Solutions to the Fall 2014 BC 3 Semester Review
1a.
1b.
2a.

x

1 x2
0
dx 
1
2

1

u
1
1
du  lim 
b   2
u
1
2



10

dx  lim 
3
a 2 
x 2


1
2
2


dx

2
x ln x 

1
  u
ln 2
sin x  2
2
1

  u
ln 2


du   lim  u
u
 b  
b
1
1
1
Since

1

10
a
du
du
 2x  dx 
dx
2x
b
1


  lim  b  1   DNE 
b




2
3
dx   lim  3  x  2
3

x 2
 a 2  2
du
2
u  1 x2 
du
1
u  ln x 
10 
a
  lim  3  4  3 3 a  22   6
 a 2  2
2


du
1
  dx  xdu
dx x
du converges ( p -integral, p  1 ),
1

dx

2
x ln x 
2
sin x  2
must also converge.
3
3
for all x  1 .
x 1
x 1
x 1
x 1 x3
 3
 sin x  2
Since
dx converges ( p -integral, p  1 ),
dx converges by the
1 x3
1
x3 1
Comparison test.

2
2
2
2c.
Limit Comparison test with f x 
dx
and g x 
1 ex 1
ex  1
ex
2
x
f x
ex
lim e  1  lim
 1 Since lim
converges to a positive number, and
2
x  g x
x 
x  e x  1
2b.
3
1
Comparison test:
dx
3


3
 



1

 
 
 
ex

dx  lim 
b  
ex
2
3





b
1


2
dx   lim  
x
x
e
 b   e
2
b
1


  lim   2    2    2 converges,
 b   e b  e 1   e 1

2
dx converges by the Limit Comparison test.
ex 1
 x 1
x 1
x
1
dx


2d.
Comparison test:
for all x  1 .
1 x x
x x x x
x
 1
 x 1
dx diverges by the Comparison
dx diverges ( p -integral, p  1 ),
Since
1 x x
1
x
test.
1



n
 

1
3a. lim an  lim     DNE  since
e
n 
n   e 
m

2
3b. lim am  lim  1  
m
m 
m  
e
2
n

r
since lim  1    e r
n
m  
3c.
lim ak  lim
4a.
k  ln
k 


k 1
n2
 4n
n 1
2
1


5a. First, consider
k 1

1
k
k 1



sin k 
k 1
m 0
2m
m
m
 2
7
5b. First, consider
m 0

k

n2
 4n
n 1

1
diverges by the nth term test.
sin k 
Since 0 
3

2
sin k 
k3
k
3

1
k
3
for all k  1 and
converges by the Comparison test.
converges absolutely.
k3

7
3

1
 0,
4
k 1
sin k 
k 1

2

converges (p-series, p  1 ),
3
Thus,
9 45

2
2
5
4n  1
sin k 
k
3
5

n2
n 


9

Since lim
1
 2k
2
2k  4  3k 
8k  6k 2
1

k
 lim
 lim
 lim
2
1
k 
k  3 1  k 2
k  3  3k 2
3
4  3k
0
k 1
3

9 
5
k 1  
5


1
0
Z
 4  3k ]

3k 1
k 1
4b.

ln 1  k
2
m
m


7
m 0
2m
m
Since 0 
m
2
converges (infinite geometric series, r   1 ),
7
m
2
m
7 m

7
m 0

m
2
7
2m
m
m
m
for all m  0 and
converges by the
Comparison test.

Thus,
7
m 0
5c.
m
converges absolutely.
n!
n  1  ! 1
n 1 

1

n 1
n 1

6b.
m


6a.
m
 2


j 2
Since
n!
n!
1
for all n  1 and


n  1 ! 1 n  1 ! n  1
1
diverges (p-series, p  1 ),
j

n!
 n  1 ! 1 diverges by the Comparison test.
n 1
n 6n

1  x 3 


1
x


cos  x 3   

2n  !
2n  !
n 0
n 0
n
2n 1
n

sin 3x  1   1  3x 
1   32n 1  x 2n

 

x
x
2
n

1
!


2n  1 !
n 0
n 0

n
2n
6c.
2
xe x  x 
x 

6d.
1x
4


1  x 4
2
7a.
 0.2

2

n 0
x 2n 1
n!

x
 0.2
1


n!
n 0
x
n
2


x   x     1 


4

n
n 0
4
n
x 4n 1
n 0
 L  cos  0.2
4!
1 1
2
3
1
1
7b. 1   2  2  L  e 2
2
2
3!

8a.

k 1
k
x  4

Use Ratio test on series
3k

k
x  4
3k
k 1
, testing for absolute convergence.
k 1
x  4
3 k  1 
lim
k
k 
x  4
x 4
 lim
k 
k 1

3 k  1 
3k
x 4
k
3k
 x 4  x 4
k  3 k  1 
 lim
3k
x  4  1  1  x  4  1  3  x  5
Now, need to test the values of x that produces value of 1 in the Ratio Test,
x = 3 and x = 5 (the endpoints of the interval).

x  3:

k
3  4 


3k
k 1

k 1
k
 1
ak 
3k
1
1
is decreasing , lim
0
3k
k  3k
so series converges by the alternating series test  include x  3 .

x 5:

k
5  4 
k 2
3k



k 1
1
1

3k 3

1
k
diverges (p-series, p  1 ),  don't include x  5
k 1
interval of convergence is [3, 5).

8b.

k 0
k
2x 

Use Ratio test on series
k!

k
2x 
k 0
k!
, testing for absolute convergence.
k 1
2x 
k  1  !
lim
k
k 
2x 
 lim
k 
2x
k 1
k  1  !

k!
2x
k
k!
interval of convergence is  ,   .
 lim
2x
k  k
1
 0  1 for all values of x.

8c.
n
 x  2
n
n 1
2

Use Ratio test on series
 3n

n 1
n
 x  2
n 2  3n
, testing for absolute convergence.
x n 1
2
lim
n  1
 32 1
 lim
xn
n 
x 2
n 
n 2  3n
n 1
n 2  3n

2
n
n 1
x 2
n

1

3
 
2
x 2
 k  x 2
 lim 



3
3
k   k  1 
x 2
 1  x  2  3  3  x  2  3  1  x  5
3
Now, need to test the values of x that produces value of 1 in the Ratio Test,
x  1 and x = 5 (the endpoints of the interval).

x  1 :

n
 1  2
n 1
n

n 2  3n
n
 3

n 1
2


 3n

n
 1
n2
n 1
an 
1
n2
is decreasing , lim
1
n  n 2
0
so series converges by the alternating series test  include x  1 .

n
x 5:
n 1
n
5  2
2
 3n


n
n 1

3n
2
 3n
1
n

n 1
2
converges (p-series, p  1 ),  include x  5
interval of convergence is  1,5  .
9.

n
 3x 
f x  


2 
n 0 

This is an infinite geometric series, so converges for r 
 2 2
interval of convergence is   ,  .
 3 3
3x
2
1 x 
2
3
f  x  
n 1

 3x 
n

2 

n 0

3x
1.
2
3
 
2
n

3
n   x n 1
2
n 0  

Derivative of a function has same radius of convergence as original function,
so just need to check endpoints for convergence.
2
x  :
3
n

n 1
3  2
n    
2
3
n 0   



n 1  3n
  1
n 1

 
 2 
n 1  3n 
 3n 
since lim    , lim  1   does not
n   2 
n 
 2 
exist,
2
so series diverges by the nth term test  don't include x   .
3
2
x  :
3

n
n 1
3 2
n   
2
3
n 0    

 don't include x 


n 1
2
.
3
 3n 
 2 
 3n 
lim     , so series diverges by the nth term test
n   2 
 2 2
interval of convergence is   ,  .
 3 3

f  x dx 
n 1

1  3x 


n 1  2 
n 0


2

3
n

 3  x n 1
 
2 n 1
n 0  

Antiderivative of a function has same radius of convergence as original function,
so just need to check endpoints for convergence.
(Note: the + C was not included since it does not affect convergence.)
n
n 1



n 1 
3  2
2
1
2
2

 an 
x  :




1
is decreasing ,


  

 3 n  1  
3
2  3
n 1
3
n

1





n 0
n 0
2
2
lim
 0 , so series converges by the alternating series test  include x   .
3
n  3 n  1 

2
x  :
3

n

n 1
3 2
   
2
3
n 0    

 don't include x 

1


n 1

2
2
1
, diverges (p-series, p  1 )
 
3
k
3
n

1


n 0
k 1

2
.
3

 2 2
interval of convergence is   ,  .
 3 3
3
10.
11.
lim
arctan 2x 
x 0
 lim
x
2x 
2x 
3
5
2x 

5
x
x 0
7
2x 

7
L
8x 2 32x 4 128x 6
 lim 2 


L
3
5
7
x 0
2
n 
 
3
0.2
0.2 
0.2 

x

x 7 x 10
x 3
xe dx 

L
x  x 4 
x
 dx 

n! 
2!
3!
0
0 
0 
 n 0






 

0.2
2
5
8

dx


11
x2 x5

 0.2   0.2   0.2   0.2  L
x8
x 11




L 

 2

5 8  2! 11  3!
2
5
8  2!
11  3!

 0
The above series is alternating, and the terms are decreasing in value toward 0,
so we need to determine the first term in it with value less than 0.00001.
5
Since

0.2
0
12.


k 0
 0.2
5
8
 0.2
 0.000064 and
2
3
xe x dx 
k
ak   x  2
 0.2
2
8  2!
 0.00000016 , we use the 1st two terms.
5

 0.2
5
 0.019936
converges if x = 7 and diverges if x = 9 gives us the following picture.
13
7
Div)
a.
b.
c.
d.

13

n 1
2
3
[ Con)
9
[Div
May be true
May be true
May be true
May be true
n 1
 x 
This series is alternating, , and the terms are decreasing in value toward 0,
n  2n
so we will use the relationship S  Sn  an 1
a.
the sixth term of this series is an upper bound for the error made in using the first
five terms of the series to approximate the function.
6 1
 x 
a6 
b.
6  26
7
 0.3

6  26
 0.000000569531
Since 0 < x < 0.3.
the (n + 1)st term of this series is an upper bound for the error made in using the
first n terms of the series to approximate the function.
n 2
an 1 
 x 
n  1  2n 1
1

1n 2
n  1  2
n 1

1
n  1  2n 1
Since 0 < x < 1.
 0.001  n  1  2n 1  1000
n  1  2
Since  6  1   261  896 and  7  1  27 1  2048 , 7 terms are necessary.
n 1
c.
the fifth term of this series is an upper bound for the error made in using the first
four terms of the series to approximate the function.
5 1
a5 
x6
 x 
5  25

x6
5  25
Since 0 < x.
 0.05  x 6  8  x  2  0  x  2  1.414
Again since 0 < x.
52
14. An upper bound for this error can be found by determining an upper bound for E5 2 ,
5
the remainder for the 10th-degree Maclaurin polynomial off  x   e x at x  2 .
11
ec
2  0

11!
E10 2  
15.
cos 2x 
for some c between 2 and 0.

11
e2
2

11!
Since 0 < c <2.

9 11
 2  0.000462
11!
Since e < 3.
2
2x 
1
2!
2x 

4!
4
6
2x 

6!
8
2x 

8!
f
(11)
x   e x 
10
2x 

10!
L
The above series is alternating, so the sixth term of this series is an upper bound for the
error made in using the first five terms of the series to approximate the function.
10
2x 
10!
10
 0.005  2x 
 18144
 2x  2.66608
 x  1.33304
 1.33304  x  1.33304
16. An upper bound for the error made when using the nth-degree Maclaurin polynomial
off  x   e x at x  1 to approximate e is the remainder, En 1 
Rn 1  
n 1
ec
1  0

n  1 !

e1
n  1 !

3
n  1 !
for some c between 1 and 0.
f
for this polynomial.
(n 1)
x   e x 
Since 0 < c < 1.
Since e < 3.
3
 0.005  n  1  !  600  n  5  six terms are necessary.
n  1 !
17.
sin 3x   3x 
3
3x 

3!
5
3x 

5!
7
3x 

7!
9
3x 

9!
L
The above series is alternating, so the fifth term of this series is an upper bound for the
error made in using the first four terms of the series to approximate the function.
9
3x 
9!
18.

9
3  0.5

9!

 0.000106
x  2  2sin   cos   
(since x  0.5 )
dx
 2cos2    2sin    2sin2  
d


y  2  2sin   sin  
dy
 2cos   sin    2cos    2cos   sin    4cos   sin    2cos  
d
dy
 0  4cos   sin    2cos    0
a.
d
1
 2cos   2sin    1  0  cos    0 or sin   
2

3

5
    2 k or  
 2 k or    2 k or  
 2 k where k  ¢
2
2
6
6
 3    
 5 
 
 Horizontal tangent occurs at  4,
 ,  1,  , and  1,
 , but NOT at  0, 
 2   6
 6 
 2



dx
is also equal to zero, which means the graph is not smooth at
d
this point, and in fact has a vertical tangent at this point (see part b).
dx
 0  2cos2    2sin    2sin2    0  2 1  sin2    2sin    2sin2    0
d
 4 sin2    2sin     0  2  0  2 sin    1 2sin    1  0
since at this point,
b.



1

7
    2 k or  
2
2
6
    7 
 Horizontal tangent occurs at  0,  ,  3,
 , and
 2  6 
 sin    1 or sin    


 2 k or  
11
 2 k , k  ¢
6
 11 
 3,
.
6 

 
dy
To see why the graph has a vertical tangent at  0,  , we need to look at
dx
 2
4 cos   sin    2cos  
2cos    2sin 2 
dy


dx 2cos2    2sin    2sin2   2sin    2cos 2 
2cos    2sin 2 
dy
 lim


 dx
 2sin    2cos 2 
lim

2

2
2cos    2sin 2 
dy
lim
 lim


 dx
 2sin    2 cos 2 

2

2
Z
0
 lim
]
Z
0

0

0


2
 lim
]
2sin    4 cos 2 

2
2cos    4 sin 2 
2sin    4 cos 2 

2cos    4 sin 2 
Z
2

]
Z
0
2
 
]
0
19.
5  4 cos   2  2cos 
6
6cos   3
4
1
2
2
cos   
2
 2 k or
3
4

 2 k where k  ¢
3

-4
-2
2
4
6
8
-2
-4
-6
Area 
1
2
2
3
2
 2 2  2cos 
d 
3
1
2

4
3
2
3
20. 3  6sin   0  6cos   3  sin  
1
Area 
2

2
0
2
3  6sin  
1
d 2
2
 
21. The rose graph r  3sin 2

2
5  4 cos 
1

5
    2 k or  
 2 k where k  ¢
2
6
6
5
6

d
2
3  6sin  
d   27 3  9  75.0398
6
has four leaves, each centered in one of the four quadrants.
The leaf in the fourth quadrant occurs at  
3
, as r is negative for this value of  .
4
dx
 6cos 2  cos    3sin 2  sin  
d
dy
y    3sin 2  sin   
 6cos 2  sin    3sin 2  cos  
d

 3  
 3  
 3 
2 3 2
x

3sin
cos

3

1












 2 
2
 4  
 2 
 4 


 3  
 3  
 3 
2
3 2
y

   3sin 
  sin 
  3  1 
2
2
 4  
 2 
 4 




x    3sin 2  cos   
dy
dx

3
4
 3
6cos 
 2

 3
6cos 
 2

 3 
 3 
 3
 sin 
  3sin 
 cos 

 4 
 2 
 4

 3 
 3 
 3
 cos 
  3sin 
 sin 

 4 
 2 
 4
 3 2

  2 1
 3 2

2


3 2
3 2
 1x 
 or y  x  3 2

2
2 

r
r
22. r t   2cos t  , tan 2t   v t   2sin t  , 2sec2 2t 
y 
2
2
2
2

2
 
   

1

2   
 v     2sin      2sec      2     2  2    65  8.06
2


6
 6  
 3 


r
r
23. a t   sin t  , 2  v t    cos t   C1,2t  C2
r
v  0   2, 0   cos  0   C1  2 and 2  0  C2  0  C1  3 and C2  0
r
r
 v t    cos t   3,2t  r t    sin t   3t  C3,t 2  C 4
r
r  0   0, 4   sin  0   3  0  C3  0 and 02  C 4  4  C3  0 and C 4  4
r
 r t    sin t   3t ,t 2  4
r
r
24. Based on information provided, v  0   20cos 25   ˆ
i  20 sin 25   ˆ
j and r  0   14ˆ
j
r
r
r
a.
a t   9.8ˆ
j  v t   9.8t  ˆ
j  v  0   20 cos 25   ˆ
i  20 sin 25   9.8t  ˆ
j
r
r
r t   20 cos 25 t  ˆi  20 sin 25 t  4.9t 2  ˆ
j  r 0 


 20 cos 25 t  ˆi  14  20 sin 25 t  4.9t   ˆ
j
 18.126t  ˆi  14  8.452t  4.9t   ˆ
j


2
2
14  20 sin 25 t  4.9t 2  0  t  2.76 seconds (since t  0)
b.
c.
20 cos 25   2.76  50 meters
d.
v 2.76 
e.
20 sin 25   9.8t  0  t  0.862
20cos 25  20sin 25  9.8 2.76
2
2
 25.97 m/s
2
14  20 sin 25  0.862  4.9  0.862  17.645 m
25a. e 2x

dy
1

 y 1  y 2dy  e 2x dx  y 1  y 2dy  e 2x dx
dx y 1  y 2

1
2


udu  e 2x dx
u  1 y2 



du
du
 2y  dy  
dy
2y
3
1 2 3 1
1
1
2
   u 2  e 2x  C   1  y 2  e 2x  C
2 3
2
3
2
dy
dy
 e x y 
 e x e y  e y dy  e x dx  e y dy  e x dx  e y  e x  C
25b.
dx
dx

1
y
x
x
x
 e   e  C  y  ln  e  C  y   ln  e  C
or y  ln 
x
  e C

dy
y
y
1
1
26. xyy   y 2  1  xy
 1  y2 
dy  dx 
dy 
dx
dx
x
x
1  y2
1  y2












1 1
du 
2 u

1
1
1
ln u  ln x  C  ln u  ln x  C  ln 1  y 2  ln x  C  ln  1  y 2   ln x  C
2
2
2



1






 x dx
 1  y2  e
ln x C
u  1  y2 
du
du
 2y  dy 
dy
2y
 
 1  y 2  Ax and since y 2  1  1  12  2A  A 
2
2
1  y2 
Therefore,
2
1 2
x or x  2  1  y 2 or y 
x 1 .
2
2
27a.
6
4
2
3
2
1
1
2
3
4
6
4
2
3
2
1
1
 
2
3
4

For y 0  7, lim y t  5
i.
t 
ii.
 
iii.
6

For y 0  4, lim y t  5
t 
For y  0  
1
, lim y t   
2 t 
27b. The line y = 1 is a horizontal
asymptote for any solution curve for
which y  0   5 since lim y t  1
4
2
t 

for these solution curves.
3
27c.
2
1
1
2
3
4
dy
dy
dy
dy
 y 6  y   5 
 6y  y 2  5 
   y  5  y  1  
 dx
dx
dx
dx
 y  5 y  1
1
 y  5 y  1

A
B
1
1

 1  A  y  1  B  y  5  A 
and B   (Do the math!)
y 5 y 1
4
4
 1
 14 
y 5
1
1
  4 
 4x  D
dy   dx  ln y  5  ln y  1  x  C  ln
y 5 y 1
4
4
y

1


y 5
y 5
y 1

 e 4x D 
 Ae 4x or
 ae 4x (could stop here, I guess.)
y 1
y 1
y 5


 y  5  yAe 4x  Ae 4x or y  1  yAe 4x  5Ae 4x
 Ae 4x  5  yAe 4x  y or 5Ae 4x  1  yAe 4x  y
y 
Ae 4x  5
5Ae 4x  1
or y 
Ae 4x  1
Ae 4x  1
28a. y   0   1  y  0.25   2   1  0.25   1.75
y   0.25   1.25  y  0.5   1.75   1.25  0.25   1.4375
y   0.5   1.5625  y  0.75   1.4375   1.5625  0.25   1.046875
y   0.75   1.953125  y 1   1.046875   1.953125  0.25   0.55859375
28b. y 1   0.34670229
28c.
dy
dy
 y 3
 dx 
dx
y 3

dy
 dx  ln y  3  x  C  y  3  e x C
y 3

 y  3  Ae x  y  Ae x  3 and sin ce y  0   2  2  Ae 0  3  A  1
 y  e x  3  y 1   3  e  0.281718
28d. The approximations determined in parts (a) and (b) are too high since y   y   y  3 , which
means y   0 for all x on the interval {0, 1}, which means the graph of y is concave down on
this interval, which means the Euler approximations will lie above the actual graph.
29a. 30,000
29b. 15,000
29c. The solution to the IVP P   0.075P 30  P  and P  0   5 (remember that P is in
thousands) is P t  
30
30
1  5e 0.225t
8
1  5e 0.225t
15
 1  5e 0.225t
4
11
5e 0.225t 
4
11
e 0.225t 
20
 11 
0.225t  ln 

 20 
.
t  2.66
So the population reaches 8000 approximately 2.66 years(?) after time t = 0..
30. Assuming that the drink obeys Newton’s law of cooling, we know that the temperature, y,
of the drink is given by y T  Ae kt , where T is the environemnt temperature, 70 here.
At time t = 0, the temperature is y  0   70  A  35  70  A  A  35 .
70  35e
35e
e
k 30 
k 30 
k 30 
 50
 20

4
7
4
30k  ln  
7
4
ln  
7
k     0.01865
30
 y  70  35e 0.01865t
a.
y  75   70  35e 0.0186575  61.6
b.
70  35e 0.01865t  65  35e 0.01865t  5  e 0.01865t 
t  104.3 minutes
1
1
 0.01865t  ln  
7
7