Section 1.4 Curves in Three Dimensions
So far, we have developed formulae for the curvature, unit tangent vector, etc., at a point
\(\vr(t)\) on a curve that lies in the
\(xy\)-plane. We now extend our discussion to curves in
\(\bbbr^3\text{.}\) Fix any
\(t\text{.}\) For
\(t'\) very close to
\(t\text{,}\) \(\vr(t')\text{,}\) will, by the Taylor expansion to second order, be very close to
\(\vr(t) + \vr'(t)\,(t'-t)
+\frac{1}{2}\vr'(t)\,(t'-t)^2\text{,}\) so that
\(\vr(t')\) almost lies in the plane through
\(\vr(t)\) that is determined by the two vectors
\(\vr'(t)\) and
\(\vr'(t)\text{.}\) Thus, if we restrict our attention to a very small part of the curve near the point of interest
\(\vr(t)\text{,}\) the curve will, to a very good approximation lie in some plane. So we can still define, for example, the osculating circle to the curve at
\(\vr(t)\) to be the circle in that plane that fits the curve best near
\(\vr(t)\text{.}\) And we still have the formulae
1
\begin{align*}
\vv&=\diff{\vr}{t}=\diff{s}{t}\,\hat\vT\\
\diff{\hat\vT}{s} &= \ka\hat\vN\\
\diff{\hat\vT}{t} &= \ka\diff{s}{t}\hat\vN\\
\va&=\difftwo{\vr}{t}=\difftwo{s}{t}\,\hat\vT
+\ka\Big(\diff{s}{t}\Big)^2\hat\vN\\
\vv\times\va &= \ka \Big(\diff{s}{t}\Big)^3\hat\vT\times\hat\vN
\end{align*}
The only
2 difference is that
\(\vv, \va, \hat\vT\) and
\(\hat\vN\) are now three component vectors rather than two component vectors.
If we are lucky and our curve happens to lie completely in a single plane, the vectors \(\hat\vT(s)\) and \(\hat\vN(s)\) are mutually perpendicular unit vectors that lie in the same plane, so that their cross product \(\hat\vB(s) = \hat\vT(s)\times\hat\vN(s)\) is a unit vector that is perpendicular to the plane. By continuity, \(\hat\vB(s)\) has to be a constant vector, i.e. be independent of \(s\text{.}\)
If, on the other hand, \(\hat\vB(s)\) is not constant, then our curve doesn't lie in a single plane, and we can use the derivative
\begin{align*}
\diff{\hat\vB}{s}
&=\diff{}{s}\big(\hat\vT\times\hat\vN\big)
=\diff{\hat\vT}{s}\times\hat\vN + \hat\vT\times \diff{\hat\vN}{s}\\
&=\hat\vT\times \diff{\hat\vN}{s}\qquad
\Big( \text{since } \diff{\hat\vT}{s} \text{ is parallel to } \hat\vN \Big)
\end{align*}
as a measure
of how badly the curve fails to lie in a plane,
i.e. how much the plane that fits the curve best near \(\vr(s)\) twists as \(s\) increases,
The cross product in \(\diff{\hat\vB}{s}=\hat\vT\times \diff{\hat\vN}{s}\) implies that \(\diff{\hat\vB}{s}\) is perpendicular to \(\hat\vT\text{.}\) In addition, \(\diff{\hat\vB}{s}\) must be perpendicular to \(\hat\vB\) because
\begin{equation*}
|\hat\vB|=1
\implies 1=\hat\vB\cdot\hat\vB
\implies 0 = \diff{}{s}\left[\hat\vB\cdot\hat\vB\right]
= 2 \hat\vB\cdot\diff{\hat\vB}{s}
\end{equation*}
So \(\diff{\hat\vB}{s}(s)\) must be parallel to \(\hat\vN(s)\text{.}\)
Definition 1.4.1.
The binormal vector at \(\vr(s)\) is \(\hat\vB(s) = \hat\vT(s)\times \hat\vN(s)\text{.}\) The normal vector \(\hat\vN(s)\) is sometimes called the unit principal normal vector to distinguish it from the binormal vector.
We define the
torsion \(\tau(s)\) by
\begin{equation*}
\diff{\hat\vB}{s}(s) = -\tau(s)\hat\vN(s)
\end{equation*}
The negative sign is included so that
\(\tau(s) \gt 0\) indicates “right handed twisting”. There will be an explanation of what this means in Example
1.4.4 below.
The
osculating plane at
\(\vr(s)\) (the plane that fits the curve best at
\(\vr(s)\)) is the plane through
\(\vr(s)\) with normal vector
\(\hat\vB(s)\text{.}\) The equation of the plane is
\begin{equation*}
\hat\vB(s)\cdot\big\{(x,y,z)-\vr(s)\big\}=0
\end{equation*}
For each
\(s\text{,}\) \(\hat\vT(s)\text{,}\) \(\hat\vN(s)\) and
\(\hat\vB(s)\) are mutually perpendicular unit vectors. They form an orthonormal basis for
\(\bbbr^3\text{,}\) just as
\(\hi\text{,}\) \(\hj\) and
\(\hk\) form an orthonormal basis for
\(\bbbr^3\text{.}\) Furthermore both
\((\hat\vT(s)\,,\,\hat\vN(s)\,,\,\hat\vB(s))\) and
\((\hi\,,\,\hj\,,\,\hk)\) are “right handed triples”
3 , meaning that
\(\hat\vB(s) = \hat\vT(s)\times\hat\vN(s)\) and
\(\hk=\hi\times\hj\text{.}\)
We have already computed \(\diff{\hat\vT}{s}\) and \(\diff{\hat\vB}{s}\text{.}\) It is now an easy matter to compute
\begin{align*}
\diff{\hat\vN}{s}
&= \diff{}{s}\big(\hat\vB(s)\times\hat\vT(s)\big)\\
&= -\tau(s)\hat\vN(s)\times\hat\vT(s)
+\hat\vB(s)\times\big(\ka(s)\hat\vN(s)\big)\\
&=\tau(s)\hat\vB(s)-\ka(s)\hat\vT(s)
\end{align*}
To see that \(\hat\vN(s)\times\hat\vT(s)=-\hat\vB(s)\) and \(\hat\vB(s)\times\hat\vN(s)=-\hat\vT(s)\text{,}\) just look at the right hand figure above.
Now suppose that we have a curve that is parametrized by \(t\) rather than \(s\text{.}\) How do we find the torsion \(\tau\text{?}\) The most obvious method is to
recall that
\(\vv\times\va = \ka \big(\diff{s}{t}\big)^3\hat\vT\times\hat\vN
= \ka \big(\diff{s}{t}\big)^3\hat\vB\) and that
\(\hat\vB(t)\) is a unit vector. So
\begin{equation*}
\hat\vB(t) = \frac{\vv(t)\times\va(t)}{|\vv(t)\times\va(t)|}
\end{equation*}
Having found
\(\vB(t)\) we can differentiate it and use
\(\diff{\hat\vB}{s}(s) = -\tau(s)\hat\vN(s)\) and the chain rule to give
\begin{equation*}
\diff{\vB}{t} = \diff{\vB}{s}\diff{s}{t} = -\tau\diff{s}{t} \hat\vN
\end{equation*}
from which we can read off
\(\tau\text{,}\) provided we know
\(\diff{s}{t}\) and
\(\hat\vN\text{.}\)
There is another, often more efficient, method to find the torsion \(\tau\) that uses
\begin{align*}
\diff{\va}{t} &= \diff{}{t}\Big(\difftwo{s}{t}\,\hat\vT
+\ka\Big(\diff{s}{t}\Big)^2\hat\vN\Big)\\
&= \ddiff{3}{s}{t}\,\hat\vT
+\difftwo{s}{t}\,\diff{s}{t}\,\ka\hat\vN
+\diff{}{t}\Big(\ka\Big(\diff{s}{t}\Big)^2\Big)\hat\vN
+\ka\Big(\diff{s}{t}\Big)^3
\big(\tau\hat\vB-\ka\hat\vT\big)
\end{align*}
While this looks a little complicated, notice that, with just one exception, namely \(\ka\big(\diff{s}{t}\big)^3\tau(s)\hat\vB(s)\text{,}\) every term on the right hand side is either in the direction \(\hat\vT\) or in the direction \(\hat\vN\) and so is perpendicular to \(\hat\vB\text{.}\) So, dotting with \(\vv\times\va = \ka \big(\diff{s}{t}\big)^3\hat\vB\) gives
\begin{gather*}
\big(\vv\times\va\big)\cdot \diff{\va}{t}
= \ka^2 \Big(\diff{s}{t}\Big)^6\,\tau
= |\vv\times\va|^2\,\tau
\end{gather*}
and hence
\begin{gather*}
\tau = \frac{\big(\vv\times\va\big)\cdot \diff{\va}{t} }{|\vv\times\va|^2}
\end{gather*}
If the curvature
4 \(\ka(s) \gt 0\) and the torsion
\(\tau(s)\) are known, then the system of equations
5
Equation 1.4.2. Frenet–Serret Formulae.
\begin{align*}
\diff{\hat\vT}{s}(s)&=\phantom{-}\ka(s)\ \hat\vN(s)\cr
\diff{\hat\vN}{s}(s)&=\phantom{-}\tau(s)\ \hat\vB(s)-\ka(s)\ \hat\vT(s)\cr
\diff{\hat\vB}{s}(s)&=-\tau(s)\ \hat\vN(s)\cr
\end{align*}
is a first order linear system of ordinary differential equations
\begin{align*}
\diff{}{s}
\left[ \begin{matrix}\hat\vT(s) \\ \hat\vN(s)\\ \hat\vB(s)\end{matrix} \right]
=\left[\begin{matrix} 0 & \ka(s) & 0 \\
-\ka(s) & 0 &\tau(s) \\
0 &-\tau(s) & 0 \end{matrix}\right]
\left[\begin{matrix}\hat\vT(s) \\ \hat\vN(s)\\ \hat\vB(s)\end{matrix}\right]
\end{align*}
for the \(9\) component vector valued function \((\hat\vT(s)\,,\,\hat\vN(s)\,,\,\hat\vB(s))\text{.}\)
Any first order linear initial value problem
\begin{equation*}
\diff{}{s}\vx(s) = M(s) \vx(s)\qquad
\vx(0)=\vx_0
\end{equation*}
where \(\vx\) is an \(n\)-component vector and \(M(s)\) is an \(n\times n\) matrix with continuous entries, has exactly one solution. If \(n=1\text{,}\) so that \(\vx(s)\) and \(M(s)\) are just functions, this is easy to see. Just let \(\cM(s)\) be the antiderivative of \(M(s)\) that obeys \(\cM(0)=0\text{.}\) Then
\begin{align*}
\diff{}{s}\vx(s) = M(s) \vx(s)
&\iff e^{-\cM(s)}\diff{}{s}\vx(s) - M(s) e^{-\cM(s)} \vx(s)=0\\
&\iff \diff{}{s}\Big(e^{-\cM(s)}\vx(s)\Big) = 0
\end{align*}
by the product rule. So \(e^{-\cM(s)}\vx(s)\) is a constant independent of \(s\text{.}\) In particular \(e^{-\cM(s)}\vx(s)=e^{-\cM(0)}\vx(0)= \vx_0\) so that \(\vx(s) = \vx_0 e^{\cM(s)}\text{.}\) This argument can be generalized to any natural number \(n\text{.}\) But that is beyond the scope of this book.
Since the Frenet-Serret formulae constitute a first order system of ordinary differential equations for the vector \((\hat\vT(s)\,,\,\hat\vN(s)\,,\,\hat\vB(s))\) and since any first order linear initial value problem has a exactly one solution,
the vector valued function \((\hat\vT(s)\,,\,\hat\vN(s)\,,\,\hat\vB(s))\) is determined by the functions \(\ka(s)\) and \(\tau(s)\) (assuming that they are continuous) together with the initial condition \((\hat\vT(0)\,,\,\hat\vN(0)\,,\,\hat\vB(0))\text{.}\)
Furthermore, once you know \(\hat\vT(s)\text{,}\) then \(\vr(s)\) is determined by \(\vr(0)\) and \(\diff{\vr}{s}(s)=\hat\vT(s)\text{.}\)
So any smooth curve \(\vr(s)\) is completely determined by \(\vr(0)\text{,}\) \((\hat\vT(0)\,,\,\hat\vN(0)\,,\,\hat\vB(0))\text{,}\) \(\ka(s)\) and \(\tau(s)\text{.}\)
That is, up to translations (you can move between any two possible choices of \(\vr(0)\) by a translation) and rotations (you can move between any two possible choices of \((\hat\vT(0)\,,\,\hat\vN(0)\,,\,\hat\vB(0))\) by a rotation) a curve is completely determined by the curvature \(\ka(s) \gt 0\) and the torsion \(\tau(s)\text{.}\) This result is called “The fundamental theorem of space curves”.
Theorem 1.4.3. The Fundamental Theorem of Space Curves.
Let \(\ka(s) \gt 0\) and \(\tau(s)\) be continuous. Then up to translations and rotations, there is a unique curve with curvature \(\ka(s)\) and torsion \(\tau(s)\text{.}\)
Example 1.4.4. Right circular helix.
The right circular helix is the curve
\begin{gather*}
\vr(t)= a\cos t\,\hi +a\sin t\,\hj + bt\,\hk
\end{gather*}
with \(a,b \gt 0\) as in the figure on the left below.
Here is why it is called a
right helix rather than a left helix. If the helix is the thread of a bolt that you are screwing into a nut, and you turn the bolt in the direction of the (curled) fingers of your right hand (as in the figure
6 on the right above), then it moves in the direction of your thumb (as in the long straight arrow of the figure on the right above).
To determine the curvature and torsion of this curve we compute
\begin{align*}
\vv(t)&= -a\sin t\,\hi +a\cos t\,\hj + b\,\hk\\
\va(t)&= -a\cos t\,\hi -a\sin t\,\hj\\
\diff{\va}{t}(t)&= a\sin t\,\hi -a\cos t\,\hj
\end{align*}
From \(\vv(t)\) we read off
\begin{align*}
\diff{s}{t}&=\sqrt{a^2+b^2}\\
\hat\vT(t)&= -\frac{a}{\sqrt{a^2+b^2}}\sin t\,\hi
+\frac{a}{\sqrt{a^2+b^2}} \cos t\,\hj
+\frac{b}{\sqrt{a^2+b^2}}\,\hk
\end{align*}
From \(\va=\difftwo{s}{t}\,\hat\vT+\ka\big(\diff{s}{t}\big)^2\hat\vN
=\ka(a^2+b^2)\hat\vN\text{,}\) we read off that
\begin{gather*}
\ka(t)=\frac{a}{a^2+b^2}\qquad
\hat\vN(t) = -\cos t\,\hi-\sin t\,\hj
\end{gather*}
From
\begin{align*}
\vv(t)\times\va(t) &= \det\left[
\begin{matrix} \hi & \hj & \hk\\
-a\sin t & a\cos t & b\\
-a\cos t &-a\sin t & 0\end{matrix} \right]
= ab\sin t\,\hi -ab\cos t\,\hj +a^2\,\hk\\
|\vv(t)\times\va(t)|^2 &=a^2b^2+a^4 = a^2(a^2+b^2)
\end{align*}
we read off
\begin{align*}
\hat\vB(t) &= \frac{\vv(t)\times\va(t)}{|\vv(t)\times\va(t)|}
= \frac{b}{\sqrt{a^2+b^2}}\sin t\,\hi
-\frac{b}{\sqrt{a^2+b^2}} \cos t\,\hj
+\frac{a}{\sqrt{a^2+b^2}}\,\hk
\end{align*}
and
\begin{align*}
\tau(t) & = \frac{\big(\vv\times\va\big)\cdot \diff{\va}{t} }{|\vv\times\va|^2}
=\frac{a^2b}{a^2(a^2+b^2)}
=\frac{b}{a^2+b^2}
\end{align*}
Note that, for the right handed helix, \(\tau \gt 0\text{.}\) Finally the centre of curvature is
\begin{align*}
\vr(t) +\frac{1}{\ka(t)}\hat\vN(t)
&=\Big(a-\frac{a^2+b^2}{a}\Big)\cos t\,\hi
+\Big(a-\frac{a^2+b^2}{a}\Big)\sin t\,\hj
+bt\,\hk\\
&=-\frac{b^2}{a}\cos t\,\hi
-\frac{b^2}{a}\sin t\,\hj
+bt\,\hk
\end{align*}
which is another helix. In the figure below, the red curve is the original helix and the blue curve is the helix traced by the centre of curvature.
Exercises Exercises
Exercise Group.
Exercises — Stage 1
1.
In the sketch below of a three-dimensional curve and its osculating circle at a point, label \(\hT\) and \(\hN\text{.}\) Will \(\hB\) be pointing out of the paper towards the reader, or into the paper away from the reader?
2.
In the formula
\begin{equation*}
\diff{s}{t}(t)=|\vv(t)|=|\vr'(t)|
\end{equation*}
does \(s\) stand for speed, or for arclength?
3.
Which curve (or curves) below have positive torsion, which have negative torsion, and which have zero torsion? The arrows indicate the direction of increasing \(t\text{.}\)
4.
Consider a curve that is parametrized by arc length \(s\text{.}\)
Show that if the curve has curvature \(\ka(s)=0\) for all \(s\text{,}\) then the curve is a straight line.
Show that if the curve has curvature \(\ka(s) \gt 0\) and torsion \(\tau(s)=0\) for all \(s\text{,}\) then the curve lies in a plane.
Show that if the curve has curvature \(\ka(s)=\ka_0\text{,}\) a strictly positive constant, and torsion \(\tau(s)=0\) for all \(s\text{,}\) then the curve is a circle.
5. (✳).
The surface \(z=x^2+y^2\) is sliced by the plane \(x=y\text{.}\) The resulting curve is oriented from \((0,0,0)\) to \((1,1,2)\text{.}\)
Sketch the curve from \((0,0,0)\) to \((1,1,2)\text{.}\)
Sketch \(\hat\vT\text{,}\) \(\hat\vN\) and \(\hat\vB\) at \(\big(\half,\half,\half\big)\text{.}\)
Find the torsion at \(\big(\half,\half,\half\big)\text{.}\)
Exercise Group.
Exercises — Stage 2
6. (✳).
Let \(C\) be the space curve
\begin{gather*}
\vr(t) = \big(e^t - e^{-t}\big)\,\hi + \big(e^t + e^{-t}\big)\,\hj +2t\,\hk
\end{gather*}
Find \(\vr'\text{,}\) \(\vr''\) and the curvature of \(C\text{.}\)
Find the length of the curve between \(\vr(0)\) and \(\vr(1)\text{.}\)
7.
Find the torsion of \(\vr(t)=(t,t^2,t^3)\) at the point \((2,4,8)\text{.}\)
8.
Find the unit tangent, unit normal and binormal vectors and the curvature and torsion of the curve
\begin{equation*}
\vr(t)=t\,\hi + \frac{t^2}{2}\,\hj + \frac{t^3}{3}\,\hk
\end{equation*}
9.
For some constant \(c\text{,}\) define \(\vr(t)=(t^3,t,e^{ct})\text{.}\) For which value(s) of \(c\) is \(\tau(5)=0\text{?}\) For each of those values of \(c\text{,}\) find an equation for the plane containing the osculating circle to the curve at \(t=5\text{.}\)
10. (✳).
Consider the parametrized space curve
\begin{equation*}
\vr(t) = \big(t^2 , t, t^3\big)
\end{equation*}
Find an equation for the plane passing through
\((1,1,1)\) with normal vector tangent to
\(\vr\) at that point.
Find the curvature of the curve from (a) as a function of the parameter \(t\text{.}\)
11. (✳).
Let \(C\) be the osculating circle to the helix \(\vr(t) =\big(\cos t\,,\,\sin t\,,\,t\big)\) at the point where \(t=\pi/6\text{.}\) Find:
the radius of curvature of \(C\)
the center of \(C\)
the unit normal to the plane of \(C\)
12. (✳).
Consider the parametrized space curve
\begin{equation*}
\vr(t) = (\cos(t), \sin(t), t^2)
\end{equation*}
Find a parametric form for the tangent line at the point corresponding to
\(t = \pi\text{.}\)
Find the tangential component \(a_T(t)\) of acceleration, as a function of \(t\text{,}\) for the parametrized space curve \(\vr(t)\text{.}\)
13. (✳).
Suppose, in terms of the time parameter \(t\) , a particle moves along the path \(\vr(t) = (\sin t - t \cos t )\,\hi + (\cos t + t \sin t )\,\hj + t^2\,\hk\text{,}\) \(1 \le t \lt \infty\text{.}\)
Find the speed of the particle at time \(t\text{.}\)
Find the tangential component of acceleration at time \(t\text{.}\)
Find the normal component of acceleration at time \(t\text{.}\)
Find the curvature of the path at time \(t\text{.}\)
14. (✳).
Assume the paraboloid \(z = x^2 + y^2\) and the plane \(2x + z = 8\) intersect in a curve \(C\text{.}\) \(C\) is traversed counter-clockwise if viewed from the positive \(z\)-axis.
Parametrize the curve \(C\text{.}\)
Find the unit tangent vector \(\hat\vT\text{,}\) the principal normal vector \(\hat\vN\text{,}\) the binormal vector \(\hat\vB\) and the curvature \(\kappa\) all at the point \((2, 0, 4)\text{.}\)
15. (✳).
Consider the curve \(C\) given by
\begin{equation*}
\vr(t) = \frac{1}{3} t^3\,\hi + \frac{1}{\sqrt{2}} t^2\,\hj + t\,\hk,\qquad
-\infty \lt t \lt \infty.
\end{equation*}
Find the unit tangent \(\hat\vT(t)\) as a function of \(t\text{.}\)
Find the curvature \(\kappa(t)\) as a function of \(t\text{.}\)
Determine the principal normal vector \(\hat\vN\) at the point \(\big(\frac{8}{3} , 2\sqrt{2}, 2\big)\text{.}\)
16. (✳).
Suppose the curve \(C\) is the intersection of the cylinder \(x^2 +y^2 = 1\) with the plane \(x+y+z = 1\text{.}\)
Find a parameterization of \(C\text{.}\)
Determine the curvature of \(C\text{.}\)
Find the points at which the curvature is maximum and determine the value of the curvature at these points.
17. (✳).
Let
\begin{gather*}
\vr(t) = t^2\,\hi + 2t\,\hj + \ln t\,\hk
\end{gather*}
Compute the unit tangent and unit normal vectors \(\hat\vT(t)\) and \(\hat\vN(t)\text{.}\) Compute the curvature \(\kappa(t)\text{.}\) Simplify whenever possible!
18. (✳).
Find the length of the curve \(\vr(t)=\big(1,\frac{t^2}{2},\frac{t^3}{3}\big)\) for \(0\le t\le 1\text{.}\)
Find the principal unit normal vector \(\hN\) to \(\vr(t) = \cos(t)\,\hi + \sin(t)\,\hj
+ t\,\hk\) at \(t =\pi/4\text{.}\)
Find the curvature of \(\vr(t) = \cos(t)\,\hi + \sin(t)\,\hj + t\,\hk\) at \(t = \pi/4\text{.}\)
19. (✳).
A particle moves along a curve with position vector given by
\begin{equation*}
\vr(t) = \big(t + 2\,,\, 1 - t\,,\, t^2 /2\big)
\end{equation*}
for \(-\infty \lt t \lt \infty\text{.}\)
Find the velocity as a function of \(t\text{.}\)
Find the speed as a function of \(t\text{.}\)
Find the acceleration as a function of \(t\text{.}\)
Find the curvature as a function of \(t\text{.}\)
Recall that the decomposition of the acceleration into tangential and normal components is given by the formula
\begin{equation*}
\vr''(t) = \difftwo{s}{t} \hat\vT(t)
+ \kappa(t)\Big(\diff{s}{t}\Big)^2\hat\vN(t)
\end{equation*}
Use this formula and your answers to the previous parts of this question to find
\(\hat\vN(t)\text{,}\) the principal unit normal vector, as a function of
\(t\text{.}\)
Find an equation for the osculating plane (the plane which best fits the curve) at the point corresponding to \(t = 0\text{.}\)
Find the centre of the osculating circle at the point corresponding to \(t = 0\text{.}\)
20. (✳).
Consider the curve \(C\) given by
\begin{equation*}
\vr(t) =\frac{t^3}{3}\,\hi + \frac{t^2}{\sqrt{2}}\,\hj + t\,\hk \qquad
-\infty \lt t \lt \infty
\end{equation*}
Find the unit tangent \(\hat\vT(t)\) as a function of \(t\text{.}\)
Find the curvature \(\kappa(t)\) as a function of \(t\text{.}\)
Evaluate \(\kappa(t)\) at \(t = 0\text{.}\)
Determine the principal normal vector \(\hat\vN(t)\) at \(t = 0\text{.}\)
Compute the binormal vector \(\hat\vB(t)\) at \(t = 0\text{.}\)
21. (✳).
A curve in \(\bbbr^3\) is given by \(\vr(t) = (t^2\,,\, t\,,\, t^3)\text{.}\)
Find the parametric equations of the tangent line to the curve at the point \((1, -1, -1)\text{.}\)
Find an equation for the osculating plane of the curve at the point \((1, 1, 1)\text{.}\)
22. (✳).
A curve in \(\bbbr^3\) is given by
\begin{equation*}
\vr(t) = (\sin t - t \cos t)\,\hi + (\cos t + t \sin t)\,\hj + t^2\,\hk,
\qquad
0 \le t \lt \infty
\end{equation*}
Find the length of the curve \(\vr(t)\) from \(\vr(0) = (0, 1, 0)\) to \(\vr(\pi) = (\pi, -1, \pi^2)\text{.}\)
Find the curvature of the curve at time \(t \gt 0\text{.}\)
23. (✳).
At time \(t=0\text{,}\) NASA launches a rocket which follows a trajectory so that its position at any time \(t\) is
\begin{equation*}
x=\frac{4\sqrt{2}}{3}t^{3/2},\ y=\frac{4\sqrt{2}}{3}t^{3/2},\ z=t(2-t)
\end{equation*}
Assuming that the flight ends when \(z=0\text{,}\) find out how far the rocket travels.
Find the unit tangent and unit normal to the trajectory at its highest point.
Also, compute the curvature of the trajectory at its highest point.
24. (✳).
Consider a particle travelling in space along the path parametrized by
\begin{equation*}
x=\cos^3t,\ y=\sin ^3t,\ z=2\sin^2 t
\end{equation*}
Calculate the arc length of this path for \(0\le t\le \pi/2\text{.}\)
Find the vectors \(\hat\vT\text{,}\) \(\hat\vN\text{,}\) \(\hat\vB\) for the particle at \(t=\pi/6\text{.}\)
25.
Suppose that the curve \(C\) is the intersection of the cylinder \(x^2 +y^2 = 1\) with the surface \(z =x^2 - y^2\text{.}\)
Find a parameterization of \(C\text{.}\)
Determine the curvature of \(C\) at the point \(\big(1/\sqrt{2}\,,1/\sqrt{2}\,,\,0\big)\text{.}\)
Find the osculating plane to \(C\) at the point \(\big(1/\sqrt{2}\,,1/\sqrt{2}\,,\,0\big)\text{.}\) In general, the osculating plane to a curve \(\vr(t)\) at the point \(\vr(t_0)\) is the plane which fits the curve best at \(\vr(t_0)\text{.}\) It passes through \(\vr(t_0)\) and has normal vector \(\hat\vB(t_0)\text{.}\)
Find the radius and the centre of the osculating circle to \(C\) at the point \(\big(1/\sqrt{2}\,,1/\sqrt{2}\,,\,0\big)\text{.}\)
Exercise Group.
Exercises — Stage 3
26. (✳).
Under the influence of a force field \(\vF\text{,}\) a particle of mass 2 kg is moving with constant speed 3 m/s along the path given as the intersection of the plane \(z = x\) and the parabolic cylinder \(z = y^2\text{,}\) in the direction of increasing \(y\text{.}\) Find \(\vF\) at the point \((1, 1, 1)\text{.}\) (Length is measured in m along the three coordinate axes.)
27. (✳).
Consider the curve \(C\) in 3 dimensions given by
\begin{equation*}
\vr(t) = 2t\hi + t^2\hj + \sqrt{3} t^2\hk
\end{equation*}
for \(t \in\bbbr\text{.}\)
Compute the unit tangent vector \(\vT(t)\text{.}\)
Compute the unit normal vector \(\vN(t)\text{.}\)
Show that the binormal vector
\(\vB\) to this curve does not depend on
\(t\) and is one of the following vectors:
\begin{equation*}
\text{(1)}\ \left[\begin{matrix}
1/2 \\
-\sqrt{3}/2 \\
0
\end{matrix}\right]\qquad
\text{(2)}\ \left[\begin{matrix}
0 \\
\sqrt{3}/2 \\
1/2
\end{matrix}\right]\qquad
\text{(3)}\ \left[\begin{matrix}
0 \\
-\sqrt{3}/2 \\
1/2
\end{matrix}\right]\qquad
\text{(4)}\ \left[\begin{matrix}
0\\
-1/2 \\
\sqrt{3}/2
\end{matrix}\right]\qquad
\end{equation*}
This implies that
\(C\) is a plane curve.
According to your choice of vector (1), (2), (3) or (4), give the equation of the plane containing \(C\text{.}\)
Compute the curvature \(\kappa(t)\) of the curve.
Are there point(s) where the curvature is maximal? If yes, give the coordinates of the point(s). If no, justify your answer.
Are there point(s) where the curvature is minimal? If yes, give the coordinates of the point(s). If no, justify your answer.
Let
\begin{equation*}
\vu := 2\,\hi,\quad
\vv := \hj + \sqrt{3}\,\hk\quad
\vw := -\sqrt{3}\,\hj + \hk
\end{equation*}
Express \(\hi\text{,}\) \(\hj\text{,}\) \(\hk\) in terms of \(\vu\text{,}\) \(\vv\text{,}\) \(\vw\text{.}\)
Using (i), write
\(\vr(t)\) in the form
\begin{equation*}
a(t)\vu + b(t)\vv + c(t)\vw
\end{equation*}
where
\(a(t)\text{,}\) \(b(t)\) and
\(c(t)\) are functions you have to determine. You should find that one of these functions is zero.
Draw the curve given by \(\big(a(t), b(t)\big)\) in the \(xy\)-plane.
Is the drawing consistent with parts (f) and (g)? Explain.
28. (✳).
Recall that if \(\hat\vT\) is the unit tangent vector to an oriented curve with arclength parameter \(s\text{,}\) then the curvature \(\ka\) and the principle normal vector \(\hat\vN\) are defined by the equation
\begin{equation*}
\diff{\hat\vT}{s} = \ka\,\hat\vN
\end{equation*}
Moreover, the torsion \(\tau\) and the binormal vector \(\hat\vB\) are defined by the equations
\begin{equation*}
\hat\vB = \hat\vT\times\hat\vN,\qquad
\diff{\hat\vB}{s} = -\tau\,\hat\vN
\end{equation*}
Show that
\begin{equation*}
\diff{\hat\vN}{s} = -\ka\,\hat\vT + \tau\,\hat\vB
\end{equation*}
29. (✳).
A skier descends the hill \(z =\sqrt{4-x^2-y^2}\) along a trail with parameterization
\begin{equation*}
x=\sin(2\theta),\qquad y=1-\cos(2\theta),\qquad z=2\cos\theta,\qquad
0\le\theta\le\frac{\pi}{2}
\end{equation*}
Let \(P\) denote the point on the trail where \(x = 1\text{.}\)
Find the vectors \(\hat\vT, \hat\vN, \hat\vB\) and the curvature \(\ka\) of the ski trail at the point \(P\text{.}\)
The skier's acceleration at \(P\) is \(\va = (-2, 3, -2\sqrt{2})\text{.}\) Find, at \(P\text{,}\)
the rate of change of the skier's speed and
the skier's velocity (a vector).
30. (✳).
A particle moves so that its position vector is given by \(\vr(t) = \big(\cos t\,,\, \sin t\,,\, c \sin t\big)\text{,}\) where \(t \gt 0\) and \(c\) is a constant.
Find the velocity \(\vv(t)\) and the acceleration \(\va(t)\) of the particle.
Find the speed \(v(t)=|\vv(t)|\) of the particle.
Find the tangential component of the acceleration of the particle.
Show that the trajectory of this particle lies in a plane.
31. (✳).
A race track between two hills is described by the parametric curve
\begin{equation*}
\vr(\theta) = \Big(4 \cos\theta\,,\, 2\sin\theta\,,\,
\tfrac{1}{4}\cos(2\theta)\Big),\qquad
0 \le \theta \le 2\pi
\end{equation*}
Compute the curvature of the track at the point \(\big(-4, 0, \frac{1}{4}\big)\text{.}\)
Compute the radius of the circle that best approximates the bend at the point\\ \(\big(-4, 0, \frac{1}{4}\big)\) (that is, the radius of the osculating circle at that point).
A car drives down the track so that its position at time \(t\) is given by \(\vr(t^2)\text{.}\) (Note the relationship between \(t\) and \(\theta\) is \(\theta = t^2\)). Compute the following quantities.
The speed at the point \(\big(-4, 0, \frac{1}{4}\big)\text{.}\)
The acceleration at the point \(\big(-4, 0, \frac{1}{4}\big)\text{.}\)
The magnitude of the normal component of the acceleration at the point
\begin{equation*}
\big(-4, 0, \tfrac{1}{4}\big)\text{.}
\end{equation*}
The arguments in the proof of Theorem
1.3.3 that we used to verify these formulae work in any plane, not just the
\(xy\)-plane. Just choose
\(\hi\) and
\(\hj\) to be any two mutually perpendicular unit vectors in the plane.
However this can be a significant difference.
We shall stick to “right handed triples” to make it easier to get various signs right.
As in two dimensions, if \(\ka(s)=0\text{,}\) then \(\hat\vN(s)\) is not defined. This makes even more sense in three dimensions than in two dimensions: if the curve is a straight line, there are infinitely many unit vectors perpendicular to it and there is no way to distinguish between them.
The equations are named after the two French mathematicians who independently discovered them: Jean Frédéric Frenet (1816–1900, the son of a wig maker), in his thesis of 1847 (actually he only gave two of the three equations), and Joseph Alfred Serret (1819–1885) in 1851.
commons.wikimedia.org/wiki/File:Right_hand_rule_simple.png
