=========================preview======================
(math204)[2008](s)final~PPSpider^_10480.pdf
Back to MATH204 Login to download
======================================================
Math 204 Final Exam
May 28, 2008

Your Name

Student Number

Number Score
1
2
3
4
5
Total

(1)
(20 points) Suppose u3 + xu2 + yu + z = 0. Find the absolute maximum and minimum of u(x, y, z) on x2 + y2 + z2 1.

(2)
(20 points) Suppose f(x, y) is second order di.erentiable at (0, 0) and satisfying f(0, 0) = 0, fy(0, 0) =.0. Prove that the function y(x) implicitly given by f(x, y)=0 near (0, 0) is also second order di.erentiable.

(3)
(20 points) Which subsets have volume? Explain.

1.
A = {(x, y, z): |x| < 1, |y| 2, |z| < 3,x2 . y z x2 + y, y z sin x4}.

2.
B = {(x, y, z): x Q,z . Q, |x| < 1, |z| < 1,y2 = x4 + xz3 + x3z + z4}.




4 33
3. C = {(x, y, z): x Q,z . Q, |x| < 1, |z| < 1,y2 x+ xz+ xz + z4}.
(4) (20 points) Compute the integrals.
1. (|x|3 + |y|3)dxdy.
x4+y41

2. (x + z 2)dy dz, where S is the parabola z = x2 + y2 1 with orientation given
S

by downward pointing normal vector.
(5) (20 points) Find continuously di.erentiable f(x), such that f(yz)xdx+f(zx)ydy +
C
f(xy)zdz depends only on the beginning and end of C. Then compute the integral from (0, 0, 0) to (1, 1, 1).
Answer to Math 204 Final, Spring 2008
not absolutely guaranteed to be correct
(1) By d(u3 + xu2 + yu + z) = (3u2 +2xu + y)du + u2dx + udy + dz, we have uz .
=0 in x2 + y2 + z2 < 1. Therefore the absolute extremes must lie on the boundary x2 + y2 + z2 =
1. The problem becomes the extremes of f(x, y, z, u)= u subject to the constraints g(x, y, z, u)= u3 + xu2 + yu + z = 0 and h(x, y, z, u)= x2 + y2 + z2 = 1. At local constrained extremes, either .h and .g are parallel (the hypersurface is not regular), which means
.g =(u 2 , u, 1, 3u 2 +2xu + y)= .h =2(x, y, z, 0)
for some , or we can use Lagrange multiplier method to get
.f = (0, 0, 0, 1) = .g + .h = (u 2 , u, 1, 3u 2 +2xu + y)+2(x, y, z, 0),
for some , . In both cases, the constraints g = 0, h = 1 are also satis.ed. In the .rst case, we have
2 32 22
x = uz, y = uz, 3u 2 +2xu + y =0,u + xu + yu + z =0,x + y + z 2 =1. It turns out that such system has no solution. In the second case, we have
32 22
x = uy, y = uz, u + xu + yu + z =0,x + y + z 2 =1.
This reduces to (u2, u, 1)u3
u 6 . u 4 . u 2 . 1=0, (x, y, z)= . .
u4 + u2 +1 Let a be the biggest root of v3 . v2 . v . 1 = 0 (in fact, the cubic equation has only one


root). Then a and . a are the absolute maximum and minimum.
(2) The di.erentiability condition means that there is a quadratic function q(x, y)= a1x + a2y + c11x 2 + c22y 2 + c12xy,
such that for any .> 0, there is > 0, such that |x| < , |y| < =.|f(x, y) . a1x . a2y . c11x 2 . c22y 2 . c12xy| .(|x| + |y|)2 .
By implicit function theorem, g(x) is di.erentiable with g(0) = 0 and g.(0) = . a1 . Write a2 a1 R(x)
g(x)= . x + R(x), where limx0 = 0, and substitute into the estimation, we get
a2 x (note that f(x, g(x)) = 0)
.2
a1