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Mathematics 309

Part I - The elementary geometry of wave motion

3. Affine functions

Two dimensions

An affine function in 2D is a scalar-valued function of the form Ax+By+C. The level curves of the function are the curves where the function takes a fixed value. For example, if the function is x+y then some of the level lines look like this:

Of course the level lines of the function Ax+By are the same as those of Ax+BY+C, but with different constants. So we may as well assume C=0, and the level lines are of the form Ax+By=k. If the coefficient B is not 0 then the equation Ax+By=k can be solved to get the equation y = (k-Ax)/B, the more familiar equation from elementary mathematics. From this equation we can figure out the slope and the y-intercept, and then plot the curve, which is actually a straight line. The level curves that escape thsi are those where B=0, or x=k/A, which are vertical. But it is unnecessary to distinguish cases. Here we want to explain more directly the geometric properties of the level curves.

  • The line Ax+By=k is perpendicular to the vector [A,B].
  • The distance of the line Ax+By=k from the origin is |k|/| [A,B] |.
  • The line Ax+By=k lies on the same side of the origin as the vector [A,B] precisely when k > 0.
Here | [A,B] | is the length of the vector [A,B], the square root of A2+B2. The first condition tells the direction of the line, the second tells how far from the origin it is. These two together narrow the possibilities down to two. The last condition fixes it uniquely. All these facts are shown here for the case [1,1].

The line x+y=1 is perpendicular to [1,1], on the same side of ther origin as [1,1], and at distance 1/ from the origin.

< Why are these things true?The first on eis easy. If P = (Px, Py) Q = (Qx, Qy) are points on the line AX+By=k, then

A Px+B Py=k
A Qx+B Qy=k
A(Px-Qx)+B(Py-Qy)=0

which means that the dot-product of [A,B] and P-Q equals 0, or that they are perpendicular. Since P-Q lies in the direction of the line, that's the first assertion.

There will be a unique scalar multiple t[A,B] which lies on the line Ax+By=k. This will be the case when

A(tA)+B(tB) = t(A2+B2) = k
t = k/(A2+B2).

The length of the vector t[A,B] is then k/| [A,B] |, and this proves the second assertion. The last one follows from the sign of t, i.e. of k.

Three dimensions

An affine function is of the form Ax+By+Cz+D. The points (x,y,z) where this has a constant value is a plane Ax+By+Cz+D=k. Here:
  • The plane Ax+By+Cz=k is perpendicular to the vector [A,B,C].
  • The distance of the plane Ax+By+Cz=k from the origin is |k|/| [A,B,C] |.
  • The line Ax+By+Cz=k lies on the same side of the origin as the vector [A,B,C] precisely when k > 0.

Exercises