Student (Print)
Section
Last, First Middle
Student (Sign)
Student ID
Instructor
MATH 152
Exam 2
Fall 2000
Test Form B
Solutions
1-10
/50
11
/10
12
/10
Part II is work out. Show all your work. Partial credit will be given.
13
/10
You may not use a calculator.
14
/10
15
/10
Part I is multiple choice. There is no partial credit.
TOTAL
Formulas:
Mn
x f x0
2
x1
f x1
2
x2
C
f x n"1 x n
2
x ¡fŸx o 2fŸx C 2fŸx n fŸx n ¢
"1
1
2
S n x ¡fŸx o 4fŸx 1 2fŸx 2 C 2fŸx n"2 4fŸx n"1 fŸx n ¢
3
KŸb " a 3
where K f UU Ÿx for all x in ¡a, b ¢
|E M | 24n 2
KŸb " a 3
where K f UU Ÿx for all x in ¡a, b ¢
|E T | 12n 2
KŸb " a 5
where K f Ÿ4 Ÿx for all x in ¡a, b ¢
|E S | 180n 4
Tn
; lnx dx
x lnx " x C
; tan x dx
" ln|cos x| C
; sec x dx
ln|sec x tan x| C
1
Part I: Multiple Choice (5 points each)
There is no partial credit. You may not use a calculator.
1. Compute ;
.
e
1 dx.
x lnx
a. 0
b. 1
e
c. 1
d. e
e. .
u
;
.
e
correctchoice
ln x
du 1x dx
1 dx ; 1 du ln|u|
u
x lnx
ln|lnx|
.
e
lim ln|ln b| " ln|lne|
bv.
.
2. After separating variables, we can solve the differential equation
a. ; 1
y
y dy
y
b. ;
2
y
c. ;Ÿy 2
dy
1
y dy
dy
dt
ty yt
by solving
; t dt
; t dt
;Ÿt 1 dt
correctchoice
2
1 1
t
;
y
dy
dt
e. None of the above, the differential equation is not separable.
d. ;
y
dy
dt
t y 1y
3. Compute nlim
v.
t
y2
y
1
y
®
y2
1
dy
®
t dt
;
y
y2
1
dy
; t dt
lnŸ2 e n 3n
a. 0
b. 1
3
c. 2
3
d. ln 2
3
correctchoice
e. .
lnŸ2 e
3n
n
Since e is much greater than 2,
Use l’Hopital’s Rule twice:
OR:
lim
nv.
n
en
2 en
lim
nv.
3
lnŸ2 e n lim
nv.
3n
1 lim e n 1 lim e n
3 nv. 2 e n
3 nv. e n
lnŸe n n 1
lim
lim
nv.
nv. 3n
3
3n
1
3
2
4
4. Which of the following gives the Trapezoid Rule approximation to ; 1
x dx with n
4?
2
1 2 1 2 1
5
7
3
8
4 4
1 1 2 1 2 1
5
7
3
4
4 2
1 1 4 2 4 1
correctchoice
5
7
4 2
3
4
1 ln 2 ln 3 ln 3 ln 7 ln 4
2
2
2
1 ln 2 2 ln 3 2 ln 3 2 ln 7 ln 4
2
4
2
a. 1
b.
c.
d.
e.
x
T4
T4
4"2 1
4
2
x fŸ2 2f
2
1 1 4 5
4 2
5 2fŸ3 2f 7
2
2
2 4 1
7
3
4
fŸ4 4
; 1x dx, which of the following is
2
the smallest upper bound on the error |E T | in the approximation?
5. If you use the Trapezoid Rule with n
1
12
1
24
1
48
1
96
1
192
a.
b.
c.
d.
e.
f UU Ÿx |E T |
4 to approximate
correctchoice
2
x3
On ¡2, 4 ¢ the maximum is K
KŸb " a 3
12n 2
1 Ÿ4 " 2 3
4
12Ÿ4 2
f UU Ÿ2 2
8
dy
dx
x sin 2x y tan x is IŸx 1
96
6. An integrating factor for the differential equation
correctchoice
a. |cos x|
b.
c.
d.
e.
1 . So the error is bounded by
4
"sec 2 x
e
e "|tan x|
"|sec x|
e |cos x|
dy
" Ÿtan x y x sin 2x
dx
sin x dx ln|cos x|
" ; tan x dx " ; cos
x
Standard form:
; Pdx
IŸ x e
; P dx
e ln|cos x|
P
" tan x
|cos x|
3
7. Find the arc length of the parametric curve x
a.
4 cos t,
y
= t t t =.
4 sin t for
3
6
=
3
2
b. =
3
c.
correctchoice
=
2
4
d. =
3
e. 3=
4
dx
dt
L
OR:
8.
;
dy
dt
"4 sin t
=/3
=/6
4 cos t
2
dx 2 dy
dt
dt
dt
=/3 " =/6 2=4 2=
3
2=
;
=/3
=/6
16 sin 2 t 16 cos 2 t dt
A tank is completely filled with water. The end is
;
=/3
=/6
4 dt
4 = " =
6
3
2=
3
5
a vertical semicircle with radius 5 m as shown.
Which integral gives the hydrostatic force on this
end of the tank?
The density of water is 1000
kg
and g
m3
5
a. 9800 ; Ÿ5 " y 2 25 " y 2 dy
9.8 m 2 .
sec
-5
0
5
correctchoice
0
5
b. 9800 ; Ÿ5 y 25 " y 2 dy
0
5
c. 9800 ; Ÿ5 " y Ÿ25 " y 2 dy
0
5
d. 9800 ; Ÿ5 y 2 25 " y 2 dy
0
5
e. 9800 ; Ÿ5 y Ÿ25 " y 2 dy
0
The slice at height y is at a distance h
x 25 " y 2
F
; >ghwdy
5 " y below the surface. Its width is w
2x where
5
9800 ; Ÿ5 " y 2 25 " y 2 dy
0
4
e t " t, y 4e t/2 for 0 t t t 2 is rotated about the x-axis. Which
integral gives the area of the surface of revolution?
HINT: Look for a perfect square.
9. The parametric curve x
2
a. ; 2=Ÿe t " t Ÿe 2t
b. ; 2=Ÿe t " t Ÿe t
0
2
c. ; 8=e t/2 Ÿe 2t
d. ; 8=e t/2 Ÿe t
0
2
e. ; 8=e t/2 e t
1 dt
correctchoice
1 dt
1 dt
1 dt
2e t
0
2
2e t
0
2
1 dt
0
dx
dt
A
dy
dt
et " 1
; 2=r ds
2
2
; 2=y
; 8=e t/2
0
0
Ÿe
2t
2e t/2
dx
dt
2
dy
dt
2
dt
2
; 2=4e t/2
Ÿe
0
" 2e t 1 Ÿ4e t dt
2
; 8=e t/2 e 2t
0
t
" 1 2 Ÿ2e t/2 2 dt
2e t
1 dt
2
; 8=e t/2 Ÿe t
0
1 dt
10. Calculate the x-component (or coordinate) of the center of mass of a plate with uniform density
> and whose shape is the quarter circle of radius 3 given by 0 t x t 3 and 0 t y t 9 " x 2 .
=
a.
4
9
b. = >
4
c. 9>
d. 8
=
4
=
e.
M
correctchoice
3
; > 9 " x 2 dx
0
My
x
> 1 =r 2
4
9= >
4
3
3/2
; >x 9 " x 2 dx "> 1 Ÿ9 " x 2 3
0
My
9> 4
4
=
M
9=>
3
0
> 1 9 3/2
3
9>
5
Part II: Work Out (10 points each)
Show all your work. Partial credit will be given.
You may not use a calculator.
11. Determine whether each of the following integrals converges or diverges. If it converges, find
its value.
Hint: Partial fractions
.
2
dx
2x x 2
Explain:
a. ;
Circle one: Converges Diverges
1
2
2x x 2
1 " 1
x
2x
Partial fractions gives
;
.
1
2
2x x
dx
2
;
.
1
1 " 1
x
2x
dx lnx " lnŸ2 x .
1
1
2
dx
2x x 2
Explain:
b. ;
;
ln
x
2x
.
1
ln 1 " ln 1
3
ln 3
Circle one: Converges Diverges
0
1
0
2
2x x
dx
2
;
1
0
1 " 1
x
2x
dx
lnx " lnŸ2 x 1
0
¡0
" ln 2 ¢ " ¡". " 0 ¢
.
12. Determine whether each of the following sequences converges or diverges. If it converges,
find the limit. Fully justify your answers.
cos n=
a. lim
nv.
Circle one: Converges Diverges
2
Explain:
cos n=
2
takes the values 0, "1, 0, 1 repeatedly. So it can never have a limit.
sin 2 n
b. lim
n
nv.
Circle one: Converges Diverges
2
Explain:
2
0 t sinn n t 1n
2
2
1
and lim
nv. 2 n
sin 2 n
0. By the sandwich theorem, nlim
v. 2 n
0.
6
13. Solve the initial value problem
x
dy
2 y cos x
x
dx
x2
dy
2xy cos x
x2
dx
x = when y 2
P
dy
dx
2
x
I
2y
e
cos x
x
; Pdx
e
yŸ= with
2
; x dx
e 2 ln x
2.
x2
d Ÿx 2 y cos x
x 2 y ; cos x dx sin x C
dx
C 2= 2
x 2 y sin x 2= 2
2= 2 sin = C
y
14. Compute the surface area of the surface obtained by rotating the curve x
1 t y t 2 about the x-axis?
; 2=r ds
2
; 2=y 1 dx
dy
1
du
2
du 8y dy
u 1 4y
8
2
3/2
2u
=
1
A
; 2= u du 4 3
8 y1
= Ÿ17 3/2 " = Ÿ5 3/2
6
6
A
2
dy
sin x 2= 2
x2
1 y 2 for
2
; 2=y 1 4y 2 dy
1
y dy
2
y1
= Ÿ1 4y 2 3/2
6
2
1
15. A tank contains 40 lb of salt mixed with 60 gal of water. Salt water containing 4 lb of salt per
gal is added to the tank at the rate of 3 gal per min. The tank is kept thoroughly mixed and
drains at the same rate.
a. Write out the differential equation and the initial condition for SŸt , the number of lbs of salt
in the tank at time t.
dS lb
dt min
4 lb 3 gal
S t lb 3 gal
" Ÿ gal min
60 gal min
dS
dt
12 " 1 S
20
SŸ0 40
b. Solve the initial value problem.
12 " 1 S
20
dS
; dt
" 20 ln 12 " 1 S t C
20
12 " 1 S
20
ln 12 " 1 S " t " C
12 " 1 S Ae "t/20
S 240 " 20Ae "t/20
20
20
20
20
SŸ0 40 240 " 20A
A 10
S 240 " 200e "t/20
dS
dt
;
c. How much salt is in the tank after 20 hours?
SŸ1200 240 " 200e "1200/20 240 " 200e "60
The problem should have said 20 min which gives
SŸ20 240 " 200e "20/20 240 " 200
e
7