M408D (54690/95/00), Final Exam Solutions: Part 1, 12/13/2010
Multiple choice questions #1.1-1.4 (20 points)
See last two pages of solutions.
Question #1.1 (20 points)
Dene the function
f x; t
(
a) Determine @f @t and @f @x.
Solution :
/
)=
t
e
x2/t ;
1/2
t> :
0
/
@f
@x
=
2
xe
@f
@t
x2/t ;
t3/2
x
2
=
1
t5/2
2
e
t3/2
x2/t
b) Consider the partial dierential equation (PDE)
@f
@t
=
1
4
@2f ;
@x2
known as the heat equation since it describes the ow of heat in a thin tube. Show that
f x; t as dened above a solution to this PDE (i.e., verify that it satises the equation).
(
)
Solution : Since
@2f @
@x2 @x
=
x
e
t
2
3/2
x
x2/ t
=
2
4
2
x2/t = 4 @f ,
e
t5/2 t3/2
@t
f satises the heat equation.
c) Let v i j be the direction of a unit vector u in the xt-plane. Find the directional
derivative Duf at the point P ; .
[Hint: Use part (a).]
Solution : In part (a) we have computed the gradient
=
(0 1)
r f h fx ; ft i
=
xe
x2/ti +
2
t3/2
=
x
2
1
e
t5/2 t3/2
2
x2/tj :
Therefore, with u p12 i j the unit vector in the direction of v, the directional derivative at P ; is
(
=
(0 1)
)
Duf (0; 1)
=
p r f h ; i
1
1
1
2
(0; 1)
p :
1
=
2
2
d) Now let x s; v; w svw and t s; v v sin s . Determine @f @s at s .
@t
cos s . Using the chain rule, we nd
Solution : First note that @x
@s vw and @s
(
)=
(
)=
=
@f @f @x @f @t
@s @x @s @t @s
=
Since x js=0
=0
+
and ts=0
=
+
(2 )
=2
=
vwx e
t3/2
2
(2 )
x
x2/t + 2 cos(2s)
v, we get that @f @ss
/
1
/
=0
=
v
=0
2
t
5/2
3/2
.
1
t
2
3/2
e
x2/t:
Question #1.2 (20 points)
Dene
f x; y
(
)=
e y y 2 x2 :
(
)
a) Find the critical points of f .
Solution : First we note that r f h xe y ; y 2 x2
y ey i. The critical points of f sat2
2
isfy r f , so that x y x
y . This implies that x and that ; and
; are the only critical points of f .
=
=0
(0
2
=(
2
(
+2
+2
)
)=0
=0
(0 0)
2)
b) Compute the Hessian
f f xx xy
H x; y
fyx fyy
(
)=
of the function. Then classify all critical points by using the second derivative test (i.e.,
determine if they are local maxima, minima, or saddle points).
Solution : Since fxx
ey, fxy fyx
xey, fyy y2 x2 y ey. Then D ;
2
det H ;
< so ; is a saddle point. However, D ;
det H ;
e 2 e 2 2 e 4 > . Since fxx ;
e 2< , ;
is a local
maximum.
=
(0 0) =
2) = (
2
2
)(
2
2
0
2
)
=
=
4
0
=
0
2
=(
+4
+ 2)
(0 0) =
(0 0)
= 4
(0
0
(0
2) =
2
0
(0
2) =
(0
2)
c) What is the absolute maximum value of f on D f x; y x ; y g
Solution : Since f is continuous and has no local maxima in the interior of D, it must
achieve its absolute maximum on the boundary of D. Testing f on each of the four
boundaries of the square D, we nd that the absolute maximum is e (achieved at ; ).
=
(
): 0
1
0
1 ?
(0 1)
Question #1.3 (20 points)
Evaluate the following integrals. Remember that iterated integrals are sometimes easier to evaluate after switching the order of integration, or after changing to dierent coordinates.
a) Determine
ZZ
R
x2 y2 dA
(4 +
)
where the region of integration is the rectangle R f x; y
x ; y g.
Solution : Writing this integral with the proper boundaries of integration we nd that
=
Z Z
1
2
(4 +
1
0
x2 y2 dydx
)
Z 1
=
4
Z 1
1
=
=
b) Consider the integral
ZZ
D
(
1
3
1
y x2 y
2
2
):
+
1
3
y3
x2 16 dx
+
x3 16 x
+
1
3
y
0
2
=2
dx
y =0
3
1
=
1
12:
x2exydA
2
over the triangular region D f x; y x ; y x g. Provide expressions for the
two possible iterated integrals (one integrating in x rst, the other in y rst) with correct
boundaries of integration. Do not evaluate these yet.
=
(
): 0
2
2
0
Solution : The two (equivalent) iterated integrals are
Z Z
2
2
x
2
y
0
2
Z Zx
2
exydxdy;
2
0
0
x2exydydx:
c) Evaluate the integral in part (b) using one of the two iterated integrals.
Solution : It is easier to evaluate the integral by rst integrating in y , then in x. Then,
Z Zx
2
2
0
0
x2exydydx
=
=
=
=
Z
Z
Z
2
xexy yy ==0x dx
]
[2
0
2
(2
0
xex2
2
x dx
)
4
eudu
e :
0
4
4
5
Question #1.4 (20 points)
We will evaluate the double integral
I
ZZ
=
sin x2
(9
R
+4
y2 dA
)
over the region R bounded by the the ellipse x2 y2 by completing the following sequence
of steps.
a) Consider the linear transformation T u; v ! x; y given by
9
+4
:(
x
Find the Jacobian J
Solution :
=
@ (x; y)
@ (u; v )
)
1
=
3
(
u;
)
y
1
=
2
v:
of the transformation.
0 @x
@ x; y det@ @u
@x
(
)
(
)
@ u; v
=1
@y
@u
@y
@v @v
=
1
A
=
0
det@
1
0
3
1
0
1
A
=
2
1
6
:
b) After the change of variables to u; v coordinates, the integral I becomes
(
)
ZZ
S
sin u2 v2 jJ jdudv
(
+
)
with S f u; v u2 v2 g. Transform to polar coordinates to express this as an iterated integral in terms of the variables r and .
Solution : With r2 u2 v 2 we write the integral as
=
(
):
+
=
1
+
I
Z Z
2
=
0
0
1
1
6
sin r2 rdrd:
c) Evaluate the answer obtained in (b).
Solution : With the substitution w r2 we nd that
=
Z1
I sin u du
=
6
(
)
0
3
=
1
6
cos u
[
(
1
1
=
0
3
)]
:
M408D (54690/95/00), Final Exam Solutions: Part 2, 12/13/2010
Multiple choice questions #2.1-2.4 (*0 points*)
Not included in exam (failed to print).
Question #2.1 (20 points)
Determine if the following series is absolutely convergent, conditionally convergent, or divergent.
You must show your work and justify the use of any test to obtain credit.
a)
1 n
X
e
n =1
[Hint: Consider limn!1 en n2.]
Solution : By L'Hospital's rule applied twice,
n2
/
lim ex x2
n!1
/
=
lim ex x
/2
n!1
=
lim ex
n!1
/2 = +
1:
Therefore, by the divergence test we have that the series diverges (since the sequence
being summed does not go to zero as n ! ).
0
b)
1
X
n =2
n ln n
(
(
1)
2/3
)
+
n
n2/3 ln n
Solution :
This problem is nearly identical to that from the rst midterm. First we test
for absolute convergence of the series. We must nd whether the series
1
X
1
2/3
n2/3ln n
n =2 n ln n
(
)
+
converges or not. To do so, note that the rst term in the denominator grows faster than
the second term and is therefore the dominant contribution as n ! 1. This can be seen
by noting that
ln n 1/3
2/3
n
lim n ln2/3
lim
n!1 n ln n
n!1 n
=0
=
(
)
by L'Hospital's rule. Therefore, by the limit comparison test we have that the series converges/diverges if the series
1
X
n =2
converges/diverges. Now note that
improper integral
Z1
2
1
x ln x
(
)
2/3
dx
P1
Z1
=
ln 2
1
n ln n 2/3
(
)
1
n =2 n (ln n)2/3
u
1
2/3
4
du
(
diverges by the integral test since the
substitution with u ln x
=
)
diverges. So
P1
1
n =2 n (ln n)2/3
diverges and the original series is not absolutely convergent.
[There are many other ways to solve this problem by comparing against
a series which we
P
1
1
know is divergent. For example, since n ( n)2/3 + n2/3 n 2n n and 2n 1 n diverges by
the integral test, we nd that the original series is not absolutely convergent.]
Now we show that
ln
1
X
n =2
ln
ln
(
1)
ln
n
n ln n 2/3 n2/3 ln n
(
)
+
is in fact conditionally convergent by the alternating series test. Indeed, it is simple to
check that
bn
=
1
n ln n
(
)
2/3
+
n2/3 ln n
satises limn!1 bn and fbng is a decreasing sequence of terms. We therefore conclude
that the series is conditionally convergent.
=0
Question #2.2 (20 points)
a) Determine the interval of convergence of the following power series centered at a
:
=1
1
X
xp n
n n
n
(
=0
1)
2
[Hint: Start by using the ratio or root test to nd the radius of convergence of the series.]
Solution : By the ratio test, we nd that the series is absolutely convergent when
x
lim n!1 x
n +1
n
1)
(
1)
(
2
pn
p
n
n
n +1
2
+1
=
1
2
jx
1
r n
jnlim
!1 n
+1
=
1
2
jx
j< :
1
1
So the radius of convergence of the series is 2 and the interval of convergence is either
; , ; , ; , or ; . Testing the endpoints we see that x
is in the
interval of convergence using the alternating series test, but x is not using the p-series
test with p
. So I
; .
(
1
3)
[
1
3)
(
1
3]
[
1
3]
=
1
=3
= 1/2
=[
1 3)
b) Find the second-degree polynomial T2 x that best approximates the function f x e
near x = 1.
Solution : We must nd the second-order Taylor polynomial centered at a . Since
(
)
(
)=
x2
=1
f0 x
(
)=
f 00 x
xe x2;
2
(
) = (4
x2
2)
e
x2
we have that
T2 x
(
)=
f
(1) +
f0
(1)(
x
1) +
1
2!
f 00
x
(1)(
2
1)
=
e
1
1
2(
x
c) Derive the Maclaurin series (i.e., Taylor series centered at 0) of e
of convergence of the series?
5
x
1) + (
x2.
1)
2
:
What is the radius
Finding the nth derivative of e
the known Maclaurin series for ex. Since
Solution :
ez
making the substitution z
e
=
=
1
X
is not particularly easy, so instead we use
x2
1 < z < 1;
n
nz ;
n =0
1
!
x2 we have that
x2 =
1
X
n
1 < x < 1:
2n
n x ;
(
1)
!
n =0
The radius of convergence of the series is innite (series converges everywhere).
Question #2.3 (20 points)
Dene the vectors
u = h4; 5; 1i
v = h1; 0; 1i
w = h3; 2; 2i:
a) Compute u v w .
Solution : First we compute v w:
(
)
v w=
i
1
0
3
Therefore, u v w
(
)=8+5+2=
j k 1
2
2
i+ j
=2
k:
2
15.
b) Determine the vector equation for the plane parallel to the vectors v and w that passes
through the point P ; ; (i.e., express in the form n r r0
by determining a
normal vector n and an intercept r0).
Solution : The plane parallel to v and w has normal vector n v w h ; ; i. Since it
passes through P ; ; , the intercept is r0 h ; ; i.
(0
1
2)
(
) = 0
=
(0 1 2)
=
= 2 1
2
0 1 2
c) Find the shortest distance between the point P ; ; and the plane given by
(0 1 2)
x
+3
y
+5
z 48
=
using the method of Lagrange multipliers. At the least, write down the proper
equations to be solved.
[Hint: Need to use equations r f r g, g const:, for an appropriate choice of function
f to be extremized under some constraint g.]
Solution : The distance between the point P ; ; and any other point Q x; y; z is
=
=
(0 1 2)
d x; y; z
(
)=
p
x2
6
+(
y
1)
(
2
z
+(
2
2)
:
)
Therefore, we seek to minimize d x; y; z subject to the constraint
(
)
g x; y; z
(
)=
x
+3
y
+5
z 48
=
(which simply says that Q must lie on the given plane). To make things simpler, note
that we can, equivalently, minimize one-half the squared distance
f x; y; z
(
)=
1
2
x2
+
1
2
(
y
1)
2
+
1
2
(
z
2)
2
subject to g x; y; z 48: We minimize this using Lagrange multipliers. That is, we need
to solve the system of equations r f r g, g 48. This yields the system of equations
(
) =
=
x
=
x =
y
z
=1+3
=2+5
+3
y
z 48:
+5
=
Substituting x; y; z in terms of into the last equation we nd that 13 35 48, i.e.,
and x; y; z
; ; p . Is is then simple to check that this corresponds to a minimal distance of d ; ;
35 between P and the plane.
+
= 1
(
=
) = (1 4 7)
(1 4 7) =
Question #2.4 (20 points)
Dene the vector-valued function r t through its derivative
( )
r 0(t) = hcos t; sin t; 0i
and suppose r
h ; ; i.
(0) = 1 1 1
a) Find r t .
Solution : Integrating, we have that
( )
r(t) = r(0) +
Zt
b) Find the unit tangent vector T t
Solution : Since jr 0(t)j =
p
( )=
0
r 0(s) ds = h1 + sin t; cos t; 0i:
r 0(t)
jr (t)j .
0
cos2t sin2t
+
, we nd that T t
=1
( )=
r 0(t) = hcos t; sin t; 0i.
c) The parametric curve traced by r t lies on the surface of the paraboloid
( )
z x2
=
+(
y
2
1)
c
for which value of c?
Solution : The parametric curve traced out by r t is a circle of radius 1 in the xy -plane
centered at ; . Therefore, the curve lies on the surface x 2 y2 , implying that
( )
(1 0)
(
c
= 2(
x y
) + 1 = 2(
7
sin t cos t
)+3
1)
:
+
=1
Note that there was an error in the statement of the question that made the problem
more complicatedthe equation for the paraboloid was originally meant to read z x
2
y2 c, in which case the answer was c . Regardless, it was possible to solve for c
by substituting the coordinates of r t for x, y, and z in the given equation.
=(
1)
+
=1
( )
d) Find the arcR length of one loop of the parametric curve, either by using the arc length
formula L ab jr 0 t jdt or by using a more direct method.
Solution : The curve returns to its starting position each time t increases by . Using the
R
arc length formula we nd L 02 dt . More simply, this is found using the formula for the length of a circle with radius 1.
=
( )
2
=
1
= 2
8
Version 001 – finalp1 – srinivasan – (54690)
This print-out should have 4 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
1
3
3
2
1
2
0
CalC15a20b
001 5.0 points
-1
1
-2
-3
Which one of the following could be the
contour map of a cone?
-3
3
5
4
3
2
1
0
-1
-2
-3
-4
-5
-6
-2
2
-1
1
0
0
4.-4
3
2
1
0
1
2
3
-1
-2
-1
-3
-2
-3
1.-4
5
4
3
2
1
0
5.
rect
5
4
3
cor-
5
4
3
2
CalC15a30a
002 5.0 points
2
1
1
0
0
-1
Which one of the following surfaces is the
graph of
-2
-3
-4
f (x, y) = 2x2 ?
2.-5
4
y
3
1.
2
1
0
2
-1
-2
4
1
0
-1
3
5 4 3 2
1
0
1
2 3 4 5
2
z
-2
1
-3
0
3.-4
-2
-1
0
1
x
2
Version 001 – finalp1 – srinivasan – (54690)
y
2.
2
Find the maximum slope on the graph of
2
1
0
f (x, y) = 3 sin(xy)
-1
-2
8
at the point P (0, 4).
6
1. max slope = 12 correct
4
z
2
2. max slope = 4π
0
-2
3. max slope = π
-1
0
1
x
correct
3.
2
4. max slope = 3
y
2
5. max slope = 1
1
0
-1
-2
4
6. max slope = 12π
2
7. max slope = 3π
z 0
-2
8. max slope = 4
-4
-2
CalC16c28s
004 5.0 points
-1
0
1
2
x
y
4.
Find the volume of the solid in the first
octant bounded by the cylinders
2
1
0
-1
x2 + y 2 = 16 ,
-2
4
y 2 + z 2 = 16 .
Hint: in the first octant the cylinders are
shown in
2
z 0
-2
z
y
-4
-2
-1
0
1
2
x
y
5.
1
0
2
-1
-2
8
6
z 4
4
2
0
-2
-1
0
4
1
x
2
x
CalC15f23s
003 5.0 points
1. volume =
112
cu. units
3
Version 001 – finalp1 – srinivasan – (54690)
2. volume =
128
cu. units correct
3
3. volume = 40 cu. units
4. volume =
116
cu. units
3
5. volume =
124
cu. units
3
3