Final-Fall 2005-Solutions - AMS 510.01 Fall 2005 Final Exam...

AMS 510.01 Fall 2005, Final Exam
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Student ID:
Score: /200
The exam is in three parts:
• The first section consists of problems involving the fundamental concepts of single and
multiple variable calculus. Do all questions from this portion. (50 points)
• The second section consists of comprehensive problems in Linear Algebra. Choose
three out of the four problems to do. (75 points)
• The third section consists of comprehensive problems in Advanced Calculus. Choose
three out of the four problems to do. (75 points)
Section 1. Fundamentals of Multivariate and Integral Calculus
1. Indicate whether each of the following statements is true or false. (2 points each)
(a) The integral F ( x ) = f ( x ) dx is unique for any given function f ( x ).
• False. F ( x ) is only unique up to an arbitrary constant.
(b) If two functions are equal everywhere in a closed interval ( f ( x ) = g ( x ) for all x in
[ a, b ]), then their derivatives must be equal everywhere on the open interval ( a, b ).
• True. However, the derivatives may not be equal (and may not even be
defined) at the points x = a and x = b .
(c) A necessary and sufficient condition for the continuity of a function f ( x, y ) at the
point ( x
o
, y
o
) is that:
lim
x → x
o
( lim
y → y
o
f ( x, y )) = lim
y → y
o
( lim
x → x
o
f ( x, y ))
• False. This is a necessary condition, but it is not sufficient.
(d)
1
0
1
0
xydxdy = 1.
• False.
1
0
1
0
xydxdy =
1
0
(
1
0
xdx ) ydy =
1
0
(
1
2
) ydy =
1
4
.
(e)
∂
2
f
∂x∂y
=
∂
2
f
∂y∂x
= for any function f ( x, y ) with continuous second order derivatives.
• True. This is a basic property of partial derivatives.
2. Consider the function f ( x ) = x cos x + sin xe cos x .
(a) Find the indefinite integral F ( x ) = f ( x ) dx . (5 points)

• Break the integral into two terms:
x cos x + sin xe
cos x
dx = x cos xdx + sin xe
cos x
dx
The first term can be solved using integration by parts. Let u = x and
v = sin x , then du = dx and dv = cos xdx . Thus:
x cos xdx = udv = uv - vdu = x sin x - sin xdx = x sin x +cos x + c
The second term can be solved by variable substitution. Let u = cos x , then
du = - sin xdx . Thus:
sin xe
cos x
dx = - e
u
du = - e
u
+ c = - e
cos x
+ c
Combining the terms:
F ( x ) = x cos x + sin xe
cos x
dx = x sin x + cos x - e
cos x
+ c
(b) Find the definite integral
π/ 2
0
f ( x ) dx . (5 points)
• The definite integral is simply the difference of the indefinite integral, evalu-
ated at each end point. Thus:
π/ 2
0
f ( x ) dx = F (
π
2
) - F (0)
= ( π
2
) sin π
2
) + cos π
2
- e cos
π
2
) - (0 sin 0 + cos 0 - e cos 0 )
= ( π
2
) + 0 - e 0 ) - (0 + 1 - e 1 )
=
π
2
- 1 - 1 + e
=
π
2
- 2 + e
3. Consider the surface in R
4
defined by:
f ( x, y, z ) = 3 x
2
+ 4 xy + y
2
+ cos(2 z )
(a) Evaluate the partial derivatives f
x
, f
y
and f
z
. (5 points)
• Partial derivatives are found simply by taking the derivative with respect to
one variable, holding all others fixed.
f
x
=
∂
∂x
(3 x
2
+ 4 xy + y
2
+ cos(2 z )) = 6 x + 4 y
f
y
=
∂
∂y
(3 x
2
+ 4 xy + y
2
+ cos(2 z )) = 4 x + 2 y
f
z
=
∂
∂z
(3 x
2
+ 4 xy + y
2
+ cos(2 z )) = - 2 sin(2 z )

(b) Identify all critical points on the surface. (5 points)
• Critical points are locations where all partial derivatives are 0, so we solve
the system of equations:
6 x + 4 y = 0
4 x + 2 y = 0
- 2 sin(2 z ) = 0
The first two equations give a unique solution of x = 0 and y = 0, while
the third gives 2 z = nπ , or z =
nπ
2
where n is any integer. Thus all points