MATH 152
Final Exam
Fall 1997
Version A
Solutions
Part I is multiple choice. There is no partial credit. You may not use a calculator.
Part II is work out. Show all your work. Partial credit will be given. You may use your
calculator.
1
Part I: Multiple Choice (5 points each)
There is no partial credit. You may not use a calculator. You have 1 hour.
1. Find the area between the curves y x 3 and y x 1/3 and within the first quadrant.
a. 1
4
b. 1
3
c. 1 ..................................................................................................................... correctchoice
2
d. 2
3
e. 3
4
x 3 x 1/3 at x 0, 1, "1. In the first quadrant, when 0 t x t 1 we have x 1/3 u x 3 . Thus
1
1
4/3
4
A ; x 1/3 " x 3 dx 3x " x
3 " 1 1
(c)
4
4 0
4
4
2
0
2. Find the plane tangent to the monkey saddle z x 3 " 3xy 2 at the point Ÿ2, 1, 2 . Its
z-intercept is z a. 2
b. 0
c. "2
d. "4..................................................................................................................... correctchoice
e. "6
fŸx, y x 3 " 3xy 2 f x Ÿx, y 3x 2 " 3y 2 f y Ÿx, y "6xy
fŸ2, 1 2
f x Ÿ2, 1 9
f y Ÿ2, 1 "12
z fŸ2, 1 f x Ÿ2, 1 Ÿx " 2 f y Ÿ2, 1 Ÿy " 1 2 9Ÿx " 2 " 12Ÿy " 1 9x " 12y " 4
So the z-intercept is "4.
(d)
3. In the partial fraction expansion
Ÿx
2
2x 1
A 2Bx C the
x"1
x x1
x 1 Ÿx " 1 coefficients are
a. A "1, B 0, C 1
b. A 1, B "1, C 0...................................................................................... correctchoice
c. A 0, B 1, C 1
d. A 0, B 0, C "1
e. A 1, B "1, C "1
2x 1 AŸx 2 x 1 ŸBx C Ÿx " 1 x1
3 AŸ3 A1
x0
1 AŸ1 CŸ"1 1 " C
C0
x "1
"1 AŸ1 Ÿ"B C Ÿ"2 1 2B
B "1
(b)
2
4. The area in the first quadrant between the hyperbola y 3
x and the line y 4 " x is
rotated about the x-axis. Find the volume of the solid swept out.
3
2
a. = ; 4 " x " 3
dx
x
1
3
2
b. 2= ; 4 " x " 3
x dx
1
3
c. 2= ; 4 " x " 3
x dx
1
3
d. = ; Ÿ4x " x 2 " 3 dx
1
3
3
x
e. = ; Ÿ4 " x 2 "
1
2
dx.................................................................................. correctchoice
4
2
3 4 " x, 3 4x " x 2 , x 2 " 4x 3 0,
x
0
1
2
3
4
Washer:
3
3
1
1
Ÿx
" 1 Ÿx " 3 0, x 1, 3
V ; =R 2 " =r 2 dx = ; Ÿ4 " x 2 "
3
x
2
dx
(e)
-2
-4
5. Which limit does not exist?
lim Ÿx 2 2y 2 a.
Ÿx,y vŸ0,0 lim
b.
Ÿx,y vŸ0,0 lim
c.
Ÿx,y vŸ0,0 d.
lim
Ÿx,y vŸ0,0 lim
e.
Ÿx,y vŸ0,0 Ÿx
2
" 2y 2 x 4 2x 2 y 2
x2 y2
x 2 " 2y 2
............................................................................................. correctchoice
x2 y2
x 4 " 2x 2 y 2
x2 y2
Try y mx in each limit. (d) gives
2
2 2
2
2
x 2 " 2y 2
lim
lim x 2 " 2m2 x2 lim 1 " 2m2 1 " 2m2
2
2
x,y v 0,0
x
x
0
0
x y
x m x
1m
1m
ymx
Ÿ
Ÿ
v
v
which depends on m. So the limit does not exist.
(d)
3
1
x3
dx x 1
a. 1 Ÿ1 " ln 2 ...................................................................................................... correctchoice
2
b. 1 Ÿ2 " ln 2 2
c. 1 Ÿ2 ln 2 2
d. 1 Ÿ1 " ln 2 e 2
e. 1 Ÿ2 " ln 2 e 2
6. ;
2
0
1 du x dx
x2 u " 1
2
1
2
3
1 du 1 ; 2 1 " 1 du 1 ¡u " lnu ¢ 2 1 ¡2 " ln 2 ¢ " 1 ¡1 " ln 1 ¢
dx 1 ; u "
; 2x
1
u
2
2
2 1 u
2 1
2
0 x 1
1
¡1 " ln 2 ¢
(a)
2
Substitute:
u x2 1
du 2x dx
7. An airplane is circling above an airport, counterclockwise as seen from above. Thus its
wings are banked with the left wing lower than the right wing. In what direction does the
binormal B point?
a. horizontally toward the center of the circle
b. vertically up.................................................................................................... correctchoice
c. vertically down
d. along the left wing
e. along the right wing
T is horizontal tangent to the circle pointing counterclockwise.
N is horizontal toward the center of the circle.
B T • N is vertical and up by the right hand rule.
(b)
8. Solve the initial value problem xy U y e x where yŸ1 e 1.
x
a. e x 1 .............................................................................................................. correctchoice
b. xŸe x " 1 c. xŸe x 1 x
d. ex 1
e. xe x " 1
x
y
;
Integrating factor: I e
Standard form: y x ex
x
xy y e x
xy ; e x dx e x C
Ÿxy e
U
U
1
x dx
e ln x x
U
Initial condition: 1Ÿe 1 e 1 C
C1
xy e x 1
x
y e x 1
(a)
4
9. A conical reservoir with the point at the top is 8 ft deep and 4 ft in radius at the base.
The liquid in the reservoir weighs 80 lb / ft 3 . How much work is done to pump the liquid
out the top of the reservoir?
a. 640= ft-lb
b. 5120 = ft-lb
3
c. 10240 = ft-lb
3
d. 4096= ft-lb
e. 20480= ft-lb ..................................................................................................... correctchoice
-4
-2
2
0
4
Measure y 0 downward (in spite of the labels in the plot.)
The dark lines are the height h y 0 and the radius r.
4
1
r
r 1y
y 8 2
2
8
8
8
y2
W ; >gh dV ; 80y =r 2 dy ; 80y = dy 20= ; y 3 dy
0
0
0
4
8
4
y
5=8 4 2 11 10= 20480=
(e)
20=
4 0
-2
-4
-6
-8
10. A nuclear power plant is producing a radioactive isotope X at the rate of 50 kg /yr. Let
NŸt kg be the amount of X at the power plant at time t. A known fact is that X
decays at the rate of 75NŸt kg /yr. Write the differential equation to be solved for NŸt .
a. dN 75N " 50
dt
dN
75N " 50t
b.
dt
c. dN 50 " 75N............................................................................................... correctchoice
dt
d. dN 50t " 75N
dt
e. dN "25N
dt
All terms must have the units of kg /yr.
dN 50 " 75N
(c)
¦
¦
dt
created
decayed
5
11. The function fŸx, y has the values given below. Estimate f Ÿ2.0, 3.1 .
fŸ2.0, 3.2 3
fŸ2.1, 3.2 7
fŸ2.2, 3.2 9
fŸ2.0, 3.1 2
fŸ2.1, 3.1 4
fŸ2.2, 3.1 8
fŸ2.0, 3.0 1
fŸ2.1, 3.0 2
fŸ2.2, 3.0 5
x
a. 2
b. 15
c. 20..................................................................................................................... correctchoice
d. 25
e. 30
f Ÿ2.0, 3.1 lim fŸ2.0 h, 3.1 " fŸ2.0, 3.1 X fŸ2.1, 3.1 " fŸ2.0, 3.1 4 " 2 20
h 0
0.1
0.1
h
x
(c)
v
x 2 s " 3t
y "1 " 2s t
12. Which line is perpendicular to the plane
z 3 " 2t
x 1 " 3t
a.
x 4 2t
y 2t
y 2"t
d.
z 3 " 2t
z "5 3t
x 2s
b.
x 3 4t
y "1 " 2s
y 2 2t
e.
z3
........................ correctchoice
z 1 " 5t
x 2t
c.
y "t
z 3t
Two tangent vectors to the plane are u ˜1, "2, 0 ™ and v ˜"3, 1, "2 ™.
So the normal is
§
§
§
1
"2
0
"3
1
"2
i
N u•v j
k
§
§
§
i Ÿ4 " 0 " j Ÿ"2 " 0 k Ÿ1 " 6 ˜4, 2, "5 ™.
This normal is tangent to the line
(e).
6
13. The legs of a right triangle are measured to be 6.0 in and 8.0 in each with a maximum
error of o0.1 in. Hence the hypotenuse is h 10.0 in o h with a maximum error of
h
a. 0.10 in
b. 0.11 in
c. 0.12 in
d. 0.13 in
e. 0.14 in .............................................................................................................. correctchoice
x 2 y 2 6 2 8 2 10
x
h X dh h x h y x
y
2
x y2
h
y
x
x y
2
y 6 0.1 8 0.1 .14
10
10
2
(e)
Part II: Work Out
Show all your work. Partial credit will be given.
You may use your calculator but only after 1 hour.
14. (10 points) Consider the curve rŸt Ÿ1 t 2 , 1 " t 2 , t . Compute each of the following:
v ˜2t, "2t, 1 ™
a. velocity
4t 2 4t 2 1 |v | b. speed (Simplify.)
a ˜2, "2, 0 ™
c. acceleration
d. curvature
§
i
v•a |v • a | §
j
§
k
2t "2t 1
2
8t 2 1
§
§
§
i Ÿ0 " "2 " j Ÿ0 " 2 k Ÿ"4t " "4t ˜2, 2, 0 ™
"2 0
440 e. tangential acceleration
f. normal acceleration
v•a
4 | 3 | 8 2 2
|v |
aT d|v |
16t
d 8t 2 1 dt
dt
2 8t 2 1
a N 4|v |2 8
8t 1
2
3
8t 2 1
2
8
8t 2 1
3
8t
8t 2 1
8
8t 1
2
7
;
15. (10 points) Compute
=/4
0
x cosŸ2x dx
ux
dv cosŸ2x dx
; x cosŸ2x dx 2x sinŸ2x " 12 ; sinŸ2x dx 2x sinŸ2x =/4
=/4
= sin
; x cosŸ2x dx 2x sinŸ2x 1 cosŸ2x 8
4
0
0
1
=
"
4
8
Use integration by parts:
du dx
v
1 cosŸ2x 4
= 1 cos =
4
2
2
sinŸ2x 2
"
0 sinŸ0 2
16. (10 points) An ant is crawling on the saddle surface z x 2 " y 2 . At time t 3, the
dy
0.2. Find dz
ant is at the position Ÿx, y, z Ÿ2, 1, 3 and satisfies dx 0.3 and
dt
dt
dt
at t 3.
dz z dx z dy 2x dx " 2y dy 2 2 0.3 " 2 1 0.2 1.2 " .4 .8
dt
dt
dt
x dt
y dt
17. (5 points) This is the direction field of a certain differential equation. Draw in the
solution y yŸx which satisfies the initial condition yŸ2 "1.
4
y
-8
-6
-4
2
0
-2
2
4
x
6
8
2
4
x
6
8
-2
-4
4
y
-8
-6
-4
-2
2
0
-2
-4
8