1
a
We nd critical points by setting
f =0
f = 4u3 4v, 4v 3 4u = 0
and so v = u3 and u = v 3 . Substituting the rst into the second, we get
u9 u = 0
which we can factor repeatedly.
u(u8 1) = u(u4 1)(u4 + 1) = u(u2 1)(u2 + 1)(u4 + 1) = u(u 1)(u + 1)(u2 + 1)
Exam 3 - Solutions
Spring 2012
1.
2. (a) Plot the region of integration in the xy-plane
2.5
2.0
1.5
+
y
x
y
=
2
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
2.5
x
(b)
u=x
y
v =x+y
u+v
2
v u
2y ! y =
2
u + v = 2x ! x =
u
v=
(c) Plot the region of integration in the uv-
Exam 2 - Solutions
Spring 2012
1. f (u, v) =
u3
3
+
v3
3
v2
2
u+2
(a) Determine the location of all critical points of f (u, v).
f = u2 1, v 2 v = 0, 0
u2 = 1 u = 1
v 2 = v v = 0, 1
Critical points are at: (1, 0); (1, 1); (1, 0); and (1, 1).
(b) Classify
Solution: APPM 2350
Exam 2
Summer 2014
1. (30 points) Suppose we want to calculate
y xy
e dA
x
R
Where R is the region in the rst quadrant bounded by the curves y =
1
4
x
, y = 2x, y = , and y = .
2
x
x
(a) Set up the integral in cartesian coordinates wit
Solution: APPM 2350
Final
Summer 2014
1. (50 points) Suppose you nd yourself in a parallel universe where the local force due to gravity takes the
x 2
y 2
z 2
improbable form F = y, x + ez , yez . Consider the surface S given by
+
+
= 1 for
2
3
4
z 0.
F d
APPM 2350
FINAL EXAM
SPRING 2014
INSTRUCTIONS: Electronic devices, books, and crib sheets are not permitted. Write your name and your instructors
name on the front of your bluebook. Work all problems. Show your work clearly. Note that a correct answer wit
APPM 2350
Spring 2012
Exam 1 Solutions
February 15, 2012
Problem 1
a
In order to determine the standard equation of the plane of intersection of two surfaces, we must nd a
relationship between x, y, and z when both surface equations are true. The easiest