Calculus IV Quadratic Approximation Exercises Spring 1999

1. Find a quadratic approximation for f(x,y) = cos x cos y near the point (0,0).

The first thing to do is work out all those derivatives and their values at (0,0):
 
f(x,y) = cos x cos y f(0,0) = cos 0 cos 0 = 1
fx(x,y) = -sin x cos y fx(0,0) = -sin 0 cos 0 = 0
fy(x,y) = -cos x sin y fy(0,0) = -cos 0 sin 0 = 0
fxx(x,y) = -cos x cos y fxx(0,0) = -cos 0 cos 0 = -1
fxy(x,y) = sin x sin y  fxy(0,0) = sin 0 sin 0 = 0
fyy(x,y) = -cos x cos y  fyy(0,0) = -cos 0 cos 0 = -1
Then if we just slot these into the regular formula for a quadratic approximation, we have:

which can be simplified to just .

2. Find a quadratic approximation for f(x,y) = cos x cos y near the point (). Think about how this compares with your answer to #1, and what the graphs of these three functions (f itself and the two approximations) look like.

To make some long computations short: The derivatives are exactly like those above, and evaluating at (pi,pi) gives exactly the same values, so the quadratic approximation look just like that above except with pi in place of some of the 0's. The result looks like .

3. Find a quadratic approximation for g(x,y) =  near (0,0).

As usual we crank out lots of derivatives:
 
g(x,y) = (1-x2-y2)1/2 g(0,0) = (1-02-02)1/2 = 1
gx(x,y) = -x(1-x2-y2)-1/2 gx(0,0) = -0(1-02-02)-1/2 = 0
gy(x,y) = -y(1-x2-y2)-1/2 gy(0,0) = -0(1-02-02)-1/2 = 0
gxx(x,y) = -(1-x2-y2)-1/2 -x2(1-x2-y2)-3/2 gxx(0,0) = -(1-02-02)-1/2 -02(1-02-02)-3/2 = -1
gxy(x,y) = -xy(1-x2-y2)-3/2 gxy(0,0) = -0(1-02-02)-3/2 = 0
gyy(x,y) = -(1-x2-y2)-1/2 -y2(1-x2-y2)-3/2 gyy(0,0) = -(1-02-02)-1/2 -02(1-02-02)-3/2 = -1
Then the quadratic approximation is , which simplifies to , just like in #1.

4. Use your approximation from #3 to estimate the value of g(0.1,0.1). Compare with the actual value of g(0.1,0.1).

p2(0.1,0.1) = 0.99, whereas g(0.1,0.1)  0.9899. That's pretty darn close.

5. Try to find a quadratic approximation for g(x,y) =  near (1,0). What goes wrong?

The derivatives are still those we got in #3, but when we try plugging in x=1 and y=0, those negative exponents produce problems with dividing by zero. The derivatives, first and second order (and beyond) are all undefined at this point.

What's happening is that we're looking at a place where the tangent plane is vertical. The graph of our function is the top half of a sphere of radius 1, and the point (1,0) is at its very edge. This is just a situation where tangent planes, or their overgrown cousins the quadratic approximations, don't work very well.

6. Find a quadratic approximation for h(x,y) = xey near (1,1).
 
h(x,y) = xey h(1,1) = 1e1 = e
hx(x,y) = e hx(1,1) = e1 = e
hy(x,y) = xey hy(1,1) = 1e1 = e
hxx(x,y) = 0 hxx(1,1) = 0
hxy(x,y) = e hxy(1,1) = e1 = e
hyy(x,y) = xey hyy(1,1) = 1e1 = e

Then plugging things in we have . Yes, you could simplify it more by distributing things out and such, but it's often better not to -- look ahead to the next problem.

7. Use your approximation from #6 to estimate h(1.1,0.9). Compare with the actual value of h(1.1,0.9).

Plugging in our values for x and y gives us e + 0.1e - 0.1e - 0.01e + .005e = 0.995e  2.70469.

This is an amazingly close fit to the true value of  2.70556.

8. Find a quadratic approximation to the plane p(x,y) = ax + by + c near the origin. Think about it.

The quadratic approximation is p2(x,y) = ax + by + c. Think about it.