MATH 152
Final Exam
Fall 1997
Version B
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 parabola y
x.
10
"
3
7
6
13
6
10
3
9 .................................................................................................................... correctchoice
2
a.
b.
c.
d.
e.
x 2 " 2x
3
A
x 2 " 2x and the line y
x 2 " 3x
x
; x " Ÿx " 2x dx
2
xŸx " 3 0
3
; Ÿ3x " x dx
0
2
0
0
So 0 t x t 3 Thus
3x 2 " x 3 3 27 " 9 9
2
3 0
2
2
2. Find the plane tangent to the hyperbolic paraboloid z
(e)
x 2 " y 2 at the point Ÿ2, 1, 3 . Its
z-intercept is z a. 1
b. 0
c. "1
d. "2
e. "3..................................................................................................................... correctchoice
fŸx, y x 2 " y 2 f x Ÿx, y 2x f y Ÿx, y "2y
fŸ2, 1 3
f x Ÿ2, 1 4 f y Ÿ2, 1 "2
z fŸ2, 1 f x Ÿ2, 1 Ÿx " 2 f y Ÿ2, 1 Ÿy " 1 3 4Ÿx " 2 " 2Ÿy " 1 So the z-intercept is "3.
(e)
3. In the partial fraction expansion
a. A
b. A
c. A
d. A
e. A
x2 4
xŸx " 2 2
A
x
B
x"2
4x " 2y " 3
C
the coefficients are
2
Ÿx " 2 "1, B 1, C 2
"1, B 0, C 2
1, B 0, C 4........................................................................................ correctchoice
0, B 1, C "4
1, B "1, C "4
x2
4
AŸx " 2 2
BxŸx " 2 Cx
x
0
4
AŸ"2 2
A
1
x
2
8
CŸ2 C
4
x
1
5
AŸ"1 2
B
0
BŸ"1 CŸ1 1"B4
(c)
2
3 and the line y
x
rotated about the y-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
4. The area in the first quadrant between the hyperbola y
4 " x is
3
d. 2= ; Ÿ4x " x 2 " 3 dx........................................................................................ correctchoice
1
3
3
x
e. 2= ; Ÿ4 " x 2 "
1
2
dx
3 4 " x,
3 4x " x 2 ,
x 2 " 4x 3 0,
x
Cylinder:
x 1, 3
Ÿx " 1 Ÿx " 3 0,
4
3
3
V
2
; 2=rh dx
1
1
3
2= ; x 4 " x " 3x
1
3
-4
-2
0
2
4
2= ; Ÿ4x " x 2 " 3 dx
dx
(d)
1
5. Which limit does not exist?
lim Ÿ2x 2 y 2 a.
Ÿx,y vŸ0,0 b.
c.
d.
e.
lim
Ÿx,y vŸ0,0 lim
Ÿx,y vŸ0,0 lim
Ÿx,y vŸ0,0 lim
Ÿx,y vŸ0,0 Try y
lim
Ÿx,y vŸ0,0 ymx
Ÿ2x
2
" y2 2x 2 " y 2
............................................................................................. correctchoice
x2 y2
2x 4 x 2 y 2
x2 y2
2x 4 " x 2 y 2
x2 y2
mx in each limit. (c) gives
2x 2 " y 2
2x 2 " m 2 x 2 lim 2 " m 2
lim
2
2
xv0 x 2 m 2 x 2
xv0 1 m 2
x y
which depends on m. So the limit does not exist.
2 " m2
1 m2
(c)
3
6. ;
2
1
a.
b.
c.
d.
e.
x 3 x 2 " 1 dx
2
15
8 ................................................................................................................... correctchoice
15
22 2 " 8
15
15
8 Ÿ2 5/4 2 3/4 15
8 Ÿ2 5/4 2 3/4 " 1 15
Substitute:
;
2
1
u
x 3 x 2 " 1 dx
1
2
2
5
2
3
x2 " 1
du
2x dx
1 ; 1 Ÿu 1 u du
2 0
1 1 8
5
3
15
1 du
2
1 ; 1 Ÿu 3/2
2 0
(b)
x dx
x2
u 1/2 du
u1
1
2
2u 5/2
5
1
2u 3/2
3
0
7. An airplane is circling above an airport, clockwise as seen from above. Thus its wings
are banked with the right wing lower than the left wing. In what direction does the
binormal B point?
a. horizontally toward the center of the circle
b. vertically up
c. vertically down............................................................................................... correctchoice
d. along the left wing
e. along the right wing
T is horizontal tangent to the circle pointing clockwise.
N is horizontal toward the center of the circle.
B T • N is vertical and down by the right hand rule.
(c)
dy
28 x y . If yŸ0 4, then yŸ1 dx
81..................................................................................................................... correctchoice
53
49
11
9
8. Solve the differential equation
a.
b.
c.
d.
e.
dy
dy
28x dx
Integrate: ;
; 28x dx
y
y
Initial condition: x 0, y 4 :
2 4 C4
2 y 14x 2 4
(a)
yŸ1 9 2 81
Separate variables:
2 y
y
14x 2
Ÿ7x
2
C
2 2
4
9. A 50 ft cable is hanging from the roof down the side of a tall building. If the cable
weighs 3 lb / ft, how much work is done to lift the cable to the roof?
a. 7500 ft-lb
b. 3750 ft-lb .......................................................................................................... correctchoice
c. 1875 ft-lb
d. 150 ft-lb
e. 75 ft-lb
0
An element of cable y ft down of length dy weighs
dF
-20
W
g> dy
3 dy and must be lifted h
50
; h dF
50
0
-40
3 50 2
2
; y g> dy
0
3750
>gy 2
2
y ft .
50
0
(b)
-60
10. A nuclear power plant is producing a radioactive isotope X at the rate of 75 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 50NŸt kg /yr. Write the differential equation to be solved for NŸt .
a. dN 50N " 75
dt
b. dN 75 " 50N............................................................................................... correctchoice
dt
c. dN 50N " 75t
dt
dN
75t " 50N
d.
dt
e. dN "25N
dt
All terms must have the units of kg /yr.
dN 75 " 50N
(b)
¦
¦
dt
created
decayed
5
11. The function fŸx, y has the values given below. Estimate f Ÿ2.1, 3.0 .
y
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
a. 2
b. 10
c. 15
d. 20..................................................................................................................... correctchoice
e. 30
f Ÿ2.1, 3.0 y
lim
hv0
fŸ2.1, 3.1 " fŸ2.1, 3.0 fŸ2.1, 3.0 H " fŸ2.1, 3.0 X
0.1
h
4"2
0.1
(d)
20
12. Which of the planes described below contain the lines:
a.
b.
c.
d.
e.
x"1 y"2 z"1
x"1 y"2 z"1 ?
and
4
2
2
4
3
3
x " 2y z "2................................................................................................ correctchoice
x " 2y z 2
2x " y z "1
2x " y z "1
x"yz 0
Line 1 has direction u
§ § §
i
u
•
v
j k
2 3 4
˜2, 3, 4 ™.
Line 2 has direction v
§
§
§
i Ÿ6 " 12 " j Ÿ4 " 16 k Ÿ6 " 12 ˜4, 3, 2 ™.
So the normal is
˜"6, 12, "6 ™ "6˜1, "2, 1 ™
4 3 2
N ˜1, "2, 1 ™
NX NP
Both lines pass through P Ÿ1, 2, 1 . So the equation is
1x " 2y 1z 1 1 " 2 2 1 1 "2
(a)
6
1 =r 2 h. If the radius is measured as r 3 o .2, and the
3
height is measured as h 4 o .1, then the volume is computed as V 12= o V
where the error in the measurement of the volume is V a. 1.1=
b. 1.7=
c. 1.9=.................................................................................................................. correctchoice
d. 2.2=
e. 2.7=
13. The volume of a cone is V
V r V h
h
r
=Ÿ1.6 .3 1.9=
V X dV
2 =rh r 1 =r 2 h
3
3
(c)
2 =3 4 .2 1 =Ÿ3 2 .1
3
3
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
a. velocity
|v |
b. speed (Simplify.)
4t 2
4t 2
1
§
i
v•a
|v • a |
§
j
8t 2
1
˜"2, 2, 0 ™
§
k
"2t 2t 1
"2
a
c. acceleration
d. curvature
˜"2t, 2t, 1 ™
§
§
§
i Ÿ0 " 2 " j Ÿ0 " "2 k Ÿ"4t " "4t ˜"2, "2, 0 ™
2 0
440
e. tangential acceleration
f. normal acceleration
8
4
2 2
aT
d|v |
dt
aN
d 8t 2
dt
4|v | 2
1
8t 2
1
3
8t 2
1
8
8t 2
16t
2 8t 2 1
8
|v • a |
|v |3
2
1
3
8t
8t 2 1
8
8t 2
1
7
;
15. (10 points) Compute
3/4
1
1 x2
0
dx
1 x 2 1 tan 2 2 sec 2
Use a tan substitution:
x tan 2
dx sec 2 2 d2
3/4
3/4
1 sec 2 2 d2 ; 3/4 sec 2 d2 ¡lnŸsec 2 tan 2 ¢ 3/4 ln 1 x 2 1
dx ;
;
x0
x0 sec 2
x0
0
1 x2
2
25 3 " lnŸ1 ln 8 ln 2
3 " ln 1 0 2 0 ln
1 3
ln
4
16
4
4
4
16. (10 points) A comet has just passed by the sun.
Its current position is Ÿx, y Ÿ3 • 10 7 mi, 4 • 10 7 mi dy
4 • 10 7 mi , 7 • 10 7 mi .
and its current velocity is v dx ,
day
day
dt dt
2
2
Hence the current distance from the sun is R x y 5 • 10 7 mi.
Find the rate at which the distance from the sun is increasing.
dR
dt
R dx
x dt
R dy
y dt
3 4 4 7
5
5
•
10 7
dx
dt
x
x2
y2
40 • 10 7
5
y
x2
y2
dy
dt
3 • 10 7 4 • 10 7
5 • 10 7
4 • 10 7 7 • 10
5 • 10 7
8 • 10 7
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 3.
4
y
-8
-6
-4
2
0
-2
2
4
x
6
8
2
4
x
6
8
-2
-4
1.
4
y
-8
-6
-4
-2
2
0
-2
-4
8