Since \(f(x)\) is the reciprocal of \(\sqrt{x}+1\text{,}\) we can use the Corollary 2.4.6:
so \(f'(1)=\dfrac{1}{2\sqrt{1}(\sqrt{1}+1)^2}=\frac{1}{8}.\)
Now, using the point \(\left(1,\frac{1}{2}\right)\) and the slope \(\frac{1}{8}\text{,}\) our tangent line has equation \(y\frac{1}{2}=\frac{1}{8}(x1)\text{.}\)
Population growth is rate of change of population. Population in year \(2000+t\) is given by \(P(t)=P_0+b(t)d(t)\text{,}\) where \(P_0\) is the initial population of the town. Then \(P'(t)\) is the expression we're looking for, and \(P'(t)=b'(t)d'(t)\text{.}\)
It is interesting to note that the initial population does not obviously show up in this calculation. It would probably affect \(b(t)\) and \(d(t)\text{,}\) but if we know these we do not need to know \(P_0\) to answer our question.
The slope of \(y=3x^2\) at \(x=a\) is \(6a\text{.}\) The tangent line to \(y=3x^2\) at \(x=a, y=3a^2\) is \(y3a^2=6a(xa)\text{.}\) This tangent line passes through \((2,9)\) if
This limit represents the derivative computed at \(x=100180\) of the function \(f(x)=\sqrt{x}\text{.}\) Since the derivative of \(f(x)\) is \(\dfrac{1}{2\sqrt{x}}\text{,}\) then its value at \(x=100180\) is exactly \(\dfrac{1}{2\sqrt{100180}}\text{.}\)
Let \(w(t)\) and \(l(t)\) be the width and length of the rectangle. Given in the problem is that \(w'(t)=2\) and \(l'(t)=5\text{.}\) Since both functions have constant slopes, both must be lines. Their slopes are given, and their intercepts are \(w(0)=l(0)=1\text{.}\) So, \(w(t)=2t+1\) and \(l(t)=5t+1\text{.}\)
The area of the rectangle is \(A(t)=w(t)\cdot l(t)\text{,}\) so using the product rule, the rate at which the area is increasing is \(A'(t)=w'(t)l(t)+w(t)l'(t)=2(5t+1)+5(2t+1)=20t+7\) square metres per second.
Using the product rule, \(f'(x)=(2x)g(x)+x^2g'(x)\text{,}\) so \(f'(0)=0\cdot g(x)+0\cdot g'(x)=0\text{.}\) (Since \(g\) is differentiable, \(g'\) exists.)
Exercise18
Answer
\begin{align*}
\\
\end{align*}
First expression, \(f(x)=\dfrac{g(x)}{h(x)}\text{:}\)
\begin{align*}
f'(x)&=\frac{h(x)g'(x)g(x)h'(x)}{h^2(x)}\\
\end{align*}
Second expresson, \(f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}\text{:}\)
\begin{align*}
f'(x)&=\left(\frac{k(x)g'(x)g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)k(x)h'(x)}{h^2(x)}\right)\\
&=\frac{k(x)g'(x)g(x)k'(x)}{k(x)h(x)}+
\frac{g(x)h(x)k'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)h(x)g(x)k'(x)}{k(x)h^2(x)}+
\frac{g(x)h(x)k'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)h(x)g(x)k'(x)+g(x)h(x)k'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)g'(x)g(x)h'(x)}{h^2(x)}\\
\end{align*}
and this is exactly what we got from differentiating the first expression.
\begin{align*}
\end{align*}
Solution
\begin{align*}
\\
\end{align*}
First expression, \(f(x)=\dfrac{g(x)}{h(x)}\text{:}\)
\begin{align*}
f'(x)&=\frac{h(x)g'(x)g(x)h'(x)}{h^2(x)}\\
\end{align*}
Second expresson, \(f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}\text{:}\)
\begin{align*}
f'(x)&=\left(\frac{k(x)g'(x)g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)k(x)h'(x)}{h^2(x)}\right)\\
&=\frac{k(x)g'(x)g(x)k'(x)}{k(x)h(x)}+
\frac{g(x)h(x)k'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)h(x)g(x)k'(x)}{k(x)h^2(x)}+
\frac{g(x)h(x)k'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)h(x)g(x)k'(x)+g(x)h(x)k'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)g'(x)g(x)h'(x)}{h^2(x)}\\
\end{align*}
and this is exactly what we got from differentiating the first expression.
\begin{align*}
\end{align*}
First, factor an \(x\) out of the derivative. What's left over looks like a quadratic equation, if you take \(x^2\) to be your variable, instead of \(x\text{.}\)
First simplify. Don't be confused by the role reversal of \(x\) and \(y\text{:}\) \(x\) is just the name of the function \(\big(2y+\tfrac{1}{y}\big)\cdot y^3\text{,}\) which is a function of the variable \(y\text{.}\) You are to differentiate with respect to \(y\text{.}\)
We could use the product rule here, but it's easier to simplify first. Don't be confused by the role reversal of \(x\) and \(y\text{:}\) \(x\) is the name of the function, and \(y\) is the variable.
We've already seen that \(\diff{}{x}\{\sqrt{x}\}=\frac{1}{2\sqrt{x}}\text{,}\) but if you forget this formula it is easy to figure out: \(\sqrt{x}=x^{1/2}\text{,}\) so \(\diff{}{x}\{\sqrt{x}\}=\frac{1}{2}x^{1/2}=\frac{1}{2\sqrt{x}}\text{.}\)
Instead of multiplying to get our usual form of this polynomial, we can use the quotient rule. If \(f_1(x)=3x^3+4x^2+x+1\) and \(f_2(x)=2x+5\text{,}\) then \(f_1'(x)=9x^2+8x+1\) and \(f_2'(x)=2\text{.}\) Then
The derivative is undefined if either \(x \lt 0\) or \(x = 0,\pm 1\) (since the squareroot is undefined for \(x \lt 0\) and the denominator is zero when \(x=0,1,1\text{.}\) Putting this together — the derivative exists for \(x \gt 0, x\neq 1\text{.}\)
The derivative is undefined if either \(x \lt 0\) or \(x = 0,\pm 1\) (since the squareroot is undefined for \(x \lt 0\) and the denominator is zero when \(x=0,1,1\text{.}\) Putting this together — the derivative exists for \(x \gt 0, x\neq 1\text{.}\)
(If you simplified differently, or used the quotient rule, you probably came up with a differentlooking answer. There is only one derivative, though, so all correct answers will look the same after sufficient algebraic manipulation.)
So our candidates for \(x\)values where \(f'(x)=0\) are \(x=0\text{,}\) \(x=5\text{,}\) and \(x=1\text{.}\) However, we need to check that \(f\) exists at these places: \(f(0)\) is undefined (and \(f'(0)\) doesn't exist). So \(f'(x)=0\) only when \(x=5\) and \(x=1\text{.}\)
Let \(m\) be the slope of such a tangent line, and let \(P_1\) and \(P_2\) be the points where the tangent line is tangent to the two curves, respectively. There are three equations \(m\) fulfils: it has the same slope as the curves at the given points, and it is the slope of the line passing through the points \(P_1\) and \(P_2\text{.}\)
Denote by \(m\) the slope of the common tangent, by \((x_1,y_1)\) the point of tangency with \(y=x^2\text{,}\) and by \((x_2,y_2)\) the point of tangency with \(y=x^22x+2\text{.}\) Then we must have
A line has equation \(y=mx+b\text{,}\) for some constants \(m\) and \(b\text{.}\) What has to be true for \(y=mb+x\) to be tangent to the first curve at the point \(x=\alpha\text{,}\) and to the second at the point \(x=\beta\text{?}\)
The line \(y=mx+b\) is tangent to \(y=x^2\) at \(x=\alpha\) if
\begin{equation*}
2\alpha=m\hbox{ and }\alpha^2=m\alpha+b
\iff m=2\alpha\hbox{ and }b=\alpha^2
\end{equation*}
The same line \(y=mx+b\) is tangent to \(y=x^2+2x5\) at \(x=\beta\) if
\begin{align*}
2\beta+2=m&\hbox{ and }\beta^2+2\beta5=m\beta+b\\
\iff m=22\beta&\hbox{ and }b=\beta^2+2\beta5(22\beta)\beta=\beta^25
\end{align*}
For the line to be simultaneously tangent to the two parabolas we need
\begin{equation*}
m=2\alpha=22\beta\hbox{ and }b=\alpha^2=\beta^25
\end{equation*}
Substituting \(\alpha=1\beta\) into \(\alpha^2=\beta^25\) gives \((1\beta)^2=\beta^25\) or \(2\beta^22\beta4=0\) or \(\beta=1,2\text{.}\) The corresponding values of the other parameters are \(\alpha=2,1\text{,}\) \(m=4,2\) and \(b=4,1\text{.}\) The two lines are {\(y=4x4\) and \(y=2x1\)}.
This limit represents the derivative computed at \(x=2015\) of the function \(f(x)=\cos(x)\text{.}\) To see this, simply use the definition of the derivative at \(a=2015\) with \(f(x)=\cos x\text{:}\)
This limit represents the derivative computed at \(x=\pi/3\) of the function \(f(x)=\cos x\text{.}\) To see this, simply use the definition of the derivative at \(a=\pi/3\) with \(f(x)=\cos x\text{:}\)
This limit represents the derivative computed at \(x=\pi\) of the function \(f(x)=\sin(x)\text{.}\) To see this, simply use the definition of the derivative at \(a=\pi\) with \(f(x)=\sin x\text{:}\)
This limit represents the derivative computed at \(x=2\) of the function \(f(x)=x^{2015}\text{.}\) To see this, simply use the definition of the derivative at \(a=2\) with \(f(x)=x^{2015}\text{:}\)
Since \(1^x=1\) for any \(x\text{,}\) we see that \((b)\) is just the constant function \(y=1\text{,}\) so D matches to \((b)\text{.}\)
Since \(2^{x}=\frac{1}{2^x}=\left(\frac{1}{2}\right)^x\text{,}\) functions \((a)\) and \((d)\) are the same. This is the only function out of the lot that grows as \(x\to\infty\) and shrinks as \(x \to \infty\text{,}\) so A matches to \((a)\) and \((d)\text{.}\)
This leaves B and C to match to \((c)\) and \((e)\text{.}\) Since \(3 \gt 2\text{,}\) when \(x \gt 0\text{,}\) \(3^x \gt 2^x\text{.}\) So, \((e)\) matches to the function that grows more quickly to the right of the \(x\)axis: B matches to \((e)\text{,}\) and C matches to \((c)\text{.}\)
First, let's consider the behaviour of exponential functions \(a^x\) based on whether \(a\) is greater or less than 1. As we know, \(\ds\lim_{x\to\infty}a^x
=\left\{\begin{array}{ll}
\infty & a \gt 1\\
0&a \lt 1
\end{array}\right.\) and \(\ds\lim_{x\to\infty}a^x=\left\{\begin{array}{ll}
0 & a \gt 1\\
\infty&a \lt 1
\end{array}\right.\text{.}\) Our function has \(\ds\lim_{x \to \infty} f(x)=\infty\) and \(\ds\lim_{x \to \infty} f(x)=0\text{,}\) so we conclude \(a \gt 1\text{:}\) thus \((d)\) and also \((b)\) hold. (We could have also seen that \((b)\) holds because \(a^x\) is defined for all real numbers.)
It remains to decide whether \(a\) is greater or less than \(e\text{.}\) (If \(a\) were equal to \(e\text{,}\) then \(f'(x)\) would be the same as \(f(x)\text{.}\)) We saw in the text that \(\diff{}{x}\{a^x\}=C(a)a^x\) for the function \(C(a)=\ds\lim_{h \to 0} \dfrac{a^h1}{h}\text{.}\) We know that \(C(e)=1\text{.}\) (Actually, we chose \(e\) to be the number that has this property.) From our graph, we see that \(f'(x) \lt f(x)\text{,}\) so \(C(a) \lt 1=C(e)\text{.}\) In other words, \(\ds\lim_{h \to 0} \dfrac{a^h1}{h} \lt \ds\lim_{h \to 0} \dfrac{e^h1}{h}\text{;}\) so, \(a \lt e\text{.}\) Thus \((e)\) holds.
The power rule tells us that \(\diff{}{x}\{x^n\}=nx^{n1}\text{.}\) In this equation, the variable is the base, and the exponent is a constant. In the function \(e^x\text{,}\) it's reversed: the variable is the exponent, and the base it a constant. So, the power rule does not apply.
\(P(t)\) is an increasing function over its domain, so the population is increasing.
There are a few ways to see that \(P(t)\) is increasing.
What we really care about is whether \(e^{0.2t}\) is increasing or decreasing, since an increasing function multiplied by 100 is still an increasing function, and a decreasing function multiplied by 100 is still a decreasing function. Since \(f(t)=e^t\) is an increasing function, we can use what we know about graphing functions to see that \(f(0.2t)=e^{0.2t}\) is also increasing.
Solution
\begin{align*}
e^{a+x}&=e^ae^x\\
\end{align*}
Since \(e^a\) is just a constant,
\begin{align*}
\diff{}{x}\{e^{a}e^{x}\}&=e^a\diff{}{x}\{e^x\}=e^ae^x=e^{a+x}
\end{align*}
If the derivative is positive, the function is increasing, so let's start by finding the derivative. We use the product rule (although Question 2.7.3.12 gives a shortcut).
Solution
\begin{align*}
e^{x}&=\frac{1}{e^x}\\
\end{align*}
Using the rule for differentiating the reciprocal:
\begin{align*}
\diff{}{x}\{e^{x}\}&=\frac{e^x}{(e^x)^2}=\frac{1}{e^x}=e^{x}
\end{align*}
We simplify the functions to get a better idea of what's going on.
(\(a\)): \(y=e^{3\log x}+1=\left(e^{\log x}\right)^3+1=x^3+1\text{.}\) This is not a line.
(\(b\)): \(2y+5=e^{3+\log x}=e^3e^{\log x}=e^3x\text{.}\) Since \(e^3\) is a constant, \(2y+5=e^3x\) is a line.
(\(c\)): There isn't a fancy simplification herethis isn't a line. If that isn't a satisfactory answer, we can check: a line is a function with a constant slope. For our function, \(y'=\diff{}{x}\{e^{2x} +4\}=\diff{}{x}\{e^{2x}\}=\diff{}{x}\left\{(e^x)^2\right\}=2e^xe^x=2e^{2x}\text{.}\) Since the derivative isn't constant, the function isn't a line.
(\(d\)): \(y=e^{\log x}3^e+\log 2=xe^3+\log 2\text{.}\) Since \(e^3\) and \(\log 2\) are constants, this is a line.
In order to be differentiable, a function should be continuous. To determine the differentiability of the function at \(x=1\text{,}\) use the definition of the derivative.
When we say a function is differentiable without specifying a range, we mean that it is differentiable over its domain. The function \(f(x)\) is differentiable when \(x \neq 1\) for any values of \(a\) and \(b\text{;}\) it is up to us to figure out which constants make it differentiable when \(x=1\text{.}\)
In order to be differentiable, a function must be continuous. The definition of continuity tells us that, for \(f\) to be continuous at \(x=1\text{,}\) we need \(\ds\lim_{x \to 1}f(x)=f(1)\text{.}\) From the definition of \(f\text{,}\) we see \(f(1)=a+b=\ds\lim_{x \to 1^}f(x)\text{,}\) so we need \(\ds\lim_{x\to 1^+}f(x)=a+b\text{.}\) Since \(\ds\lim_{x \to 1^+}f(x)=e^1=e\text{,}\) we specifically need
\begin{equation*}
e=a+b.
\end{equation*}
Now, let's consider differentiability of \(f\) at \(x=1\text{.}\) We need the following limit to exist:
\begin{align*}
\lim_{h \to 0} \frac{f(1+h)f(1)}{h}&\\
\end{align*}
In particular, we need the onesided limits to exist and be equal:
\begin{align*}
\textcolor{red}{\lim_{h \to 0^}\frac{f(1+h)f(1)}{h}}&=\textcolor{blue}{\lim_{h \to 0^+}\frac{f(1+h)f(1)}{h}}\\
\end{align*}
If \(h \lt 0\text{,}\) then \(1+h \lt 1\text{,}\) so \(f(1+h)=a(1+h)^2+b\text{.}\) If \(h \gt 0\text{,}\) then \(1+h \gt 1\text{,}\) so \(f(1+h)=e^{1+h}\text{.}\) With this in mind, we begin to evaluate the onesided limits:
\begin{align*}
\color{red}\lim_{h \to 0^}\frac{f(1+h)f(1)}{h}&\color{red}=
\lim_{h \to 0^}\frac{[a(1+h)^2+b][a+b]}{h}\\
&\color{red}=\lim_{h \to 0^}\frac{ah^2+2ah}{h}=2a\\
\color{blue}\lim_{h \to 0^+}\frac{f(1+h)f(1)}{h}&\color{blue}=
\lim_{h \to 0^+}\frac{e^{1+h}(a+b)}{h}\\
\end{align*}
Since we take \(a+b\) to be equal to \(e\) (to ensure continuity):
\begin{align*}
&\color{blue}=
\lim_{h \to 0^+}\frac{e^{1+h}e^1}{h}\\
&\color{blue}=\left.\diff{}{x}\{e^x\}\right_{x=1}=e^1=e
\end{align*}
The graph \(f(x)=\sin x\) has horizontal tangent lines precisely at those points where \(\cos x=0\text{.}\) This must be true, since \(\diff{}{x}\{\sin x\}=\cos x\text{:}\) where the derivative of sine is zero, cosine itself is zero.
The graph \(f(x)=\sin x\) has maximum slope at those points where \(\cos x\) has a maximum. This makes sense, because \(f'(x)=\cos x\text{:}\) the maximum values of the slop of sine correspond to the maximum values of cosine.
\(f'(x)=\cos x  \sin x\text{,}\) so \(f'(x)=0\) precisely when \(\sin x = \cos x\text{.}\) This happens at \(\pi/4\text{,}\) but it also happens at \(\pi/4\) and \(5\pi/4\text{.}\) By looking at the unit circle, it is clear that \(\sin x = \cos x\) whenever \(x = \frac{\pi}{4}+\pi n\) for some integer \(n\text{.}\)
Solution 1: \(f(x)=\sin^2x+\cos^2x=1\text{,}\) so \(f'(x)=\diff{}{x}\{1\}=0\text{.}\)
Solution 2: Using the formula for the derivative of a squared function,
\begin{gather*}
f'(x)=2\sin x \cos x + 2\cos x( \sin x)=2\sin x \cos x  2 \sin x \cos x =0.
\end{gather*}
\begin{align*}
f'(x)&=\dfrac{(\cos x + \tan x)(2\cos x + 3 \sec^2 x)(2\sin x+3\tan x)(\sin x + \sec^2 x)}{(\cos x + \tan x)^2}\\
&=\frac{2\cos^2x+3\cos x \sec^2 x + 2 \cos x \tan x + 3 \tan x \sec^2 x}{(\cos x + \tan x)^2}\\
&\qquad
+\frac{2\sin^2x2\sin x \sec^2x+3\sin x \tan x 3\tan x \sec^2 x}{(\cos x + \tan x)^2}\\
&=\frac{2+3 \sec x + 2 \sin x 2\tan x \sec x+3\sin x \tan x }{(\cos x + \tan x)^2}
\end{align*}
We don't know how to differentiate this function as it is written, but an identity helps us. Since \(\sin\left(\frac{\pi}{2}\theta \right)=\cos \theta\text{,}\) we see \(f'(\theta)=\diff{}{\theta}\{\cos \theta\}=\sin(\theta)\text{.}\)
The only spot to worry about is when \(x=0\text{.}\) For \(f(x)\) to be differentiable, it must be continuous, so first find the value of \(b\) that makes \(f\) continuous at \(x=0\text{.}\) Then, find the value of \(a\) that makes the derivatives from the left and right of \(x=0\) equal to each other.
In order for \(f\) to be differentiable at \(x=0\text{,}\) it must also be continuous at \(x=0\text{.}\) This forces \(\cos(x)\big_{x=0}=\big[ax+b\big]_{x=0}\) or \(b=1\text{.}\) In order for \(f\) to be differentiable at \(x=0\text{,}\) the left hand derivative of \(ax+b\) at \(x=0\text{,}\) which is \(a\text{,}\) must be the same as the right hand derivative of \(\cos(x)\) at \(x=0\text{,}\) which is \(\sin(x)\big_{x=0}=0\text{.}\) So, we need \(a=0\) and \(b=1\text{.}\)
We compute the derivative of \(\cos(x)+2x\) as being \(\sin(x)+2\text{,}\) which evaluated at \(x=\frac{\pi}{2}\) yields \(1+2=1\text{.}\) Since we also compute \(\cos(\pi/2)+2(\pi/2)=0+\pi\text{,}\) then the equation of the tangent line is
\begin{gather*}
y  \pi = 1\cdot (x\pi/2).
\end{gather*}
This limit represents the derivative computed at \(x=\pi/3\) of the function \(f(x)=\cos x\text{.}\) Since the derivative of \(f(x)\) is \(\sin x\text{,}\) then its value at \(x=\pi/3\) is exactly \(\sqrt{3}/2\text{.}\)
In order for the function \(f(x)\) to be continuous at \(x=0\text{,}\) the left half formula \(ax+b\) and the right half formula \(\dfrac{6\cos x}{2+\sin x+\cos x}\) must match up at \(x=0\text{.}\) This forces
In order for the derivative \(f'(x)\) to exist at \(x=0\text{,}\) the limit \(\ds\lim_{h \rightarrow 0} \dfrac{f(h)f(0)}{h}\) must exist. In particular, the limits \(\ds\lim_{h \rightarrow 0^} \dfrac{f(h)f(0)}{h}\) and \(\ds\lim_{h \rightarrow 0^+} \dfrac{f(h)f(0)}{h}\) must exist and be equal to each other.
When \(h \to 0^\text{,}\) this means \(h \lt 0\text{,}\) so \(f(h)=ah+b=ah+2\text{.}\) So:
In order for \(f'(x)\) to exist, \(f(x)\) has to exist. We already know that \(\tan x \) does not exist whenever \(x=\frac{\pi}{2}+n\pi\) for any integer \(n\text{.}\) If we look a little deeper, since \(\tan x = \frac{\sin x}{\cos x}\text{,}\) the points where tangent does not exist correspond exactly to the points where cosine doesn't exist.
From its graph, tangent looks like a smooth curve over its domain, so we might guess that everywhere tangent is defined, its derivative is defined. We can check this: \(f'(x) = \sec^2 x = \left(\frac{1}{\cos x}\right)^2\text{.}\) Indeed, wherever \(\cos x\) is nonzero, \(f'\) exists.
So, \(f'(x)\) exists for all values of \(x\) except when \(x=\frac{\pi}{2}+n\pi\) for some integer \(n\text{.}\)
which is welldefined unless \(x^2+x6=0\text{.}\) We solve \(x^2+x6=(x2)(x+3)=0,\) and get \(x=2\) and \(x=3\text{.}\) So, the function is differentiable for all real values \(x\) except for \(x=2\) and for \(x=3\text{.}\)
which is welldefined unless \(x^2+x6=0\text{.}\) We solve \(x^2+x6=(x2)(x+3)=0,\) and get \(x=2\) and \(x=3\text{.}\) So, the function is differentiable for all real values \(x\) except for \(x=2\) and for \(x=3\text{.}\)
which is welldefined unless \(\sin x = 0\text{.}\) This happens when \(x\) is an integer multiple of \(\pi\text{.}\) So, the function is differentiable for all real values \(x\) except \(x=n\pi,\text{,}\) where \(n\) is any integer.
which is welldefined unless \(\sin x = 0\text{.}\) This happens when \(x\) is an integer multiple of \(\pi\text{.}\) So, the function is differentiable for all real values \(x\) except \(x=n\pi,\text{,}\) where \(n\) is any integer.
We compute the derivative of \(\tan(x)\) as being \(\sec^2(x)\text{,}\) which evaluated at \(x=\frac{\pi}{4}\) yields \(2\text{.}\) Since we also compute \(\tan(\pi/4)=1\text{,}\) then the equation of the tangent line is
\begin{gather*}
y  1 = 2\cdot (x\pi/4).
\end{gather*}
We compute the derivative \(y' = \cos(x)\sin(x)+e^x\text{,}\) which evaluated at \(x=0\) yields \(10+1 = 2\text{.}\) Since we also compute \(y(0)=0+1+1=2\text{,}\) the equation of the tangent line is
We are asked to solve \(f'(x)=0\text{.}\) That is, \(e^x[\sin x + \cos x]=0\text{.}\) Since \(e^x\) is always positive, that means we need to find all points where \(\sin x + \cos x =0\text{.}\) That is, we need to find all values of \(x\) where \(\sin x =  \cos x\text{.}\) Looking at the unit circle, we see this happens whenever \(x = \frac{3\pi}{4}+n\pi\) for any integer \(n\text{.}\)
Recall \(x=\left\{\begin{array}{rl}
x&x\ge 0\\
x&x \lt 0
\end{array}\right.\text{.}\) To determine whether \(h(x)\) is differentiable at \(x=0\text{,}\) use the definition of the derivative.
As usual, when dealing with the absolute value function, we can make things a little clearer by splitting it up into two pieces.
\begin{align*}
x&=\left\{\begin{array}{rl}
x&x\ge 0\\
x&x \lt 0
\end{array}\right.\\
\end{align*}
So,
\begin{align*}
\sinx&=\left\{\begin{array}{rl}
\sin x&x\ge 0\\
\sin(x)&x \lt 0
\end{array}\right.
=\left\{\begin{array}{rl}
\sin x&x\ge 0\\
\sin x&x \lt 0
\end{array}\right.\\
\end{align*}
where we used the identity \(\sin(x)=\sin x\text{.}\) From here, it's easy to see \(h'(x)\) when \(x\) is anything other than zero.
\begin{align*}
\diff{}{x}\{\sinx\}&=\left\{\begin{array}{rl}
\cos x&x \gt 0\\
??&x=0\\
\cos x&x \lt 0
\end{array}\right.\\
\end{align*}
To decide whether \(h(x)\) is differentiable at \(x=0\text{,}\) we use the definition of the derivative. One word of explanation: usually in the definition of the derivative, \(h\) is the tiny “change in \(x\)” that is going to zero. Since \(h\) is the name of our function, we need another letter to stand for the tiny change in \(x\text{,}\) the size of which is tending to zero. We chose \(t\text{.}\)
\begin{align*}
\lim_{t \to 0} \frac{h(t+0)h(0)}{t}&=\lim_{t \to 0}\frac{\sint}{t}\\
\end{align*}
We consider the behaviour of this function to the left and right of \(t=0\text{:}\)
\begin{align*}
\frac{\sin t}{t}&=\left\{\begin{array}{ll}
\frac{\sin t}{t} & t\ge 0\\
\frac{\sin (t)}{t} & t \lt 0
\end{array}\right.
=\left\{\begin{array}{ll}
\frac{\sin t}{t} & t\ge 0\\
\frac{\sin t}{t} & t \lt 0
\end{array}\right.\\
\end{align*}
Since we're evaluating the limit as \(t\) goes to zero, we need the fact that \(\ds\lim_{t \to 0} \dfrac{\sin t}{t}=1\text{.}\) We saw this in Section 2.7, but also we know enough now to evaluate it another way. Using the definition of the derivative:
\begin{align*}
\lim_{t \to 0}\frac{\sin t}{t}&=\lim_{t \to 0}\frac{\sin (t+0)\sin (0)}{t}=\left.\diff{}{x}\{\sin x\}\right_{t=0}=\cos 0=1\\
\end{align*}
At any rate, since we know \(\ds\lim_{t \to 0\dfrac{\sin t}{t}=1\text{,}\) then:
\begin{align*}
\lim_{t \to 0^+} \frac{h(t+0)h(0)}{t}&=\lim_{t \to 0^+}\frac{\sin t}{t}=1\qquad
\lim_{t \to 0^} \frac{h(t+0)h(0)}{t}=\lim_{t \to 0^}\frac{\sin t}{t}=1\\
\end{align*}
So, since the onesided limits disagree,
\begin{align*}
\lim_{t \to 0} \frac{h(t+0)h(0)}{t}&=DNE\\
\end{align*}
so \(h(x)\) is not differentiable at \(x=0\text{.}\) Therefore,
\begin{align*}
h'(x)&=\left\{\begin{array}{rl}
\cos x&x \gt 0\\
\cos x&x \lt 0
\end{array}\right.
\end{align*}
Statement 2.8.8.26.i is false, since \(f(0)=0\text{.}\) Statement 2.8.8.26.iv cannot hold, since a function that is differentiable is also continuous.
Since \(\ds\lim_{x\rightarrow 0+}\frac{\sin x}{x}=1\) (we saw this in Section 2.8 ),
\begin{align*}
\lim_{x\rightarrow 0+}f(x)&=\lim_{x\rightarrow 0+}\frac{\sin x}{\sqrt{x}}\\
&=\lim_{x\rightarrow 0+}\sqrt{x}\frac{\sin x}{x}\\
&=0\cdot 1=0\\
\end{align*}
So \(f\) is continuous at \(x=0\text{,}\) and so Statement 2.8.8.26.ii does not hold. Now let's consider \(f'(x)\)
\begin{align*}
\lim_{x\rightarrow 0+}\frac{f(x)f(0)}{x}
&=\lim_{x\rightarrow 0+}\frac{\frac{\sin x}{\sqrt{x}}0}{x}\\
&=\lim_{x\rightarrow 0+}\frac{1}{\sqrt{x}}\frac{\sin x}{x}=+\infty\\
\end{align*}
Therefore, using the definition of the derivative,
\begin{align*}
f'(0)&=\lim_{x \to 0}\frac{f(x)f(0)}{x}\quad\mbox{ if it exists, but}\\
\lim_{x \to 0}\frac{f(x)f(0)}{x}&=DNE
\end{align*}
since one of the onesided limits does not exist. So \(f\) is continuous but not differentiable at \(x=0\text{.}\) The correct statement is 2.8.8.26.iii.
In this chapter, we learned \(\ds\lim_{x \to 0}\dfrac{\sin x}{x}=1\text{.}\) If you divide the numerator and denominator by \(x^5\text{,}\) you can make use of this knowledge.
Recall that \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x} =1\text{.}\) In order to take advantage of this knowledge, we divide the numerator and denominator by \(x^5\) (because \(5\) is the power of sine in the denominator, and a denominator that goes to zero generally makes a limit harder).
\begin{align*}
\lim_{x\rightarrow 0} \dfrac{\sin x^{27}+2x^5 e^{x^{99}}}{\sin^5 x}
&=\lim_{x\rightarrow 0} \dfrac{\dfrac{\sin x^{27}}{x^{5}}+2 e^{x^{99}}}
{\left(\dfrac{\sin x}{x}\right)^5}\\
\end{align*}
Now the denominator goes to 1, which is nice, but we need to take care of the fraction \(\dfrac{\sin^{27}x}{x^5}\) in the numerator. This fraction isn't very familiar, but we know what happens to the fraction \(\dfrac{\sin^{27}x}{x^{27}}=\left(\dfrac{\sin x}{x}\right)^{27}\text{:}\)
\begin{align*}
&=\lim_{x\rightarrow 0} \dfrac{x^{22}\dfrac{\sin x^{27}}{x^{27}}+2 e^{x^{99}}}
{\left(\dfrac{\sin x}{x}\right)^5}
=\dfrac{0\times 1+2\times 1}{1^5}
=2
\end{align*}
2.9.4.1.a More urchins means less kelp, and fewer urchins means more kelp. This means kelp and urchins are negatively correlated, so \(\diff{K}{U} \lt 0\text{.}\)
If you aren't sure why that is, we give a more detailed explanation here, using the definition of the derivative. When \(h\) is a positive number, \(U+h\) is greater than \(U\text{,}\) so \(K(U+h)\) is less than \(U\text{,}\) hence \(K(U+h)K(U) \lt 0\text{.}\) Therefore:
2.9.4.1.b More otters means fewer urchins, and fewer otters means more urchins. So, otters and urchins are negatively correlated: \(\diff{U}{O} \lt 0\text{.}\)
2.9.4.1.c Using the chain rule, \(\diff{K}{O} = \diff{K}{U}\cdot\diff{U}{O}\text{.}\) Parts 2.9.4.1.a and 2.9.4.1.b tell us both these derivatives are negative, so their product is positive: \(\diff{K}{O} \gt 0\text{.}\)
We can also see that \(\diff{K}{O} \gt 0\) by thinking about the relationships as described. When the otter population increases, the urchin population decreases, so the kelp population increases. That means when the otter population increases, the kelp population also increases, so kelp and otters are positively correlated. The chain rule is a formal version of this kind of reasoning.
If \(g(x)=\cos x\) and \(h(x)=5x+3\text{,}\) then \(f(x)=g(h(x))\text{.}\) So we apply the chain rule, with “outside” function \(\cos x\) and “inside” function \(5x+3\text{.}\)
You can expand this into a polynomial, but it's easier to use the chain rule. If \(g(x)=x^5\text{,}\) and \(h(x)=x^2+2\text{,}\) then \(f(x)=g(h(x))\text{.}\)
You can expand this into a polynomial, but it's easier to use the chain rule. If \(g(k)=k^{17}\text{,}\) and \(h(k)=4k^4+2k^2+1\text{,}\) then \(T(k)=g(h(k))\text{.}\)
If we define \(g(x)=\sqrt{x}\) and \(h(x)=\dfrac{x^2+1}{x^21}\text{,}\) then \(f(x)=g(h(x))\text{.}\) To differentiate the square root function: \(\ds\diff{}{x}\{\sqrt{x}\}=\ds\diff{}{x}\left\{x^{1/2}\right\}=\dfrac{1}{2}x^{1/2}=\dfrac{1}{2\sqrt{x}}\text{.}\)
If we let \(g(x)=e^x\) and \(h(x)=\cos(x^2)\text{,}\) then \(f(x)=g(h(x))\text{,}\) so \(f'(x)=g'(h(x))\cdot h'(x)\text{.}\)
\begin{align*}
f'(x)&=e^{\textcolor{red}{\cos(x^2)}}\cdot \diff{}{x}\{\textcolor{red}{\cos(x^2)}\}\\
\end{align*}
In order to evaluate \(\diff{}{x}\{{\cos(x^2)}\}\text{,}\) we'll need the chain rule again.
\begin{align*}
&=e^{\cos(x^2)}\cdot [\sin(\textcolor{orange}{x^2})]\cdot\diff{}{x}\{\textcolor{orange}{x^2}\}\\
&=e^{\cos(x^2)}\cdot \sin(x^2)\cdot 2x
\end{align*}
First, we manipulate our function to make it easier to differentiate:
\begin{align*}
f(x)&=x^{2}+(x^21)^{1/2}\\
\end{align*}
Now, we can use the power rule to differentiate \(\dfrac{1}{x^2}\text{.}\) This will be easier than differentiating \(\dfrac{1}{x^2}\) using quotient rule, but if you prefer, quotient rule will also work.
\begin{align*}
f'(x)&=2x^{3}+\frac{1}{2}(\textcolor{red}{x^21})^{1/2}\cdot\diff{}{x}\{\textcolor{red}{x^21}\}\\
&=2x^{3}+\frac{1}{2}({x^21})^{1/2}(2x)\\
&=\frac{2}{x^3}+\frac{x}{\sqrt{x^21}}
\end{align*}
The function \(f(x)\) is only defined when \(x \neq 0\) and when \(x^21 \geq 0\text{.}\) That is, when \(x\) is in \((\infty,1] \cup [1,\infty)\text{.}\) We have an added restriction on the domain of \(f'(x)\text{:}\) \(x^21\) must not be zero. So, the domain of \(f'(x)\) is \((\infty,1)\cup(1,\infty)\text{.}\)
If we let \(g(x)=\sec x\) and \({h(x)}=e^{2x+7}\text{,}\) then \(f(x)=g(h(x))\text{,}\) so by the chain rule, \(f'(x)=g'(h(x))\cdot h'(x)\text{.}\) However, in order to evaluate \(h'(x)\text{,}\) we'll need to use the chain rule again.
If we let \(g(x)=\sec x\) and \(h(x)=e^{2x+7}\text{,}\) then \(f(x)=g(h(x))\text{,}\) so by the chain rule, \(f'(x)=g'(h(x))\cdot h'(x)\text{.}\) Since \(g'(x)=\sec x \tan x\text{:}\)
\begin{align*}
f'(x)&=g'(h(x))\cdot h'(x)\\
&=\sec(h(x))\tan(h(x)) \cdot h'(x)\\
&=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \diff{}{x}\left\{e^{2x+7}\right\}\\
\end{align*}
Here, we need the chain rule again:
\begin{align*}
&=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \left[e^{\textcolor{red}{2x+7}}\cdot \diff{}{x}\{\textcolor{red}{2x+7}\}\right]\\
&=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \left[e^{2x+7}\cdot2\right]\\
&=2e^{2x+7}\sec(e^{2x+7})\tan(e^{2x+7})
\end{align*}
It is possible to start in on this problem with the product rule and then the chain rule, but it's easier if we simplify first. Since \(\tan^2x+1=\sec^2 x=\frac{1}{\cos^2x}\text{,}\) we see
for all values of \(x\) for which \(\cos x\) is nonzero. That is, \(f(x)=1\) for every \(x\) that is not an integer multiple of \(\pi/2\) (and \(f(x)\) is not defined when \(x\) is an integer multiple of \(\pi/2\)). Therefore, \(f'(x)=0\) for every \(x\) on which \(f\) exists, and in particular \(f'(\pi/4)=0\text{.}\) Also, \(f(\pi/4)=1\text{,}\) so the tangent line to \(f\) at \(x=\pi/4\) is the line with slope 0, passing through the point \((\pi/4,1)\text{:}\)
Velocity is the derivative of position with respect to time. So, the velocity of the particle is given by \(s'(t)\text{.}\) We need to find \(s'(t)\text{,}\) and determine when it is zero.
To differentiate, we us the chain rule.
\begin{align*}
s'(t)&=e^{\textcolor{red}{t^37t^2+8t}}\cdot\diff{}{t}\{\textcolor{red}{t^37t^2+8t}\}\\
&=e^{{t^37t^2+8t}}\cdot(3t^214t+8)\\
\end{align*}
To determine where this function is zero, we factor:
\begin{align*}
&=e^{{t^37t^2+8t}}\cdot(3t2)(t4)
\end{align*}
So, the velocity is zero when \(e^{{t^37t^2+8t}}=0\text{,}\) when \(3t2=0\text{,}\) and when \(t4=0\text{.}\) Since \(e^{{t^37t^2+8t}}\) is never zero, this tells us that the velocity is zero precisely when \(t=\frac{2}{3}\) or \(t=4\text{.}\)
The slope of the tangent line is the derivative. If we let \(f(x)=\tan x\) and \(g(x)=e^{x^2}\text{,}\) then \(f(g(x))=\tan(e^{x^2})\text{,}\) so \(y'=f'(g(x)) \cdot g'(x)\text{:}\)
\begin{align*}
y'&=\sec^2(\textcolor{red}{e^{x^2}})\cdot\diff{}{x}\{\textcolor{red}{e^{x^2}}\}\\
\end{align*}
We find ourselves once more in need of the chain rule:
\begin{align*}
&=\sec^2({e^{x^2}})\cdot{e^{\textcolor{red}{x^2}}}\diff{}{x}\{\textcolor{red}{x^2}\}\\
&=\sec^2(e^{x^2})\cdot e^{x^2}\cdot 2x\\
\end{align*}
Finally, we evaluate this derivative at the point \(x=1\text{:}\)
\begin{align*}
y'(1)&=\sec^2(e)\cdot e \cdot 2\\
&=2e\sec^2e
\end{align*}
\begin{align*}
y'&=\sin\big(\textcolor{red}{x^2+\sqrt{x^2+1}}\big)\cdot\diff{}{x}\left\{\textcolor{red}{x^2+\sqrt{x^2+1}}\right\}\\
&=\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\diff{}{x}\left\{\sqrt{x^2+1}\right\}\right)\\
\end{align*}
and find ourselves in need of chain rule a second time:
\begin{align*}
&=\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{1}{2\sqrt{\textcolor{red}{x^2+1}}}\cdot\diff{}{x}\left\{\textcolor{red}{x^2+1}\right\}\right)\\
&=\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{2x}{2\sqrt{x^2+1}}\right)
\end{align*}
Solution
\begin{align*}
y&=(1+x^2)\cos^2 x\\
\end{align*}
Using the product rule,
\begin{align*}
y'&=(2x)\cos^2 x + (1+x^2)\diff{}{x}\{\cos^2 x\}\\
\end{align*}
Here, we'll need to use the chain rule. Remember \(\cos^2 x = [\cos x]^2\text{.}\)
\begin{align*}
&=2x\cos^2x+(1+x^2) 2\textcolor{red}{\cos x} \cdot \diff{}{x}\{\textcolor{red}{\cos x}\}\\
&=2x\cos^2x+(1+x^2) 2\cos x \cdot (\sin x)\\
&=2x\cos^2x2(1+x^2) \sin x\cos x
\end{align*}
Let \(f(x)=xe^{(x^21)/2}=xe^{(1x^2)/2}\text{.}\) Then, using the product rule,
\begin{align*}
f'(x)&=e^{(1x^2)/2}+x\cdot\diff{}{x}\left\{e^{(1x^2)/2}\right\}\\
\end{align*}
Here, we need the chain rule:
\begin{align*}
&=e^{(1x^2)/2}+x\cdot e^{\textcolor{red}{(1x^2)/2}}\diff{}{x}\left\{\textcolor{red}{\frac{1}{2}(1x^2)}\right\}\\
&=e^{(1x^2)/2}+x\cdot e^{{(1x^2)/2}}\cdot (x)\\
&=(1x^2)e^{(1x^2)/2}
\end{align*}
There is no power of \(e\) that is equal to zero; so if the product above is zero, it must be that \(1x^2=0\text{.}\) This happens for \(x=\pm 1\text{.}\) On the curve, when \(x=1\text{,}\) \(y=1\text{,}\) and when \(x=1\text{,}\) \(y=1\text{.}\) So the points are \((1,1)\) and \((1,1)\text{.}\)
When \(t\ge1\text{,}\) \(\frac{1}{t}\) is between 0 and 1. Since \(\cos \theta\) is positive for \(0 \leq \theta \lt \pi/2\text{,}\) and \(\pi/2 \gt 1\text{,}\) we see that \(\cos\left(\frac{1}{t}\right)\) is positive for the entire domain of \(s(t)\text{.}\) Also, \(\frac{1}{t^2}\) is negative for the entire domain of the function. We conclude that \(s'(t)\) is negative for the entire domain of \(s(t)\text{,}\) so the particle is always moving in the negative direction.
The notation \(\cos^3(5x7)\) means \(\left[\cos(5x7)\right]^3\text{.}\) So, if \(g(x)=x^3\) and \(h(x)=\cos(5x7)\text{,}\) then \(g(h(x))=\left[\cos(5x+7)\right]^3=\cos^3(5x+7)\text{.}\)
We present two solutions: one where we dive right in and use the quotient rule, and another where we simplify first and use the product rule.
Solution 1: We begin with the quotient rule:
\begin{align*}
f'(x) &= \frac{\cos^3(5x7)\diff{}{x}\{e^x\}e^x\diff{}{x}\{\cos^3(5x7)\}}{\cos^6(5x7)}\\
&= \frac{\cos^3(5x7)e^xe^x\diff{}{x}\{\cos^3(5x7)\}}{\cos^6(5x7)}\\
\end{align*}
Now, we use the chain rule. Since \(\cos^3(5x7)=[\cos(5x7)]^3\text{,}\) our “outside” function is \(g(x)=x^3\text{,}\) and our “inside” function is \(h(x)=\cos(5x1)\text{.}\)
\begin{align*}
&= \frac{\cos^3(5x7)e^xe^x\cdot3\textcolor{red}{\cos}^2\textcolor{red}{(5x7)} \cdot \diff{}{x}\{\textcolor{red}{\cos(5x7)}\}}{\cos^6(5x7)}\\
\end{align*}
We need the chain rule again!
\begin{align*}
&= \frac{\cos^3(5x7)e^xe^x\cdot3{\cos}^2{(5x7)} \cdot[{\sin(\textcolor{red}{5x7})\cdot \diff{}{x}\{\textcolor{red}{5x7}\}}]}{\cos^6(5x7)}\\
&= \frac{\cos^3(5x7)e^xe^x\cdot3{\cos}^2{(5x7)} \cdot[{\sin({5x7})\cdot5}]}{\cos^6(5x7)}\\
\end{align*}
We finish by simplifying:
\begin{align*}
&= \frac{e^x\cos^2(5x7)\left(\cos(5x7)+15\sin(5x7)\right)}{\cos^6(5x7)}\\
&=e^x \frac{\cos(5x7)+15\sin(5x7)}{\cos^4(5x7)}\\
&=e^x(\sec^3(5x7)+15\tan(5x7)\sec^3(5x7))\\
&=e^x\sec^3(5x7)(1+15\tan(5x7))
\end{align*}
Solution 2: We simplify to avoid the quotient rule:
\begin{align*}
f(x)&=\dfrac{e^{x}}{\cos^3 (5x7)}\\
&=e^x\sec^3(5x7)\\
\end{align*}
Now we use the product rule to differentiate:
\begin{align*}
f'(x)&=e^x\sec^3(5x7)+e^x\diff{}{x}\{\sec^3(5x7)\}\\
\end{align*}
Here, we'll need the chain rule. Since \(\sec^3(5x7)=[\sec (5x7)]^3\text{,}\) our “outside” function is \(g(x)=x^3\) and our “inside” function is \(h(x)=\sec(5x7)\text{,}\) so that \(g(h(x))=[\sec(5x7)]^3=\sec^3(5x7)\text{.}\)
\begin{align*}
&=e^x\sec^3(5x7)+e^x\cdot3\;\textcolor{red}{\sec}^2\textcolor{red}{(5x7)} \cdot \diff{}{x}\{\textcolor{red}{\sec(5x7)}\}\\
\end{align*}
We need the chain rule again! Recall \(\diff{}{x}\{\sec x\}=\sec x \tan x\text{.}\)
\begin{align*}
&=e^x\sec^3(5x7)+e^x\cdot3\;{\sec}^2{(5x7)} \cdot {\sec(\textcolor{orange}{5x7})\tan(\textcolor{orange}{5x7})\cdot\diff{}{x}\{\textcolor{orange}{5x7}\}}\\
&=e^x\sec^3(5x7)+e^x\cdot3\;{\sec}^2{(5x7)} \cdot {\sec({5x7})\tan({5x7})\cdot 5}\\
\end{align*}
We finish by simplifying:
\begin{align*}
&=e^x\sec^3(5x7)(1+15\tan({5x7}))
\end{align*}
At time \(t\text{,}\) the particle is at the point \(\big(x(t),y(t)\big)\text{,}\) with \(x(t)=\cos t\) and \(y(t)=\sin t\text{.}\) Over time, the particle traces out a curve; let's call that curve \(y=f(x)\text{.}\) Then \(y(t) = f\big(x(t)\big)\text{,}\) so the slope of the curve at the point \(\big(x(t),y(t)\big)\) is \(f'\big(x(t)\big)\text{.}\) You are to determine the values of \(t\) for which \(f'\big(x(t)\big)=1\text{.}\)
At time \(t\text{,}\) the particle is at the point \(\big(x(t),y(t)\big)\text{,}\) with \(x(t)=\cos t\) and \(y(t)=\sin t\text{.}\) Over time, the particle traces out a curve; let's call that curve \(y=f(x)\text{.}\) Then \(y(t) = f\big(x(t)\big)\text{,}\) so the slope of the curve at the point \(\big(x(t),y(t)\big)\) is \(f'\big(x(t)\big)\text{.}\) You are to determine the values of \(t\) for which \(f'\big(x(t)\big)=1\text{.}\)
is \(1\) precisely when \(\sin t = \cos t\text{.}\) This happens whenever \(t = \frac{\pi}{4}\text{.}\)
Remark: the path traced by the particle is a semicircle. You can think about the point on the unit circle with angle t, or you can notice that \(x^2 + y^2 = \sin^2t + \cos^2t = 1\text{.}\)
Since \(f(0)=g(0)\text{,}\) and \(f'(x) \gt g'(x)\) for all \(x \gt 0\text{,}\) that means \(f\) and \(g\) start at the same place, but \(f\) always grows faster. Therefore, \(f(x) \gt g(x)\) for all \(x \gt 0\text{.}\)
Since \(f(0)=g(0)\text{,}\) and \(f'(x) \gt g'(x)\) for all \(x \gt 0\text{,}\) that means \(f\) and \(g\) start at the same place, but \(f\) always grows faster. Therefore, \(f(x) \gt g(x)\) for all \(x \gt 0\text{.}\)
This is a long, nasty problem, but it doesn't use anything you haven't seen before. Be methodical, and break the question into as many parts as you have to. At the end, be proud of yourself for your problemsolving abilities and tenaciousness!
Answer
\begin{align*}
f'(x)&=
\frac{1}{3}\left(
\dfrac{ \sqrt{x^39} \tan x }{e^{\csc x^2}}
\right)^{\frac{2}{3}}\cdot\\
&\left(\frac{ \sqrt{x^39}\tan x {(2x)e^{\csc x^2}\csc(x^2)\cot(x^2)}e^{\csc x^2}{\left(\frac{3x^2\tan x}{2\sqrt{{x^39}}}+\sqrt{x^39}\sec^2 x\right)}}{(\tan^2 x)(x^39) }\right)
\end{align*}
Solution
\begin{align*}
f(x)&=\sqrt[3]{\dfrac{e^{\csc x^2}}{ \sqrt{x^39} \tan x }}\\
&=\left({\dfrac{e^{\csc x^2}}{ \sqrt{x^39} \tan x }}\right)^{\frac{1}{3}}\\
\end{align*}
To begin the differentiation, we can choose our “outside” function to be \(g(x)=x^{\frac{1}{3}}\text{,}\) and our “inside” function to be \(h(x)=\dfrac{e^{\csc x^2}}{ \sqrt{x^39} \tan x }\text{.}\) Then \(f(x)=g(h(x))\text{,}\) so \(f'(x)=g'(h(x))\cdot h'(x)=\frac{1}{3}(h(x))^{\frac{2}{3}}h'(x)\text{:}\)
\begin{align*}
f'(x)&=\frac{1}{3}\left(\textcolor{red}{\dfrac{e^{\csc x^2}}{ \sqrt{x^39} \tan x }}\right)^{\frac{2}{3}}\cdot\diff{}{x}\left\{\textcolor{red}{\dfrac{e^{\csc x^2}}{ \sqrt{x^39} \tan x }}\right\}\\
&=\frac{1}{3}
\left(
\dfrac{ \sqrt{x^39} \tan x }{e^{\csc x^2}}
\right)^{\frac{2}{3}}
\cdot
\diff{}{x}\left\{\textcolor{red}{\dfrac{e^{\csc x^2}}{ \sqrt{x^39} \tan x }}\right\}\\
\end{align*}
This leads us to use the quotient rule:
\begin{align*}
&=\frac{1}{3}\left(
\dfrac{ \sqrt{x^39} \tan x }{e^{\csc x^2}}
\right)^{\frac{2}{3}}
\left(\frac{ \sqrt{x^39}\tan x \diff{}{x}\left\{e^{\csc x^2}\right\}e^{\csc x^2}\diff{}{x}\left\{\sqrt{x^39}\tan x \right\}}{(\tan^2 x)(x^39) }\right)\\
\end{align*}
Let's figure out those two derivatives on their own, then plug them in. Using the chain rule twice:
\begin{align*}
\diff{}{x}\left\{e^{\csc x^2}\right\}&=e^{\textcolor{red}{\csc x^2}}\diff{}{x}\left\{\textcolor{red}{\csc x^2}\right\}=e^{\csc x^2}\cdot (\csc(\textcolor{orange}{x^2})\cot(\textcolor{orange}{x^2}))\cdot\diff{}{x}\{\textcolor{orange}{x^2}\}\\
&=2xe^{\csc x^2}\csc(x^2)\cot(x^2)\\
\end{align*}
For the other derivative, we start with the product rule, then chain:
\begin{align*}
\diff{}{x}\left\{ \sqrt{x^39} \tan x \right\}&=
\diff{}{x}\left\{\sqrt{x^39}\right\}\cdot\tan x+\sqrt{x^39}\sec^2 x\\
&=\frac{1}{2\sqrt{\textcolor{red}{x^39}}}\diff{}{x}\left\{\textcolor{red}{x^39}\right\}\cdot \tan x+\sqrt{x^39}\sec^2 x\\
&=\frac{3x^2 \tan x}{2\sqrt{{x^39}}}+\sqrt{x^39}\sec^2 x\\
\end{align*}
Now, we plug these into our equation for \(f'(x)\text{:}\)
\begin{align*}
f'(x)&=\frac{1}{3}\left(
\dfrac{ \sqrt{x^39} \tan x }{e^{\csc x^2}}
\right)^{\frac{2}{3}}
\left(\frac{ \sqrt{x^39}\tan x \textcolor{blue}{\diff{}{x}\left\{e^{\csc x^2}\right\}}e^{\csc x^2}\textcolor{blue}{\diff{}{x}\left\{\sqrt{x^39}\tan x \right\}}}{(\tan^2 x)(x^39) }\right)\\
&=\frac{1}{3}\left(
\dfrac{ \sqrt{x^39} \tan x }{e^{\csc x^2}}
\right)^{\frac{2}{3}}\cdot\\
&\left(\frac{ \sqrt{x^39}\tan x \textcolor{blue}{(2x)e^{\csc x^2}\csc(x^2)\cot(x^2)}e^{\csc x^2}\textcolor{blue}{\left(\frac{3x^2\tan x}{2\sqrt{{x^39}}}+\sqrt{x^39}\sec^2 x\right)}}{(\tan^2 x)(x^39) }\right)
\end{align*}
The particle traces the curve \(y=1x^2\) restricted to domain \([1,1]\text{.}\) At \(t=0\text{,}\) the particle is at the top of the curve, \((1,0)\text{.}\) Then it moves to the right, and goes back and forth along the curve, repeating its path every \(2\pi\) units of time.
2.9.4.35.a The table below gives us a number of points on our graph, and the times they occur.
\(t\)
\((\sin t,\cos^2 t)\)
\hline \(0\)
\((0,1)\)
\(\pi/4\)
\((\frac{1}{\sqrt{2}},\frac{1}{2})\)
\(\pi/2\)
\((1,0)\)
\(3\pi/4\)
\((\frac{1}{\sqrt{2}},\frac{1}{2})\)
\(\pi\)
\((0,1)\)
\(5\pi/4\)
\((\frac{1}{\sqrt{2}},\frac{1}{2})\)
\(3\pi/2\)
\((1,0)\)
\(7\pi/4\)
\((\frac{1}{\sqrt{2}},\frac{1}{2})\)
\(2\pi\)
\((0,1)\)
These points will repeat with a period of \(2\pi\text{.}\) With this information, we have a pretty good idea of the particle's motion:
The particle traces out an arc, pointing down. It starts at \(t=0\) at the top part of the graph at \((1,0)\text{,}\) then is moves to the right until it hits \((1,0)\) at time \(t=\pi/2\text{.}\) From there it reverses direction and moves along the curve to the left, hitting the top at time \(t=\pi\) and reaching \((1,0)\) at time \(t=3\pi/2\text{.}\) Then it returns to the top at \(t=2\pi\) and starts again.
So, it starts at the top of the curve, then moves back for forth along the length of the curve. If goes right first, and repeats its cycle every \(2\pi\) units of time.
2.9.4.35.b Let \(y=f(x)\) be the curve the particle traces in the \(xy\)plane. Since \(x\) is a function of \(t\text{,}\) \(y(t)=f(x(t))\text{.}\) What we want to find is \(\ds\diff{f}{x}\) when \(t=\left(\dfrac{10\pi}{3}\right)\text{.}\) Since \(\ds\diff{f}{x}\) is a function of \(x\text{,}\) we note that when \(t=\left(\dfrac{10\pi}{3}\right)\text{,}\) \(x=\sin\left(\dfrac{10\pi}{3}\right)=\sin\left(\dfrac{4\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\text{.}\) So, the quantity we want to find (the slope of the tangent line to the curve \(y=f(x)\) traced by the particle at the time \(t=\left(\dfrac{10\pi}{3}\right)\) is given by \(\ds\diff{f}{x}\left(\dfrac{\sqrt{3}}{2}\right)\text{.}\)
Using the chain rule:
\begin{align*}
y(t)&=f(x(t))\\
\diff{y}{t}=\diff{}{t}\left\{f(x(t))\right\}&=\diff{f}{x}\cdot\diff{x}{t}\\
\mbox{so, }\qquad\diff{f}{x}&=\diff{y}{t}\div \diff{x}{t}\\
\end{align*}
Using \(y(t)=\cos^2 t\) and \(x(t)=\sin t\text{:}\)
\begin{align*}
\diff{f}{x}&=\left(2\cos t \sin t \right)\div\left( \cos t \right)=2\sin t=2x\\
\end{align*}
So, when \(t=\dfrac{10\pi}{3}\) and \(x=\dfrac{\sqrt{3}}{2},\)
\begin{align*}
\diff{f}{x}\left(\dfrac{\sqrt{3}}{2}\right)&=2\cdot\frac{\sqrt{3}}{2}=\sqrt{3}.
\end{align*}
Remark: The standard way to write this problem is to omit the notation \(f(x)\text{,}\) and let the variable \(y\) stand for two functions. When \(t\) is the variable, \(y(t)=\cos^2t\) gives the \(y\)coordinate of the particle at time \(t\text{.}\) When \(x\) is the variable, \(y(x)\) gives the \(y\)coordinate of the particle given its position along the \(x\)axis. This is an abuse of notation, because if we write \(y(1)\text{,}\) it is not clear whether we are referring to the \(y\)coordinate of the particle when \(t=1\) (in this case, \(y=\cos^2 (1) \approx 0.3\)), or the \(y\)coordinate of the particle when \(x=1\) (in this case, looking at our table of values, \(y=0\)). Although this notation is not strictly “correct,” it is very commonly used. So, you might see a solution that looks like this:
The slope of the curve is \(\ds\diff{y}{x}\text{.}\) To find \(\ds\diff{y}{x}\text{,}\) we use the chain rule:
\begin{align*}
\diff{y}{t}&=\diff{y}{x}\cdot\diff{x}{t}\\
\diff{}{t}\left\{\cos^2 t\right\}&=\diff{y}{x}\cdot\diff{}{t}\{\sin t\}\\
2\cos t \sin t &= \diff{y}{x} \cdot \cos t\\
\diff{y}{x}&=2\sin t\\
\end{align*}
So, when \(t=\dfrac{10\pi}{3}\text{,}\)
\begin{align*}
\diff{y}{x}&=2\sin\left(\frac{10\pi}{3}\right)=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}.
\end{align*}
In this case, it is up to the reader to understand when \(y\) is used as a function of \(t\text{,}\) and when it is used as a function of \(x\text{.}\) This notation (using \(y\) to be two functions, \(y(t)\) and \(y(x)\)) is actually the accepted standard, so you should be able to understand it.
Each speaker produces 3dB of noise, so if \(P\) is the power of one speaker, \(3=V(P)=10\log_{10}\left(\frac{P}{S}\right)\text{.}\) Use this to find \(V(10P)\) and \(V(100P)\text{.}\)
We are given that one speaker produces 3dB. So if \(P\) is the power of one speaker,
\begin{align*}
3=V(P)&=10\log_{10}\left(\frac{P}{S}\right).\\
\end{align*}
So, for ten speakers:
\begin{align*}
V(10P)&=10\log_{10}\left(\frac{10P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(10\right)\\
&=3+10(1)=13 \mathrm{dB}\\
\end{align*}
and for one hundred speakers:
\begin{align*}
V(100P)&=10\log_{10}\left(\frac{100P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(100\right)\\
&=3+10(2)=23 \mathrm{dB}
\end{align*}
From our logarithm rules, we know that when \(y\) is positive, \(\log (y^2)=2\log y\text{.}\) However, the expression \(\cos x\) does not always take on positive values, so (a) is not correct. (For instance, when \(x=\pi\text{,}\) \(\log(\cos^2 x)=\log(\cos^2\pi)=\log\left((1)^2\right) = \log(1)=0\text{,}\) while \(2\log (\cos \pi)=2\log(1)\text{,}\) which does not exist.)
Because \(\cos^2 x\) is never negative, we notice that \(\cos^2 x = \cos x^2\text{.}\) When \(\cos x\) is nonzero, \(\cos x\) is positive, so our logarithm rules tell us \(\log\left(\cos x^2\right) =2\log\cos x \text{.}\) When \(\cos x\) is exactly zero, then both \(\log(\cos^2x)\) and \(2\log\cos x\) do not exist. So, \(\log(\cos^2x) = 2\log\cos x\text{.}\)
Don't be fooled by a common mistake: \(\log(x^2+x)\) is not the same as \(\log(x^2)+\log x\text{.}\) We differentiate using the chain rule: \(\ds\diff{}{x}\left\{\log(x^2+x)\right\}=\dfrac{1}{x^2+x}\cdot(2x+1) = \dfrac{2x+1}{x^2+x}\text{.}\)
We know the derivative of the natural logarithm (base \(e\)), so we use the basechange formula:
\begin{align*}
f(x)=\log_{10}x&=\frac{\log x}{\log 10}\\
\end{align*}
Since \(\log 10\) is a constant:
\begin{align*}
f'(x)&=\frac{1}{x\log 10}.
\end{align*}
Solution 1: Using the quotient rule,
\begin{equation*}
y'=\frac{x^3\frac{1}{x}(\log x)\cdot 3x^2}{x^6}=
\frac{x^23x^2\log x}{x^6}=\frac{13\log x}{x^4}.
\end{equation*}
Solution 2: Using the product rule with \(y=\log x \cdot x^{3}\text{,}\)
\begin{equation*}
y'=\frac{1}{x}x^{3}+\log x \cdot(3)x^{4}=x^{4}(13\log x)
\end{equation*}
Remark: the domain of the function \(\log(\sec \theta)\) is those values of \(\theta\) for which \(\sec\theta\) is positive: so, the intervals \(\left(\left(2n\frac{1}{2}\right)\pi,\left(2n+\frac{1}{2}\right)\pi\right)\) where \(n\) is any integer. Certainly the tangent function has a larger domain than this, but outside the domain of \(\log(\sec \theta)\text{,}\) \(\tan \theta\) is not the derivative of \(\log(\sec \theta)\text{.}\)
Remark: Although we have a logarithm in the exponent, we can't cancel. The expression \(e^{\cos (\log x)}\) is not the same as the expression \(x^{\cos x}\text{,}\) or \(\cos x\text{.}\)
Solution
\begin{align*}
y&=\log(x^2+\sqrt{x^4+1})\\
\end{align*}
So, we'll need the chain rule:
\begin{align*}
y'&=\frac{\diff{}{x}\left\{\textcolor{red}{x^2+\sqrt{x^4+1}}\right\}}{\textcolor{red}{x^2+\sqrt{x^4+1}}}\\
&=\frac{2x+\diff{}{x}\left\{\sqrt{x^4+1}\right\}}{x^2+\sqrt{x^4+1}}\\
\end{align*}
We need the chain rule again:
\begin{align*}
&=\frac{2x+\frac{\diff{}{x}\left\{\textcolor{red}{x^4+1}\right\}}{2\sqrt{\textcolor{red}{x^4+1}}}}{x^2+\sqrt{x^4+1}}\\
&=\frac{2x+\frac{4x^3}{2\sqrt{x^4+1}}}{x^2+\sqrt{x^4+1}}.
\end{align*}
Remark: it looks strange to see a negative sign in the argument of a square root. Since the cosine function always gives values that are at most 1, \(\log(\cos x)\) is always negative or zero over its domain. So, \(\sqrt{\log(\cos x)}\) is only defined for the points where \(\cos x=1\) (and so \(\log(\cos x) = 0\)this isn't a very interesting function! In contrast, \(\log(\cos x)\) is always positive or zero over its domainand therefore we can always take its square root.
In the text, we saw that \(\ds\diff{}{x}\left\{a^x\right\}=a^x\log a\) for any constant \(a\text{.}\) So, \(\ds\diff{}{x}\left\{\pi^x\right\}=\pi^x\log \pi\text{.}\)
By the power rule, \(\ds\diff{}{x}\left\{x^{\pi}\right\}=\pi x^{\pi1}\text{.}\)
Remark: we had to use two different rules for the two different terms in \(g(x)\text{.}\) Although the functions \(\pi^x\) and \(x^\pi\) look superficially the same, they behave differently, as do their derivatives. A function of the form \((\mbox{constant})^{x}\) is an exponential function and not eligible for the power rule, while a function of the form \(x^{\mbox{constant}}\) is exactly the class of function the power rule applies to.
You'll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
We have the power rule to tell us the derivative of functions of the form \(x^n\text{,}\) where \(n\) is a constant. However, here our exponent is not a constant. Similarly, in this section we learned the derivative of functions of the form \(a^x\text{,}\) where \(a\) is a constant, but again, our base is not a constant! Although the result \(\ds\diff{}{x} a^x=a^x\log a\) is not what we need, the method used to differentiate \(a^x\) will tell us the derivative of \(x^x\text{.}\)
We'll set \(g(x)=\log(x^x)\text{,}\) because now we can use logarithm rules to simplify:
\begin{align*}
g(x)=\log(f(x))&=x\log x\\
\end{align*}
Now, we can use the product rule to differentiate the right side, and the chain rule to differentiate \(\log(f(x))\text{:}\)
\begin{align*}
g'(x)=\frac{f'(x)}{f(x)}&=\log x +x\frac{1}{x}=\log x +1\\
\end{align*}
Finally, we solve for \(f'(x)\text{:}\)
\begin{align*}
f'(x)&=f(x)(\log x + 1) = x^x(\log x + 1)
\end{align*}
In Question 2.10.3.19, we saw \(\ds\diff{}{x}\left\{x^x\right\}=x^x(\log x+1)\text{.}\) Using the basechange formula, \(\log_{10}(x)=\dfrac{\log x}{\log 10}\text{.}\) Since \(\log_{10}\) is a constant,
To make this easier, use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
Rather than set in with a terrible chain rule problem, we'll use logarithmic differentiation. Instead of differentiating \(f(x)\text{,}\) we differentiate a new function \(\log(f(x))\text{,}\) after simplifying.
\begin{align*}
\log(f(x))&=\log\sqrt[4]{\dfrac{(x^4+12)(x^4x^2+2)}{x^3}}\\
&=\frac{1}{4}\log\left(\frac{(x^4+12)(x^4x^2+2)}{x^3}\right)\\
&=\frac{1}{4}\left(\log(x^4+12)+\log(x^4x^2+2)3\log x\right)\\
\end{align*}
Now that we've simplified, we can efficiently differentiate both sides. It is important to remember that we aren't differentiating \(f(x)\) directlywe're differentiating \(\log(f(x))\text{.}\)
\begin{align*}
\frac{f'(x)}{f(x)}&=\frac{1}{4}\left(\frac{4x^3}{x^4+12}+\frac{4x^32x}{x^4x^2+2}\frac{3}{x}\right)\\
\end{align*}
Our final step is to solve for \(f'(x)\text{:}\)
\begin{align*}
{f'(x)}&=f(x)\frac{1}{4}\left(\frac{4x^3}{x^4+12}+\frac{4x^32x}{x^4x^2+2}\frac{3}{x}\right)\\
&=\frac{1}{4}\left(
{\sqrt[4]{\dfrac{(x^4+12)(x^4x^2+2)}{x^3}}}\right)\left(\frac{4x^3}{x^4+12}+\frac{4x^32x}{x^4x^2+2}\frac{3}{x}\right)
\end{align*}
It was possible to differentiate this function without logarithms, but the logarithms make it more efficient.
To make this easier, use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
It's possible to do this using the product rule a number of times, but it's easier to use logarithmic differentiation. Set
\begin{align*}
g(x)=\log(f(x))&=\log\left[(x+1)(x^2+1)^2(x^3+1)^3(x^4+1)^4(x^5+1)^5\right]\\
\end{align*}
Now we can use logarithm rules to change \(g(x)\) into a form that is friendlier to differentiate:
\begin{align*}
&=\log(x+1)+\log(x^2+1)^2+\log(x^3+1)^3+\log(x^4+1)^4+\log(x^5+1)^5\\
&=\log(x+1)+2\log(x^2+1)+3\log(x^3+1)+4\log(x^4+1)+5\log(x^5+1)\\
\end{align*}
Now, we differentiate \(g(x)\) using the chain rule:
\begin{align*}
g'(x)=\frac{f'(x)}{f(x)}&=\frac{1}{x+1}+\frac{4x}{x^2+1}
+\frac{9x^2}{x^3+1}+\frac{16x^3}{x^4+1}+\frac{25x^4}{x^5+1}\\
\end{align*}
Finally, we solve for \(f'(x)\text{:}\)
\begin{align*}
f'(x)&=f(x)\left[\frac{1}{x+1}+\frac{4x}{x^2+1}
+\frac{9x^2}{x^3+1}+\frac{16x^3}{x^4+1}+\frac{25x^4}{x^5+1}\right]\\
&=(x+1)(x^2+1)^2(x^3+1)^3(x^4+1)^4(x^5+1)^5\\
&\;\;\;\;\;\; \cdot\left[\frac{1}{x+1}+\frac{4x}{x^2+1}
+\frac{9x^2}{x^3+1}+\frac{16x^3}{x^4+1}+\frac{25x^4}{x^5+1}\right]
\end{align*}
We could do this with quotient and product rules, but it would be pretty painful. Insteady, let's use a logarithm.
\begin{align*}
f(x) &= \left(\dfrac{5x^2+10x+15}{3x^4+4x^3+5}\right)\left(\dfrac{1}{10(x+1)}\right)
= \left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{2(x+1)}\right)\\
\log(f(x)) &= \log\left[\left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{2(x+1)}\right)\right]\\
&=\log\left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)
+
\log\left(\dfrac{1}{2(x+1)}\right)\\
&=\log\left({x^2+2x+3}\right)
\log\left({3x^4+4x^3+5}\right)
\log(x+1)\log(2)\\
\end{align*}
Now we have a function that we can differentiate more cleanly than our original function.
\begin{align*}
\diff{}{x}\left\{\log(f(x))\right\}&=\diff{}{x}\left\{
\log\left({x^2+2x+3}\right)
\log\left({3x^4+4x^3+5}\right)

\log\left({x+1}\right)
\log\left({2}\right)
\right\}\\
\frac{f'(x)}{f(x)}&=\frac{2x+2}{x^2+2x+3}\frac{12x^3+12x^2}{3x^4+4x^3+5}\frac{1}{x+1}\\
&=\frac{2(x+1)}{x^2+2x+3}\frac{12x^2(x+1)}{3x^4+4x^3+5}\frac{1}{x+1}\\
\end{align*}
Finally, we solve for \(f(x)\text{:}\)
\begin{align*}
f'(x)&=f(x)\left(\frac{2(x+1)}{x^2+2x+3}\frac{12x^2(x+1)}{3x^4+4x^3+5}\frac{1}{x+1}\right)\\
&= \left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{2(x+1)}\right)\left(\frac{2(x+1)}{x^2+2x+3}\frac{12x^2(x+1)}{3x^4+4x^3+5}\frac{1}{x+1}\right)\\
&= \left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\frac{1}{x^2+2x+3}\frac{6x^2}{3x^4+4x^3+5}\frac{1}{2(x+1)^2}\right)
\end{align*}
You'll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
Since \(f(x)\) has the form of a function raised to a functional power, we will use logarithmic differentiation.
\begin{align*}
\log(f(x))&=\log\left( (\cos x)^{\sin x}\right)=\sin x \cdot \log (\cos x)\\
\end{align*}
Logarithm rules allowed us to simplify. Now, we differentiate both sides of this equation:
\begin{align*}
\frac{f'(x)}{f(x)}&=(\cos x ) \log(\cos x)+ \sin x \cdot \frac{\sin x}{\cos x}\\
&=(\cos x) \log (\cos x)  \sin x \tan x\\
\end{align*}
Finally, we solve for \(f'(x)\text{:}\)
\begin{align*}
f'(x)&=f(x)\left[(\cos x) \log (\cos x)  \sin x \tan x\right]\\
&= (\cos x)^{\sin x}\left[(\cos x) \log (\cos x)  \sin x \tan x\right]
\end{align*}
Remark: negative numbers behave in a complicated manner when they are the base of an exponential expression. For example, the expression \((1)^x\) is defined when \(x\) is the reciprocal of an odd number (like \(x=\frac{1}{5}\) or \(x=\frac{1}{7}\)), but not when \(x\) is the reciprocal of an even number (like \(x=\frac{1}{2}\)). Since the domain of \(f(x)\) was restricted to \((0,\tfrac{\pi}{2})\text{,}\) \(\cos x\) is always positive, and we avoid these complications.
You'll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
Since \(f(x)\) has the form of a function raised to a functional power, we will use logarithmic differentiation. We take the logarithm of the function, and make use of logarithm rules:
\begin{align*}
\log\left((\tan x)^x\right)&=x\log(\tan x)\\
\end{align*}
Now, we can differentiate:
\begin{align*}
\frac{\diff{}{x}\left\{(\tan x)^x\right\}}{(\tan x)^x}&=\log(\tan x) + x\cdot\frac{\sec^2 x}{\tan x}\\
&=\log(\tan x) + \frac{x}{\sin x \cos x}\\
\end{align*}
Finally, we solve for the derivative we want, \(\ds\diff{}{x}\{(\tan x)^x\}\text{:}\)
\begin{align*}
{\diff{}{x}\left\{(\tan x)^x\right\}}&={(\tan x)^x}\left(\log(\tan x) + \frac{x}{\sin x \cos x}\right)
\end{align*}
Remark: the restricted domain \((0,\pi/2)\) ensures that \(\tan x\) is a positive number, so we avoid the problems that arise by raising a negative number to a variety of powers.
You'll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
You'll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
You'll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
Remark: negative numbers behave in a complicated manner when they are the base of an exponential expression. For example, the expression \((1)^x\) is defined when \(x\) is the reciprocal of an odd number (like \(x=\frac{1}{5}\) or \(x=\frac{1}{7}\)), but not when \(x\) is the reciprocal of an even number (like \(x=\frac{1}{2}\)). Since the domain of \(f(x)\) was restricted so that \(x\) is always positive, we avoid these complications.
You'll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
We will use logarithmic differentiation. First, we take the logarithm of our function, so we can use logarithm rules.
\begin{align*}
\log\left([f(x)]^{g(x)}\right)&=g(x)\log(f(x))\\
\end{align*}
Now, we differentiate. On the left side we use the chain rule, and on the right side we use product and chain rules.
\begin{align*}
\ds\diff{}{x}\left\{\log\left([f(x)]^{g(x)}\right)\right\}&=\ds\diff{}{x}\left\{g(x)\log(f(x))\right\}\\
\frac{\diff{}{x}\{[f(x)]^{g(x)}\}}{[f(x)]^{g(x)}}&=
g'(x)\log(f(x))+g(x)\cdot\frac{f'(x)}{f(x)}\\
\end{align*}
Finally, we solve for the derivative of our original function.
\begin{align*}
{\diff{}{x}\{[f(x)]^{g(x)}\}}&={[f(x)]^{g(x)}}\left(
g'(x)\log(f(x))+g(x)\cdot\frac{f'(x)}{f(x)}\right)
\end{align*}
Remark: in this section, we have differentiated problems of this type several timesfor example, Questions 2.10.3.24
Let \(g(x):=\log(f(x))\text{.}\) Notice \(g'(x)=\frac{f'(x)}{f(x)}\text{.}\)
In order to show that the two curves have horizontal tangent lines at the same values of \(x\text{,}\) we will show two things: first, that if \(f(x)\) has a horizontal tangent line at some value of \(x\text{,}\) then also \(g(x)\) has a horizontal tangent line at that value of \(x\text{.}\) Second, we will show that if \(g(x)\) has a horizontal tangent line at some value of \(x\text{,}\) then also \(f(x)\) has a horizontal tangent line at that value of \(x\text{.}\)
Suppose \(f(x)\) has a horizontal tangent line where \(x=x_0\) for some point \(x_0\text{.}\) This means \(f'(x_0)=0\text{.}\) Then \(g'(x_0)=\frac{f'(x_0)}{f(x_0)}\text{.}\) Since \(f(x_0) \neq 0\text{,}\) \(\frac{f'(x_0)}{f(x_0)}=\frac{0}{f(x_0)}=0\text{,}\) so \(g(x)\) also has a horizontal tangent line when \(x=x_0\text{.}\) This shows that whenever \(f\) has a horizontal tangent line, \(g\) has one too.
Now suppose \(g(x)\) has a horizontal tangent line where \(x=x_0\) for some point \(x_0\text{.}\) This means \(g'(x_0)=0\text{.}\) Then \(g'(x_0)=\frac{f'(x_0)}{f(x_0)}=0\text{,}\) so \(f'(x_0)\) exists and is equal to zero. Therefore, \(f(x)\) also has a horizontal tangent line when \(x=x_0\text{.}\) This shows that whenever \(g\) has a horizontal tangent line, \(f\) has one too.
Let \(g(x):=\log(f(x))\text{.}\) Notice \(g'(x)=\frac{f'(x)}{f(x)}\text{.}\)
In order to show that the two curves have horizontal tangent lines at the same values of \(x\text{,}\) we will show two things: first, that if \(f(x)\) has a horizontal tangent line at some value of \(x\text{,}\) then also \(g(x)\) has a horizontal tangent line at that value of \(x\text{.}\) Second, we will show that if \(g(x)\) has a horizontal tangent line at some value of \(x\text{,}\) then also \(f(x)\) has a horizontal tangent line at that value of \(x\text{.}\)
Suppose \(f(x)\) has a horizontal tangent line where \(x=x_0\) for some point \(x_0\text{.}\) This means \(f'(x_0)=0\text{.}\) Then \(g'(x_0)=\frac{f'(x_0)}{f(x_0)}\text{.}\) Since \(f(x_0) \neq 0\text{,}\) \(\frac{f'(x_0)}{f(x_0)}=\frac{0}{f(x_0)}=0\text{,}\) so \(g(x)\) also has a horizontal tangent line when \(x=x_0\text{.}\) This shows that whenever \(f\) has a horizontal tangent line, \(g\) has one too.
Now suppose \(g(x)\) has a horizontal tangent line where \(x=x_0\) for some point \(x_0\text{.}\) This means \(g'(x_0)=0\text{.}\) Then \(g'(x_0)=\frac{f'(x_0)}{f(x_0)}=0\text{,}\) so \(f'(x_0)\) exists and is equal to zero. Therefore, \(f(x)\) also has a horizontal tangent line when \(x=x_0\text{.}\) This shows that whenever \(g\) has a horizontal tangent line, \(f\) has one too.
Remark: if we were not told that \(f(x)\) gives only positive numbers, it would not necessarily be true that \(f(x)\) and \(\log(f(x))\) have horizontal tangent lines at the same values of \(x\text{.}\) If \(f(x)\) had a horizontal tangent line at an \(x\)value where \(f(x)\) were negative, then \(\log(f(x))\) would not exist there, let alone have a horizontal tangent line.
We use the power rule (a) and the chain rule (b): the power rule tells us to “bring down the 2”, and the chain rule tells us to multiply by \(y'\text{.}\)
There is no need for the quotient rule here, as there are no quotients. Exponential functions have the form \((\mbox{constant})^{\mbox{function}}\text{,}\) but our function has the form \((\mbox{function})^{\mbox{constant}}\text{,}\) so we did not use (d).
At \((0,4)\) and \((0,4)\text{,}\) the curve looks to be horizontal, if you zoom in: a tangent line here would have derivative zero. At the origin, the curve looks like its tangent line is vertical, so \(\ds\diff{y}{x}\) does not exist.
(a) no \qquad (b) no \(\ds\diff{y}{x}=\dfrac{x}{y}\text{.}\) It is not possible to write \(\ds\diff{y}{x}\) as a function of \(x\text{,}\) because (as stated in (b)) one value of \(x\) may give two values of \(\ds\diff{y}{x}\text{.}\) For instance, when \(x=\pi/4\text{,}\) at the point \(\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)\) the circle has slope \(\ds\diff{y}{x}=1\text{,}\) while at the point \(\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)\) the circle has slope \(\ds\diff{y}{x}=1\text{.}\)
(a) No. A function must pass the vertical line test: one input cannot result in two (or more) outputs. Since one value of \(x\) sometimes corresponds to two values of \(y\) (for example, when \(x=\pi/4\text{,}\) \(y\) is \(\pm 1/\sqrt{2}\)), there is no function \(f(x)\) so that \(y=f(x)\) captures every point on the circle.
Remark: \(y=\pm\sqrt{1x^2}\) does capture every point on the unit circle. However, since one input \(x\) sometimes results in two outputs \(y\text{,}\) this expression is not a function.
(b) No, for the same reasons as (a). If \(f'(x)\) is a function, then it can give at most one slope corresponding to one value of \(x\text{.}\) Since one value of \(x\) can correspond to two points on the circle with different slopes, \(f'(x)\) cannot give the slope of every point on the circle. For example, fix any \(0 \lt a \lt 1\text{.}\) There are two points on the circle with \(x\)coordinate equal to \(a\text{.}\) At the upper one, the slope is strictly negative. At the lower one, the slope is strictly positive.
(c) We differentiate:
\begin{align*}
2x+2y\diff{y}{x}&=0\\
\end{align*}
and solve for \(\ds\diff{y}{x}\)
\begin{align*}
\diff{y}{x}&=\frac{x}{y}
\end{align*}
But there is a \(y\) in the righthand side of this equation, and it's not clear how to get it out. Our answer in (b) tells us that, actually, we can't get it out, if we want the righthand side to be a function of \(x\text{.}\) The derivative cannot be expressed as a function of \(x\text{,}\) because one value of \(x\) corresponds to multiple points on the circle.
Remark: since \(y=\pm\sqrt{1x^2}\text{,}\) we could try writing
but this is not a function of \(x\text{.}\) Again, in a function, one input leads to at most one output, but here one value of \(x\) will usually lead to two values of \(\diff{y}{x}\text{.}\)
Remember that \(y\) is a function of \(x\text{.}\) Use implicit differentiation, then collect all the terms containing \(\ds\diff{y}{x}\) on one side of the equation to solve for \(\ds\diff{y}{x}\text{.}\)
Differentiate both sides of the equation with respect to \(x\text{:}\)
\begin{align*}
e^y\diff{y}{x}&=x\cdot2y\diff{y}{x}+y^2+1\\
\end{align*}
Now, get the derivative on one side and solve
\begin{align*}
e^y\diff{y}{x}2xy\diff{y}{x}&=y^2+1\\
\diff{y}{x}\left(e^y2xy\right)&=y^2+1\\
\diff{y}{x}&=\frac{y^2+1}{e^y2xy}
\end{align*}
First we find the \(x\)coordinates where \(y=1\text{.}\)
\begin{align*}
x^2\tan\left(\frac{\pi}{4}\right)+2x\log(1) &= 16 \\
x^2\cdot 1 +2x\cdot 0 &=16\\
x^2 &= 16\\
\end{align*}
So \(x=\pm 4\text{.}\)
Now we use implicit differentiation to get \(y'\) in terms of \(x,y\text{:}\)
\begin{align*}
x^2\tan(\pi y/4)+2x\log(y) &= 16\\
2x\tan(\pi y/4) + x^2 \frac{\pi}{4}\sec^2(\pi y/4)\cdot y' + 2\log(y) + \frac{2x}{y} \cdot y' &= 0\,.\\
\end{align*}
Now set \(y=1\) and use \(\tan(\pi/4)=1\,, \sec(\pi/4)=\sqrt{2}\) to get
\begin{align*}
2x\tan(\pi/4) + x^2 \frac{\pi}{4} \sec^2(\pi/4)y' + 2\log(1) + 2x\cdot y' &= 0\\
2x + \frac{\pi}{2} x^2 y' +2x y' &= 0 \\
y' = \frac{2x}{x^2 \pi/2 + 2x} &= \frac{4}{\pi x + 4} \\
\end{align*}
So at \((x,y)=(4,1)\) we have \(y' = \dfrac{4}{4\pi+4} =
\dfrac{1}{\pi + 1}\)
and at \((x,y)=(4,1)\) we have \(y' = \dfrac{1}{\pi1}\)
First we find the \(x\)coordinates where \(y=0\text{.}\)
\begin{align*}
x^2e^0+4x\cos(0) &= 5 \\
x^2 +4x  5 &=0\\
(x+5)(x1)&=0
\end{align*}
So \(x=1,5\text{.}\)
Now we use implicit differentiation to get \(y'\) in terms of \(x,y\text{:}\)
\begin{align*}
x^2e^y+4x\cos(y) &= 5 & \text{differentiate both sides} \\
x^2 \cdot e^y \cdot y' + 2x e^y + 4x(\sin(y)) \cdot y' + 4\cos(y) &= 0
\end{align*}
Now set \(y=0\) to get
\begin{align*}
x^2 \cdot e^0 \cdot y' + 2x e^0 + 4x(\sin(0)) \cdot y' + 4\cos(0) &= 0 \\
x^2y' + 2x + 4 &=0 \\
y' &=  \frac{4+2x}{x^2}.
\end{align*}
So at \((x,y)=(1,0)\) we have \(y' = 6\text{,}\)
and at \((x,y)=(5,0)\) we have \(y' = \frac{6}{25}\text{.}\)
First we find the \(x\)coordinates where \(y=0\text{.}\)
\begin{align*}
x^2\cos(0)+2xe^0 &= 8 \\
x^2 +2x  8 &=0\\
(x+4)(x2)&=0
\end{align*}
So \(x=2,4\text{.}\)
Now we use implicit differentiation to get \(y'\) in terms of \(x,y\text{:}\)
\begin{align*}
x^2\cos(y)+2xe^y &= 8 & \text{differentiate both sides} \\
x^2 \cdot (\sin y) \cdot y' + 2x \cos y + 2xe^y \cdot y' + 2e^y &= 0
\end{align*}
The question asks at which points on the ellipse \(\ds\diff{y}x{}=1\text{.}\) So, we begin by differentiating, implicitly:
\begin{align*}
2x+6y\diff{y}{x}&=0\\
\end{align*}
We could solve for \(\ds\diff{y}{x}\) at this point, but it's not necessary. We want to know when \(\ds\diff{y}{x}\) is equal to one:
\begin{align*}
2x+6y(1)&=0\\
x&=3y\\
\end{align*}
That is, \(\ds\diff{y}{x}=1\) at those points along the ellipse where \(x=3y\text{.}\) We plug this into the equation of the ellipse to find the coordinates of these points.
\begin{align*}
\left(3y\right)^2+3y^2&=1\\
12y^2&=1\\
y=\pm\frac{1}{\sqrt{12}}=\pm\frac{1}{2\sqrt{3}}
\end{align*}
So, the points along the ellipse where the tangent line is parallel to the line \(y=x\) occur when \(y=\dfrac{1}{2\sqrt{3}}\) and \(x=3y\text{,}\) and when \(y=\dfrac{1}{2\sqrt{3}}\) and \(x=3y\text{.}\) That is, the points \(\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2\sqrt{3}}\right)\) and \(\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2\sqrt{3}}\right)\text{.}\)
Implicitly differentiating \(x^2y(x)^2+x\sin(y(x))=4\) with respect to \(x\) gives
\begin{align*}
2xy^2&+2x^2yy'+\sin y+xy'\cos y=0\\
\end{align*}
Then we gather the terms containing \(y'\) on one side, so we can solve for \(y'\text{:}\)
\begin{align*}
2x^2yy'&+xy'\cos y=2xy^2\sin y\\
y'(2x^2y&+x\cos y)=2xy^2\sin y\\
y'&=\frac{2xy^2+\sin y}{2x^2y+x\cos y}
\end{align*}
First we find the \(x\)ordinates where \(y=0\text{.}\)
\begin{align*}
x^2+(1)e^0 &= 5 \\
x^2 +1 &=5\\
x^2&=4
\end{align*}
So \(x=2,2\text{.}\)
Now we use implicit differentiation to get \(y'\) in terms of \(x,y\text{:}\)
\begin{align*}
2x+(y+1)e^y\diff{y}{x}+e^y\diff{y}{x}&=0
\end{align*}
Now set \(y=0\) to get
\begin{align*}
2x+(0+1)e^0\diff{y}{x}+e^0\diff{y}{x}&=0\\
2x+\diff{y}{x}+\diff{y}{x}&=0\\
2x&=2\diff{y}{x}\\
x&=\diff{y}{x}
\end{align*}
So at \((x,y)=(2,0)\) we have \(y' = 2\text{,}\)
and at \((x,y)=(2,0)\) we have \(y' = 2\text{.}\)
The slope of the tangent line is, of course, given by the derivative, so let's start by finding \(\diff{y}{x}\) of both shapes.
\begin{align*}
\\
\end{align*}
For the circle, we differentiate implicitly
\begin{align*}
2x+2y\diff{y}{x}&=0\\
\end{align*}
and solve for \(\ds\diff{y}{x}\)
\begin{align*}
\diff{y}{x}&=\frac{x}{y}\\
\end{align*}
For the ellipse, we also differentiate implicitly:
\begin{align*}
2x+6y\diff{y}{x}&=0\\
\end{align*}
and solve for \(\ds\diff{y}{x}\)
\begin{align*}
\diff{y}{x}&=\frac{x}{3y}\\
\end{align*}
What we want is a value of \(x\) where both derivatives are equal. However, they might have different values of \(y\text{,}\) so let's let \(y_1\) be the \(y\)values associated with \(x\) on the circle, and let \(y_2\) be the \(y\)values associated with \(x\) on the ellipse. That is, \(x^2+ y_1^2=1\) and \(x^2+3y_2^2=1\text{.}\) For the slopes at \((x,y_1)\) on the circle and \((x,y_2)\) on the ellipse to be equal, we need:
\begin{align*}
\frac{x}{y_1}&=\frac{x}{3y_2}\\
x\left(\frac{1}{y_1}\frac{1}{3y_2}\right)&=0\\
\end{align*}
So \(x=0\) or \(y_1=3y_2\text{.}\) Let's think about which \(x\)values will have a \(y\)coordinate of the circle be three times as large as a \(y\)coordinate of the ellipse. If \(y_1=3y_2\text{,}\) \((x,y_1)\) is on the circle, and \((x,y_2)\) is on the ellipse, then \(x^2+y_1^2 =x^2+(3y_2)^2=1\) and \(x^2+3y_2^2=1\text{.}\) In this case:
\begin{align*}
x^2+9y_2^2&=x^2+3y_2^2\\
9y_2^2&=3y_2^2\\
y_2&=0\\
x&=\pm 1
\end{align*}
We need to be a tiny bit careful here: when \(y=0\text{,}\) \(y'\) is not defined for either curve. For both curves, when \(y=0\text{,}\) the tangent lines are vertical (and so have no realvalued slope!). Two vertical lines are indeed parallel.
So, for \(x=0\) and for \(x=\pm1\text{,}\) the two curves have parallel tangent lines.
(a) We can plug any number into the cosine function, and it will return a number in \([1,1]\text{.}\) The domain of \(\arcsin x\) is \([1,1]\text{,}\) so any number we plug into cosine will give us a valid number to plug into arcsine. So, the domain of \(f(x)\) is all real numbers.
(b) We can plug any number into the cosine function, and it will return a number in \([1,1]\text{.}\) The domain of \(\arccsc x\) is \((\infty,1] \cup [1,\infty)\text{,}\) so in order to have a valid number to plug into arccosecant, we need \(\cos x = \pm 1\text{.}\) That is, the domain of \(g(x)\) is all values \(x=n\pi\) for some integer \(n\text{.}\)
(c) The domain of arccosine is \([1,1]\text{.}\) The domain of sine is all real numbers, so no matter what number arccosine spits out, we can safely plug it into sine. So, the domain of \(h(x)\) is \([1,1]\text{.}\)
False: \(\cos t=1\) for infinitely many values of \(t\text{;}\) arccosine gives only the single value \(t=0\) for which \(\cos t=1\) and \(0 \leq t \leq \pi\text{.}\) The particle does not start moving until \(t=10\text{,}\) so \(t=0\) is not in the domain of the function describing its motion.
The particle will have height \(1\) at time \(2\pi n\text{,}\) for any integer \(n \geq 2\text{.}\)
First, we restrict the domain of \(f\) to force it to be onetoone. There are many intervals we could choose over which \(f\) is onetoone, but the question asks us to contain \(x=0\) and be as large as possible; this leaves us with the following restricted function:
The inverse of a function swaps the role of the input and output; so if the graph of \(y=f(x)\) contains the point \((a,b)\text{,}\) then the graph of \(Y=f^{1}(X)\) contains the point \((b,a)\text{.}\) That is, the graph of \(Y=f^{1}(X)\) is the graph of \(y=f(x)\) with the \(x\)coordinates and \(y\)coordinates swapped. (So, since \(y=f(x)\) crosses the \(y\)axis at \(y=1\text{,}\) then \(Y=f^{1}(X)\) crosses the \(X\)axis at \(X=1\text{.}\)) This swapping is equivalent to reflecting the curve \(y=f(x)\) over the line \(y=x\text{.}\)
Remark: while you're getting accustomed to inverse functions, it is sometimes clearer to consider \(y=f(x)\) and \(Y=f^{1}(X)\text{:}\) using slightly different notations for \(x\) (the input of \(f\text{,}\) hence the output of \(f^{1}\)) and \(X\) (the input of \(f^{1}\text{,}\) which comes from the output of \(f\)). However, the convention is to use \(x\) for the inputs of both functions, and \(y\) as the outputs of both functions, as is written on the graph above.
If \(a \gt 1\text{,}\) there is no point where the curve has horizontal tangent line.
If \(a=1\text{,}\) the curve has a horizontal tangent line where \(x=2\pi n + \dfrac{a\pi}{2}\) for any integer \(n\text{.}\)
If \(a \lt 1\text{,}\) the curve has a horizontal tangent line where \(x=2\pi n+\arcsin(a)\) or \(x=(2 n +1) \pi  \arcsin (a)\) for any integer \(n\text{.}\)
The tangent line is horizontal when \(0=y'=a\sin x\text{.}\) That is, when \(a=\sin x\text{.}\)
If \(a \gt 1\text{,}\) then there is no value of \(x\) for which \(a=\sin x\text{,}\) so the curve has no horizontal tangent lines.
If \(a = 1\text{,}\) then there are infinitely many solutions to \(a=\sin x\text{,}\) but only one solution in the interval \([\pi,\pi]\text{:}\) \(x=\arcsin(a)=\arcsin(\pm1)=\pm\frac{\pi}{2}\text{.}\) Then the values of \(x\) for which \(a=\sin x\) are \(x=2\pi n +a \frac{\pi}{2}\) for any integer \(n\text{.}\)
If \(a \lt 1\text{,}\) then there are infinitely many solutions to \(a=\sin x\text{.}\) The solution in the interval \(\left(\frac{\pi}{2},\frac{\pi}{2}\right)\) is given by \(x=\arcsin(a)\text{.}\) The other solution in the interval \(\left(\pi,\pi\right)\) is given by \(x=\pi\arcsin(a)\text{,}\) as shown in the unit circles below.
So, the values of \(x\) for which \(x=\sin a\) are \(x=2\pi n+\arcsin(a)\) and \(x=2\pi n + \pi  \arcsin (a)\) for any integer \(n\text{.}\)
Remark: when \(a=1\text{,}\) then
\begin{equation*}
2\pi n+\arcsin(a) = 2\pi n + \dfrac{\pi}{2}=2\pi n +\pi \left(\dfrac {\pi}{2}\right)=2\pi n+\pi\arcsin(a).
\end{equation*}
So, if we try to use the descriptions in the third bullet point to describe points where the tangent line is horizontal when \(a=1\text{,}\) we get the correct points but each point is listed twice. This is why we separated the case \(a=1\) from the case \(a \lt 1\text{.}\)
The function \(\arcsin x\) is only defined for \(x \leq 1\text{,}\) and the function \(\arccsc x\) is only defined for \(x \geq 1\text{,}\) so \(f(x)\) has domain \(x=1\text{.}\) That is, \(x=\pm1\text{.}\)
In order for \(f(x)\) to be differentiable at a point, it must exist in an open interval around that point. (See Definition 2.2.1.) Since our function does not exist over any open interval, \(f(x)\) is not differentiable anywhere.
So, actually, \(f(x)\) is a pretty boring function, which we can entirely describe as: \(f(1)=\pi\) and \(f(1)=\pi\text{.}\)
Since the domain of arcsine is \([1,1]\text{,}\) and we are plugging in \(\dfrac{x}{3}\) to arcsine, the values of \(x\) that we can plug in are those that satisfy \(1 \le \dfrac{x}{3} \leq 1\text{,}\) or \(3\leq x \leq 3\text{.}\) So the domain of \(f\) is \([3,3]\text{.}\)
The domain of arccosine is \([1,1]\text{,}\) and since \(t^21\) is in the denominator, the domain of \(f\) requires \(t^21 \neq 0\text{,}\) that is, \(t \neq \pm 1\text{.}\) So the domain of \(f(t)\) is \((1,1)\text{.}\)
The domain of \(\arcsec x\) is \(x \geq 1\text{:}\) that is, we can plug into arcsecant only values with absolute value greater than or equal to one. Since \(x^22 \leq 2\text{,}\) every real value of \(x\) gives us an acceptable value to plug into arcsecant. So, the domain of \(f(x)\) is all real numbers.
To differentiate, we use the chain rule. Remember \(\ds\diff{}{x}\left\{\arcsec x\right\} = \dfrac{1}{x\sqrt{x^21}}\text{.}\)
We differentiate using the \textcolor{blue}{product} and \textcolor{red}{chain} rules.
\begin{align*}
\diff{}{x}\left\{\textcolor{blue}{x\arcsin x} + \textcolor{red}{\sqrt{1x^2}}\right\}&=
\textcolor{blue}{\arcsin x + \frac{x}{\sqrt{1x^2}}}+\textcolor{red}{\frac{2x}{2\sqrt{1x^2}}}\\
&=\arcsin x
\end{align*}
The domain of \(\arcsin x\) is \([1,1]\text{,}\) and the domain of \(\sqrt{1x^2}\) is all values of \(x\) so that \(1x^2 \geq 0\text{,}\) so \(x\) in \([1,1]\text{.}\) Therefore, the domain of \(f(x)\) is \([1,1]\text{.}\)
Using formulas you should memorize from this section,
\begin{equation*}
\diff{}{x}\{\arcsin x + \arccos x\}=\frac{1}{\sqrt{1x^2}}+\frac{1}{\sqrt{1x^2}}=0
\end{equation*}
Remark: the only functions with derivative equal to zero everywhere are constant functions, so \(\arcsin x + \arccos x\) should be a constant. Since \(\sin \theta = \cos \left(\frac{\pi}{2}\theta\right)\text{,}\) we can set
\begin{align*}
\sin\theta&=x & \cos\left(\frac{\pi}{2}\theta\right)&=x\\
\end{align*}
where \(x\) and \(\theta\) are the same in both expressions, and \(\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{.}\) Then
\begin{align*}
\arcsin x &=\theta & \arccos x &= \frac{\pi}{2}\theta\\
\end{align*}
We note here that arcsine is the inverse of the sine function restricted to \(\left[\frac{\pi}{2}, \frac{\pi}{2}\right]\text{.}\) So, since we restricted \(\theta\) to this domain, \(\sin \theta=x\) really does imply \(\arcsin x = \theta\text{.}\) (For an example of why this matters, note \(\sin(2\pi)=0\text{,}\) but \(\arcsin (0)=0 \neq 2\pi\text{.}\)) Similarly, arccosine is the inverse of the cosine function restricted to \([0,\pi]\text{.}\) Since \(\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{,}\) then \(0 \leq (\frac{\pi}{2}\theta) \leq \pi\text{,}\) so \(\cos\left(\frac{\pi}{2} \theta\right) =x\) really does imply \(\arccos x=\frac{\pi}{2}\theta\text{.}\)
\begin{align*}
\end{align*}
So,
\begin{equation*}
\arcsin x+\arccos x =\theta+\frac{\pi}{2}\theta =\frac{\pi}{2}
\end{equation*}
which means the derivative we were calculating was actually just \(\ds\diff{}{x}\left\{\dfrac{\pi}{2}\right\}=0\text{.}\)
You can simplify the expression before you differentiate to remove the trigonometric functions. If \(\arctan x =\theta\text{,}\) then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:
With the sides labeled, you can figure out \(\sin\left(\arctan x\right)=\sin\left(\theta\right)\text{.}\)
Let \(\theta = \arctan x\text{.}\) Then \(\theta\) is the angle of a right triangle that gives \(\tan \theta = x\text{.}\) In particular, the ratio of the opposite side to the adjacent side is \(x\text{.}\) So, we have a triangle that looks like this:
where the length of the hypotenuse came from the Pythagorean Theorem. Now,
Let \(\theta = \arctan x\text{.}\) Then \(\theta\) is the angle of a right triangle that gives \(\tan \theta = x\text{.}\) In particular, the ratio of the opposite side to the adjacent side is \(x\text{.}\) So, we have a triangle that looks like this:
where the length of the hypotenuse came from the Pythagorean Theorem. Now,
Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression \(\ds\diff{}{x}\left\{\sin\left(\arctan x\right)\right\}=\dfrac{\cos(\arctan x)}{1+x^2}\text{.}\)
You can simplify the expression before you differentiate to remove the trigonometric functions. If \(\arcsin x =\theta\text{,}\) then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:
With the sides labeled, you can figure out \(\cot\left(\arcsin x\right)=\cot\left(\theta\right)\text{.}\)
Let \(\theta = \arcsin x\text{.}\) Then \(\theta\) is the angle of a right triangle that gives \(\sin \theta = x\text{.}\) In particular, the ratio of the opposite side to the hypotenuse is \(x\text{.}\) So, we have a triangle that looks like this:
where the length of the adjacent side came from the Pythagorean Theorem. Now,
Let \(\theta = \arcsin x\text{.}\) Then \(\theta\) is the angle of a right triangle that gives \(\sin \theta = x\text{.}\) In particular, the ratio of the opposite side to the hypotenuse is \(x\text{.}\) So, we have a triangle that looks like this:
where the length of the adjacent side came from the Pythagorean Theorem. Now,
Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression \(\ds\diff{}{x}\left\{\cot\left(\arcsin x\right)\right\}=\frac{\csc^2(\arcsin x)}{\sqrt{1x^2}}\text{.}\)
The line \(y=2x+9\) has slope \(2\text{,}\) so we must find all values of \(x\) between \(1\) and \(1\) (\(\arcsin x\) is only defined for these values of \(x\)) for which \(\diff{}{x}\{\arcsin x\}=2\text{.}\) Evaluating the derivative:
So if \(f'(x)=0\text{,}\) then \(\cos x=0\text{,}\) and this happens when \(x=\dfrac{(2n+1)\pi}{2}\) for any integer \(n\text{.}\) We should check that these points are in the domain of \(f\text{.}\) Arctangent is defined for all real numbers, so we only need to check the domain of cosecant; when \(x=\dfrac{(2n+1)\pi}{2}\text{,}\) then \(\sin x=\pm1 \neq 0\text{,}\) so \(\csc x = \dfrac{1}{\sin x}\) exists.
\begin{align*}
f(g(y))&=f\left(f^{1}(y)\right)=y\\
\end{align*}
Now, we can differentiate with respect to \(y\) using the chain rule.
\begin{align*}
\diff{}{y}\left\{f(g(y))\right\}&=\diff{}{y}\{y\}\\
f'(g(y))\cdot g'(y)&=1\\
g'(y)&=\frac{1}{f'(g(y))}=\frac{1}{1\sin g(y)}
\end{align*}
Write \(g(y)=f^{1}(y)\text{.}\) Then \(g(f(x))=x\text{,}\) so differentiating both sides (using the chain rule), we see
\begin{align*}
g'(f(x))\cdot f'(x)=1\\
\end{align*}
What we want is \(g'(\pi1)\text{,}\) so we need to figure out which value of \(x\) gives \(f(x)=\pi1\text{.}\) A little trial and error leads us to \(x=\frac{\pi}{2}\text{.}\)
\begin{align*}
g'(\pi1)\cdot f'\left(\frac{\pi}{2}\right)&=1\\
\end{align*}
Since \(f'(x)=2\cos(x)\text{,}\) \(f'\left(\frac{\pi}{2}\right)=20=2\text{:}\)
\begin{align*}
g'(\pi1)\cdot 2&=1\\
g'(\pi1)=\frac{1}{2}
\end{align*}
Write \(g(y)=f^{1}(y)\text{.}\) Then \(g(f(x))=x\text{,}\) so differentiating both sides (using the chain rule), we see
\begin{align*}
g'(f(x))f'(x)&=1\\
\end{align*}
What we want is \(g'(e+1)\text{,}\) so we need to figure out which value of \(x\) gives \(f(x)=e+1\text{.}\) A little trial and error leads us to \(x=1\text{.}\)
\begin{align*}
g'(f(1))f'(1)&=1\\
g'(e+1)\cdot f'(1)&=1\\
g'(e+1) &= \frac{1}{f'(1)}\\
\end{align*}
It remains only to note that \(f'(x)=e^x+1\text{,}\) so \(f'(1)=e+1\)
\begin{align*}
g'(e+1)&=\frac{1}{e+1}
\end{align*}
\(f'(x)=[\sin x +2]^{\arcsec x}\left(\dfrac{\log[\sin x +2]}{x\sqrt{x^21}}+ \dfrac{\arcsec x \cdot\cos x}{\sin x +2}\right)\text{.}\) The domain of \(f(x)\) is \(x\ge 1\text{.}\)
We use logarithmic differentiation, our standard method of differentiating an expression of the form \((\mbox{function})^{\mbox{function}}\text{.}\)
\begin{align*}
f(x)&=[\sin x +2]^{\arcsec x}\\
\log(f(x))&=\arcsec x \cdot \log[\sin x +2]\\
\frac{f'(x)}{f(x)}&=\frac{1}{x\sqrt{x^21}}\log[\sin x +2]+\arcsec x \cdot \frac{\cos x}{\sin x +2}\\
f'(x)&=[\sin x +2]^{\arcsec x}\left(\frac{\log[\sin x +2]}{x\sqrt{x^21}}+ \frac{\arcsec x \cdot\cos x}{\sin x +2}\right)
\end{align*}
The domain of \(\arcsec x\) is \(x \geq 1\text{.}\) For any \(x\text{,}\) \(\sin x +2\) is positive, and a positive number can be raised to any power. (Recall negative numbers cannot be raised to any powerfor example, \((1)^{1/2}=\sqrt{1}\) is not a real number.) So, the domain of \(f(x)\) is \(x \geq 1\text{.}\)
The function \(\dfrac{1}{\sqrt{x^21}}\) exists only for those values of \(x\) with \(x^21 \gt 0\text{:}\) that is, the domain of \(\dfrac{1}{\sqrt{x^21}}\) is \(x \gt 1\text{.}\) However, the domain of arcsine is \(x \leq 1\text{.}\) So, there is not one single value of \(x\) where \(\arcsin x\) and \(\dfrac{1}{\sqrt{x^21}}\) are both defined.
If the derivative of \(\arcsin(x)\) were given by \(\dfrac{1}{\sqrt{x^21}}\text{,}\) then the derivative of \(\arcsin(x)\) would not exist anywhere, so we would probably just write “derivative does not exist,” instead of making up a function with a mismatched domain. Also, the function \(f(x)=\arcsin(x)\) is a smooth curveits derivative exists at every point strictly inside its domain. (Remember not all curves are like this: for instance, \(g(x)=x\) does not have a derivative at \(x=0\text{,}\) but \(x=0\) is strictly inside its domain.) So, it's a pretty good bet that the derivative of arcsine is not \(\dfrac{1}{\sqrt{x^21}}\text{.}\)
The function \(\dfrac{1}{\sqrt{x^21}}\) exists only for those values of \(x\) with \(x^21 \gt 0\text{:}\) that is, the domain of \(\dfrac{1}{\sqrt{x^21}}\) is \(x \gt 1\text{.}\) However, the domain of arcsine is \(x \leq 1\text{.}\) So, there is not one single value of \(x\) where \(\arcsin x\) and \(\dfrac{1}{\sqrt{x^21}}\) are both defined.
If the derivative of \(\arcsin(x)\) were given by \(\dfrac{1}{\sqrt{x^21}}\text{,}\) then the derivative of \(\arcsin(x)\) would not exist anywhere, so we would probably just write “derivative does not exist,” instead of making up a function with a mismatched domain. Also, the function \(f(x)=\arcsin(x)\) is a smooth curveits derivative exists at every point strictly inside its domain. (Remember not all curves are like this: for instance, \(g(x)=x\) does not have a derivative at \(x=0\text{,}\) but \(x=0\) is strictly inside its domain.) So, it's a pretty good bet that the derivative of arcsine is not \(\dfrac{1}{\sqrt{x^21}}\text{.}\)
This limit represents the derivative computed at \(x=1\) of the function \(f(x)=\arctan x\text{.}\) To see this, simply use the definition of the derivative at \(a=1\text{:}\)
First, let's interpret the given information: when the input of our function is \(2x+1\) for some \(x\text{,}\) then its output is \(\dfrac{5x9}{3x+7}\text{,}\) for that same \(x\text{.}\) We're asked to evaluate \(f^{1}(7)\text{,}\) which is the number \(y\) with the property that \(f(y)=7\text{.}\) If the output of our function is 7, that means
\begin{align*}
7&=\frac{5x9}{3x+7}\\
\end{align*}
and so
\begin{align*}
7(3x+7)&=5x9\\
x&=\frac{29}{8}\\
\end{align*}
So, when \(x=\dfrac{29}{8}\text{,}\) our equation \(f(2x+1)=\dfrac{5x9}{3x+7}\) becomes:
\begin{align*}
f\left(2\cdot\frac{29}{8}+1\right)&=\dfrac{5\cdot\frac{29}{8}9}{3\cdot\frac{29}{8}+7}\\
\end{align*}
Or, equivalently:
\begin{align*}
f\left(\frac{25}{4}\right)&=7
\end{align*}
If \(f^{1}(y)=0\text{,}\) that means \(f(0)=y\text{.}\) So, we want to find out what we plug into \(f^{1}\) to get 0. Since we only know \(f^{1}\) in terms of a variable \(x\text{,}\) let's figure out what \(x\) gives us an output of 0:
Solution 1: We begin by differentiating implicitly. Following the usual convention, we use \(y'\) to mean \(y'(x)\text{.}\)
\begin{align*}
\arcsin(x+2y)&=x^2+y^2 \qquad\mbox{Using the chain rule:}\\
\frac{1+2y'}{\sqrt{1(x+2y)^2}}&=2x+2yy'\\
\frac{1}{\sqrt{1(x+2y)^2}}+\frac{2y'}{\sqrt{1(x+2y)^2}}&=2x+2yy'\\
\frac{2y'}{\sqrt{1(x+2y)^2}}2yy'&=2x\frac{1}{\sqrt{1(x+2y)^2}}\\
y'\left(\frac{2}{\sqrt{1(x+2y)^2}}2y\right)&=2x\frac{1}{\sqrt{1(x+2y)^2}}\\
y'&=\frac{2x\frac{1}{\sqrt{1(x+2y)^2}}}{\frac{2}{\sqrt{1(x+2y)^2}}2y}
\left(\frac{\sqrt{1(x+2y)^2}}{\sqrt{1(x+2y)^2}}\right)\\
y'&=\frac{2x\sqrt{1(x+2y)^2}1}{22y\sqrt{1(x+2y)^2}}
\end{align*}
Solution 2: We begin by taking the sine of both sides of the equation.
\begin{align*}
\arcsin(x+2y)&=x^2+y^2\\
x+2y&=\sin(x^2+y^2)\\
\end{align*}
Now, we differentiate implicitly.
\begin{align*}
1+2y'&=\cos(x^2+y^2)\cdot(2x+2yy')\\
1+2y'&=2x\cos(x^2+y^2)+2yy'\cos(x^2+y^2)\\
2y'2yy'\cos(x^2+y^2)&=2x\cos(x^2+y^2)1\\
y'\left(22y\cos(x^2+y^2)\right)&=2x\cos(x^2+y^2)1\\
y'&=\frac{2x\cos(x^2+y^2)1}{22y\cos(x^2+y^2)}
\end{align*}
We used two different methods, and got two answers that look pretty different. However, the answers ought to be equivalent. To see this, we remember that for all values of \(x\) and \(y\) that we care about (those pairs \((x,y)\) in the domain of our curve), the equality
We know the top speed of the caribou, so we can use this to give the minimum possible number of hours the caribou spent travelling during its migration. If the caribou travels at 70kph, it will take \((5000 \mathrm{km})\left(\dfrac{1 \mathrm{hr}}{70 \mathrm{km}}\right) \approx 71.4 \mathrm{hrs}\) to travel 5000 kilometres. Probably the caribou wasn't sprinting the whole time, so probably it took it longer than that, but we can only say for sure that the caribou spent at least about 71.4 hours migrating.
If \(f(x)\) is the position of the crane at time \(x\text{,}\) measured in hours, then (if we let \(x=0\) be the beginning of the day) we know that \(f(24)f(0)=240\text{.}\) Since \(f(x)\) is the position of the bird, \(f(x)\) is continuous and differentiable. So, the MVT says there is a \(c\) in \((0,24)\) such that \(f'(x)=\dfrac{f(24)f(0)}{240}=\dfrac{240}{24}=10\text{.}\) That is, at some point \(c\) during the day, the speed of the crane was exactly 10 kph.
The MVT guarantees there is some point \(c\) strictly between \(a\) and \(b\) where the tangent line to \(f(x)\) at \(x=c\) has the same slope as the secant line of \(f(x)\) from \(x=a\) to \(x=b\text{.}\) So, let's start by drawing in the secant line.
What we're looking for is a point on the curve where the tangent line is parallel to this secant line. In fact, there are two.
So, either of the two values \(c_1\) and \(c_2\) marked below can serve as the point guaranteed by the MVT:
Since \(f(x)\) is differentiable for all \(x \in (0,10)\text{,}\) then \(f(x)\) is also continuous for all \(x \in (0,10)\text{.}\) If \(f(x)\) were continuous on the closed interval \([0,10]\text{,}\) then the MVT would guarantee \(f'(x)=\dfrac{f(10)f(0)}{100}=1\) for some \(c \in (0,10)\text{;}\) however, this is not the case. So, it must be that \(f(x)\) is continuous for all \(x \in (0,10)\text{,}\) but not for all \(x \in [0,10]\text{.}\)
Since \(f'(c)=0\) for \(c \in (0,10)\text{,}\) that means \(f\) is constant on that interval. So, \(f(x)\) is a function like this:
where the height of the constant function can be anything.
So, one possible answer is \(f(x) = \left\{\begin{array}{lr}
0&x \neq 10\\
10&x=10
\end{array}\right.\text{.}\)
The function \(f(x)\) is continuous over all real numbers, but it is only differentiable when \(x \neq 0\text{.}\) So, if we want to apply the MVT, our interval must consist of only positive numbers or only negative numbers: the interval \((4,13)\) is not valid.
It is possible to use the mean value theorem to prove what we want: if \(a=1\) and \(b=144\text{,}\) then \(f(x)\) is differentiable over the interval \((1,144)\) (since 0 is not contained in that interval), and \(f(x)\) is continuous everywhere, so by the mean value theorem there exists some point \(c\) where \(f'(x)=\dfrac{\sqrt{144}\sqrt{1}}{1441}=\dfrac{11}{143}=\dfrac{1}{13}\text{.}\)
That being said, an easier way to prove that a point exists is to simply find itwithout using the MVT. When \(x \gt 0\text{,}\) \(f(x)=\sqrt{x}\text{,}\) so \(f'(x)=\dfrac{1}{2\sqrt {x}}\text{.}\) Then \(f'\left(\dfrac{169}{4}\right)=\dfrac{1}{13}\text{.}\)
The function \(f(x)\) is continuous over all real numbers, but it is only differentiable when \(x \neq 0\text{.}\) So, if we want to apply the MVT, our interval must consist of only positive numbers or only negative numbers: the interval \((4,13)\) is not valid.
It is possible to use the mean value theorem to prove what we want: if \(a=1\) and \(b=144\text{,}\) then \(f(x)\) is differentiable over the interval \((1,144)\) (since 0 is not contained in that interval), and \(f(x)\) is continuous everywhere, so by the mean value theorem there exists some point \(c\) where \(f'(x)=\dfrac{\sqrt{144}\sqrt{1}}{1441}=\dfrac{11}{143}=\dfrac{1}{13}\text{.}\)
That being said, an easier way to prove that a point exists is to simply find itwithout using the MVT. When \(x \gt 0\text{,}\) \(f(x)=\sqrt{x}\text{,}\) so \(f'(x)=\dfrac{1}{2\sqrt {x}}\text{.}\) Then \(f'\left(\dfrac{169}{4}\right)=\dfrac{1}{13}\text{.}\)
To use Rolle's Theorem, you will want two values where the function is zero. If you're stuck finding one of them, think about when \(x^22\pi x\) is equal to zero.
We note that \(f(0)=f(2\pi)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we conclude that there exists \(c\) in \((0,2\pi)\) such that
We note that \(f(0)=f(2\pi)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we conclude that there exists \(c\) in \((0,2\pi)\) such that
We note that \(f(0)=f(1)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists \(c\in (0,1)\) such that
We note that \(f(0)=f(1)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists \(c\in (0,1)\) such that
We note that \(f(0)=f(2\pi)=\sqrt{3} + \pi^2\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers since \(3+\sin x \gt 0\)), we get that there exists \(c\in (0,2\pi)\) such that
We note that \(f(0)=f(2\pi)=\sqrt{3} + \pi^2\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers since \(3+\sin x \gt 0\)), we get that there exists \(c\in (0,2\pi)\) such that
We note that \(f(0)=0\) and \(f(\pi/4)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists \(c\in
(0,\pi/4)\) such that
We note that \(f(0)=0\) and \(f(\pi/4)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists \(c\in
(0,\pi/4)\) such that
By inspection, we see \(x=0\) is a root of \(f(x)\text{.}\) The question now is whether there could possibly be other roots. Since \(f(x)\) is differentiable over all real numbers, if there is another root \(a\text{,}\) then by Rolle's Theorem, \(f'(c)=0\) for some \(c\) strictly between \(0\) and \(a\text{.}\) However, \(f'(x)=3\cos x\) is never zero. So, there is no second root: \(f(x)\) has precisely one root.
To show that there are exactly \(n\) roots, you need to not only show that \(n\) exist, but also that there are not more than \(n\text{.}\) If you can't explicitly find the root(s), you can use the intermediate value theorem to show they exist.
The function \(f(x)\) is continuous and differentiable over all real numbers. If \(a\) and \(b\) are distinct roots of \(f(x)\text{,}\) then \(f'(c)=0\) for some \(c\) strictly between \(a\) and \(b\) (Rolle's Theorem). So, let's think about \(f'(x)\text{.}\)
Hence \(f'(c)=0\) only when \(c=\dfrac{1}{2}\text{.}\) That the derivative only has a single zero is very useful. It means (via Rolle's theorem) that if \(f(x)\) has distinct roots \(a\) and \(b\) with \(a \lt b\text{,}\) then we must have \(a \lt \dfrac{1}{2} \lt b\text{.}\) This also means that \(f(x)\) cannot have 3 distinct roots \(a,b\) and (say) \(q\) with \(a \lt b \lt q\text{,}\) because then Rolle's theorem would imply that \(f'(x)\) would have two zeros — one between \(a\) and \(b\) and another between \(b\) and \(q\text{.}\)
We've learned that \(f(x)\) has at most two roots, and we've learned something about where those roots can exist, if there are indeed two of them. But that means \(f(x)\) could have 0, 1, or 2 roots.
It's not easy to find a root of \(f(x)\) by inspection. But we can get a good enough picture of the graph of \(y=f(x)\) to tell exactly how many roots there are, just by exploiting the following properties of \(f(x)\) and \(f'(x)\text{.}\)
As \(x\) tends to \(\pm \infty\text{,}\) \(f(x)\) tends to \(+\infty\text{.}\)
The derivative \(f'(x)=(4x+1)^3+1\) is negative for \(x \lt \frac{1}{2}\) and is positive for \(x \gt \frac{1}{2}\text{.}\) That is, \(f(x)\) is decreasing for \(x \lt \frac{1}{2}\) and increasing for \(x \gt \frac{1}{2}\text{.}\)
This means the function must look something like the graph below:
except we are unsure of the locations of the \(x\)intercepts. At present we only know that one is to the left of \(x=1/2\) and one is to the right.
Note what happens to \(f(x)\) as \(x\) increases from strongly negative values to strongly positive values.
When \(x\) is large and negative, \(f(x) \gt 0\text{.}\)
As \(x\) increases, \(f(x)\) decreases continuously until \(x=\frac{1}{2}\text{,}\) where \(f(x) \lt 0\text{.}\) In particular, since \(f(1)=\frac{65}{16} \gt 0\) and \(f\big(\frac{1}{2}\big) \lt 0\) and \(f(x)\) is continuous, the intermediate value theorem guarantees that \(f(x)\) takes the value zero for some \(x\) between \(\frac{1}{2}\) and \(1\text{.}\) More descriptively put, as \(x\) increases from hugely negative numbers to \(\frac{1}{2}\text{,}\) \(f(x)\) passes through zero exactly once.
As \(x\) increases beyond \(\frac{1}{2}\text{,}\) \(f(x)\) increases continuously, starting negative and becoming very large and positive when \(x\) becomes large and positive. In particular, since \(f(0)=\frac{1}{16} \gt 0\) and \(f\big(\frac{1}{2}\big) \lt 0\) and \(f(x)\) is continuous, the intermediate value theorem guarantees that \(f(x)\) takes the value zero for some \(x\) between \(\frac{1}{2}\) and \(0\text{.}\) So, as \(x\) increases from \(\frac{1}{2}\) to near \(+\infty\text{,}\) \(f(x)\) again passes through zero exactly once.
So \(f(x)\) must have exactly two roots, one with \(x \lt \frac{1}{2}\) and one with \(x \gt \frac{1}{2}\text{.}\)
If \(f(x)=0\text{,}\) then \(x^3=\left\sin\left(x^5\right)\right \leq 1\text{.}\) When \(x \lt 1\text{,}\) is \(\cos(x^5)\) positive or negative?
We can see by inspection that \(f(0)=0\text{,}\) so there is at least one root.
To hunt for other roots, notice than when \(f(x)=0\text{,}\) \(x^3=\sin\left(x^5\right)\text{,}\) and since the sine function only takes on values in \([1,1]\text{,}\) this means \(1 \leq x^3 \leq 1\text{.}\) That is, all roots of \(f(x)\) are in the interval \([1,1]\text{.}\) (An equivalent way to see this is to notice that for all \(x \gt 1\text{,}\) \(f(x) \gt 0\text{,}\) and for all \(x \lt 1\text{,}\) \(f(x) \lt 0\text{.}\))
If there is a root \(a \neq 0\text{,}\) then by Rolle's Theorem (since \(f(x)\) is continuous and differentiable for all real numbers) \(f'(c)=0\) for some \(c\) strictly between 0 and \(a\text{.}\) In particular, since we already know any roots \(a\) will be between \(1\) and \(1\text{,}\) if \(f(x)\) has two roots then \(f'(c)=0\) for some \(c \in (1,0) \cup (0,1)\text{.}\)
\begin{equation*}
f'(x)=3x^2+5x^4\cos\left(x^5\right)=x^2\left(3+5x^2\cos\left(x^5\right)\right)
\end{equation*}
So, if \(f'(x)=0\text{,}\) then \(x=0\) or \(\left(3+5x^2\cos\left(x^5\right)\right)=0\text{.}\) If \(x \in (1,0) \cup (0,1)\text{,}\) then \(x \lt 1\) and so \(x^5 \lt 1 \lt \frac{\pi}{2}\text{,}\) so \(\cos\left(x^5\right) \gt 0\text{.}\) Then \(3+5x^2\cos\left(x^5\right) \gt 3\text{,}\) so \(f'(x) \neq 0\text{.}\) That is, there is no \(c \in (1,0) \cup (0,1)\) with \(f'(c)=0\text{.}\) Therefore, following our last bullet point, \(f(x)\) has only one root.
Note here that \(f'(x)\) has many zeroesinfinitely many, in fact. However, \(x=0\) is the only root of \(f'(x)\) in the interval \((1,1)\text{.}\)
We are to find the number of positive solutions to the equation \(e^x = 4\cos(2x)\text{.}\) The figure below contains the graphs of \(y=e^x\) and \(y= 4\cos(2x)\text{.}\) The solutions to \(e^x = 4\cos(2x)\) are precisely the \(x\)'s where \(y=e^x\) and \(y= 4\cos(2x)\) cross.
It sure looks like there is exactly one crossing with \(x\ge 0\) and that one crossing is somewhere between \(x=0\) and \(x=1\text{.}\) Indeed since \(\big[e^x 4\cos(2x)\big]_{x=0} = 3 \lt 0\) and \(\big[e^x 4\cos(2x)\big]_{x=1} \gt e \gt 0\) and \(f(x)=e^x 4\cos(2x)\) is continuous, the intermediate value theorem guarantees that there is at least one root with \(0 \lt x \lt 1\text{.}\)
We still have to show that there is no second root — even if our graphs are not accurate.
Recall that the range of the cosine function is \([1,1]\text{.}\) If \(e^x=4\cos(2x)\text{,}\) then \(e^x \leq 4\text{,}\) so \(x \leq \log(4)\approx 1.39\text{.}\) So, we only need to search for roots of \(f(x)\) on the interval \((0,1.4)\text{:}\) we are guaranteed there are no roots elsewhere. Over this interval, \(2x \in (0,2.8)\text{,}\) so \(\sin(2x) \gt 0\text{,}\) and thus \(f'(x)=e^x+8\sin(2 x) \gt 0\text{.}\) Since \(f'(x)\) has no roots in \((0,1.4)\text{,}\) we conclude by Rolle's Theorem that \(f(x)\) has at most one root in \((0,1.4)\) (and so at most one positive root total). Since we've already found that a root of \(f(x)\) exists in \((0,1)\text{,}\) we conclude \(e^x=4\cos(2x)\) has precisely one positivevalued solution.
The derivative is nonnegative everywhere. The only values of \(x\) for which \(f'(x)=0\) are \(1\) and \(1\text{,}\) so \(f'(x) \gt 0\) for every \(x\) in \((1,1)\text{.}\)
2.13.5.15.b If \(f(x)\) has two roots \(a\) and \(b\) in \([1,1]\text{,}\) then by Rolle's Theorem, \(f'(c)=0\) for some \(x\) strictly between \(a\) and \(b\text{.}\) But since \(a\) and \(b\) are in \([1,1]\text{,}\) and \(c\) is between \(a\) and \(b\text{,}\) that means \(c\) is in \((1,1)\text{;}\) however, we know for every \(c\) in \((1,1)\text{,}\) \(f'(c) \gt 0\text{,}\) so this can't happen. Therefore, \(f(x)\) does not have two roots \(a\) and \(b\) in \([1,1]\text{.}\) This means \(f(x)\) has at most one root in \([1,1]\text{.}\)
The derivative is nonnegative everywhere. The only values of \(x\) for which \(f'(x)=0\) are \(1\) and \(1\text{,}\) so \(f'(x) \gt 0\) for every \(x\) in \((1,1)\text{.}\)
2.13.5.15.b If \(f(x)\) has two roots \(a\) and \(b\) in \([1,1]\text{,}\) then by Rolle's Theorem, \(f'(c)=0\) for some \(x\) strictly between \(a\) and \(b\text{.}\) But since \(a\) and \(b\) are in \([1,1]\text{,}\) and \(c\) is between \(a\) and \(b\text{,}\) that means \(c\) is in \((1,1)\text{;}\) however, we know for every \(c\) in \((1,1)\text{,}\) \(f'(c) \gt 0\text{,}\) so this can't happen. Therefore, \(f(x)\) does not have two roots \(a\) and \(b\) in \([1,1]\text{.}\) This means \(f(x)\) has at most one root in \([1,1]\text{.}\)
Write \(f(x)=e^x\text{.}\) Since \(f(x)\) is continuous and differentiable, the Mean Value Theorem asserts that there exists some \(c\) between \(0\) and \(T\) such that
\begin{align*}
f'(c)&=\frac{f(T)f(0)}{T0}\\
\end{align*}
The problem asks us to find this value of \(c\text{.}\) Solving:
\begin{align*}
e^c&=\frac{e^Te^0}{T}\\
e^c&=\frac{e^T1}{T}\\
c&=\log\left(\frac{e^T1}{T}\right)
\end{align*}
The domains of \(\arcsec x\) and \(C\arcsec x\) are the same: \(x \geq 1\text{.}\) Define \(f(x)=\arcsec x + \arccsc x\text{,}\) and note the domain of \(f(x)\) is also \(x \geq 1\text{.}\) Using Theorem 2.12.8,
By Corollary 2.13.12, this means \(f(x)\) is constant. That is, \(f(x)=C\) for some constant \(C\text{,}\) so \(C=\arcsec x + \arccsc x\text{,}\) hence
\begin{equation*}
\arcsec x = C\arccsc x
\end{equation*}
The domains of \(\arcsec x\) and \(C\arccsc x\) are the same: \(x \geq 1\text{.}\) Define \(f(x)=\arcsec x + \arccsc x\text{,}\) and note the domain of \(f(x)\) is also \(x \geq 1\text{.}\) Using Theorem 2.12.8,
By Corollary 2.13.12, this means \(f(x)\) is constant. That is, \(f(x)=C\) for some constant \(C\text{,}\) so \(C=\arcsec x + \arccsc x\text{,}\) hence
\begin{equation*}
\arcsec x = C\arccsc x
\end{equation*}
for all \(x \geq 1\text{.}\)
In order to find \(C\text{,}\) we can use any value of \(x\) where we know cosecant and arcsecant. For instance, \(x=1\text{:}\) \(\arcsec(1)=\arccos\left(\frac{1}{1}\right)=0\text{,}\) and \(\arccsc(1)=\arcsin\left(\frac{1}{1}\right)=\frac{\pi}{2}\text{.}\) So,
Show that \(f\) is differentiable by showing that \(f'(x)\) exists for every \(x\text{.}\) Then, the Mean Value Theorem applies. What is the largest \(f'(x)\) can be, for any \(x\text{?}\) If \(f(100) \lt 100\text{,}\) what does the MVT tell you must be true of \(f'(c)\) for some \(c\text{?}\)
Since \(f'(x)\) exists for every \(x\text{,}\) we see that \(f\) is differentiable, so the Mean Value Theorem applies. If \(f(100)\) is greater than or equal to 100, then by the Mean Value Theorem, there would have to be some \(c\) between \(0\) and \(100\) such that
Since \(f'(x) \leq 1\) for every \(x\text{,}\) there is no value of \(c\) as described. Therefore, it is not possible that \(f(100) \geq 100\text{.}\) So, \(f(100) \lt 100\text{.}\)
Since \(f'(x)\) exists for every \(x\text{,}\) we see that \(f\) is differentiable, so the Mean Value Theorem applies. If \(f(100)\) is greater than or equal to 100, then by the Mean Value Theorem, there would have to be some \(c\) between \(0\) and \(100\) such that
Since \(f'(x) \lt 1\) for every \(x\text{,}\) there is no value of \(c\) as described. Therefore, it is not possible that \(f(100) \geq 100\text{.}\) So, \(f(100) \lt 100\text{.}\)
If \(2x+\sin x\) is onetoone over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if \(f(a)=f(b)\) for distinct \(a\) and \(b\text{,}\) then \(f'(c)=0\) for some \(c\) between \(a\) and \(b\text{.}\) However, \(f'(x)=2+\cos x\text{,}\) which is never zero. In fact, \(f'(x)\ge 1\) for all \(x\text{,}\) so \(f(x)\) is strictly increasing over its entire domain. Therefore, our function \(f\) never takes the same value twice, so it is onetoone over all the real numbers, \((\infty,\infty)\text{.}\)
When we define the inverse function \(f^{1}(x)\text{,}\) the domain of \(f\) is the range of \(f^{1}\text{,}\) and viceversa. In general, we might have to restrict the domain of \(f\) (and hence the range of \(f^{1}\)) to an interval where \(f\) is onetoone, but in our case, this isn't necessary. So, the range of \(f^{1}\) is \((\infty,\infty)\) and the domain of \(f^{1}\) is the range of \(f\text{:}\) \((\infty,\infty)\text{.}\)
Onetoone interval, and range of \(f^{1}\text{:}\) \(\left[\frac{2\pi}{3},\frac{2\pi}{3}\right]\) Domain of \(f^{1}\text{:}\) \(\left[\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]\)
If \(f(x)=\dfrac{x}{2}+\sin x\) is onetoone over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if \(f(a)=f(b)\) for distinct \(a\) and \(b\text{,}\) then \(f'(c)=0\) for some \(c\) between \(a\) and \(b\text{.}\) Since \(f'(x)=\frac{1}{2}+\cos x\text{,}\) \(f'(x)=0\) when \(x=2n\pi\pm\frac{2\pi}{3}\) for some integer \(n\text{.}\) So, in particular, if \(a\) and \(b\) are distinct numbers in the interval \(\left[\frac{2\pi}{3},\frac{2\pi}{3}\right]\text{,}\) then for every \(c\) strictly between \(a\) and \(b\text{,}\) \(f'(c) \neq 0\text{,}\) so by Rolle's Theorem \(f(a) \neq f(b)\text{.}\) Therefore \(f(x)\) is onetoone on the interval \(\left[\frac{2\pi}{3},\frac{2\pi}{3}\right]\text{.}\)
We should also show that the interval \(\left[\frac{2\pi}{3},\frac{2\pi}{3}\right]\) cannot be extended to a larger interval over which \(f(x)\) is still onetoone. Consider the derivative \(f'(x) = \frac{1}{2}+\cos x\text{.}\) For all \(\frac{2\pi}{3} \lt x \lt \frac{2\pi}{3}\text{,}\) we have \(\cos x \gt \frac{1}{2}\) (sketch the graph of \(\cos x\) yourself) so that \(f'(x) \gt 0\) and \(f(x)\) is increasing. But at \(x=\frac{2\pi}{3}\text{,}\) \(f'(x)=0\text{,}\) and then for \(x\) a bit bigger than \(\frac{2\pi}{3}\) we have \(\cos x \lt \frac{1}{2}\) so that \(f'(x) \lt 0\) and \(f(x)\) is decreasing. So the graph “reverses direction”, and \(f(x)\) repeats values. (See the graph of \(y=f(x)\) below.) The same is true for \(x\) a little smaller than \(\frac{2\pi}{3}.\)
When we define the inverse function \(f^{1}(x)\text{,}\) first we restrict \(f\) to \(\left[\frac{2\pi}{3},\frac{2\pi}{3}\right]\text{.}\) Then the range of \(f^{1}\) is also \(\left[\frac{2\pi}{3},\frac{2\pi}{3}\right]\text{.}\) The domain of \(f^{1}\) is the range of \(f\) over this interval, so \(\left[\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]\text{.}\)
Define \(h(x)=f(x)g(x)\text{,}\) and notice \(h(a)=f(a)g(a) \lt 0\) and \(h(b)=f(b)g(b) \gt 0\text{.}\) Since \(h\) is the difference of two functions that are continuous over \([a,b]\) and differentiable over \((a,b)\text{,}\) also \(h\) is continuous over \([a,b]\) and differentiable over \((a,b)\text{.}\) So, by the Mean Value Theorem, there exists some \(c \in (a,b)\) with
Since \((a,b)\) is an interval, \(b \gt a\text{,}\) so the denominator of the above expression is positive; since \(h(b) \gt 0 \gt h(a)\text{,}\) also the numerator of the above expression is positive. So, \(h'(c) \gt 0\) for some \(c \in (a,b)\text{.}\) Since \(h'(c)=f'(c)g'(c)\text{,}\) we conclude \(f'(c) \gt g'(c)\) for some \(c \in (a,b)\text{.}\)
Define \(h(x)=f(x)g(x)\text{,}\) and notice \(h(a)=f(a)g(a) \lt 0\) and \(h(b)=f(b)g(b) \gt 0\text{.}\) Since \(h\) is the difference of two functions that are continuous over \([a,b]\) and differentiable over \((a,b)\text{,}\) also \(h\) is continuous over \([a,b]\) and differentiable over \((a,b)\text{.}\) So, by the Mean Value Theorem, there exists some \(c \in (a,b)\) with
Since \((a,b)\) is an interval, \(b \gt a\text{,}\) so the denominator of the above expression is positive; since \(h(b) \gt 0 \gt h(a)\text{,}\) also the numerator of the above expression is positive. So, \(h'(c) \gt 0\) for some \(c \in (a,b)\text{.}\) Since \(h'(c)=f'(c)g'(c)\text{,}\) we conclude \(f'(c) \gt g'(c)\) for some \(c \in (a,b)\text{.}\)
Since \(f(x)\) is differentiable over all real numbers, it is also continuous over all real numbers. We claim that \(f(x)\) cannot have four or more distinct roots. For every two distinct roots \(a \lt b\text{,}\) Rolle's Theorem tells us there is a \(c \in (a,b)\) such that \(f'(c)=0\text{:}\) that is, \(c\) is a root of \(f'\text{.}\) Since \(f'\) has only two distinct roots, \(f\) can have at most three distinct roots.
We are asked to find the number of solutions to the equation \(x^2+5x+1 = \sin x\text{.}\) The figure below contains the graphs of \(y=x^2+5x+1\) and \(y=\sin x\text{.}\) The solutions to \(x^2+5x+1 = \sin x\) are precisely the \(x\)'s where \(y=x^2+5x+1\) and \(y=\sin x\) cross.
From the figure, it sure looks like there are two crossings. Since the function \(\sin x\) has range \([1,1]\text{,}\) if the two functions cross, then also \(1 \leq x^2+5x+1\leq 1 \text{.}\) This portion of the quadratic function is highlighted in blue in the figure. The \(x\) coordinates of the end points of the blue arcs are found by solving \(x^2+5x+1 = \pm1\text{,}\) i.e. \(x=0\text{,}\) \(5\text{,}\) and (using the quadratic equation) \(x= \frac{5\pm\sqrt{17}}{2}\text{.}\)
We are now in a position to exploit the intuition that we have built using the above figure to write a concise argument showing that \(f(x)\) has exactly two roots. Remember that, in general, if we want to show that a function has \(n\) roots, we need to show that there exist \(n\) distinct roots somewhere, and that there do not exist \(n+1\) distinct roots. This argument is given below, in blue text. \color{blue}
\(f(x)\) is continuous over all real numbers
\(f(x)=0\) only when \(\sin x=x^2+5x+1\text{,}\) which only happens when \(x^2+5x+1 \leq 1\text{.}\) Thus, \(f(x)\) only has roots in the intervals \(\left[5,\frac{5\sqrt{17}}{2}\right]\) and \(\left[\frac{5+\sqrt{17}}{2},0\right]\text{.}\)
\(f(5)=\sin(5)+1 \gt 0\text{,}\) and \(f\left(\frac{5\sqrt{17}}{2}\right)=\sin\left(\frac{5\sqrt{17}}{2}\right)1 \lt 0\text{.}\) So, by the IVT, \(f(c)=0\) for some \(c \in \left(5,\frac{5\sqrt{17}}{2}\right)\text{.}\)
\(f(0)=1 \gt 0\text{,}\) and \(f\left(\frac{5+\sqrt{17}}{2}\right)=\sin\left(\frac{5+\sqrt{17}}{2}\right)1 \lt 0\text{.}\) So, by the IVT, \(f(c)=0\) for some \(c \in \left(\frac{5+\sqrt{17}}{2},0\right)\text{.}\)
\(f'(x)=\cos x + 2x + 5\text{.}\) If \(f'(x)=0\text{,}\) then \(2x+5=\cos x\text{,}\) so \(2x+5 \leq 1\text{.}\) So, the only interval that can contain roots of \(f'(x)\) is \([3,2]\text{.}\)
Suppose \(f(x)\) has more than two roots. Then it has two roots in the interval \(\left[5,\frac{5\sqrt{17}}{2}\right]\) OR it has two roots in the interval \(\left[\frac{5+\sqrt{17}}{2},0\right]\text{.}\) Since \(f(x)\) is differentiable for all real numbers, Rolle's Theorem tells us that \(f'(x)\) has a root in \(\left(5,\frac{5\sqrt{17}}{2}\right)\) or in \(\left(\frac{5+\sqrt{17}}{2},0\right)\text{.}\) However, since all roots of \(f'(x)\) are in the interval \([3,2]\text{,}\) and this interval shares no points with \(\left(5,\frac{5\sqrt{17}}{2}\right)\) or \(\left(\frac{5+\sqrt{17}}{2},0\right)\text{,}\) this cannot be the case. Therefore \(f(x)\) does not have more than two roots.
Since \(f(x)\) has at least two roots, and not more than two roots, \(f(x)\) has exactly two roots.
The derivative of \(e^x\) is \(e^x\text{:}\) taking derivatives leaves the function unchanged, even if we do it 180 times. So \(f^{(180)}=e^x\text{.}\)
Since \(f'(x) \gt 0\) over \((a,b)\text{,}\) we know from Corollary 2.13.12 that \(f(x)\) is increasing over \((a,b)\text{,}\) so 2.14.2.2.ii holds. Since \(f''(x) \gt 0\text{,}\) and \(f''(x)\) is the derivative of \(f'(x)\text{,}\) by the same reasoning we see that \(f'(x)\) is increasing. Since \(f'(x)\) is the rate at which \(f(x)\) is increasing, that means that the rate at which \(f'(x)\) is increasing is itself increasing: this is, 2.14.2.2.iv holds (and not2.14.2.2.iii).
There is no reason to think 2.14.2.2.i or 2.14.2.2.v holds, but to be thorough we will give an example showing that they do not need to be true. If \(f(x)=x^210\) and \((a,b)=(0,1)\text{,}\) then \(f'(x)=2x \gt 0\) over \((0,1)\text{,}\) and \(f''(x)=2 \gt 0\) everywhere, but \(f(x) \lt 0\) for all \(x \in (0,1)\text{,}\) so 2.14.2.2.i does not hold. Also, \(f'''(x)=0\) everywhere, so 2.14.2.2.v does not hold either.
Every time we differentiate \(f(x)\text{,}\) the constant out front gets multiplied by an everdecreasing constant, while the power decreases by one. As in Example 2.14.2, \(\ds\ddiff{{15}}{}{x}ax^{15}=a\cdot 15!\text{.}\) So, if \(a\cdot 15!=3\text{,}\) then \(a=\dfrac{3}{15!}\text{.}\)
The derivative \(\ds\diff{y}{x}\) is \(\dfrac{11}{4}\) only at the point \((1,3)\text{:}\) it is not constantly \(\dfrac{11}{4}\text{,}\) so it is wrong to differentiate the constant \(\dfrac{11}{4}\) to find \(\ds\ddiff{2}{y}{x}\text{.}\) Below is a correct solution.
\begin{align*}
28x+2y+2xy'+2yy'&=0\\
\end{align*}
Plugging in \(x=1\text{,}\) \(y=3\text{:}\)
\begin{align*}
28+6+2y'+6y'&=0\\
y'&=\frac{11}{4} \quad\mbox{\textcolor{red}{ at the point $(1,3)$}}\\
\end{align*}
Differentiating \textcolor{red}{the equation $28x+2y+2xy'+2yy'=0$:}
\begin{align*}
28+2y'+2y'+2xy''+2y'y'+2yy''&=0\\
4y'+2(y')^2+2xy''+2yy''&=28\\
\end{align*}
At the point \((1,3)\text{,}\) \(y'=\dfrac{11}{4}\text{.}\) Plugging in:
\begin{align*}
4\left(\frac{11}{4}\right)+2\left(\frac{11}{4}\right)^2+2(1)y''+2(3)y''&=28\\
y''&=\frac{15}{64}
\end{align*}
Solution
The derivative \(\ds\diff{y}{x}\) is \(\dfrac{11}{4}\) only at the point \((1,3)\text{:}\) it is not constantly \(\dfrac{11}{4}\text{,}\) so it is wrong to differentiate the constant \(\dfrac{11}{4}\) to find \(\ds\ddiff{2}{y}{x}\text{.}\) Below is a correct solution.
\begin{align*}
28x+2y+2xy'+2yy'&=0\\
\end{align*}
Plugging in \(x=1\text{,}\) \(y=3\text{:}\)
\begin{align*}
28+6+2y'+6y'&=0\\
y'&=\frac{11}{4} \quad\mbox{\textcolor{red}{ at the point $(1,3)$}}\\
\end{align*}
Differentiating \textcolor{red}{the equation $28x+2y+2xy'+2yy'=0$}:
\begin{align*}
28+2y'+2y'+2xy''+2y'y'+2yy''&=0\\
4y'+2(y')^2+2xy''+2yy''&=28\\
\end{align*}
At the point \((1,3)\text{,}\) \(y'=\dfrac{11}{4}\text{.}\) Plugging in:
\begin{align*}
4\left(\frac{11}{4}\right)+2\left(\frac{11}{4}\right)^2+2(1)y''+2(3)y''&=28\\
y''&=\frac{15}{64}
\end{align*}
\begin{align*}
2x+2yy'&=0\\
2+(2y)y''+(2y')y'&=0\\
y''&=\frac{(y')^2+1}{y}\\
\end{align*}
So, we need an expression for \(y'\text{.}\) We use the equation \(2x+2yy'=0\) to conclude \(y'=\dfrac{x}{y}\text{:}\)
\begin{align*}
y''&=\frac{\left(\frac{x}{y}\right)^2+1}{y}\\
&=\frac{\frac{x^2}{y^2}+1}{y}\\
&=\frac{x^2+y^2}{y^3}\\
&=\frac{1}{y^3}
\end{align*}
\begin{align*}
\diff{}{x}\{\log(5x^212)\}&=\frac{10x}{5x^212}\\
\end{align*}
Using the quotient rule:
\begin{align*}
\ddiff{2}{}{x}\{\log(5x^212)\}&=\diff{}{x}\left\{\frac{10x}{5x^212}\right\}\\
&=\frac{(5x^212)(10)10x(10x)}{(5x^212)^2}\\
&=\frac{10(5x^2+12)}{(5x^212)^2}\\
\end{align*}
Using the quotient rule one last time:
\begin{align*}
\ddiff{3}{}{x}\{\log(5x^212)\}&=\diff{}{x}\left\{
\frac{10(5x^2+12)}{(5x^212)^2}
\right\}\\
&=\frac{(5x^212)^2(10)(10x)+10(5x^2+12)(2)(5x^212)(10x)}{(5x^212)^4}\\
&=\frac{(5x^212)(100x)+(200x)(5x^2+12)}{(5x^212)^3}\\
&=\frac{100x(5x^2+12+10x^2+24)}{(5x^212)^3}\\
&=\frac{100x(5x^2+36)}{(5x^212)^3}
\end{align*}
The velocity of the particle is given by \(h'(t)=\sin t\text{.}\) Note \(0 \lt 1 \lt \pi\text{,}\) so \(h'(1) \gt 0\)the particle is rising (moving in the positive direction, in this case “up”). The acceleration of the particle is \(h''(t)=\cos t\text{.}\) Since \(0 \lt 1 \lt \frac{\pi}{2}\text{,}\) \(h''(t) \gt 0\text{,}\) so \(h'(t)\) is increasing: the particle is moving up, and it's doing so at an increasing rate. So, the particle is speeding up.
\(h'(t)\) gives the velocity of the particle, and \(h''(t)\) gives its accelerationthe rate the velocity is changing. Be wary of signsas in legends, they may be misleading.
For this problem, remember that velocity has a sign indicating direction, while speed does not.
The velocity of the particle is given by \(h'(t)=3t^22t5\text{.}\) At \(t=1\text{,}\) the velocity of the particle is \(4\text{,}\) so the particle is moving downwards with a speed of 4 units per second. The acceleration of the particle is \(h''(t)=6t2\text{,}\) so when \(t=1\text{,}\) the acceleration is (positive) \(4\) units per second per second. That means the velocity (currently \(4\) units per second) is becoming a bigger numbersince the velocity is negative, a bigger number is closer to zero, so the speed of the particle is getting smaller. (For instance, a velocity of \(3\) represents a slower motion than a velocity of \(4\text{.}\)) So, the particle is slowing down at \(t=1\text{.}\)
Solution
\begin{align*}
x^2+x+y&=\sin(xy)\\
\end{align*}
We differentiate implicitly. For ease of notation, we write \(y'\) for \(\ds\diff{y}{x}\text{.}\)
\begin{align*}
2x+1+y'&=\cos(xy)(y+xy')\\
\end{align*}
We're interested in \(y''\text{,}\) so we implicitly differentiate again.
\begin{align*}
2+y''&=\sin(xy)(y+xy')^2+\cos(xy)(2y'+xy'')\\
\end{align*}
We want to know what \(y''\) is when \(x=y=0\text{.}\) Plugging these in yields the following:
\begin{align*}
2+y''&=2y'\\
\end{align*}
So, we need to know what \(y'\) is when \(x=y=0\text{.}\) We can get this from the equation \(2x+1+y'=\cos(xy)(y+xy')\text{,}\) which becomes \(1+y'=0\) when \(x=y=0\text{.}\) So, at the origin, \(y'=1\text{,}\) and
\begin{align*}
2+y''&=2(1)\\
y''&=4
\end{align*}
Remark: a common mistake is to stop at the equation \(2x+1+y'=\cos(xy)(y+xy')\text{,}\) plug in \(x=y=0\text{,}\) find \(y'=1\text{,}\) and decide \(y''=\ds\diff{}{x}\{1\}=0\text{.}\) This is due to a slight sloppiness in the usual notation. When we wrote \(y'=1\text{,}\) what we meant is that at the point \((0,0)\text{,}\) \(\ds\diff{y}{x}=1\text{.}\) More properly written: \(\left.\ds\diff{y}{x}\right_{x=0, y=0}=1\text{.}\) This is not the same as saying \(y'=1\) everywhere (in which case, indeed, \(y''\) would be 0 everywhere).
\begin{align*}
\diff{}{x} \sin x &= \cos x\\
\diff{}{x} \cos x &= \sin x\\
\diff{}{x} \{\sin x\} &= \cos x\\
\diff{}{x} \{\cos x\} &= \sin x\\
\diff{}{x} \sin x &= \cos x\\
\end{align*}
The fourth derivative is \(\sin x\) is \(\sin x\text{,}\) and the fourth derivative of \(\cos x\) is \(\cos x\text{,}\) so (a) and (b) are true.
\begin{align*}
\diff{}{x}\tan x &=\sec^2 x\\
\diff{}{x}\sec^2 x &=2\sec x (\sec x \tan x)=2\sec^2x\tan x\\
\diff{}{x}\{2\sec^2x\tan x\}&=(4\sec x \cdot \sec x \tan x)\tan x+2\sec^2x\sec^2x\\
&=4\sec^2x\tan^2x+2\sec^4x\\
\diff{}{x}\{4\sec^2x\tan^2x+2\sec^4x\}&=(8\sec x \cdot \sec x \tan x)\tan^2x+4\sec^2x(2\tan x \cdot\sec^2x)\\
&\quad+8\sec^3x\cdot\sec x \tan x\\
&=8\sec^2x\tan^3x+16\sec^4x\tan x
\end{align*}
So, \(\ds\ddiff{4}{}{x} \tan x =8\sec^2x\tan^3x+16\sec^4x\tan x\text{.}\) It certainly seems like this is not the same as \(\tan x\text{,}\) but remember that sometimes trig identities can fool you: \(\tan^2x+1=\sec^2x\text{,}\) and so on. So, to be absolutely sure that these are not equal, we need to find a value of \(x\) so that the output of one is not the same as the output of the other. When \(x=\frac{\pi}{4}\text{:}\)
\begin{equation*}
8\sec^2x\tan^3x+16\sec^4x\tan x = 8\left({\sqrt{2}}\right)^2(1)^3+16\left({\sqrt{2}}\right)^4(1)=80\neq 1=\tan x.
\end{equation*}
Since \(f'(x) \lt 0\text{,}\) we need a decreasing function. This only applies to (ii), (iii), and (v). Since \(f''(x) \gt 0\text{,}\) that means \(f'(x)\) is increasing, so the slope of the function must be increasing. In (v), the slope is constant, so \(f''(x)=0\)therefore, it's not (v). In (iii), the slope is decreasing, because near \(a\) the curve is quite flat (\(f'(x)\) near zero) but near \(b\) the curve is very steeply decreasing (\(f'(x)\) is a large negative number), so (iii) has a negative second derivative. By contrast, in (ii), the line starts out as steeply decreasing (\(f'(x)\) is a strongly negative number) and becomes flatter and flatter (\(f'(x)\) nears 0), so \(f'(x)\) is increasingin other words, \(f''(x) \gt 0\text{.}\) So, (ii) is the only curve that has \(f'(x) \lt 0\) and \(f''(x) \gt 0\text{.}\)
\begin{align*}
\diff{}{x}\{2^x\}&=2^x\log 2\\
\ddiff{2}{}{x}\{2^x\}&=2^x\log2 \cdot \log 2 = 2^x(\log2)^2\\
\ddiff{3}{}{x}\{2^x\}&=2^x(\log2)^2 \cdot \log 2 = 2^x(\log2)^3\\
\end{align*}
Every time we differentiate, we multiply the original function by another factor of \(\log 2\text{.}\) So, the \(n\)th derivative is given by:
\begin{align*}
\ddiff{n}{}{x}\{2^x\}&=2^x(\log2)^n
\end{align*}
In the above work, remember that \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are all constants. Since they are nonzero constants, \(\ds\ddiff{3}{f}{x} =6a\neq 0\text{.}\) So, the fourth derivative is the first derivative to be identically zero: \(n=4\text{.}\)
2.14.2.17.c \(f\) and \(h\) “start at the same place”, since \(f(0)=h(0)\text{.}\) Also \(f'(0)=h'(0)\text{,}\) and \(f''(x)=(4x^2+4x+3)e^{x+x^2} \gt 3e^{x+x^2} \gt 3=h''(x)\) when \(x \gt 0\text{.}\) Since \(f'(0)=h'(0)\text{,}\) and since \(f'\) grows faster than \(h'\) for positive \(x\text{,}\) we conclude \(f'(x) \gt h'(x)\) for all positive \(x\text{.}\) Now we can conclude that (since \(f(0)=h(0)\) and \(f\) grows faster than \(h\) when \(x \gt 0\)) also \(f(x) \gt h(x)\) for all positive \(x\text{.}\)
2.14.2.17.c \(f\) and \(h\) “start at the same place”, since \(f(0)=h(0)\text{.}\) If it were clear that \(f'(x)\) were greater than \(h'(x)\) for \(x \gt 0\text{,}\) then we would know that \(f\) grows faster than \(h\text{,}\) so we could conclude that \(f(x) \gt h(x)\text{,}\) as desired. Unfortunately, it is not obvious whether \((1+2x)e^{x+x^2}\) is always greater than \(1+3x\) for positive \(x\text{.}\) So, we look to the second derivative. \(f'(0)=h'(0)\text{,}\) and \(f''(x)=(4x^2+4x+3)e^{x+x^2} \gt 3e^{x+x^2} \gt 3=h''(x)\) when \(x \gt 0\text{.}\) Since \(f'(0)=h'(0)\text{,}\) and since \(f'\) grows faster than \(h'\) for positive \(x\text{,}\) we conclude \(f'(x) \gt h'(x)\) for all positive \(x\text{.}\) Now we can conclude that (since \(f(0)=h(0)\) and \(f\) grows faster than \(h\) when \(x \gt 0\)) also \(f(x) \gt h(x)\) for all positive \(x\text{.}\)
For 2.14.2.18.b, you know a point where the curve and tangent line intersect, and you know what the tangent line looks like. What do the derivatives tell you about the shape of the curve?
We differentiate implicitly.
\begin{align*}
x^3y(x)+y(x)^3&=10 x\\
3x^2y(x)+x^3y'(x)+3y(x)^2y'(x)&=10\\
\end{align*}
Subbing in \(x=1\) and \(y(1)=2\) gives
\begin{align*}
(3)(1)( 2)+(1)y'(1)+(3)(4) y'(1)&=10\\
13y'(1)&=4\\
y'(1)&=\frac{4}{13}
\end{align*}
From part 2.14.2.18.a, the slope of the curve at \(x=1,\ y=2\) is \(\dfrac{4}{13}\text{,}\) so the curve is increasing, but fairly slowly. The angle of the tangent line is \(\tan^{1}\left(\frac{4}{13}\right)\approx 17^\circ\text{.}\) We are also told that \(y''(1) \lt 0\text{.}\) So the slope of the curve is decreasing as \(x\) passes through 1. That is, the line is more steeply increasing to the left of \(x=1\text{,}\) and its slope is decreasing (getting less sleep, then possibly the slope even becomes negative) as we move past \(x=1\text{.}\)
2.14.2.19.c We notice that the coefficients of the derivatives of \(f\) correspond to the entries in the rows of Pascal's Triangle.
In the first derivative of \(g\text{,}\) the coefficients of \(f\) and \(f'\) correspond to the entries in the second row of Pascal's Triangle.
In the second derivative of \(g\text{,}\) the coefficients of \(f\text{,}\) \(f'\text{,}\) and \(f''\) correspond to the entries in the third row of Pascal's Triangle.
In the third derivative of \(g\text{,}\) the coefficients of \(f\text{,}\) \(f'\text{,}\) \(f''\text{,}\) and \(f'''\) correspond to the entries in the fourth row of Pascal's Triangle.
We guess that, in the fourth derivative of \(g\text{,}\) the coefficients of \(f\text{,}\) \(f'\text{,}\) \(f''\text{,}\) \(f'''\text{,}\) and \(f^{(4)}\) will correspond to the entries in the fifth row of Pascal's Triangle.
Rolle's Theorem relates the roots of a function to the roots of its derivative. So, the fifth derivative tells us something about the fourth, the fourth derivative tells us something about the third, and so on.
Since \(f(x)\) is differentiable over all real numbers, it is also continuous over all real numbers. Similarly, \(f'(x)\) is differentiable over all real numbers, so it is also continuous over all real numbers, and so on for the first \(n\) derivatives of \(f(x)\text{.}\)
Rolle's Theorem tells us that if \(a\) and \(b\) are distinct roots of a function \(g\text{,}\) then \(g'(x)=0\) for some \(c\) in \((a,b)\text{.}\) That is, \(g'\) has a root strictly between \(a\) and \(b\text{.}\) Expanding this idea, if \(g\) has \(m+1\) distinct roots, then \(g'\) must have at least \(m\) distinct roots, as in the sketch below.
So, if \(f^{(n)}(x)\) has only \(m\) roots, then \(f^{(n1)}(x)\) has at most \(m+1\) roots. Similarly, since \(f^{(n1)}(x)\) has at most \(m+1\) roots, \(f^{(n2)}(x)\) has at most \(m+2\) roots. Continuing in this way, we see \(f(x)=f^{(nn)}(x)\) has at most \(m+n\) distinct roots.
Let's begin by noticing that the domain of \(f(x)\) is \((1,\infty)\text{.}\)
By inspection, \(f(0)=0\text{,}\) so \(f(x)\) has at least one root.
If \(x\in(1,0)\text{,}\) then \((x+1)\) is positive, \(\log(x+1)\) is negative, \(\sin(x)\) is negative, and \(x^2\) is negative. Therefore, if \(x \lt 0\) is in the domain of \(f\text{,}\) then \(f(x) \lt 0\text{.}\) So, \(f(x)\) has no negative roots. We focus our attention on the case \(x \gt 0\text{.}\)
\(f'(x)=12x+\log(x+1)+\cos x\text{.}\) We would like to know how many positive roots \(f'(x)\) has, but it isn't obvious. So, let's differentiate again.
\(f''(x)=2+\frac{1}{x+1}\sin x\text{.}\) When \(x \gt 0\text{,}\) \(\frac{1}{x+1} \lt 1\text{,}\) so \(f''(x) \lt 1\sin(x) \leq 0\text{,}\) so \(f''(x)\) has no positive roots. Since \(f'(x)\) is continuous and differentiable over \((0,\infty)\text{,}\) and since \(f''(x) \neq 0\) for all \(x \in (0, \infty)\text{,}\) by Rolle's Theorem, \(f'(x)\) has at most one root in \([0,\infty)\text{.}\)
Since \(f(x)\) is continuous and differentiable over \([0,\infty)\text{,}\) and \(f'(x)\) has at most one root in \((0,\infty)\text{,}\) by Rolle's Theorem \(f(x)\) has at most two distinct roots in \([0,\infty)\text{.}\) (Otherwise, \(f(a)=f(b)=f(c)=0\) for some values \(0\le a \lt b \lt c\text{,}\) so \(f'(d)=f'(e)=0\) for some \(d\in(a,b)\) and some \(e \in (b,c)\text{,}\) but since \(f'(x)\) has at most one root, this is impossible.)
We know \(f(0)=0\text{,}\) so the remaining question is whether or not \(f(x)\) has a second root (which would have to be positive). As usual, we can show another root exists using the intermediate value theorem. We see that for large values of \(x\text{,}\) \(f(x)\) is negative, for example:
For positive values of \(x\) closer to zero, we hope to find a positive value of \(f(x)\text{.}\) However, it's quite difficult to get a number \(c\) that obviously gives \(f(c) \gt 0\text{.}\) It suffices to observe that \(f(0)=0\) and \(f'(0)=2 \gt 0\text{.}\) From the definition of the derivative, we can conclude \(f(x) \gt 0\) for some \(x \gt 0\text{.}\) (If it is not true that \(f(x) \gt 0\) for some \(x \gt 0\text{,}\) then \(f(x)\le 0\) for all \(x \gt 0\text{.}\) The definition of the derivative tells us that [since \(f'(0)\) exists] \(f'(0)=\ds\lim_{h \to 0^+}\frac{f(h)f(0)}{h}=\ds\lim_{h \to 0^+}\frac{f(h)}{h}\text{;}\) the denominator is positive, so if the numerator were always less than or equal to zero, the limit would be less than or equal to zero as well. However, the derivative is positive, so \(f(x) \gt 0\) for some \(x \gt 0\text{.}\)) Therefore, \(f(x)\) has a second root, so \(f(x)\) has precisely two roots.
You can rewrite this function as a piecewise function, with branches \(x \ge 0\) and \(x \lt 0\text{.}\) To figure out the derivatives at \(x = 0\text{,}\) use the definition of a derivative.
2.14.2.22.a In order to make \(f(x)\) a little more tractable, let's change the format. Since \(x=\left\{\begin{array}{rl}
x&x \geq 0\\
x&x \lt 0
\end{array}\right.\text{,}\) then:
Now, we turn to the definition of the derivative to figure out whether \(f'(0)\) exists.
\begin{align*}
f'(0)&=\lim_{h \to 0} \frac{f(0+h)f(0)}{h}=\lim_{h \to 0}\frac{f(h)0}{h} =\lim_{h \to 0}\frac{f(h)}{h}\qquad\mbox{if it exists.}\\
\end{align*}
Since \(f\) looks different to the left and right of 0, in order to evaluate this limit, we look at the corresponding onesided limits. Note that when \(h\) approaches 0 from the right, \(h \gt 0\) so \(f(h)=h^2\text{.}\) By contrast, when \(h\) approaches 0 from the left, \(h \lt 0\) so \(f(h)=h^2\text{.}\)
\begin{align*}
& \lim_{h \to 0^+} \frac{f(h)}{h}=\lim_{h \to 0^+}\frac{h^2}{h}=\lim_{h \to 0^+}h=0\\
& \lim_{h \to 0^} \frac{f(h)}{h}=\lim_{h \to 0^}\frac{h^2}{h}=\lim_{h \to 0^}h=0\\
\end{align*}
Since both onesided limits exist and are equal to 0,
\begin{align*}
& \lim_{h \to 0} \frac{f(0+h)f(0)}{h}=0
\end{align*}
and so \(f\) is differentiable at \(x=0\) and \(f'(0)=0\text{.}\)
So, whenever \(x \neq 0\text{,}\) \(f''(x)\) exists. To investigate the differentiability of \(f'(x)\) when \(x=0\text{,}\) again we turn to the definition of a derivative. If
\begin{align*}
&\lim_{h \to 0}\frac{f'(0+h)f'(0)}{h}\\
\end{align*}
exists, then \(f''(0)\) exists.
\begin{align*}
\lim_{h \to 0}\frac{f'(0+h)f'(0)}{h}&=\lim_{h \to 0} \frac{f'(h)0}{h}=\lim_{h \to 0}\frac{f'(h)}{h}\\
\end{align*}
Since \(f(h)\) behaves differently when \(h\) is greater than or less than zero, we look at the onesided limits.
\begin{align*}
\lim_{h \to 0^+}\frac{f'(h)}{h}&=\lim_{h \to 0^+}\frac{2h}{h}=2\\
\lim_{h \to 0^}\frac{f'(h)}{h}&=\lim_{h \to 0^}\frac{2h}{h}=2\\
\end{align*}
Since the onesided limits do not agree,
\begin{align*}
\lim_{h \to 0}\frac{f'(0+h)f'(0)}{h}&=DNE
\end{align*}
So, \(f''(0)\) does not exist. Now we have a complete picture of \(f''(x)\text{:}\)
2.14.2.22.a In order to make \(f(x)\) a little more tractable, let's change the format. Since \(x=\left\{\begin{array}{rl}
x&x \geq 0\\
x&x \lt 0
\end{array}\right.\text{,}\) then:
Now, we turn to the definition of the derivative to figure out whether \(f'(0)\) exists.
\begin{align*}
f'(0)&=\lim_{h \to 0} \frac{f(0+h)f(0)}{h}=\lim_{h \to 0}\frac{f(h)0}{h} =\lim_{h \to 0}\frac{f(h)}{h}\qquad\mbox{if it exists.}\\
\end{align*}
Since \(f\) looks different to the left and right of 0, in order to evaluate this limit, we look at the corresponding onesided limits. Note that when \(h\) approaches 0 from the right, \(h \gt 0\) so \(f(h)=h^2\text{.}\) By contrast, when \(h\) approaches 0 from the left, \(h \lt 0\) so \(f(h)=h^2\text{.}\)
\begin{align*}
&\;\; \lim_{h \to 0^+} \frac{f(h)}{h}=\lim_{h \to 0^+}\frac{h^2}{h}=\lim_{h \to 0^+}h=0\\
&\;\; \lim_{h \to 0^} \frac{f(h)}{h}=\lim_{h \to 0^}\frac{h^2}{h}=\lim_{h \to 0^}h=0\\
\end{align*}
Since both onesided limits exist and are equal to 0,
\begin{align*}
&\;\; \lim_{h \to 0} \frac{f(0+h)f(0)}{h}=0
\end{align*}
and so \(f\) is differentiable at \(x=0\) and \(f'(0)=0\text{.}\)
So, whenever \(x \neq 0\text{,}\) \(f''(x)\) exists. To investigate the differentiability of \(f'(x)\) when \(x=0\text{,}\) again we turn to the definition of a derivative. If
\begin{align*}
&\lim_{h \to 0}\frac{f'(0+h)f'(0)}{h}\\
\end{align*}
exists, then \(f''(0)\) exists.
\begin{align*}
\lim_{h \to 0}\frac{f'(0+h)f'(0)}{h}&=\lim_{h \to 0} \frac{f'(h)0}{h}=\lim_{h \to 0}\frac{f'(h)}{h}\\
\end{align*}
Since \(f(h)\) behaves differently when \(h\) is greater than or less than zero, we look at the onesided limits.
\begin{align*}
\lim_{h \to 0^+}\frac{f'(h)}{h}&=\lim_{h \to 0^+}\frac{2h}{h}=2\\
\lim_{h \to 0^}\frac{f'(h)}{h}&=\lim_{h \to 0^}\frac{2h}{h}=2\\
\end{align*}
Since the onesided limits do not agree,
\begin{align*}
\lim_{h \to 0}\frac{f'(0+h)f'(0)}{h}&=DNE
\end{align*}
So, \(f''(0)\) does not exist. Now we have a complete picture of \(f''(x)\text{:}\)
False. The acceleration of the ball is given by \(h''(t)=9.8\text{.}\) This is constant throughout its trajectory (and is due to gravity).
Remark: the velocity of the ball at \(t=2\) is zero, since \(h'(2)=9.8(2)+19.6=0\text{,}\) but the velocity is only zero for an instant. Since the velocity is changing, the acceleration is nonzero.
It takes 10 seconds to accelerate from \(2\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(3\;\frac{\mathrm{m}}{\mathrm{s}}\text{,}\) and \(100\) seconds to accelerate from \(3\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(13\;\frac{\mathrm{m}}{\mathrm{s}}\text{.}\)
The acceleration is constant, which means the rate of change of the velocity is constant. So, since it took 10 seconds for the velocity to increase by 1 metre per second (from \(1\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(2\;\frac{\mathrm{m}}{\mathrm{s}}\)), then it always takes 10 seconds for the velocity to increase by 1 metre per second.
So, it takes 10 seconds to accelerate from \(2\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(3\;\frac{\mathrm{m}}{\mathrm{s}}\text{.}\) To accelerate from \(3\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(13\;\frac{\mathrm{m}}{\mathrm{s}}\) (that is, to change its velocity by 10 metres per second), it takes \(10\times 10=100\) seconds.
Let \(v(a) = s'(a)\) be the velocity of the particle. If \(s''(a) \gt 0\text{,}\) then \(v'(a) \gt 0\) — so the velocity of the particle is increasing. However, that does not mean that its speed (the absolute value of velocity) is increasing as well. For example, if a velocity is increasing from \(4\) kph to \(3\) kph, the speed is decreasing from \(4\) kph to \(3\) kph. So, the statement is false in general.
Since \(s'(a) \gt 0\text{,}\) \(s'(a)=s'(a)\text{:}\) that is, the speed and velocity of the particle are the same. (This means the particle is moving in the positive direction.) If \(s''(a) \gt 0\text{,}\) then the velocity (and hence speed) of the particle is increasing. So, the statement is true.
The equation of an object falling from rest on the earth is derived in Example 3.1.2. It would be difficult to use exactly the version given for \(s(t)\text{,}\) but using the same logic, you can find an equation for the height of the flower pot at time \(t\text{.}\)
From Example 3.1.2, we know that an object falling from rest on the Earth is subject to the acceleration due to gravity, \(9.8\;\frac{\mathrm{m}}{\mathrm{s}^2}\text{.}\) So, if \(h(t)\) is the height of the flower pot \(t\) seconds after it rolls out the window, then \(h''(t)=9.8\text{.}\) (We make the acceleration negative, since the measure “height” has “up” as the positive direction, while gravity pulls the pot in the negative direction, “down.”)
Then \(h'(t)\) is a function whose derivative is the constant \(9.8\) and with \(h'(0)=0\) (since the object fell, instead of being thrown up or down), so \(h'(t)=9.8t\text{.}\)
What we want to know is \(h'(t)\) at the time \(t\) when the pot hits the ground. We don't know yet exactly what time that happens, so we go a little farther and find an expression for \(h(t)\text{.}\) The function \(h(t)\) has derivative \(9.8t\) and \(h(0)=10\text{,}\) so (again following the ideas in Example 3.1.2)
Now, we can find the time when the pot hits the ground: it is the time when \(h(t)=0\) (and \(t \gt 0\)).
\begin{align*}
0&=\frac{9.8}{2}t^2+10\\
\frac{9.8}{2}t^2&=10\\
t^2&=\frac{20}{9.8}\\
t&=+\sqrt{\frac{20}{9.8}}\approx1.4 \mathrm{sec}\\
\end{align*}
The velocity of the pot at this time is
\begin{align*}
h'\left(\sqrt{\frac{20}{9.8}}\right)&=9.8\left(\sqrt{\frac{20}{9.8}}\right)=\sqrt{20\cdot 9.8}
=14 \frac{\mathrm{m}}{\mathrm{s}}
\end{align*}
So, the pot is falling at 14 metres per second, just as it hits the ground.
Let \(s(t)\) be the distance the stone has fallen \(t\) seconds after dropping it. Since the acceleration due to gravity is \(9.8\;\frac{\mathrm{m}}{\mathrm{s}^2}\text{,}\) \(s''(t)=9.8\text{.}\) (We don't make this negative, because \(s(t)\) measures how far the stone has fallen, which means the positive direction in our coordinate system is “down”, which is exactly the way gravity is pulling.)
Then \(s'(t)\) has a constant derivative of 9.8, so \(s'(t)=9.8t+c\) for some constant \(c\text{.}\) Notice \(s'(0)=c\text{,}\) so \(c\) is the velocity of the stone at the very instant you dropped it, which is zero. Therefore, \(s'(t)=9.8t\text{.}\)
So, \(s(t)\) is a function with derivative \(9.8t\text{.}\) It's not too hard to figure out by guessing and checking that \(s(t)=\frac{9.8}{2}t^2+d\) for some constant \(d\text{.}\) Notice \(s(0)=d\text{,}\) so \(d\) is the distance the rock has travelled at the instant you dropped it, which is zero. So, \(s(t)=\frac{9.8}{2}t^2=4.9t^2\text{.}\)
Remark: this is exactly the formula found in Example 3.1.2. You may, in general, use that formula without proof, but you need to know where it comes from and be able to apply it in other circumstances where it might be slightly differentlike part (b) below.
The rock falls for \(x\) seconds, so the distance fallen is
\begin{equation*}
4.9x^2
\end{equation*}
Remark: this is a decent (if imperfect) way to figure out how deep a well is, or how tall a cliff is, when you're out and about. Drop a rock, square the time, multiply by 5.
(b) We'll go through a similar process as before.
Again, let \(s(t)\) be the distance the rock has fallen \(t\) seconds after it is let go. Then \(s''(t)=9.8\text{,}\) so \(s'(t)=9.8t+c\text{.}\) In this case, since the initial speed of the rock is \(1\) metre per second, \(1=s'(0)=c\text{,}\) so \(s'(t)=9.8t+1\text{.}\)
Then, \(s(t)\) is a function whose derivative is \(9.8t+1\text{,}\) so \(s(t)=\frac{9.8}{2}t^2+t+d\) for some constant \(d\text{.}\) Since \(0=s(0)=d\text{,}\) we see \(s(t)=4.9t^2+t\text{.}\)
So, if the rock falls for \(x\) seconds, the distance fallen is
\begin{equation*}
4.9x^2+x
\end{equation*}
Remark: This means there is an error of \(x\) metres in your estimation of the depth of the well.
Acceleration is constant, so finding a formula for the distance your keys have travelled is a similar problem to finding a formula for something falling.
Let \(s(t)\) be the distance your keys have travelled since they left your hand. The rate at which they are travelling, \(s'(t)\text{,}\) is decreasing by 0.25 metres per second. That is, \(s''(t)=0.25\text{.}\) Therefore, \(s'(t)=0.25t+c\) for some constant \(c\text{.}\) Since \(c=s'(0)=2\text{,}\) we see
\begin{equation*}
s'(t)=20.25t
\end{equation*}
Then \(s(t)\) has \(20.25t\) as its derivative, so \(s(t)=2t\frac{1}{8}t^2+d\) for some constant \(d\text{.}\) At time \(t=0\text{,}\) the keys have not yet gone anywhere, so \(0=s(0)=d\text{.}\) Therefore,
We need to figure out which of these values of \(t\) is really the time when the keys reach your friend. The keys travel this way from \(t=0\) to the time they reach your friend. (Then \(s(t)\) no longer describes their motion.) So, we need to find the first value of \(t\) that is positive with \(s(t)=2\text{.}\) Since \(84\sqrt{3} \gt 0\text{,}\) this is the first time \(s(t)=2\) and \(t \gt 0\text{.}\) So, the keys take
We proceed with the technique of Example 3.1.3 in mind.
Let \(v(t)\) be the velocity (in kph) of the car at time \(t\text{,}\) where \(t\) is measured in hours and \(t=0\) is the instant the brakes are applied. Then \(v(0)=100\) and \(v'(t)=50000\text{.}\) Since \(v'(t)\) is constant, \(v(t)\) is a line with slope \(50000\) and intercept \((0,100)\text{,}\) so
\begin{equation*}
v(t)=10050000t
\end{equation*}
The car comes to a complete stop when \(v(t)=0\text{,}\) which occurs at \(t=\frac{100}{50000}=\frac{1}{500}\) hours. This is a confusing measure, so we convert it to seconds:
Suppose the deceleration provided by the brakes is \(d\;\frac{\mathrm{km}}{\mathrm{hr}^2}\text{.}\) Then if \(v(t)\) is the velocity of the car, \(v(t)=120dt\) (at \(t=0\text{,}\) the velocity is 120, and it decreases by \(d\) kph per hour). The car stops when \(0=v(t)\text{,}\) so \(t=\frac{120}{d}\) hours.
Let \(s(t)\) be the distance the car has travelled \(t\) hours after applying the brakes. Then \(s'(t)=v(t)\text{,}\) so \(s(t)=120t\frac{d}{2}t^2+c\) for some constant \(c\text{.}\) Since \(0=s(0)=c\text{,}\)
Since your deceleration is constant, your speed decreases smoothly from 100 kph to 0 kph. So, one second before your stop, you only have \(\frac{1}{7}\) of our speed left: you're going \(\frac{100}{7}\) kph.
A less direct way to solve this problem is to note that \(v(t)=100dt\) is the velocity of car \(t\) hours after braking, if \(d\) is its deceleration. Since it stops in 7 seconds (or \(\frac{7}{3600}\) hours), \(0=v\left(\frac{7}{3600}\right)=100\frac{7}{3600}d\text{,}\) so \(d=\frac{360000}{7}\text{.}\) Then
We know that the acceleration of the ball will be constant. If the height of the ball is given by \(h(t)\) while it is in the air, \(h''(t)=9.8\text{.}\) (The negative indicates that the velocity is decreasing: the ball starts at its largest velocity, moving in the positive direction, then the velocity decreases to zero and then to a negative number as the ball falls.) As in Example 3.1.2, we need a function \(h(t)\) with \(h''(t)=9.8\text{.}\) Since this is a constant, \(h'(t)\) is a line with slope \(9.8\text{,}\) so it has the form
\begin{equation*}
h'(t)=9.8t+a
\end{equation*}
for some constant \(a\text{.}\) Notice when \(t=0\text{,}\) \(h'(0)=a\text{,}\) so in fact \(a\) is the initial velocity of the ballthe quantity we want to solve for.
Again, as in Example 3.1.2, we need a function \(h(t)\) with \(h'(t)=9.8t+a\text{.}\) Such a function must have the form
for some constant \(b\text{.}\) You can find this by guessing and checking, or simply remember it from the text. (In Section 4.1, you'll learn more about figuring out which functions have a particular derivative.) Notice when \(t=0\text{,}\) \(h(0)=b\text{,}\) so \(b\) is the initial height of the baseball, which is 0.
So, \(h(t)=4.9t^2+at = t(4.9t+a)\text{.}\) The baseball is at height zero when it is pitched (\(t=0\)) and when it hits the ground (which we want to be \(t=10\)). So, we want \((4.9)(10)+a=0\text{.}\) That is, \(a=49\text{.}\) So, the initial pitch should be at \(49\) metres per second.
Incidentally, this is on par with the fastest pitch in baseball, as recorded by Guiness World Records: \small\url{http://www.guinnessworldrecords.com/worldrecords/fastestbaseballpitch(male)}
Be very careful with units. The acceleration of gravity you're used to is \(9.8\) metres per second squared, so you might want to convert \(325\) kpm to metres per second.
The acceleration of a falling object due to gravity is \(9.8\) metres per second squared. So, the object's velocity \(t\) seconds after being dropped is
Since gravity alone brings your cannon ball down, its acceleration is a constant \(9.8\;\frac{\mathrm{m}}{\mathrm{s}^2}\text{.}\) So, \(v(t)=v_09.8t\) and thus its height is given by \(s(t)=v_0t4.9t^2\) (if we set \(s(0)=0\)).
We want to know what value of \(v_0\) makes the maximum height 100 metres. The maximum height is reached when \(v(t)=0\text{,}\) which is at time \(t=\frac{v_0}{9.8}\text{.}\) So, we solve:
First, find an equation for \(a(t)\text{,}\) the acceleration of the car, noting that \(a'(t)\) is constant. Then, use this to find an equation for the velocity of the car. Be careful about seconds versus hours.
The derivative of acceleration is constant, so the acceleration \(a(t)\) has the form \(mt+b\text{.}\) We know \(a(0)=50000\) and \(a\left(\frac{3}{3600}\right)=60000\) (where we note that \(3\) seconds is \(\frac{3}{3600}\) hours). So, the slope of \(a(t)\) is \(\frac{60000+50000}{\frac{3}{3600}}=12000000\text{,}\) which leads us to
Remark: When acceleration is constant, the position function is a quadratic function, but we don't want you to get the idea that position functions are always quadratic functionsin the example you just did, it was the velocity function that was quadratic. Position, velocity, and acceleration functions don't have to be polynomial at allit's only in this section, where we're dealing with the simplest cases, that they seem that way.
We recommend using two different functions to describe your height: \(h_1(t)\) while you are in the air, not yet touching the trampoline, and \(h_2(t)\) while you are in the trampoline, going down.
Both \(h_1(t)\) and \(h_2(t)\) are quadratic equations, since your acceleration is constant over both intervals, but be very careful about signs.
Different forces are acting on you (1) after you jump but before you land on the trampoline, and (2) while you are falling into the trampoline. In both instances, the acceleration is constant, so both height functions are quadratic, of the form \(\frac{a}{2}t^2+vt+h\text{,}\) where \(a\) is the acceleration, \(v\) is the velocity when \(t=0\text{,}\) and \(h\) is the initial height.
Let's consider (1) first, the time during your jump before your feet touch the trampoline. Let \(t=0\) be the moment you jump, and let the rim of the trampoline be height \(0\text{.}\) Then, since your initial velocity was (positive) \(1\) meter per second, your height is given by
Notice that, because your acceleration is working against your positive velocity, it has a negative sign.
We'll need to know your velocity when your feet first touch the trampoline on your fall. The time your feet first first touch the trampoline after your jump is precisely when \(h_1(t)=0\) and \(t \gt 0\text{.}\) That is, when \(t=\frac{2}{9.8}\text{.}\) Now, since \(h'(t)=9.8t+1\text{,}\) \(h'\left(\frac{2}{9.8}\right)=9.8\left(\frac{2}{9.8}\right)+1=1\text{.}\) So, you are descending at a rate of 1 metre per second at the instant your feet touch the trampoline.
Remark: it is not only coincidence that this was your initial speed. Think about the symmetries of parabolas, and conservation of energy.
Now we need to think about your height as the trampoline is slowing your fall. One thing to remember about our general equation \(\frac{a}{2}t^2+vt+h\) is that \(v\) is the velocity when \(t=0\text{.}\) But, you don't hit the trampoline at \(t=0\text{,}\) you hit it at \(t=\frac{2}{9.8}\text{.}\) In order to keep things simple, let's use a different time scale for this second part of your journey. Let's let \(h_2(T)\) be your height at time \(T\text{,}\) from the moment your feet touch the trampoline skin (\(T=0\)) to the bottom of your fall. Now, we can use the fact that your initial velocity is \(1\) metres per second (negative, since your height is decreasing) and your acceleration is \(4.9\) metres per second per second (positive, since your velocity is increasing from a negative number to zero):
where still the height of the rim of the trampoline is taken to be zero.
Remark: if it seems very confusing that your freefalling acceleration is negative, while your acceleration in the trampoline is positive, remember that gravity is pushing you down, but the trampoline is pushing you up.
How long were you falling in the trampoline? The equation \(h_2(T)\) tells you your height only as long as the trampoline is slowing your fall. You reach the bottom of your fall when your velocity is zero.
\begin{equation*}
h_2'(T)=4.9T1
\end{equation*}
so you reach the bottom of your fall at \(T=\frac{1}{4.9}\text{.}\) Be careful: this is \(\frac{1}{4.9}\) seconds after you entered the trampoline, not after the peak of your fall, or after you jumped.
The last piece of the puzzle is how long it took you to fall from the peak of your jump to the surface of the trampoline. We know the equation of your motion during that time: \(h_1(t)=\frac{9.8}{2}t^2+t\text{.}\) You reached the peak when your velocity was zero:
So, you fell from your peak at \(t=\frac{1}{9.8}\) and reached the level of the trampoline rim at \(t=\frac{2}{1.98}\text{,}\) which means the fall took \(\frac{1}{9.8}\) seconds.
Remark: by the symmetry mentioned early, the time it took to fall from the peak of your jump to the surface of the trampoline is the same as onehalf the time from the moment you jumped off the rim to the moment you're back on the surface of the trampoline.
So, your time falling from the peak of your jump to its bottom was \(\frac{1}{9.8}+\frac{1}{4.9}\approx 0.3\) seconds.
Let \(v(t)\) be the velocity of the object. From the given information:
\(v(0)\) is some value, call it \(v_0\text{,}\)
\(v(1)=2v_0\) (since the speed doubled in the first second),
\(v(2)=2(2)v_0\) (since the speed doubled in the second second),
\(v(3)=2(2)(2)v_0\text{,}\) and so on.
So, for general \(t\text{:}\)
\begin{equation*}
v(t)=2^tv(0)
\end{equation*}
To find its acceleration, we simply differentiate. Recall \(\ds\diff{}{x}\{2^x\}=2^x\log 2\text{,}\) where \(\log\) denotes logarithm base \(e\text{.}\)
We have an equation relating \(P\) and \(Q\text{:}\)
\begin{align*}
P&=Q^3\\
\end{align*}
We differentiate implicitly with respect to a third variable, \(t\text{:}\)
\begin{align*}
\diff{P}{t}&=3Q^2\cdot\diff{Q}{t}
\end{align*}
If we know two of the three quantities \(\ds\diff{P}{t}\text{,}\) \(Q\text{,}\) and \(\ds\diff{Q}{t}\text{,}\) then we can find the third. Therefore, ii is a question we can solve. If we know \(P\text{,}\) then we also know \(Q\) (it's just the cube root of \(P\)), so also we can solve iv. However, if we know neither \(P\) nor \(Q\text{,}\) then we can't find \(\ds\diff{P}{t}\) based only off \(\ds\diff{Q}{t}\text{,}\) and we can't find \(\ds\diff{Q}{t}\) based only off \(\ds\diff{P}{t}\text{.}\) So we can't solve i or iii.
Suppose that at time \(t\text{,}\) the point is at \(\big(x(t),y(t)\big)\text{.}\) Then \(x(t)^2+y(t)^2=1\) so that \(2x(t)x'(t)+2y(t)y'(t)=0\text{.}\) We are told that at some time \(t_0\text{,}\) \(x(t_0)=2/\sqrt{5}\text{,}\) \(y(t_0)=1/\sqrt{5}\) and \(y'(t_0)=3\text{.}\) Then
The distance \(z(t)\) between the particles at any moment in time is
\begin{align*}
z^2(t)&= x(t)^2+y(t)^2,
\end{align*}
where \(x(t)\) is the position on the \(x\)axis of the particle A at time \(t\) (measured in seconds) and \(y(t)\) is the position on the \(y\)axis of the particle B at the same time \(t\text{.}\)
We differentiate the above equation with respect to \(t\) and get
\begin{align*}
2z \cdot z' &= 2x \cdot x' + 2 y \cdot y',
\end{align*}
We are told that \(x'=2\) and \(y'=3\text{.}\) (The values are negative because \(x\) and \(y\) are decreasing.) It will take \(3\) seconds for particle \(A\) to reach \(x=4\text{,}\) and in this time particle \(B\) will reach \(y=3\text{.}\)
At this point \(z = \sqrt{x^2+y^2}=\sqrt{3^2+4^2}=5\text{.}\)
Hence
\begin{align*}
10z' &= 8\cdot(2)+6\cdot(3) = 34 \\
z' &= \frac{34}{10} = \frac{17}{5} \text{ units per second}.
\end{align*}
We compute the distance \(z(t)\) between the two particles after \(t\) seconds as
\begin{align*}
z^2(t)&= 3^2 + (y_A(t)y_B(t))^2,
\end{align*}
where \(y_A(t)\) and \(y_B(t)\) are the \(y\)coordinates of particles \(A\) and \(B\) after \(t\) seconds, and the horizontal distance between the two particles is always 3 units.
We are told the distance between the particles is 5 units, this happens when
\begin{align*}
(y_Ay_B)^2 &= 5^23^2 = 16\\
y_Ay_B&=4
\end{align*}
That is, when the difference in \(y\)coordinates is \(4\text{.}\) This happens when \(t=4\text{.}\)
We differentiate the distance equation (from the first bullet point) with respect to \(t\) and get
\begin{align*}
2z \cdot z' &= 2 ({y_A}'  {y_B}') (y_A  y_B),
\end{align*}
We know that \((y_Ay_B)=4\text{,}\) and we are told that \(z=5\text{,}\) \(y_A'=3\text{,}\) and \(y_B'=2\text{.}\) Hence
\begin{align*}
10z'(4) &= 2 \times1\times4 = 8
\end{align*}
Therefore
\begin{align*}
z'(4) &= \frac{8}{10} = \frac{4}{5} \text{ units per second}.
\end{align*}
If \(x(t),\ y(t)\) and \(z(t)\) are the distances shown, at time \(t\text{,}\) then
\begin{align*}
x(t)^2+y(t)^2&=z(t)^2\\
\end{align*}
Differentiating with respect to \(t\text{,}\)
\begin{align*}
2x(t)x'(t)+2y(t)y'(t)&=2z(t)z'(t)\\
x(t)x'(t)+y(t)y'(t)&=z(t)z'(t)\\
\end{align*}
At the specified time, \(x(t)\) is decreasing, so \(x'(t)\) is negative.
\begin{align*}
(300)(15)+(400)(20) &= \sqrt{300^2+400^2}z'(t)\\
500 z'(t)&=3500\\
z'(t)&=7 \mbox{ mph}
\end{align*}
We compute the distance \(d(t)\) between the two snails after \(t\) minutes as
\begin{align*}
d^2(t)&= 30^2 + (y_1(t)y_2(t))^2,
\end{align*}
where \(y_1(t)\) is the altitude of the first snail, and \(y_2(t)\) the altitude of the second snail after \(t\) minutes.
We differentiate the above equation with respect to \(t\) and get
\begin{align*}
2d \cdot d' &= 2 ({y_1}'  {y_2}') (y_1  y_2)\\
d \cdot d' &= ({y_1}'  {y_2}') (y_1  y_2)
\end{align*}
We are told that \({y_1}'=25\) and \({y_2}'=15\text{.}\) It will take \(4\) minutes for the first snail to reach \(y_1=100\text{,}\) and in this time the second snail will reach \(y_2=60\text{.}\)
At this point \(d^2 = 30^2 + (10060)^2 = 900 + 1600 = 2500\text{,}\) hence \(d = 50\text{.}\)
Therefore
\begin{align*}
50 d' &= (25  15) \times (100  60) \\
d' &= \frac{400}{50} = 8 \text{ cm per minute}.
\end{align*}
If a trapezoid has height \(h\) and (parallel) bases \(b_1\) and \(b_2\text{,}\) then its area is \(h\left(\frac{b_1+b_2}{2}\right)\text{.}\) To figure out how wide the top of the water is when the water is at height \(h\text{,}\) you can cut the trapezoid up into a rectangle and two triangles, and make use of similar triangles.
What we're given is \(\ds\diff{V}{t}\) (where \(V\) is volume of water in the trough, and \(t\) is time), and what we are asked for is \(\ds\diff{h}{t}\) (where \(h\) is the height of the water). So, we need an equation relating \(V\) and \(h\text{.}\) First, let's get everything in the same units: centimetres.
We can calculate the volume of water in the trough by multiplying the area of its trapezoidal cross section by 200 cm. A trapezoid with height \(h\) and bases \(b_1\) and \(b_2\) has area \(h\left(\frac{b_1+b_2}{2}\right)\text{.}\) (To see why this is so, draw the trapezoid as a rectangle flanked by two triangles.) So, using \(w\) as the width of the top of the water (as in the diagram above), the area of the cross section of the water in the trough is
We need a formula for \(w\) in terms of \(h\text{.}\) If we draw lines straight up from the bottom corners of the trapezoid, we break it into rectangles and triangles.
Using similar triangles, \(\dfrac{a}{h}=\dfrac{20}{50}\text{,}\) so \(a=\dfrac{2}{5}h\text{.}\) Then
\begin{align*}
w&=60+2a\\
&=60+2\left(\frac{2}{5}h\right)=60+\frac{4}{5}h\\
\end{align*}
so
\begin{align*}
V&=100h(60+w)\\
&=100h(120+\frac{4}{5}h)\\
&=80h^2+12000h\\
\end{align*}
This is the equation we need, relating \(V\) and \(h\text{.}\) Differentiating implicitly with respect to \(t\text{:}\)
\begin{align*}
\diff{V}{t}
&=2\cdot80h\cdot\diff{h}{t}+12000\diff{h}{t}\\
&=\left(160h+12000\right)\diff{h}{t}\\
\end{align*}
We are given that \(h=25\) and \(\ds\diff{V}{t}=3\) litres per minute. Converting to cubic centimetres, \(\ds\diff{V}{t} = 3000\) cubic centimetres per minute. So:
\begin{align*}
3000&=\left(160\cdot25+12000\right)\diff{h}{t}\\
\diff{h}{t}&=\frac{3}{16}=.1875\;\frac{\mathrm{cm}}{\mathrm{min}}
\end{align*}
So, the water level is dropping at \(\dfrac{3}{16}\) centimetres per minute.
If \(V\) is the volume of the water in the tank, and \(t\) is time, then we are given \(\ds\diff{V}{t}\text{.}\) What we want to know is \(\ds\diff{h}{t}\text{,}\) where \(h\) is the height of the water in the tank. A reasonable plan is to find an equation relating \(V\) and \(h\text{,}\) and differentiate it implicitly with respect to \(t\text{.}\)
Let's be a little careful about units. The volume of water in the tank is
\begin{equation*}
\text{(area of cross section of water)$\times$(length of tank)}
\end{equation*}
If we measure these values in metres (area in square metres, length in metres), then the volume is going to be in cubic metres. So, when we differentiate with respect to time, our units will be cubic metres per second. The water is flowing in at one litre per second, or \(1000\) cubic centimetres per second. So, we either have to measure our areas and distances in centimetres, or convert litres to cubic metres. We'll do the latter, but both are fine.
If we imagine one cubic metre as a cube, with each side of length 1 metre, then it's easy to see the volume inside is \((100)^3=10^6\) cubic centimetres: it's the volume of a cube with each side of length 100 cm. Since a litre is \(10^3\) cubic centimetres, and a cubic metre is \(10^{6}\) cubic centimetres, one litre is \(10^{3}\) cubic metres. So, \textcolor{red}{\(\ds\diff{V}{t}=\frac{1}{10^3}\)} cubic metres per second.
Let \(h\) be the height of the water (in metres). We can figure out the area of the cross section by breaking it into three pieces: a triangle on the left, a rectangle in the middle, and a trapezoid on the right.
The triangle on the left has height \(h\) metres. Let its base be \(a\) metres. It forms a similar triangle with the triangle whose height is 1.25 metres and width is 1 metre, so:
\begin{align*}
\frac{a}{h}&=\frac{1}{1.25}\\
a&=\frac{4}{5}h\\
\end{align*}
So, the area of the triangle on the left is
\begin{align*}
\frac{1}{2}ah&=\frac{2}{5}h^2
\end{align*}
The rectangle in the middle has length 3 metres and height \(h\) metres, so its area is \(3h\) square metres.
The trapezoid on the right is a portion of a triangle with base 3 metres and height 1.25 metres. So, its area is
\begin{align*}
&\underbrace{\left(\frac{1}{2}(3)(1.25)\right) }_{\mbox{area of big triangle}} \underbrace{\left(\frac{1}{2}(b)(1.25h)\right)}_{\mbox{area of little triangle}}\\
\end{align*}
The little triangle (of base \(b\) and height \(1.25h\)) is formed by the air on the right side of the tank. It is a similar triangle to the triangle of base 3 and height 1.25, so
\begin{align*}
\frac{b}{1.25h}&=\frac{3}{1.25}\\
b&=\frac{3}{1.25}(1.25h)\\
\end{align*}
So, the area of the trapezoid on the right is
\begin{align*}
&\frac{1}{2}(3)(1.25)\frac{1}{2}\left(\frac{3}{1.25}\right)\left(1.25h\right)(1.25h)\\
&=3h\frac{6}{5}h^2
\end{align*}
So, the area \(A\) of the cross section of the water is
\begin{align*}
A&=\underbrace{\frac{2}{5}h^2}_{\mbox{triangle}}+
\underbrace{3h}_{\mbox{rectangle}}+
\underbrace{3h\frac{6}{5}h^2}_{\mbox{trapezoid}}\\
&=6h\frac{4}{5}h^2\\
\end{align*}
So, the volume of water is
\begin{align*}
V&=5\left(6h\frac{4}{5}h^2\right)=30h4h^2\\
\end{align*}
Differentiating with respect to time, \(t\text{:}\)
\begin{align*}
\diff{V}{t}&=30\diff{h}{t}8h\diff{h}{t}\\
\end{align*}
When \(h=\dfrac{1}{10}\) metre, and \(\ds\diff{V}{t}=\frac{1}{10^3}\) cubic metres per second,
\begin{align*}
\frac{1}{10^3}&=30\diff{h}{t}8\left(\frac{1}{10}\right)\diff{h}{t}\\
\diff{h}{t}&=\frac{1}{29200} \mbox{ metres per second}
\end{align*}
This is about 1 centimetre every five minutes. You might want a bigger hose.
Let \(\theta\) be the angle of your head, where \(\theta=0\) means you are looking straight ahead, and \(\theta=\dfrac{\pi}{2}\) means you are looking straight up. We are interested in \(\ds\diff{\theta}{t}\text{,}\) but we only have information about \(h\text{.}\) So, a reasonable plan is to find an equation relating \(h\) and \(\theta\text{,}\) and differentiate with respect to time.
The right triangle formed by you, the rocket, and the rocket's original position has adjacent side (to \(\theta\)) length 2km, and opposite side (to \(\theta\)) length \(h(t)\) kilometres, so
\begin{align*}
\tan\theta&=\frac{h}{2}\\
\end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*}
\sec^2\theta \cdot \diff{\theta}{t}&=\frac{1}{2}\diff{h}{t}\\
\diff{\theta}{t}&=\frac{1}{2}\cos^2\theta\cdot\diff{h}{t}
\end{align*}
We know \(\tan\theta= \frac{h}{2}\text{.}\) We draw a right triangle with angle \(\theta\) (filling in the sides using SOH CAH TOA and the Pythagorean theorem) to figure out \(\cos\theta\text{:}\)
Using the triangle, \(\cos\theta = \dfrac{2}{\sqrt{h^2+4}}\text{,}\) so
\begin{align*}
\diff{\theta}{t}&=\frac{1}{2}\left(\frac{2}{\sqrt{h^2+4}}\right)^2\cdot\diff{h}{t}\\
&=\left(\frac{2}{h^2+4}\right)\diff{h}{t}\\
\end{align*}
So, the quantities we need to know one minute after liftoff (that is, when \(t = \dfrac{1}{60}\)) are \(h\left(\dfrac{1}{60}\right)\) and \(\ds\diff{h}{t}\left(\frac{1}{60}\right)\text{.}\) Recall \(h(t)=61750t^2\text{.}\)
\begin{align*}
h\left(\frac{1}{60}\right)&=\frac{61750}{3600}=\frac{1235}{72}\\
\diff{h}{t}&=2(61750)t\\
\diff{h}{t}\left(\frac{1}{60}\right)&=\frac{2(61750)}{60}=\frac{6175}{3}\\
\end{align*}
Returning to the equation \(\ds\diff{\theta}{t}=\left(\dfrac{2}{h^2+4}\right)\ds\diff{h}{t}
\text{:}\)
\begin{align*}
\diff{\theta}{t}\left(\frac{1}{60}\right)&=\left(\frac{2}{\left(\frac{1235}{72}\right)^2+4}\right)\left(\frac{6175}{3}
\right)\approx 13.8\;\frac{\mathrm{rad}}{\mathrm{hour}}\approx 0.0038\;\frac{\mathrm{rad}}{\mathrm{sec}}
\end{align*}
3.2.2.13.a Let \(x(t)\) be the distance of the train along the track at time \(t\text{,}\) measured from the point on the track nearest the camera. Let \(z(t)\) be the distance from the camera to the train at time \(t\text{.}\)
Then \textcolor{red}{\(x'(t)=2\)} and at the time in question, \textcolor{red}{\(z(t)=1.3\)} km and \(\textcolor{red}{x(t)}=\sqrt{1.3^20.5^2}=\textcolor{red}{1.2}\) km. So
3.2.2.13.b Let \(\theta(t)\) be the angle shown at time \(t\text{.}\) Then
\begin{align*}
\sin\left(\theta(t)\right)&=\frac{x(t)}{z(t)}\\
\end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*}
\theta'(t)\cos\left(\theta(t)\right)
&=\frac{x'(t)z(t)x(t)z'(t)}{z(t)^2}\\
\theta'(t)&=\frac{x'(t)z(t)x(t)z'(t)}{z(t)^2\cos\left(\theta(t)\right)}\\
\end{align*}
From our diagram, we see \(\cos\left(\theta(t)\right)=\dfrac{0.5}{z(t)}\text{,}\) so:
\begin{align*}
&=2\frac{x'(t)z(t)x(t)z'(t)}{z(t)}\\
\end{align*}
Substituting in \(x'(t)=2\text{,}\) \(z(t)=1.3\text{,}\) \(x(t)=1.2\text{,}\) and \(z'(t)=\dfrac{2\times1.2}{1.3}\text{:}\)
\begin{align*}
\theta'(t)
&=2\frac{2\times 1.31.2\times\frac{2\times 1.2}{1.3}}{1.3}
\approx .592 \mbox{ radians/min}
\end{align*}
In the diagram, there are two right triangles: one formed by the light, the ball, and the ball's original position, and one formed by the ball, the tip of its shadow, and the ball's eventual position on the ground. The two angles marked below have the same measure, \(\alpha\text{.}\) (Whenever two straignt lines cross, the opposite angles formed have the same measure.)
Since the two triangles share two angles in common (a right angle, and an angle of measure \(\alpha\)), they are similar triangles. Let's call \(s(t)\) the distance from the shadow to the point on the ground directly underneath the ball. Since the triangles are similar,
Let \(\theta\) be the angle between the two hands. Using the Law of Cosines, you can get an expression for \(D\) in terms of \(\theta\text{.}\) To find \(\ds\diff{\theta}{t}\text{,}\) use what you know about how fast clock hands move.
Let \(\theta\) be the angle between the two hands.
The Law of Cosines (Appendix ) tells us that
\begin{align*}
D^2&=5^2+10^22\cdot5\cdot10\cdot\cos\theta\\
D^2&=125100\cos\theta\\
\end{align*}
Differentiating with respect to time \(t\text{,}\)
\begin{align*}
2D\diff{D}{t}&=100\sin\theta\cdot\diff{\theta}{t}\\
\end{align*}
Our tasks now are to find \(D\text{,}\) \(\theta\) and \(\ds\diff{\theta}{t}\) when the time is 4:00. At 4:00, the minute hand is straight up, and the hour hand is \(\dfrac{4}{12}=\dfrac{1}{3}\) of the way around the clock, so \textcolor{red}{\(\theta = \dfrac{1}{3}(2\pi)=\dfrac{2\pi}{3}\)} at 4:00. Then \(D^2=125100\cos\left(\frac{2\pi}{3}\right)=125100\left(\frac{1}{2}\right)=175\text{,}\) so \textcolor{red}{\(D=\sqrt{175}=5\sqrt{7}\)} at 4:00.
To calculate \(\ds\diff{\theta}{t}\text{,}\) remember that both hands are moving. The hour hand makes a full rotation every 12 hours, so its rotational speed is \(\dfrac{2\pi}{12}=\dfrac{\pi}{6}\) radians per hour. The hour hand is being chased by the minute hand. The minute hand makes a full rotation every hour, so its rotational speed is \(\dfrac{2\pi}{1}=2\pi\) radians per hour. Therefore, the angle \(\theta\) between the two hands is changing at a rate of
The area at time \(t\) is the area of the outer circle minus the area of the inner circle:
\begin{align*}
A(t)&=\pi\big(R(t)^2r(t)^2\big)\\
\mbox{So, }\;A'(t)&=2\pi\big(R(t)R'(t)r(t)r'(t)\big)\\
\end{align*}
Plugging in the given data,
\begin{align*}
A'&=2\pi\big(3\cdot 21\cdot 7\big)=2\pi
\end{align*}
So the area is shrinking at a rate of \(2\pi\;\dfrac{\mathrm{cm}^2}{\mathrm{s}}\text{.}\)
We differentiate \(R=10+2t\) and \(r=6t\) to find \(\ds\diff{R}{t}=2\) and \(\ds\diff{r}{t}=6\text{.}\) When \(R=2r\text{,}\) \(10+2t=2(6t)\text{,}\) so \(t=1\text{.}\) When \(t=1\text{,}\) \(R=12\) and \(r=6\text{.}\) So:
So the volume between the two spheres is increasing at 288 cubic units per unit time.
Remark: when the radius of the inner sphere increases, we are “subtracting” more area. Since the radius of the inner sphere grows faster than the radius of the outer sphere, we might expect the area between the spheres to be decreasing. Although the radius of the outer sphere grows more slowly, a small increase in the radius of the outer sphere results in a larger change in volume than the same increase in the radius of the inner sphere. So, a result showing that the volume between the spheres is increasing is not unreasonable.
We know something about the rate of change of the height \(h\) of the triangle, and we want to know something about the rate of change of its area, \(A\text{.}\) A reasonable plan is to find an equation relating \(A\) and \(h\text{,}\) and differentiate implicitly with respect to \(t\text{.}\) The area of a triangle with height \(h\) and base \(b\) is
\begin{align*}
A&=\frac{1}{2}bh\\
\end{align*}
Note, \(b\) will change with time as well as \(h\text{.}\) So, differentiating with respect to time, \(t\text{:}\)
\begin{align*}
\diff{A}{t}&=\frac{1}{2}\left(\diff{b}{t}\cdot h+b\cdot\diff{h}{t}\right)
\end{align*}
We are given \(\diff{h}{t}\) and \(h\text{,}\) but those \(b\)'s are a mystery. We need to relate them to \(h\text{.}\) We can do this by breaking our triangle into two right triangles and using the Pythagorean Theorem:
So, the base of the triangle is
\begin{align*}
b&=\sqrt{150^2h^2}+\sqrt{200^2h^2}\\
\end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*}
\diff{b}{t}&=\frac{2h\diff{h}{t}}{2\sqrt{150^2h^2}}+
\frac{2h\diff{h}{t}}{2\sqrt{200^2h^2}}\\
&=\frac{h\diff{h}{t}}{\sqrt{150^2h^2}}+
\frac{h\diff{h}{t}}{\sqrt{200^2h^2}}\\
\end{align*}
Using \(\ds\diff{h}{t}=3\) centimetres per minute:}
\begin{align*}
\diff{b}{t}&=\frac{3h}{\sqrt{150^2h^2}}+
\frac{3h}{\sqrt{200^2h^2}}\\
\end{align*}
When \(h=120\text{,}\) \(\sqrt{150^2h^2}=90\) and \(\sqrt{200^2h^2}=160\text{.}\) So, at this moment in time:
\begin{align*}
b&=90+160=250\\
\diff{b}{t}&=\frac{3(120)}{90}+\frac{3(120)}{160}
=
4+\frac{9}{4}=\frac{25}{4}\\
\end{align*}
We return to our equation relating the derivatives of \(A\text{,}\) \(b\text{,}\) and \(h\text{.}\)
\begin{align*}
\diff{A}{t}&=\frac{1}{2}\left(\diff{b}{t}\cdot h + b \cdot \diff{h}{t}\right)\\
\end{align*}
When \(h=120\) cm, \(b=250\text{,}\) \(\ds\diff{h{t}=3\text{,}\) and \(\ds\diff{b}{t}=\dfrac{25}{4}\text{:}\)
\begin{align*}
\diff{A}{t}&=\frac{1}{2}\left(\frac{25}{4}(120)+250(3)\right)\\
&=0
\end{align*}
Remark: What does it mean that \(\left.\ds\diff{A}{t}\right_{h=120}=0\text{?}\) Certainly, as the height changes, the area changes as well. As the height sinks to 120 cm, the area is increasing, but after it sinks past 120 cm, the area is decreasing. So, at the instant when the height is exactly 120 cm, the area is neither increasing nor decreasing: it is at a local maximum. You'll learn more about this kind of problem in Section 3.5.
The easiest way to figure out the area of the sector of an annulus (or a circle) is to figure out the area of the entire annulus, then multiply by what proportion of the entire annulus the sector is. For example, if your sector is \(\frac{1}{10}\) of the entire annulus, then its area is \(\frac{1}{10}\) of the area of the entire annulus. (See Section to see how this works out for circles.)
Let \(S\) be the flow of salt (in cubic centimetres per second). We want to know \(\ds\diff{S}{t}\text{:}\) how fast the flow is changing at time \(t\text{.}\) We are given an equation for \(S\text{:}\)
\begin{equation*}
S=\frac{1}{5}A
\end{equation*}
where \(A\) is the uncovered area of the cutout. So,
If we can find \(\ds\diff{A}{t}\text{,}\) then we can find \(\ds\diff{S}{t}\text{.}\) We are given information about how quickly the door is rotating. If we let \(\theta\) be the angle made by the leading edge of the door and the far edge of the cutout (shown below), then \(\ds\diff{\theta}{t}=\dfrac{\pi}{6}\) radians per second. (Since the door is covering more and more of the cutout, \(\theta\) is getting smaller, so \(\ds\diff{\theta}{t}\) is negative.)
Since we know \(\ds\diff{\theta}{t}\text{,}\) and we want to know \(\ds\diff{A}{t}\) (in order to get \(\ds\diff{S}{t}\)), it is reasonable to look for an equation relating \(A\) and \(\theta\text{,}\) and differentiate it implicitly with respect to \(t\) to get an equation relating \(\ds\diff{A}{t}\) and \(\ds\diff{\theta}{t}\text{.}\)
The area of an annulus with outer radius 6 cm and inner radius 1 cm is \(\pi\cdot 6^2\pi\cdot1^2=35\pi\) square centimetres. A sector of that same annulus with angle \(\theta\) has area \(\left(\frac{\theta}{2\pi}\right)(35\pi)\text{,}\) since \(\frac{\theta}{2\pi}\) is the ratio of the sector to the entire annulus. (For example, if \(\theta=\pi\text{,}\) then the sector is half of the entire annulus, so its area is \((1/2)35\pi\text{.}\))
So, when \(0 \leq \theta \leq \frac{\pi}{2}\text{,}\) the area of the cutout that is open is
\begin{align*}
A&=\frac{\theta}{2\pi}(35\pi)=\frac{35}{2}\theta\\
\end{align*}
This is the formula we wanted, relating \(A\) and \(\theta\text{.}\) Differentiating with respect to \(t\text{,}\)
\begin{align*}
\diff{A}{t}&=\frac{35}{2}\diff{\theta}{t}=\frac{35}{2}\left(\frac{\pi}{6}\right)=\frac{35\pi}{12}\\
\end{align*}
Since \(\ds\diff{S}{t}=\frac{1}{5}\ds\diff{A}{t}\text{,}\)
\begin{align*}
\diff{S}{t}&=\frac{1}{5}\frac{35\pi}{12}=\frac{7\pi}{12}\approx 1.8\;\frac{\mathrm{cm}^3}{\mathrm{sec}^2}
\end{align*}
Remark: the change in flow of salt is constant while the door covers more and more of the cutout, so we never used the fact that precisely half of the cutout was open. We also never used the radius of the lid, which is immaterial to the flow of salt.
Let \(F\) be the flow of water through the pipe, so \(F=\dfrac{1}{5}A\text{.}\) We want to know \(\ds\diff{F}{t}\text{,}\) so differentiating implicitly with respect to \(t\text{,}\) we find
If we can find \(\ds\diff{A}{t}\text{,}\) then we can find \(\ds\diff{F}{t}\text{.}\) We know something about the shape of the uncovered area of the pipe; a reasonable plan is to find an equation relating the height of the door with the uncovered area of the pipe. Let \(h\) be the distance from the top of the pipe to the bottom of the door, measured in metres.
Since the radius of the pipe is 1 metre, the orange line has length \(1h\) metres, and the blue line has length 1 metre. Using the Pythagorean Theorem, the green line has length \(\sqrt{1^2(1h)^2}=\sqrt{2hh^2}\) metres.
The uncovered area of the pipe can be broken up into a triangle (of height \textcolor{orange}{\(1h\)} and base \textcolor{green}{\(2\sqrt{2hh^2}\)}) and a sector of a circle (with angle \(2\pi2\theta\)). The area of the triangle is
Remember: what we want is to find \(\ds\diff{A}{t}\text{,}\) and what we know is \(\ds\diff{h}{t}=0.01\) metres per second. If we find \(\theta\) in terms of \(h\text{,}\) we find \(A\) in terms of \(h\text{,}\) and then differentiate with respect to \(t\text{.}\)
Since \(\theta\) is an angle in a right triangle with hypotenuse \(1\) and adjacent side length \(1h\text{,}\) \(\cos\theta = \frac{1h}{1} =1h\text{.}\) We want to conclude that \(\theta = \arccos(1h)\text{,}\) but let's be a little careful: remember that the range of the arccosine function is angles in \([0,\pi]\text{.}\) We must be confident that \(0\le\theta\le\pi\) in order to conclude \(\theta = \arccos(1h)\)but clearly, \(\theta\) is in this range. (Remark: we could also have said \(\sin\theta=\frac{\sqrt{2hh^2}}{1}\text{,}\) and so \(\theta = \arcsin\left(\sqrt{2hh^2}\right)\text{.}\) This would require \(\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}\text{,}\) which is true when \(h \lt 1\text{,}\) but false for \(h \gt 1\text{.}\) Since our problem asks about \(h=0.25\text{,}\) we could also use arcsine.)
Now, we know the area of the open pipe in terms of \(h\text{.}\)
\begin{align*}
A&=(\mbox{area of triangle})+(\mbox{area of sector})\\
&=(1h)\sqrt{2hh^2}+(\pi\theta)\\
&=(1h)\sqrt{2hh^2}+\pi\arccos\left(1h\right)\\
\end{align*}
We want to differentiate with respect to \(t\text{.}\) Using the chain rule:
\begin{align*}
\diff{A}{t}&=\diff{A}{h}\cdot\diff{h}{t}\\
\diff{A}{t}&=\left((1h)\frac{22h}{2\sqrt{2hh^2}}+(1)\sqrt{2hh^2}
+\frac{1}{\sqrt{1(1h)^2}}\right)\diff{h}{t}\\
&=\left(\frac{(1h)^2}{\sqrt{2hh^2}}\sqrt{2hh^2}\frac{1}{\sqrt{2hh^2}}\right)\diff{h}{t}\\
&=\left(\frac{(1h)^21}{\sqrt{2hh^2}}\sqrt{2hh^2}\right)\diff{h}{t}\\
&=\left(\frac{(2hh^2)}{\sqrt{2hh^2}}\sqrt{2hh^2}\right)\diff{h}{t}\\
&=\left(\sqrt{2hh^2}\sqrt{2hh^2}\right)\diff{h}{t}\\
&=2\sqrt{2hh^2}\diff{h}{t}\\
\end{align*}
We note here that the negative sign makes sense: as the door lowers, \(h\) increases and \(A\) decreases, so \(\ds\diff{h}{t}\) and \(\ds\diff{A}{t}\) should have opposite signs.
\begin{align*}
\\
\end{align*}
When \(h=\dfrac{1}{4}\) metres, and \(\ds\diff{h}{t}=\frac{1}{100}\) metres per second:
\begin{align*}
\diff{A}{t}&=2\sqrt{\frac{2}{4}\frac{1}{4^2}}\left(\frac{1}{100}\right)=\frac{\sqrt{7}}{200}\;\frac{\mathrm{cm}^2}{\mathrm{s}}\\
\end{align*}
Since \(\ds\diff{F}{t}=\frac{1}{5}\ds\diff{A}{t}\text{:}\)
\begin{align*}
\diff{F}{t}&=\frac{\sqrt{7}}{1000}\;\frac{\mathrm{m}^3}{\mathrm{sec^2}}
\end{align*}
That is, the flow is decreasing at a rate of \(\dfrac{\sqrt{7}}{1000}\;\dfrac{\mathrm{m}^3}{\mathrm{sec}^2}\text{.}\)
We are given the rate of change of the volume of liquid, and are asked for the rate of change of the height of the liquid. So, we need an equation relating volume and height.
The volume \(V\) of a cone with height \(h\) and radius \(r\) is \(\frac{1}{3}\pi r^2h\text{.}\) Since we know \(\ds\diff{V}{t}\text{,}\) and want to know \(\ds\diff{h}{t}\text{,}\) we need to find a way to deal with the unwanted variable \(r\text{.}\) We can find \(r\) in terms of \(h\) by using similar triangles. Viewed from the side, the conical glass is an equilateral triangle, as is the water in it. Using the Pythagorean Theorem, the cone has height \(5\sqrt{3}\text{.}\)
Using similar triangles, \(\dfrac{r}{h}=\dfrac{5}{5\sqrt{3}}\text{,}\) so \(r=\dfrac{h}{\sqrt{3}}\text{.}\) (Remark: we could also use the fact that the water forms a cone that looks like an equilateral triangle when viewed from the side to conclude \(r=\dfrac{h}{\sqrt{3}}\text{.}\))
Now, we can write the volume of water in the cone in terms of \(h\text{,}\) and no other variables.
\begin{align*}
V&=\frac{1}{3}\pi r^2h\\
&=\frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2h\\
&=\frac{\pi}{9}h^3\\
\end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*}
\diff{V}{t}&=\frac{\pi}{3}h^2\diff{h}{t}\\
\end{align*}
When \(h=7\) cm and \(\ds\diff{V}{t}=5\) mL per minute,
\begin{align*}
5&=\frac{\pi}{3}(49)\diff{h}{t}\\
\diff{h}{t}&=\frac{15}{49\pi}\approx 0.097 \mbox{ cm per minute}
\end{align*}
If you were to install the buoy, how would you choose the length of rope? For which values of \(\theta\) do \(\sin\theta\) and \(\cos\theta\) have different signs? How would those values of \(\theta\) look on the diagram?
As is so often the case, we use a right triangle in this problem to relate the quantities.
\begin{align*}
\sin\theta&=\frac{D}{2}\\
D&=2\sin\theta\\
\end{align*}
Using the chain rule, we differentiate both sides with respect to time, \(t\text{.}\)
\begin{align*}
\diff{D}{t}&=2\cos\theta\cdot\diff{\theta}{t}\\
\end{align*}
So, if \(\ds\diff{\theta}{t}=0.25\) radians per hour and \(\theta = \dfrac{\pi}{4}\) radians, then
\begin{align*}
(a)\qquad\diff{D}{t}&=2\cos\left(\frac{\pi}{4}\right)\cdot 0.25=2\left(\frac{1}{\sqrt{2}}\right)\frac{1}{4}=\frac{1}{2\sqrt{2}}\;\mbox{metres per hour}.
\end{align*}
Setting aside part (b) for a moment, let's think about (c). If \(\ds\diff{\theta}{t}\) and \(\ds\diff{D}{t}\) have different signs, then because \(\ds\diff{D}{t}=2\cos\theta\cdot\ds\diff{\theta}{t}\text{,}\) that means \(\cos\theta \lt 0\text{.}\) We have to have a nonnegative depth, so \(D \gt 0\) and \(D=2\sin\theta\) implies \(\sin \theta \gt 0\text{.}\) If \(\sin\theta\ge0\) and \(\cos \theta \lt 0\text{,}\) then \(\theta\in(\pi/2,\pi]\text{.}\) On the diagram, that looks like this:
That is: the water has reversed direction. This happens, for instance, when a river empties into the ocean and the tide is high. Skookumchuck Narrows provincial park, in the Sunshine Coast, has reversing rapids.
Now, let's return to (b). If the rope is only 2 metres long, and the river rises higher than 2 metres, then our equation \(D=2\sin\theta\) doesn't work any more: the buoy might be stationary underwater while the water rises or falls (but stays at or above \(2\) metres deep).
At both points of interest, the point is moving along a straight line. From the diagram, you can figure out the equation of that line.
For the question “How fast is the point moving?” in part (b), remember that the velocity of an object can be found by differentiating (with respect to time) the equation that gives the position of the object. The complicating factors in this case are that (1) the position of our object is not given as a function of time, and (2) the position of our object is given in two dimensions, not one.
(a) When the point is at \((0,2)\text{,}\) its \(y\)coordinate is not changing, because it is moving along a horizontal line. So, the rate at which the particle moves is simply \(\ds\diff{x}{t}\text{.}\) Let \(\theta\) be the angle an observer would be looking at, in order to watch the point. Since we know \(\ds\diff{\theta}{t}\text{,}\) a reasonable plan is to find an equation relating \(\theta\) and \(x\text{,}\) and then differentiate implicitly with respect to \(t\text{.}\) To do this, let's return to our diagram.
When the point is a little to the right of \((0,2)\text{,}\) then we can make a triangle with the origin, as shown. If we let \(\theta\) be the indicated angle, then \(\ds\diff{\theta}{t}=1\) radian per second. (It is given that the observer is turning one radian per second, so this is how fast \(\theta\) is increasing.) From the right triangle in the diagram, we see
Now, we have to take care of a subtle point. The diagram we drew only makes sense for the point when it is at a position a little to the right of \((0,2)\text{.}\) So, right now, we've only made a setup that will find the derivative from the right. But, with a little more thought, we see that even when \(x\) is negative (that is, when the point is a little to the left of \((0,2)\)), our equation holds if we are careful about how we define \(\theta\text{.}\) Let \(\theta\) be the angle between the line connecting the point and the origin, and the \(y\)axis, where \(\theta\) is negative when the point is to the left of the \(y\)axis.
Since \(x\) and \(\theta\) are both negative when the point is to the left of the \(y\)axis,
\begin{align*}
\tan\theta&=\frac{2}{x}\\
\tan(\theta) &= \frac{x}{2}\\
\end{align*}
So, since \(\tan(\theta)=\tan(\theta)\text{:}\)
\begin{align*}
\tan\theta&=\frac{x}{2}
\end{align*}
So, we've shown that the relationship \(\tan\theta=\dfrac{x}{2}\) holds when our point is at \((x,2)\text{,}\) regardless of the sign of \(x\text{.}\)
Moving on, since we are given \(\ds\diff{\theta}{t}\) and asked for \(\ds\diff{x}{t}\text{,}\) we differentiate with respect to \(t\text{:}\)
\begin{align*}
\sec^2\theta \cdot \diff{\theta}{t}&=\frac{1}{2}\cdot\diff{x}{t}\\
\end{align*}
When the point is at \((0,2)\text{,}\) since the observer is turning at one radian per second, also \(\ds\diff{\theta}{t}=1\text{.}\) Also, looking at the diagram, \(\theta=0\text{.}\) Plugging in these values:
\begin{align*}
\sec^2\left(0\right)\cdot(1)&=\frac{1}{2}\cdot\diff{x}{t}\\
1&=\frac{1}{2}\cdot\diff{x}{t}\\
\diff{x}{t}&=2
\end{align*}
So, the particle is moving at 2 units per second.
(b) When the point is at \((0,2)\text{,}\) it is moving along a line with slope \(\frac{1}{2}\) and \(y\)intercept \(2\text{.}\) So, it is on the line
That is, at time \(t\text{,}\) if the point is at \((x(t),y(t))\text{,}\) then \(x(t)\) and \(y(t)\) satisfy \(y(t)=2\frac{1}{2}x(t)\text{.}\) Implicitly differentiating with respect to \(t\text{:}\)
So, when \(\ds\diff{x}{t}=1\text{,}\) \(\ds\diff{y}{t}=\dfrac{1}{2}\text{.}\) That is, its \(y\)coordinate is decreasing at \(\dfrac{1}{2}\) unit per second.
For the question “How fast is the point moving?”, remember that the velocity of an object can be found by differentiating (with respect to time) the equation that gives the position of the object. The complicating factors in this case are that (1) the position of our object is not given as a function of time, and (2) the position of our object is given in two dimensions (an \(x\) coordinate and a \(y\) coordinate), not one.
\textcolor{blue}{Remark: the solution below is actually pretty complicated. It is within your abilities to figure it out, but later on in your mathematical career you will learn an easier way, using vectors. For now, take this as a relatively tough exercise, and a motivation to keep learning: your intuition that there must be an easier way is well founded!}
The point is moving along a straight line. So, to take care of complication (2), we can give its position as a point on the line. We can take the line as a sort of axis. We'll need to choose a point on the axis to be the “origin”: \((2,1)\) is a convenient point. Let \(D\) be the point's (signed) distance along the “axis” from \((2,1)\text{.}\) When the point is a distance of one unit to the left of \((2,1)\text{,}\) we'll have \(D=1\text{,}\) and when the point is a distance of one unit to the right of \((2,1)\text{,}\) we'll have \(D=1\text{.}\) Then \(D\) changes with respect to time, and \(\ds\diff{D}{t}\) is the velocity of the point. Since we know \(\ds\diff{x}{t}\) and \(\ds\diff{y}{t}\text{,}\) a reasonable plan is to find an equation relating \(x\text{,}\) \(y\text{,}\) and \(D\text{,}\) and differentiate implicitly with respect to \(t\text{.}\) (This implicit differentiation takes care of complication (1).) Using the Pythagorean Theorem:
\begin{align*}
D^2&=(x2)^2+(y1)^2\\
\end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*}
2D\cdot\diff{D}{t}&=2(x2)\cdot\diff{x}{t}+2(y1)\cdot\diff{y}{t}\\
\end{align*}
We plug in \(x=0\text{,}\) \(y=2\text{,}\) \(\diff{x}{t}=1\text{,}\) \(\diff{y}{t}=\frac{1}{2}\text{,}\) and \(D=\sqrt{(02)^2+(21)^2}=\sqrt{5}\) (negative because the point is to the left of \((2,1)\)):
\begin{align*}
2\sqrt{5}\cdot\diff{D}{t}&=2(2)(1)+2(1)\left(\frac{1}{2}\right)\\
\diff{D}{t}&=\frac{\sqrt{5}}{2}\;\mbox{units per second}
\end{align*}
(a) Since the perimeter of the cross section of the bottle does not change, \(p\) (the perimeter of the ellipse) is the same as the perimeter of the circle of radius 5.
(b) The volume of the bottle will be the area of its cross section times its height. This is always the case when you have some twodimensional shape, and turn it into a threedimensional object by “pulling” the shape straight up. (For example, you can think of a cylinder as a circle that has been “pulled” straight up. To understand why this formula works, think about what is means to measure the area of a shape in square centimetres, and the volume of an object in cubic centimetres.)
(c) You can use what you know about \(a\) and the formula from (a) to find \(b\) and \(\ds\diff{b}{t}\text{.}\) Then use the formula from \((b)\text{.}\)
(a) \(10\pi=\pi\left[3(a+b)+\sqrt{(a+3b)(3a+b)}\right]\) or equivalently, \(10=3(a+b)+\sqrt{(a+3b)(3a+b)}\)
(b) \(20\pi a b\)
(c) The water is spilling out at about 36.64 cubic centimetres per second. The exact amount is \(\dfrac{200\pi}{9+\sqrt{35}}\left(12\left(\dfrac{3\sqrt{35}+11}{3\sqrt{35}+13}\right)\right)\;\dfrac{\mathrm{cm}^3}{\mathrm{sec}}\text{.}\)
(a) Since the perimeter of the bottle is unchanged (you aren't stretching the plastic), it is always the same as the perimeter before it was smooshed, which is the circumference of a circle of radius 5, or \(2\pi(5)=10\pi\text{.}\) So, using our approximation for the perimeter of an ellipse,
(b) The area of the base of the bottle is \(\pi a b\) (see Section ), and its height is 20 cm, so the volume of the bottle is
\begin{equation*}
V=20\pi a b
\end{equation*}
(c) As you smoosh the bottle, its volume decreases, so the water spills out. (If it turns out that the volume is increasing, then no water is spilling outbut life experience suggests, and our calculations verify, that this is not the case.) The water will spill out at a rate of \(\ds\diff{V}{t}\) cubic centimetres per second, where \(V\) is the volume inside the bottle. We know something about \(a\) and \(\ds\diff{a}{t}\text{,}\) so a reasonable plan is to differentiate the equation from (b) (relating \(V\) and \(a\)) with respect to \(t\text{.}\)
Using the product rule, we differentiate the equation in (b) implicitly with respect to \(t\) and get
So, we need to find the values of \textcolor{red}{\(a\)}, \textcolor{red}{\(b\)}, \textcolor{red}{\(\ds\diff{a}{t}\)}, and \textcolor{red}{\(\ds\diff{b}{t}\)} at the moment when \(a=2b\text{.}\)
The equation from (a) tells us \(10=3(a+b)+\sqrt{(a+3b)(3a+b)}\text{.}\) So, when \(a=2b\text{,}\)
\begin{align*}
10&=3(2b+b)+\sqrt{(2b+3b)(6b+b)}\\
10&=9b+\sqrt{(5b)(7b)}=b\left(9+\sqrt{35}\right)\\
\color{red}b&\color{red}=\frac{10}{9+\sqrt{35}}\\
\end{align*}
where we use the fact that \(b\) is a positive number, so \(\sqrt{b^2}=b=b\text{.}\)
\begin{align*}
\\
\end{align*}
Since \(a=2b\text{,}\)
\begin{align*}
\color{red}a&\color{red}=\frac{20}{9+\sqrt{35}}\\
\end{align*}
Now we know \(a\) and \(b\) at the moment when \(a=2b\text{.}\) We still need to know \(\ds\diff{a}{t}\) and \(\ds\diff{b}{t}\) at that moment. Since \(a=5+t\text{,}\) always \textcolor{red}{\(\ds\diff{a}{t}=1\)}. The equation from (a) relates \(a\) and \(b\text{,}\) so differentiating both sides with respect to \(t\) will give us an equation relating \(\ds\diff{a}{t}\) and \(\ds\diff{b}{t}\text{.}\) When differentiating the portion with a square root, be careful not to forget the chain rule.
\begin{align*}
0&=3\left(\diff{a}{t}+\diff{b}{t}\right)+\frac{\left(\diff{a}{t}+3\diff{b}{t}\right)(3a+b)+(a+3b)\left(3\diff{a}{t}+\diff{b}{t}\right)}{2\sqrt{(a+3b)(3a+b)}}\\
\end{align*}
Since \(\ds\diff{a}{t}=1\text{:}\)
\begin{align*}
0&=3\left(1+\diff{b}{t}\right)+\frac{\left(1+3\diff{b}{t}\right)(3a+b)+(a+3b)\left(3+\diff{b}{t}\right)}{2\sqrt{(a+3b)(3a+b)}}\\
\end{align*}
At this point, we could plug in the values we know for \(a\) and \(b\) at the moment when \(a=2b\text{.}\) However, the algebra goes a little smoother if we start by plugging in \(a=2b\text{:}\)
\begin{align*}
0&=3\left(1+\diff{b}{t}\right)+\frac{\left(1+3\diff{b}{t}\right)(7b)+(5b)\left(3+\diff{b}{t}\right)}{2\sqrt{(5b)(7b)}}\\
0&=3\left(1+\diff{b}{t}\right)+\frac{b\left(7+21\diff{b}{t}+15+5\diff{b}{t}\right)}{2b\sqrt{35}}\\
0&=3\left(1+\diff{b}{t}\right)+\frac{22+26\diff{b}{t}}{2\sqrt{35}}\\
0&=3+3\diff{b}{t}+\frac{11}{\sqrt{35}}+\frac{13}{\sqrt{35}}\diff{b}{t}\\
3\frac{11}{\sqrt{35}}&=\left(3+\frac{13}{\sqrt{35}}\right)\diff{b}{t}\\
\color{red}\diff{b}{t}&=\frac{3\frac{11}{\sqrt{35}}}{3+\frac{13}{\sqrt{35}}}=\color{red}\frac{3\sqrt{35}11}{3\sqrt{35}+13}\\
\end{align*}
Now, we can calculate \(\ds\diff{V}{t}\) at the moment when \(a=2b\text{.}\) We already found
\begin{align*}
\diff{V}{t}&=20\pi\left(\diff{a}{t}b+a\diff{b}{t}\right)\\
\end{align*}
So, plugging in the values of \(a\text{,}\) \(b\text{,}\) \(\ds\diff{a}{t}\text{,}\) and \(\ds\diff{b}{t}\) at the moment when \(a=2b\text{:}\)
\begin{align*}
\diff{V}{t}&=20\pi\left((1)\left(\frac{10}{9+\sqrt{35}}\right)+\left(\frac{20}{9+\sqrt{35}}\right)\left(\frac{3\sqrt{35}11}{3\sqrt{35}+13}\right)\right)\\
&=\frac{200\pi}{9+\sqrt{35}}\left(12\left(\frac{3\sqrt{35}+11}{3\sqrt{35}+13}\right)\right)\\
&\approx 36.64\;\frac{\mathrm{cm}^3}{\mathrm{sec}}
\end{align*}
So the water is spilling out of the cup at about 36.64 cubic centimetres per second.
Remark: the algebra in this problem got a little nasty, but the method behind its solution is no more difficult than most of the problems in this section. One of the reasons why calculus is so widely taught in universities is to give you lots of practice with problemsolving: taking a big problem, breaking it into pieces you can manage, solving the pieces, and getting a solution.
A problem like this can sometimes derail people. Breaking it up into pieces isn't so hard, but when you actually do those pieces, you can get confused and forget why you are doing the calculations you're doing. If you find yourself in this situation, look back a few steps to remind yourself why you started the calculation you just did. It can also be helpful to write notes, like “We are trying to find \(\diff{V}{t}\text{.}\) We already know that \(\diff{V}{t} = ...\text{.}\) We still need to find \(a\text{,}\) \(b\text{,}\) \(\diff{a}{t}\) and \(\diff{b}{t}\text{.}\)”
Since \(A=0\text{,}\) the equation relating the variables tells us:
\begin{align*}
0&=\log\left(C^2+D^2+1\right)\\
1&=C^2+D^2+1\\
0&=C^2+D^2\\
0&=C=D\\
\end{align*}
This will probably be useful information. Since we're also given the value of a derivative, let's differentiate the equation relating the variables implicitly with respect to \(t\text{.}\) For ease of notation, we will write \(\ds\diff{A}{t}=A'\text{,}\) etc.
\begin{align*}
A'B+AB'&=\frac{2CC'+2DD'}{C^2+D^2+1}\\
\end{align*}
At \(t=10\text{,}\) \(A=C=D=0\text{:}\)
\begin{align*}
A'B+0&=\frac{0+0}{0+0+1}\\
A'B&=0\\
\end{align*}
at \(t=10\text{,}\) \(A'=2\) units per second:
\begin{align*}
2B&=0\\
B&=0
\end{align*}
In the beginning of this section, the text says “A differential equation is an equation for an unknown function that involves the derivative of the unknown function.” Our unknown function is \(y\text{,}\) so a differential equation is an equation that relates \(y\) and \(\ds\diff{y}{x}\text{.}\) This applies to (a) and (b), but not (c), (d), or (e).
Note that \(\ds\diff{x}{x}=1\text{:}\) this is the derivative of \(x\) with respect to \(x\text{.}\)
Theorem 3.3.2 tells us that a function is a solution to the differential equation \(\ds\diff{Q}{t}=kQ(t)\) if and only if the function has the form \(Q(t)=Ce^{kt}\) for some constant \(C\text{.}\) In our case, we want \(Q(t)=5\ds\diff{Q}{t}\text{,}\) so \(\ds\diff{Q}{t}=\dfrac{1}{5}Q(t)\text{.}\) So, the theorem tells us that the solutions are the functions of the form \(Q(t)=Ce^{t/5}\text{.}\) This applies to (a) (with \(C=0\)) and (d) (with \(C=1\)), but none of the other functions.
We don't actually need a theorem to answer this question, though: we can just test every option.
[(a)] \(\ds\diff{Q}{t}=0\text{,}\) so \(Q(t)=0=5\cdot 0 = 5\ds\diff{Q}{t}\text{,}\) so (a) is a solution.
[(b)] \(\ds\diff{Q}{t}=5e^t=Q(t)\text{,}\) so \(Q(t)=\ds\diff{Q}{t} \neq 5 \ds\diff{Q}{t}\text{,}\) so (b) is not a solution.
[(c)] \(\ds\diff{Q}{t}=5e^{5t}=5Q(t)\text{,}\) so \(Q(t)=\dfrac{1}{5}\ds\diff{Q}{t}\neq5\ds\diff{Q}{t}\text{,}\) so (c) is not a solution.
[(d)] \(\ds\diff{Q}{t}=\frac{1}{5}e^{t/5}=\frac{1}{5}Q(t)\text{,}\) so \(Q(t)=5\ds\diff{Q}{t}\text{,}\) so (d) is a solution.
[(e)] \(\ds\diff{Q}{t}=\frac{1}{5}e^{t/5}=\frac{1}{5}\left(Q(t)1\right)\text{,}\) so \(Q(t)=5\ds\diff{Q}{t}+1\text{,}\) so (e) is not a solution.
\begin{align*}
Q(t)&=0\\
\end{align*}
That is,
\begin{align*}
Ce^{kt}&=0
\end{align*}
If \(C=0\text{,}\) then this is the case for all \(t\text{.}\) There was no isotope to begin with, and there will continue not being any undecayed isotope forever.
If \(C \gt 0\text{,}\) then since \(e^{kt} \gt 0\text{,}\) also \(Q(t) \gt 0\text{:}\) so \(Q(t)\) is never 0 for any value of \(t\text{.}\) (But as \(t\) gets bigger and bigger, \(Q(t)\) gets closer and closer to 0.)
Remark: The last result is somewhat disturbing: surely at some point the last atom has decayed. The differential equation we use is a model that assumes \(Q\) runs continuously. This is a good approximation only when there is a very large number of atoms. In practice, that is almost always the case.
In Theorem 3.3.2, we saw that if \(y\) is a function of \(t\text{,}\) and \(\ds\diff{y}{t}=ky\text{,}\) then \(y=Ce^{kt}\) for some constant \(C\text{.}\)
Our equation \(y\) satisfies \(\ds\diff{y}{t}=3y\text{,}\) so the theorem tells us \(y=Ce^{3t}\) for some constant \(C\text{.}\)
We are also told that \(y(1)=2\text{.}\) So, \(2=Ce^{3 \times 1}\) tells us \(C=2e^3\text{.}\) Then:
From the text, we see the halflife of Carbon14 is 5730 years. A microgram (\(\mu\)g) is onemillionth of a gram, but you don't need to know that to solve this problem.
The amount of Carbon14 in the sample \(t\) years after the animal died will be
\begin{equation*}
Q(t)=5e^{kt}
\end{equation*}
for some constant \(k\) (where 5 is the amount of Carbon14 in the sample at time \(t=0\)). So, the answer we're looking for is \(Q(10000)\text{.}\) We need to replace \(k\) with an actual number to evaluate \(Q(10000)\text{,}\) and the key to doing this is the halflife. The text tells us that the halflife of Carbon14 is 5730 years, so we know:
\begin{align*}
Q(5730)&=\frac{5}{2}\\
5e^{k\cdot5730}&=\frac{5}{2}\\
\left(e^{k}\right)^{5730}&=\frac{1}{2}\\
e^{k}&=\sqrt[5730]{\frac{1}{2}}=2^{\tfrac{1}{5730}}\\
\end{align*}
So:
\begin{align*}
Q(t)&=5\left(e^{k}\right)^t\\
&=5\cdot2^{\tfrac{t}{5730}}\\
\end{align*}
Now, we can evaluate:
\begin{align*}
Q(10000)&=5\cdot 2^{\tfrac{10000}{5730}}\approx 1.5\;\mu g
\end{align*}
Remark: after \(2(5730)=11,460\) years, the sample will have been sitting for two halflives, so its remaining Carbon14 will be a quarter of its original amount, or \(1.25\) \(\mu\)g. It makes sense that at 10,000 years, the sample will contain slightly more Carbon14 than at 11,460 years. Indeed, 1.5 is slightly larger than 1.25, so our answer seems plausible.
It's a good habit to look for ways to quickly check whether your answer seems plausible, since a small algebra error can easily turn into a big error in your solution.
The quantity of Radium226 in the sample at time \(t\) will be \(Q(t)=Ce^{kt}\) for some positive constants \(C\) and \(k\text{.}\) You can use the given information to find \(C\) and \(e^{k}\text{.}\)
In the following work, remember we use \(\log\) to mean natural logarithm, \(\log_e\text{.}\)
Let 100 years ago be the time \(t=0\text{.}\) Then if \(Q(t)\) is the amount of Radium226 in the sample, \(Q(0)=1\text{,}\) and
\begin{align*}
Q(t)&=e^{kt}\\
\end{align*}
for some positive constant \(k\text{.}\) When \(t=100\text{,}\) the amount of Radium226 left is 0.9576 grams, so
\begin{align*}
0.9576=Q(100)&=e^{k\cdot 100}=\left(e^{k}\right)^{100}\\
e^{k}&=0.9576^{\tfrac{1}{100}}\\
\end{align*}
This tells us
\begin{align*}
Q(t)&=0.9576^{\tfrac{t}{100}}\\
\end{align*}
So, if half the original amount of Radium226 is left,
\begin{align*}
\frac{1}{2}&=0.9576^{\tfrac{t}{100}}\\
\log\left(\frac{1}{2}\right)&=\log\left(0.9576^{\tfrac{t}{100}}\right)\\
\log 2&=\frac{t}{100}\log(0.9576)\\
t&=100\frac{\log 2}{\log 0.9576}\approx 1600
\end{align*}
So, the half life of Radium226 is about 1600 years.
The fact that the mass of the sample decreases at a rate proportional to its mass tells us that, if \(Q(t)\) is the mass of Polonium201, the following differential equation holds:
Let \(Q(t)\) denote the mass at time \(t\text{.}\) Then \(\ds\diff{Q}{t}\) is the rate at which the mass is changing. Since the rate the mass is decreasing is proportional to the mass remaining, we know \(\ds\diff{Q}{t}=kQ(t)\text{,}\) where \(k\) is a positive constant. (Remark: since \(Q\) is decreasing, \(\ds\diff{Q}{t}\) is negative. Since we cannot have a negative mass, if we choose \(k\) to be positive, then \(k\) and \(Q\) are both positivethis is why we added the negative sign.)
for some constant \(k \gt 0\text{.}\) By Theorem 3.3.2, we know
\begin{equation*}
Q(t)=Ce^{kt}
\end{equation*}
for some constant \(C\text{.}\) Since \(Q(0)=Ce^{0}=C\text{,}\) the given information tells us \(6=C\text{.}\) (This is the initial mass of our sample.) So, \(Q(t)=6e^{kt}\text{.}\) To get the full picture of the behaviour of \(Q\text{,}\) we should find \(k\text{.}\) We do this using the given information \(Q(1)=1\text{:}\)
\begin{align*}
1&=Q(1)=6e^{k(1)}\\
6^{1}=\frac{1}{6}&=e^{k}\\
\end{align*}
So, all together,
\begin{align*}
Q(t)&=6\left(e^{k}\right)^t=6\cdot \left(6^{1}\right)^t=6^{1t}
\end{align*}
The question asks us to determine the time \(t_{h}\) which obeys \(Q(t_{h})=\dfrac{6}{2}=3\text{.}\) Now that we know the equation for \(Q(t)\text{,}\) we simply solve:
The halflife of Polonium210 is \(\dfrac{\log 2}{\log 6}\) years, or about 141 days.
Remark: The actual halflife of Polonium210 is closer to 138 days. The numbers in the question are made to work out nicely, at the expense of some accuracy.
The amount of Radium221 in a sample at time \(t\) will be \(Q(t)=Ce^{kt}\) for some positive constants \(C\) and \(k\text{.}\) You can leave \(C\) as a variableit's the original amount in the sample, which isn't specified. What you want to find is the value of \(t\) such that \(Q(t)=0.0001Q(0)=0.0001C\text{.}\)
\begin{align*}
Q(t)&=Ce^{kt}\\
\end{align*}
where \(C\) is the amount in the sample at time \(t=0\text{,}\) and \(k\) is some positive constant. We know the halflife of the isotope, so we can find \(e^{k}\text{:}\)
\begin{align*}
\frac{C}{2}=Q(30)&=Ce^{k\cdot 30}\\
\frac{1}{2}&=\left(e^{k}\right)^{30}\\
2^{\tfrac{1}{30}}&=e^{k}\\
\end{align*}
So,
\begin{align*}
Q(t)&=C\left(e^{k}\right)^t=C\cdot2^{\tfrac{t}{30}}\\
\end{align*}
When only 0.01% of the original sample is left, \(Q(t)=0.0001C\text{:}\)
\begin{align*}
0.0001C=Q(t)&=C\cdot 2^{\tfrac{t}{30}}\\
0.0001&=2^{\tfrac{t}{30}}\\
\log(0.0001)&=\log\left(2^{\tfrac{t}{30}}\right)\\
\log\left(10^{4}\right)&=\frac{t}{30}\log2\\
4\log 10&=\frac{t}{30}\log2\\
t&=120\cdot\frac{\log10}{\log2}\approx 398.6
\end{align*}
It takes about 398.6 seconds (that is, roughly 6 and a half minutes) for all but \(0.01\)% of the sample to decay.
Remark: we can do another reality check here. The halflife is 30 seconds. 6 and a half minutes represents 13 halflives. So, the sample is halved 13 times: \(\left(\tfrac{1}{2}\right)^{13}\approx 0.00012=0.012\%\text{.}\) So these 13 halflives should reduce the sample to about 0.01% of its original amount, as desired.
You don't need to know the original amount of Polonium210 in order to answer this question: you can leave it as some constant \(C\text{,}\) or you can call it 100%.
We know that the amount of Polonium210 in a sample after \(t\) days is given by
\begin{align*}
Q(t)&=Ce^{kt}\\
\end{align*}
where \(C\) is the original amount of the sample, and \(k\) is some positive constant.
\begin{align*}
\\
\end{align*}
The question asks us what percentage of the sample decays in a day. Since \(t\) is measured in days, the amount that decays in a day is \(Q(t)Q(t+1)\text{.}\) The percentage of \(Q(t)\) that this represents is \(100\dfrac{Q(t)Q(t+1)}{Q(t)}\text{.}\) (For example, if there were two grams at time \(t\text{,}\) and one gram at time \(t+1\text{,}\) then \(100\dfrac{21}{1}=50\text{:}\) 50% of the sample decayed in a day.)
\begin{align*}
\\
\end{align*}
In order to simplify, we should figure out a better expression for \(Q(t)\text{.}\) As usual, we make use of the halflife.
\begin{align*}
Q(138)&=\frac{C}{2}\\
Ce^{k\cdot138}&=\frac{C}{2}\\
\left(e^{k}\right)^{138}&=\frac{1}{2}=2^{1}\\
e^{k}&=2^{\tfrac{1}{138}}\\
\end{align*}
Now, we have a better formula for \(Q(t)\text{:}\)
\begin{align*}
Q(t)&=C\left(e^{k}\right)^t\\
Q(t)&=C\cdot2^{\tfrac{t}{138}}\\
\end{align*}
Finally, we can evaluate what percentage of the sample decays in a day.
\begin{align*}
100\frac{Q(t)Q(t+1)}{Q(t)}&=100\frac{C\cdot2^{\tfrac{t}{138}}C\cdot2^{\tfrac{t+1}{138}}}{C\cdot2^{\tfrac{t}{138}}}\left(\frac{\frac{1}{C}}{\frac{1}{C}}\right)\\
&=100\frac{2^{\tfrac{t}{138}}2^{\tfrac{t+1}{138}}}{2^{\tfrac{t}{138}}}\\
&=100\left(2^{\tfrac{t}{138}}2^{\tfrac{t+1}{138}}\right)2^{\tfrac{t}{138}}\\
&=100\left(12^{\tfrac{1}{138}}\right)\approx 0.5
\end{align*}
About 0.5% of the sample decays in a day.
Remark: when we say that half a percent of the sample decays in a day, we don't mean half a percent of the original sample. If a day starts out with, say, 1 microgram, then what decays in the next 24 hours is about half a percent of that 1 microgram, regardless of what the “original” sample (at some time \(t=0\)) held.
In particular, since the sample is getting smaller and smaller, that half of a percent that decays every day represents fewer and fewer actual atoms decaying. That's why we can't say that half of the sample (50%) will decay after about 100 days, even though 0.5% decays every day and \(100\times 0.5 = 50\text{.}\)
The amount of Uranium232 in the sample of ore at time \(t\) will be
\begin{align*}
Q(t)&=Q(0)e^{kt}\\
\end{align*}
where \(6.9 \leq Q(0) \leq 7.5\text{.}\) We don't exactly know \(Q(0)\text{,}\) and we don't exactly know the halflife, so we also won't exactly know \(Q(10)\text{:}\) we can only say that is it between two numbers. Our strategy is to find the highest and lowest possible values of \(Q(10)\text{,}\) given the information in the problem.
\begin{align*}
\\
\end{align*}
In order for the most possible Uranium232 to be in the sample after 10 years, we should start with the most and have the longest halflife (since this represents the slowest decay). So, we take \(Q(0)=7.5\) and \(Q(70)=\frac{1}{2}(7.5)\text{.}\)
\begin{align*}
Q(t)&=7.5e^{kt}\\
\frac{1}{2}(7.5)=Q(70)&=7.5e^{k(70)}\\
\frac{1}{2}&=\left(e^{k}\right)^{70}\\
2^{\tfrac{1}{70}}&=e^{k}\\
\end{align*}
So, in this secenario,
\begin{align*}
Q(t)&=7.5\cdot 2^{\tfrac{t}{70}}\\
\end{align*}
After ten years,
\begin{align*}
Q(10)&=7.5\cdot 2^{\tfrac{10}{70}}\approx 6.79\\
\end{align*}
So after ten years, the sample contains at most 6.8 \(\mu\)g.
\begin{align*}
\\
\end{align*}
Now, let's think about the least possible amount of Uranium232 that could be left after 10 years. We should start with as little as possible, so take \(Q(0)=6.9\text{,}\) and the sample should decay quickly, so take the halflife to be 68.8 years.
\begin{align*}
Q(t)&=6.9e^{kt}\\
\frac{1}{2}6.9=Q(68.8)&=6.9e^{k(68.8)}\\
\frac{1}{2}&=\left(e^{k}\right)^{68.8}\\
2^{\tfrac{1}{68.8}}&=e^{k}\\
\end{align*}
In this scenario,
\begin{align*}
Q(t)&=6.9\cdot 2^{\tfrac{t}{68.8}}\\
\end{align*}
After ten years,
\begin{align*}
Q(10)&=6.9\cdot 2^{\tfrac{10}{68.8}}\approx 6.24\\
\end{align*}
So after ten years, the sample contains at least 6.2 \(\mu\)g.
\begin{align*}
\end{align*}
After ten years, the sample contains between 6.2 and 6.8 \(\mu\)g of Uranium232.
You can refer to Corollary 3.3.8, but you can also just differentiate the various proposed functions and see whether, in fact, \(\ds\diff{T}{t}\) is the same as \(5[T20]\text{.}\)
for some constant \(T(0)\text{.}\) This fits (a) (with \(T(0)=20\)), (c) (with \(T(0)=21)\text{,}\) and (d) (with \(T(0)=40\)), but not (b) (since the constant has the wrong sign).
Instead of using the corollary, we can also just check each function for ourselves.
[(a)] \(\ds\diff{T}{t}=0=5\cdot0=5[T(t)20]\text{,}\) so (a) gives a solution to the differential equation.
[(b)] \(\ds\diff{T}{t}=5[20e^{5t}]=5[T+20] \neq 5[T20]\text{,}\) so (b) does not give a solution to the differential equation.
[(c)] \(\ds\diff{T}{t}=5[e^{5t}]=5[T20]\text{,}\) so (c) gives a solution to the differential equation.
[(d)] \(\ds\diff{T}{t}=5[20e^{5t}]=5[T20]\text{,}\) so (d) gives a solution to the differential equation.
where \(A\) is the ambient temperature (the temperature of the room), \(T(0)\) is the initial temperature of the copper, and \(K\) is some constant. So, the ambient temperaturethe temperature of the room is \(10\) degrees. Since the coefficient of the exponential part of the function is positive, the temperature of the object is higher than the temperature of the room.
As \(t\) grows very large, \(T(t)\) approaches \(A\text{.}\) That is:
\begin{align*}
\ds\lim_{t \to \infty}T(t)&=A\\
\lim_{t \to \infty}[T(0)A]e^{Kt}+A&=A\\
\lim_{t \to \infty}[T(0)A]e^{Kt}&=0\\
\end{align*}
Since the object is warmer than the room, \(T(0)A\) is a nonzero constant. So,
\begin{align*}
\lim_{t \to \infty}e^{Kt}&=0
\end{align*}
This tells us that \(K\) is a negative number. So, \(K\) must be negativenot zero, and not positive.
Remark: in our work, we used the fact that the object and the room have different temperatures (but it didn't matter which one was hotter). If not, then \(T(0)=A\text{,}\) and \(T(t)=A\text{:}\) that is, the temperature of the object is constant. In this case, our usual form for the temperature of the object looks like this:
\begin{equation*}
T(t)=0e^{Kt}+A
\end{equation*}
Keeping the exponential piece in there is overkill: the temperature isn't changing, the function is simply constant. If the object and the room have the same temperature, \(K\) could be any real number since we're multiplying \(e^{Kt}\) by zero.
If the object has a different initial temperature than its surroundings, then \(T(t)\) is never equal to \(A\text{.}\) (But as time goes on, it gets closer and closer.)
If the object starts out with the same temperature as its surrounding, then \(T(t)=A\) for all values of \(t\text{.}\)
\begin{align*}
[T(0)A]e^{kt}+A&=A\\
\end{align*}
That is, when
\begin{align*}
[T(0)A]e^{kt}&=0\\
\end{align*}
Since \(e^{kt} \gt 0\) for all values of \(k\) and \(t\text{,}\) this happens exactly when
\begin{align*}
T(0)A=0
\end{align*}
So: if the initial temperature of the object is not the same as the ambient temperature, then according to this model, it never will be! (However, as \(t\) gets larger and larger, \(T(t)\) gets closer and closer to \(A\)it just never exactly reaches there.)
If the initial temperature of the object starts out the same as the ambient temperature, then \(T(t)=A\) for all values of \(t\text{.}\)
where \(A\) is the ambient temperature, \(T(0)\) is the initial temperature of the copper, and \(K\) is some constant. Use the given information to find an expression for \(T(t)\) not involving any unknown constants.
where \(A\) is the ambient temperature (100\(^\circ\)), \(T(0)\) is the temperature of the copper at time 0 (let's make this the instant it was dumped in the water, so \(T(0)=25^\circ\)), and \(K\) is some constant. That is:
\begin{align*}
T(t)&=[25100]e^{Kt}+100\\
&=75e^{Kt}+100\\
\end{align*}
The information given tells us that \(T(10)=90\text{,}\) so
\begin{align*}
90&=75e^{10K}+100\\
75\left(e^K\right)^{10}&=10\\
\left(e^K\right)^{10}&=\frac{2}{15}\\
e^K&=\left(\frac{2}{15}\right)^{\tfrac{1}{10}}\\
\end{align*}
This lets us describe \(T(t)\) without any unknown constants.
\begin{align*}
T(t)&=75\left(e^K\right)^{t}+100\\
&=75\left(\frac{2}{15}\right)^{\tfrac{t}{10}}+100\\
\end{align*}
The question asks what value of \(t\) gives \(T(t)=99.9\text{.}\)
\begin{align*}
99.9&=75\left(\frac{2}{15}\right)^{\tfrac{t}{10}}+100\\
75\left(\frac{2}{15}\right)^{\tfrac{t}{10}}&=0.1\\
\left(\frac{2}{15}\right)^{\tfrac{t}{10}}&=\frac{1}{750}\\
\log\left(\left(\frac{2}{15}\right)^{\tfrac{t}{10}}\right)&=\log\left(\frac{1}{750}\right)\\
\frac{t}{10}\log\left(\frac{2}{15}\right)&=\log(750)\\
t&=\frac{10\log(750)}{\log\left(\tfrac{2}{15}\right)}\approx 32.9
\end{align*}
The temperature of the stone \(t\) minutes after taking it from the bonfire is
\begin{align*}
T(t)&=[T(0)A]e^{Kt}+A\\
&=[5000]e^{Kt}+0\\
&=500e^{Kt}\\
\end{align*}
for some constant \(K\text{.}\) We are given that \(T(10)=100\text{.}\)
\begin{align*}
100=T(10)&=500e^{10K}\\
e^{10K}&=\frac{1}{5}\\
e^K&=5^{\tfrac{1}{10}}\\
\end{align*}
This gives us the more complete picture for the temperature of the stone.
\begin{align*}
T(t)&=500\left(e^K\right)^t=500\cdot 5^{\tfrac{t}{10}}\\
\end{align*}
If \(T(t)=50:\)
\begin{align*}
50=T(t)&=500\cdot 5^{\tfrac{t}{10}}\\
\frac{1}{10}=10^{1}&=5^{\tfrac{t}{10}}\\
10&=5^{\tfrac{t}{10}}\\
\log(10)&=\frac{t}{10}\log(5)\\
t&=10\frac{\log(10)}{\log(5)}\approx 14.3
\end{align*}
So the stone has been out of the fire for about 14.3 minutes.
First scenario: At time \(0\text{,}\) Newton mixes 9 parts coffee at temperature \(95^\circ\) C with 1 part cream at temperature \(5^\circ\) C. The resulting mixture has temperature
\begin{equation*}
\frac{9\times 95+1\times 5}{9+1}=86^\circ
\end{equation*}
The mixture cools according to Newton's Law of Cooling, with initial temperature 86\(^\circ\) and ambient temperature 22\(^\circ\text{:}\)
\begin{align*}
T(t)&=[8622]e^{kt}+22\\
T(t)&=64e^{kt}+22\\
\end{align*}
After 10 minutes,
\begin{align*}
\textcolor{red}{54}=T(10)&=22+64e^{10 k}\\
e^{10 k}&=\frac{5422}{64}=\frac{1}{2}
\end{align*}
We could compute \(k\) from this, but we don't need it.
Second scenario: At time \(0\text{,}\) Newton gets hot coffee at temperature \(95^\circ\) C. It cools according to Newton's Law of Cooling
\begin{gather*}
T(t)=[T(0)22]e^{kt}+22
\end{gather*}
In this second scenario, \(T(0)=95\text{,}\) so
\begin{equation*}
T(t)=[9522]e^{kt}+22=73e^{kt}+22
\end{equation*}
The value of \(k\) is the same as in the first scenario, so after 10 minutes
\begin{equation*}
T(10)=22+73e^{10k}=22+73\frac{1}{2}=58.5
\end{equation*}
This cooled coffee is mixed with cold cream to yield a mixture of temperature
\begin{equation*}
\frac{9\times 58.5+1\times 5}{9+1}=\color{red}53.15
\end{equation*}
Under the second (add cream just before drinking) scenario, the coffee ends up {cooler by \(0.85^\circ\) C}\(\,\text{.}\)
By Newton's law of cooling, the rate of change of temperature is proportional to the difference between \(T(t)\) and the ambient temperature, which in this case is \(30^\circ\text{.}\) Thus
for some constant of proportionality \(k\text{.}\) To answer part (a), all we have to do is find \(k\text{.}\)
Under Newton's Law of Cooling, the temperature at time \(t\) will be given by
\begin{align*}
T(t)&=[T(0)A]e^{kt}+A\\
&=[530]e^{kt}+30\\
&=25e^{kt}+30\\
\end{align*}
We are told \(T(5)=10\text{:}\)
\begin{align*}
10&=25e^{5k}+30\\
25e^{5k}&=20\\
e^{5k}&=\frac{4}{5}\\
5k&=\log(4/5)\\
k&=\tfrac{1}{5}\log(4/5)\\
\end{align*}
So, the differential equation is
\begin{align*}
\diff{T}{t}(t)&=\frac{1}{5}\log(4/5)[T(t)30]
\end{align*}
Since \(T(t)=3025e^{kt}\text{,}\) the temperature of the tea is \(20^\circ\) when
\begin{align*}
3025 e^{kt}&=20\\
e^{kt}&=\frac{10}{25}\\
kt&=\log\left(\frac{10}{25}\right)\\
t&=\frac{1}{k}\log\frac{2}{5}\\
&=\frac{5\log(2/5)}{\log(4/5)}\\
&\approx{20.53\;\mathrm{ min}}
\end{align*}
As time goes on, temperatures that follow Newton's Law of Cooling get closer and closer to the ambient temperature. So, \(\ds\lim_{t \to \infty} T(t)\) exists. In particular, \(\ds\lim_{t \to \infty} 0.8^{kt}\) exists.
If \(k \lt 0\text{,}\) then \(\ds\lim_{t \to \infty}0.8^{kt} = \infty\text{,}\) since \(0.8 \lt 1\text{.}\) So, \(k \geq 0\text{.}\)
If \(k=0\text{,}\) then \(T(t)=16\) for all values of \(t\text{.}\) But, in the statement of the question, the object is changing temperature. So, \(k \gt 0\text{.}\)
Therefore, \(k\) is positive.
Remark: contrast this to Question 3.3.4.3. The reason we get a different answer is that our base in this question (0.8) is less than one, while the base in Question 3.3.4.3 (\(e\)) is greater than one.
Although the given equation \(T(t)\) does not exactly look like the Newton's Law equations we're used to, it is equivalent. Remembering \(e^{\log(0.8)}=0.8\text{:}\)
Since \(b\) is a positive constant, \(\ds\lim_{t \to \infty}e^{bt}=\infty\text{.}\) Therefore:
\begin{align*}
\lim_{t \to \infty}P(t)&=\lim_{t \to \infty} P(0)e^{bt}
=\left\{\begin{array}{cc}
0&\mbox{ if }P(0)=0\\
\infty&\mbox{ if }P(0) \gt 0
\end{array}\right.
\end{align*}
If \(P(0)=0\text{,}\) then the model simply says “a population that starts with no individuals continues to have no individuals indefinitely,” which certainly makes sense. If \(P(0) \neq 0\text{,}\) then (since we can't have a negative population) \(P(0) \gt 0\text{,}\) and the model says “a population that starts out with some individuals will end up with any gigantically huge number you can think of, given enough time.” This one doesn't make so much sense. Populations only grow to a certain finite amount, due to scarcity of resources and such. In the derivation of the Malthusian model, we assume a constant net birth ratethat the birth and death rates (per individual) don't depend on the population, which is not a reasonable assumption longterm.
The assumption that the animals grow according to the Malthusian model tells us that their population \(t\) years after 2015 is given by \(P(t)=121e^{bt}\) for some constant \(b\text{.}\)
The assumption that the animals grow according to the Malthusian model tells us that their population \(t\) years after 2015 is given by \(P(t)=121e^{bt}\) for some constant \(b\text{,}\) since \(121=P(0)\text{,}\) the population 0 years after 2015. The information about 2016 tells us
\begin{align*}
136=P(1)&=121e^{b}\\
\frac{136}{121}&=e^b\\
\end{align*}
This gives us a better idea of \(P(t)\text{:}\)
\begin{align*}
P(t)&=121e^{bt}=121\left(\frac{136}{121}\right)^t\\
\end{align*}
2020 is 5 years after 2015, so in 2020 (assuming the population keeps growing according to the Malthusian model) the population will be
\begin{align*}
P(5)&=121\left(\frac{136}{121}\right)^5\approx217
\end{align*}
In 2020, the Malthusian model predicts the herd will number 217 individuals.
The Malthusian model says that the population of bacteria \(t\) hours after being placed in the dish will be \(P(t)=1000e^{bt}\) for some constant \(b\text{.}\)
Since the initial population of bacteria is 1000 individuals, the Malthusian model says that the population of bacteria \(t\) hours after being placed in the dish will be \(P(t)=1000e^{bt}\) for some constant \(b\text{.}\) Since \(P(1)=2000\text{,}\)
\begin{align*}
2000=P(1)&=1000e^{b}\\
2&=e^b\\
\end{align*}
So, the population at time \(t\) is
\begin{align*}
P(t)&=1000\cdot 2^t\\
\end{align*}
We want to know at what time the population triples, to 3,000 individuals.
\begin{align*}
3000&=1000\cdot 2^t\\
3&=2^t\\
\log(3)&=\log\left(2^t\right)=t\log(2)\\
t&=\frac{\log(3)}{\log(2)}\approx 1.6
\end{align*}
According to the Malthusian Model, if the ship wrecked at year \(t=0\) and 2 rats washed up on the island, then \(t\) years after the wreck, the population of rats will be
\begin{align*}
P(t)&=2e^{bt}\\
\end{align*}
for some constant \(b\text{.}\) We want to get rid of this extraneous variable \(b\text{,}\) so we use the given information. If 1928 is \(a\) years after the wreck:
\begin{align*}
1000=P(a)&=2e^{ba}\\
1500=P(a+1)&=2e^{b(a+1)}=2e^{ba}e^b\\
\end{align*}
So,
\begin{align*}
(1000)\left(e^b\right)&=\left(2e^{ba}\right)\left(e^b\right)=1500\\
\end{align*}
Which tells us
\begin{align*}
e^b&=\frac{1500}{1000}=1.5\\
\end{align*}
Now, our model is complete:
\begin{align*}
P(t)&=2\left(e^b\right)^t=2\cdot1.5^t\\
\end{align*}
Since \(P(a)=1000\text{,}\) we can find \(a\text{:}\)
\begin{align*}
1000=P(a)&=2\cdot1.5^a\\
500&=1.5^a\\
\log(500)&=\log\left(1.5^a\right)=a\log(1.5)\\
a&=\frac{\log(500)}{\log(1.5)}\approx 15.3
\end{align*}
So, the year 1928 was about 15.3 years after the shipwreck. Since we aren't given a month when the rats reached exactly 1000 in number, that puts the shipwreck at the year 1912 or 1913.
If the population has a net birthrate per individual per unit time of \(b\text{,}\) then the Malthusian model predicts that the number of individuals at time \(t\) will be \(P(t)=P(0)e^{bt}\text{.}\) You can use the test population to find \(e^b\text{.}\)
The Malthusian model suggests that, if we start with \(P(0)\) cochineals, their population after 3 months will be
\begin{align*}
P(t)&=P(0)e^{bt}\\
\end{align*}
for some constant \(b\text{.}\) The constant \(b\) is the net birthrate per population member per unit time. Assuming that the net birthrate for a larger population will be the same as the test population, we can use the data from the test to find \(e^b\text{.}\) Let \(Q(t)\) be the number of individuals in the test population at time \(t\text{.}\)
\begin{align*}
Q(t)&=Q(0)e^{bt}=200e^{bt}\\
1000&=Q(3)=200e^{3b}\\
5&=e^{3b}\\
e^b&=5^{1/3}\\
\end{align*}
Now that we have an idea of the birthrate, we predict
\begin{align*}
P(t)&=P(0)\left(e^{b}\right)^t=P(0)\cdot5^{\tfrac{t}{3}}\\
\end{align*}
We want \(P(12)=10^6+P(0)\text{.}\)
\begin{align*}
10^6+P(0)=P(12)&=P(0)\cdot 5^{\tfrac{12}{3}}=P(0)\cdot 5^4\\
10^6&=P(0)\cdot 5^4P(0)=P(0)\left[5^41\right]\\
P(0)&=\frac{10^6}{5^41}\approx 1603
\end{align*}
The farmer should use an initial population of (at least) about 1603 individuals.
Remark: if we hadn't specified that we need to save \(P(0)\) individuals to start next year's population, the number of individual cochineals we would want to start with to get a million in a year would be 1600almost the same!
[(a)] Since \(f(t)\) gives the amount of the radioactive isotope in the sample at time \(t\text{,}\) the amount of the radioactive isotope in the sample when \(t=0\) is \(f(0)=100e^0=100\) units. Since the sample is decaying, \(f(t)\) is decreasing, so \(f'(t)\) is negative. Differentiating, \(f'(t)=k(100e^{kt})\text{.}\) Since \(100e^{kt}\) is positive and \(f'(t)\) is negative, \(k\) is negative.
[(b)] Since \(f(t)\) gives the size of the population at time \(t\text{,}\) the number of individuals in the population when \(t=0\) is \(f(0)=100e^0=100\text{.}\) Since the population is growing, \(f(t)\) is increasing, so \(f'(t)\) is positive. Differentiating, \(f'(t)=k(100e^{kt})\text{.}\) Since \(100e^{kt}\) is positive and \(f'(t)\) is positive, \(k\) is also positive.
[(c)] Newton's Law of Cooling give the temperature of an object at time \(t\) as \(f(t)=[f(0)A]e^{kt}+A\text{,}\) where \(A\) is the ambient temperature surrounding the object. In our case, the ambient temperature is 0 degrees. In an object whose temperature is being modelled by Newton's Law of Cooling, it doesn't matter whether the object is heating or cooling, \(k\) is negative. We saw this in Question 3.3.4.3 of Section 3.3.2, but it bears repeating. Since \(f(t)\) approaches the ambient temperature (in this case, 0) as \(t\) goes to infinty:
\begin{equation*}
\lim_{t \to \infty}100e^{kt}=0
\end{equation*}
so \(k\) is negative.
The first piece of information given tells us \(\ds\diff{f}{x}=\pi f(x)\text{.}\) Then by Theorem 3.3.2,
\begin{align*}
f(x)&=Ce^{\pi x}\\
\end{align*}
for some constant \(C\text{.}\) The second piece of given information tells us \(f(0)=2\text{.}\) Using this, we can solve for \(C\text{:}\)
\begin{align*}
2&=f(0)=Ce^{0}=C\\
\end{align*}
Now, we know \(f(x)\) entirely:
\begin{align*}
f(x)&=2e^{\pi t}\\
\end{align*}
So, we can evaluate \(f(2)\)
\begin{align*}
f(2)&=2e^{2\pi}
\end{align*}