Selected Solutions to SP 2013 BC 1,2 Semester Review
1a.
1b.
1c.
1d.
1e.
3
3
6
x2  3 0  3
1
0 2
2
x  1  2x 3
x  1
2
x3
2
x
1  cos 3x 
1  cos 3x  3
lim
 lim
  0 3  0
x
3x
1
x 0
x 0
sin 3x 
sin 3x  cos 2x 
sin 3x 
cos 2x  3
2x
3 3
lim
 lim

 lim


  1 1 1  
1
1
2
2 2
x 0 tan 2x 
x 0
sin 2x  x 0 3x
sin 2x 
lim
3x  6x
1

x
3
 lim
4
5

4
4x  5x
x x3 0  0
lim
 x  lim

0
1
1
20
x  2x 4  1
x 
2
x4
x4
x 5
x 5
lim
 lim 
 1
x 5
x 5 5  x
x 5
1
3
1f.
x  4 x  3  lim x  4   3  4   7
x 2  x  12
lim
 lim
2x  6
2
2
2
x 3
x 3
x 3
2  x  3
1g.
x 4
 
x 2 2  x
1h.
1i.
2.
lim
lim
x 3
x 2
lim
 0
x 2  2x
sin3  4 
3
x 3

x 2 x  x  2 
 lim
3
 sin  4  
sin3  4  64
  64  13  64
 lim

 64 lim 
3
1
4 
 0 64
 0 


Only point where f might not be continuous is at x = 1.
(1)
Does f(1) exist? Yes, f(1) = 2
(2)
Does lim f (x ) exist? Yes
x 1
lim f (x )  lim x 2  3  12  3  2
x 1
x 1
 
lim f (x )  lim 2x  4  2  4  2
x 1
x 1
Since lim f (x )  lim f x , lim f (x ) exists.
x 1
(3)
x 1

x 1

Does lim f (x )  f 1 . No, since lim f (x )  2  2  f 1 .
x 1
x 1
Since f is not continuous is at x = 1, f is not continuous.
BC 1,2 Semester Review Solutions p. 1
3.
Evaluating f(1) = 14 – 6 + 2 = -3, and f(2) = 24 – 12 + 2 = 6, we see that f(1) < 0, f(2) > 0.
Since f is continuous on the closed interval [1, 2], it satisfies the hypotheses of the IVT,
which then guarantees that f hits every value between –3 and 6, including 0.
4.
Determine values of x near x = 2 for which x 3  3  5  0.1 or x 3  8  0.1 .


x 3  8  0.1  0.1  x 3  8  0.1  7.9  x 3  8.1  1.992  x  2.008


So, if x  1.992,2.008  , then x 3  3  5  0.1 .
5.
8
6
4
2
1.0
0.5
0.5
1.0
1.5
2.0
Determine endpoints of segments that form approximation for y, starting with y(–1) = –5.
First step: 1  x  0
Second step: 0  x  1
Third step: 1  x  2
ynew  yold  m  x
ynew  yold  m  x
ynew  yold  m  x
 5  8  1
3
 3  31
6
BC 1,2 Semester Review Solutions p. 2
 6  0 1
6
6
(1, 6)
(2, 6)
4
2
1.0
(0, 3)
0.5
0.5
1.0
1.5
2
4
(–1, –5)
f a  h   f a 
6.
f   a   lim
7a.
f  x   x 2  3x  1
h 0
h
OR f  a   lim
x a
f  x   f a 
x a
2

 x  h   3  x  h   1  x 2  3x  1
f   x   lim
h
h 0
x 2  2xh  h 2  3x  3h  1  x 2  3x  1
 

 lim 
h
h 0
2
2xh  h  3h
 lim
h
h 0
 lim 2x  h  3
h 0
 2x  3
7b.
g x   x  2
BC 1,2 Semester Review Solutions p. 3
2.0
x h 2  x 2 x h 2  x 2

h
x h 2  x 2
 x  h  2   x  2
g   x   lim
h 0
 lim
h 0 h
 lim
h 0 h
 lim
h 0

7c.
h x  
1
 x h 2  x 2
h
 x h 2  x 2
1
x h 2  x 2
2 x 2
1
x 2
1
1

h   x   lim x  h  2 x  2
h
h 0
1  x  2   x  h  2
 lim 
h 0 h
 x  h  2 x  2
1
h

h  0 h  x  h  2  x  2 
 lim
1
h  0  x  h  2  x  2 
 lim

8a.
1
2
 x  2


1
3


2

1 2
f  x   x  3x  4  x  3x  4
x  3x  4 3  2x  3 .
3
f is not differentiable at values of x for which f  is undefined, which occur when:
3
2
2
 f  x  
x 2  3x  4  0
 x  4  x  1  0
x  4 or x  1
8b.
g  x   x 2  3x  10 is not differentiable at values of x for which g  x   0 , since at
these values of x, the graph of g will have corners.
BC 1,2 Semester Review Solutions p. 4
x 2  3x  10  0
 x  5  x  2  0
x  5 or x  2
9.
Many answers are possible. The three types, with one example of each, are given below.
Corner
Cusp
y   x  2
y  x 2
Vertical Tangent
2
3
y   x  2
1
3
2
1.5
1
1.5
1.25
0.5
1
1
0.75
1
0.5
0.5
2
-0.5
0.25
-1
1
10a.
10b.
2
3
4
1
2
3
4
 2
  0

 3  
 3

s 2   s  0   2  1
 0 1
1
Average velocity over first 2 seconds:

20
2
3
Instantaneous velocity at t = 2 seconds is s  2  .
s  t  
1  t  1   t  1 
2
t  1
1

1
2
t  1
 s  2  
10c.
Since s  t  
11.
Use Nderiv on the calculator, f  1   0.173 .
2
t  1
Can also calculate
1
2
2  1 

1
9
can never be equal to 0, the particle is never at rest.
f 1  h   f  1 
h
for small values of h.
BC 1,2 Semester Review Solutions p. 5
3
4
h  0.01 :
f  1  h   f 1 
h
h  0.001 :
h
 sin 1.001

h
1
 0.166
0.001
0.001
f 1  h   f 1 
 sin 1.0001

1
0.0001
 0.172
1
0.0001
 0.173
1
2
  
 
dy 1
 sec x  sin e     sec x tan x  2 sin e  cos e  e 
dx 2 
y  sec x  sin2 e x  sec x  sin2 e x
2
13.
0.01
0.01
f  1  h   f 1 
h  0.0001 :
12.
 sin 1.01

x


1
x
2

x

f  x   ln x 2  3x  f 2   ln 22  3  2  ln 10 
f  x  
1
x 2  3x
 2x  3  f  2 
1
22  3  2
 2  2  3 
Therefore, the equation of the normal line is:
14a.
x
7
10
y  ln 10   
10
 x  2
7
14b.
15a.
f has a local maximum at x = 0.15 since f ′ changes from positive to negative at this point.
15b.
f is decreasing on the interval [0.15, 0.84] since f ′ is negative (nonpositive) on this
interval.
BC 1,2 Semester Review Solutions p. 6
15c.
f is concave down on the intervals (–0.3, 0.5) and (1.4, 2) since f ′ is decreasing (and thus f
′′ is negative) on these intervals.
15d.
f has inflection points at x = –0.3, x = –0.5, x = 1.4, and x = 2, since f ′ has local maximums
or minimums (and thus f ′′ changes sign) at these point.
BC 1,2 Semester Review Solutions p. 7
15e
.

0.3
0.15
0.5
0.84
1.4
2
16.
Critical points are points where the derivative of a function is either zero or undefined.
17.
g  x   ln x 2  3x  4

g  x  
1
2
x  3x  4
2x  3

 2x  3
For 0,3 ,
3
9 9

7
g    ln    4   ln  
2
4 2

4
g  0   ln  4 
0
x 2  3x  4
2x  3  0
x 
g 3  ln  9  9  4   ln  4 
3
2
7
Global (absolute) maximum value of g: ln(4); Global (absolute) maximum value of g: ln  
4
18.
f  x   3sin x  2 sin3 x
f   x   3cos x  6 sin2 x cos x


3
2
or x 
 3cos x 1  2 sin2 x
x 

2
or x 
3cos x  0
cos x  0

4
or x 
or 1  2 sin2 x  0
or
sin x  
3
5
7
or x 
or x 
on 0,2 
4
4
4
BC 1,2 Semester Review Solutions p. 8
1
2
0
f
f
+


4
2
-
3
4
5
4
+
inc dec inc
-
3
2
+
dec
7
4
-
2
+
inc dec inc
5
7
.
,x 
2
4
4

3
3
Local maxima occur at x  , x 
.
,x 
4
4
2
Now, check values at these points an at endpoints to determine global maxs/mins.
 
 
 3 
 5 
 7 
 3 
f    1; f    f 
  2; f 
 f 
   2; f 
  1;f  0   f 2   0
2
4
 4 
 4 
 4 
 2 
Local minima occur at x 

,x 
5
7
and x 
.
4
4

3
2 occurs at x 
and x 
.
4
4
Global minimum value of  2 occurs at x 
Global maximum value of
19.
1
f  x   x 1  x  x 1  x 2
1
1

1
x
f   x   1  x  2  x  1  x  2   1   1  x 
2 1x
2

 f has a critical point at x  1 since f  is undefined at this value of x .
1x 
x
2
3
0
2 1x
2 1  x   x  0
1
2  3x  0
f
+
-
2
3
f
inc
dec
x 
Since f  is positive for all x 
2
2
and negative for all  x  1 ,
3
3
2 2 1
2
2 3
f 

or
is the maximum value of f.
3
3
3
9
3
3
 
20.
y  x 3  sin x  y   3x 2  cos x  3x 2  cos x  0  x  0.5354 or x  0.5354
21.
f  x   x 4  6x 3  7x  3
BC 1,2 Semester Review Solutions p. 9
f   x   4x 3  18x 2  7
f   x   24x  36  f   0   36 and f  3  36
f   x   12x  36x
2
12x  36x  0
Since f  is negative at x  0, f  is concave down at
this point, meaning f  has a local max at x  0. Since
x  0 or x  2
f  is positive at x  3, f  is concave up at this point,
meaning f  has a local min at x  3.
2
12x  x  3  0
22.
g  f 1  g   a  
 g  9 
23.
1

f  f 1  a 

1
f f
1
 g  9 
1
f  5 
 g  9 
1
3
g  f 1  g   a  
 g  9 

9
Since f 5  9,f
1

f  f 1  a 
1

f  f 1  9 
1
9  5


To find f 1  9  , first we need to solve 2x 5  3x  1  9 for x .
2x 5  3x  1  9  x  f 1  9   1.175


f   x   10x 4  3  f  f 1  9   22.05
 g   9   0.454
24.
9x 2  6 y  x 2y  17
BC 1,2 Semester Review Solutions p. 10
9x 2  6 y  x 2y  17  18x  6 
dy
dy
 2xy  x 2 
dx
2 y dx
1

dy
dy
 x2 
 2xy  18x
dx
y dx
3



dy  3

 x 2   2x  y  9 

dx  y


2x  y  9  y
dy 2x  y  9 


or
3
dx
3x2 y
x2
y

a)
2
2
9 1  6 y  1  y  17

y 6 y 80
y 2


dy
dx
y 4 0
y  2 or
y 4
y  4 or y  16
1,4 

2 1  4  9  4
2
3  1 
4
dy
dx
1,16
 20

2 1 16  9  16
2
3  1 
16
 56
Equations of tangent lines are y  4  20  x  1  and y  16  56  x  1  .
b)
To have horizontal tangent line, must have 2x  y  9   0.
x  0 or y  9
2
2
9  0   6 y   0  y  17 or 9x 2  6 9  x 2  9   17
17
or 9x 2  18  9x 2  17
6
289
y 
or 18  17 no solution! 
36
 289 
Horizontal tangent line occurs only at  0,
.
36


y 
25a.
The function f(x) is not differentiable on any interval containing x = –1 or x = 2, so the
MVT cannot be applied here, meaning it can tell us nothing about f(x).
25b
Since the function f is continuous on a closed, bounded interval, the EVT guarantees that
f(x) has both a minimum and a maximum value of [-3, 5]. But since (-3, 5) is not a closed
interval, EVT guarantees nothing, and in fact the function does not have a maximum on
that interval.
BC 1,2 Semester Review Solutions p. 11
26.
Average value of h on [-1, 3] is
h 3  h  1 
3   1 

47   5 
4
 13 .
h   x   3x 2  6
3c 2  6  13  c  
7
3
Only the positive square root is on [–1, 3], so answer is c 
27.
The graph at right works for all three parts. It is not
continuous, so fails all the hypotheses. But it clearly hits
every y value between its endpoints (IVT), has an
instantaneous rate of change at least one point in the
interval that is equal to the average of change of the
function over the interval (MVT), and has a minimum and
maximum value (EVT),
28.
V 

3
r 2h and h  3r V   r 3 
We know
7
or
3
21
.
3
dV
dr
 3 r 2
dt
dt
dV
 12 and when h = 4, r = 4/3.
dt
2
 4  dr
dr
9
 12  3  


dt 4
 3  dt
 The radius is increasing at
29.
9
ft/sec.
4
In the diagram, the 300 is fixed, but the x and the y
dx
change. In fact, we are given that
is 8 ft/s.
dt
300
dy
dx
2
2
2
y  300  x  2y
 2x
dt
dt
dy
 2  500 
 2  400  8
dt
dy 32


dt
5
32
ft/sec .
Thus, the rate at which string is being let out is
5
BC 1,2 Semester Review Solutions p. 12
x
y
Now, if  is the angle of elevation to the ground, we have:
300
d
300 dx
tan   
 sec2  

x
dt
x 2 dt
2
 500  d 
300


8

4002
 400  dt
d
6


dt 625
Thus, the angle of elevation is decreasing at
30.
V   r 2h 
dV
dr
dh
 2 r
h  r 2
dt
dt
dt
Since the volume of the cylinder is fixed,
2
dr
2     5   0.25 

dt
25
dr

 20
4
dt
dr
5


dt 16
5
ft/sec .
Thus, the radius is changing at
16
6
or about 0.096 rad/sec. .
625
dV
 0.
dt
 0  2  5 
31.
Linearization, L, of g at x = 4:
L  x   g   4  x  4   g  4   L  x   3  x  4   6 or L  x   3x  18
L  4.2   3  4.2   18  5.4  g  4.2   5.4
Since g   x   5  0 for 2  x  6, g is concave up for 4  x  4.2.
Thus, the linearization of g at x  4, which is the tangent line to g at x  4,
will lie below the graph of y  g  x  , meaning that the estimate will be less than
the actual value of g  4.2  .
32.
1  f   x   4 and f 2   5  5  1  x  2   f  x   5  4  x  2 
 5  1 7  2  f 7   5  4 7  2
 10  f  7   25
BC 1,2 Semester Review Solutions p. 13
33.
Suppose the printed part of the poster has dimensions x by y, with x going from left to
50
right and y going from top to bottom. Thus, xy  50 or y 
. . Since the side margins
x
are 2 inches, the width of the whole poster is going to be x + 4, and since the top and
bottom margins are 4 inches, the height will be y + 8. We need to minimize area.
 50

400
A   x  4  y  8  A  
 4   y  8  A 
 4y  32 .
y
 y

Differentiating, we find that A  
400
4.
y2
Setting this to zero yields y = 10, which gives a minimum because A 
800
y3
, which is
greater than 0 when y = 12. Then x is 50/10 = 5, the width is 5 + 4 = 9 inches, and the
height is 10 + 8 = 18 inches.
34.
(a)
The critical points of f occur at values of x for which f  is 0 or undefined.
f   5x 4  15ax 2
5x 4  15ax 2  0


5x 2 x 2  3a  0
x 2  0 or x 2  3a
x  0 or x   3a
x  0 : f 0  b
x  3a : f
 

3a 
3a

5
 5a

3a

3
b
 9a 2 3a  15a 2 3a  b
 6a 2 3a  b

 
x   3a : f  3a   3a

5

 5a  3a

3
b
 9a 2 3a  15a 2 3a  b
 6a 2 3a  b


Critical points are  0, b  ,  3a , 6a 2 3a  b ,
34.
(b)

3a , 6a 2 3a  b .
Since the critical points of f occur at x  0, x  3a , x   3a ,
to have exactly one, either a = 0, so that all three critical points occur
at x = 0, or a > 0, so that the critical points at x  3a and x   3a
BC 1,2 Semester Review Solutions p. 14
do not exist. In either case, the coordinates of the critical point will be  0, b  .
can
To determine if this critical point a local maximum, local minimum, or neither, we
look again at the first derivative of f.
f   5x 4  15ax 2
 5x
2
x
2
 3a
0

f   x   0 for x  0 and a  0
f   x   0 for x  0 and a  0
f
+
f
inc
+
inc
Since f  DOES NOT change sign at x = 0, the critical point at x = 0 is neither
a local maximum, nor a local minimum.
34.
(c)
Since the critical points of f occur at x  0, x  3a , x   3a ,
to have exactly three, we must have a < 0 so that 3a is defined but not.
equal to 0. The coordinates of the three critical points will then be
 0, b  ,


3a , 6a 2 3a  b ,

3a , 6a 2 3a  b .
In part (b), we determined that the critical point at x = 0 is neither a local
maximum,
nor a local minimum. To determine if the other two critical points are local maxima,
local minima, or neither, we can look at the second derivative of f.
f   5x 4  15ax 2
f   20x 3  30ax





 60a  3a   30a  3a 
 30a  3a 

f   3a  20  3a
3
 30a  3a
f 


3a  20

3a

 30a 
 60a

3
 30a

3a 

3a  30a
3a


3a

Since a  0  30a 3a  0 , f  is negative at x   3a , which means the graph
of f
is concave down at x   3a , making this critical point a local maximum.
Since a  0  30a 3a  0 , f  is positive at x  3a , which means the graph
of f
is concave up at x  3a , making this critical point a local minimum.
BC 1,2 Semester Review Solutions p. 15
35.
36.
There are many such graphs. Critical
points are where the derivative is
zero or undefined. This can happen
when the function has a flat spot (x2).
It can also happen with the function
still being continuous with things like
corners (absolute value), cusps (x2/3),
and vertical tangents (x1/3). At right
is an example with each of these
types. There are yet others (like x
sin(1/x) ). Any graph that contains
three different such points will work
here.
1.0
0.5
5
4
3
2
1
1
2
3
4
5
0.5
1.0
 24  3x 
V x   x 
 16  2x 
2


V  x   3x 3  48x 2  192x
V   x   9x 2  96x  192
9x 2  96x  192  0
3  x  8 3x  8   0
x  8 or x 
8
3
Of course, x  8 is not possible.
Why?
8
V   x   18x  96 V     48
3
8
Since V  is negative at x  , V is concave down at
3
8
this point, meaning V is maximal at x  .
3
37.
Let h represent the height of the rectangle, and w represent the width. If the rectangle
has perimeter 36, h + w = 18, so h = 18 – w. If the rectangle is rotated around its height
h,
BC 1,2 Semester Review Solutions p. 16
then the cylinder swept out has height h and radius w, so the volume, V, is
V   w 2 18  w  .
To maximize the volume, we need to determine values of w for which the derivative of
the
volume function changes sign from positive to negative.
V w    w 2 18  w 
Of course, w  0 is not possible.
V  w   36  6 w V  12   36
 18 w 2   w 3
V  w   36 w  3 w 2
Since V  is negative at w  12, V is concave down at
this point, meaning V is maximal at w  8.
Thus, the maximum possible volume for the cylinder
36 w  3 w 2  0
3 w 12  w   0
2
is  12  18  12  or 864 in3 .
w  0 or w  12
38.
3x  y  216  y 
F  6x  4y
x
72
x
288
x
288
F  x   6x 
F  x   6 
F   x  
6
288
x
2
576
y
x
y
x
y
y
x2
x3
0
6
Why?
288
x2
x 2  48
x
 
F  4 3 
576
4 3
3
x
x
0
Since F  is positive at x  4 3, F is concave up at
this point, meaning F has a minimum value at x  4 3.
x 4 3
Choose the positive value of x
since x represents a dimension.
 
Dimensions: 3 4 3 or 12 3 m in direction with
2 fences,
72
4 3
or 6 3 m in direction with 4 fences.
Total fencing: 6  4 3  4  6 3  48 3 m .
38.
Revenue = (# of passengers)  (fare per passenger).
BC 1,2 Semester Review Solutions p. 17
If z is the number of 5 cent increases (or decreases) added onto the original fare
of $1.50, then the fare will become 1.50 + 0.05z and the number of passengers will
be 600 – 40z. To maximize revenue, R, we need to determine values of z for which the
derivative of the revenue function changes sign from positive to negative.
R  z    600  4z 1.50  0.05z 
R   z   4  R   7.5   4
 900  30z  2z 2
Since R  is negative at z  7.5,
R   z   30  4z
R is concave down at this point,
0  30  4z
meaning R is maximal at z  7.5.
z  7.5
So, if you assume not just 5 cent changes in the fare are possible, the maximal revenue is
generated when the fare is 1.50 + (0.05) (-7.5) or $1.12 and a half cent (well, there are
half-pennies in Great Britain!).
If you decide to assume only 5 cent changes in the fare are possible, you'd need to test
whether 7 or 8 five cent decreases produce the highest revenue.


R  7   600  40  7  1.50  0.05  7 
  880 1.15 

 1012


R  8   600  40  8  1.50  0.05  8 
  920 1.10 

 1012
Thus, either fare of $1.15 or $1.10 produces the maximum revenue of $1012 under the
assumption that only 5 cent changes in the fare are possible.
Note: The value of z that generates maximum revenue can be determined without using
calculus by using what we know about the location of the vertex of a quadratic function,
but since this IS a calculus class, one should be using calculus to solve the problem.
39.
a
lim
x 0
sin 3x 
sin 1 2x 
Z0
3cos 3x 
 lim
x 0
] 0
1
2
2
1  2x 
 lim
x 0
31
1
2
1  2  0 

2
3
2
BC 1,2 Semester Review Solutions p. 18
39.
39.
b 
c

ln  x  1   x
lim 
x 0  x  e x  1



Z0




]
0
Z0

1
1

x 1
 lim 
x 0  e x  1  x  e x







]

1

2
x  1


 lim 
x 0  e x  e x  x  e x
1

2







0
1
tan  
1
x 
lim x tan    lim
1
x 
 x  x 
x
1  1 
sec2     

x   x2  .
 lim
1
x 

x2
1
 lim sec2  
x 
x 
1
39.
(d)
This is an indeterminate form of type 1 . We have to use the logarithm method to
calculate this type of limit, remember?
BC 1,2 Semester Review Solutions p. 19
2x

3
a  lim  1  
x
x  
2x

3
ln  a   lim ln  1  
x
x  

3
 lim 2x ln  1  
x
x 

Z0

3
2ln  1  
x

 lim
1
x 
x
]
2
1
0
3


3  x2
1  
x

 lim
1
x 

x2
6
 lim
x  
3
1  
x

6
2x

3
ln  a   6  a  lim  1  
x
x  
40.
a
dy
dx
 et and
 8e 2t
dt
dt

40.
b 
 e6
dy 8e 2t

dx
et
dy
 8et or 8x
dx
 
dy
dx
 sin t   t  cos t  and
 2t sin t 2
dt
dt
 
2t sin t 2
dy


dx
sin t   t  cos t 
BC 1,2 Semester Review Solutions p. 20
41.
a
2 sin  4 
dy
dy
dx
 2cos   and
 4 sin  4  

d
d
dx
cos  
dy
dx
 
6
 2 
3
2 sin 

2
3 

2  2


 
3
cos  
2
6
 
 
 
 2
x    2 sin    1 and y    cos 
6
6
6
 3
1
3
y   2  x  1  or y  2x 
2
2
41.
b 

1

2

2
2
 dy 
 dx 
Speed  

 
d  
d  
2
2


  
 2  
  2cos      4 sin 

 6 
 3 



 3    2 3 
2
2
 15
41.
c
y  cos  4 
y  2 cos2 2   1

y  2 1  2 sin2  

2
1
2
2

 x  

y  2 1  2 
1

2 



2

x2 
y  2 1 
 1


2


y  1  2x 2 
42.
x4
2
1

dy
dx
 a   cos t   and
 a sin t 
dt
dt
2



BC 1,2 Semester Review Solutions p. 21
(a)
This curve has a horizontal tangent when


a sin t   0
dy
 0.
dt
sin t   0
t  k ,k  ¢
(b)
 k

k
x a
 sin  k    x  a 
,k  ¢
2
 2

dx
This curve has a vertical tangent when
0.
dt
1

a   cos t    0
2

cos t  
1
2
t 

3
 2 k , k  ¢


  2 k






3
x  a  3
 sin   2 k    x  a  
 k ,k  ¢
6 2

2
3




OR
 

   2 k

 




3
x  a  3
 sin    2 k    x  a   
 k ,k  ¢
 6

2
2
 3




43.
(a)
In the figure at the right, which is nowhere
close to being drawn to scale, x represents the
distance from the radar station to the rocket and y
represents the height of the rocket. At the
dx
 2000 mph and y = 4
specific time in question,
dt
miles, which means that x  41 miles due to the
Pythagorean Theorem.
x
y
θ
5
BC 1,2 Semester Review Solutions p. 22
52  y 2
dy
2y
dt
dy
4
dt
dy
dt
 x2
 2x
dx
dt
 41 2000 
 500 41
So, at this time, the rocket is rising at 500 41 or approximately 3202 mph.
43.
(b)
In the figure above at the right, θ represents the radar angle of elevation.
x
sec   
5
d  1 dx
sec   tan  

dt 5 dt
41 4 d  1

 2000 
5 5 dt 5
d  2500

dt
41
So, at this time, the radar angle of elevation is increasing at a rate of
approximately 390 radians per hour, which is equivalent to
approximately 0.108 radians per second.
44.
(a)
6 41
41
or
or
In the figure below at the right, x represents the thickness of the ice coating the
spherical ball. At the specific time in question, x = 2 and
is
5
2500
dV
in3
 10
, where V
dt
min .
the volume of the ice coating the spherical ball. Note that V is equal to (volume of
ice-coated spherical ball) minus (volume of spherical ball).
BC 1,2 Semester Review Solutions p. 23
3
3
4
4
 4  x    4
3
3
4
V   64  48x  12x 2  x 3  64
3
4
V   48x  12x 2  x 3
3
dV
4  dx
dx
dx 
   48
 24x
 3x 2

dt
3 
dt
dt
dt 
V 




x in
4 in
2 dx
4 
  48  24 2   3 2  
3 
 dt
4
dx
10   108 
3
dt
dx
5

dt
72
10 
So, at this time, the thickness of the ice is decreasing at a rate of
approximately 0.022 inches per minute.
44.
b 
5
or
72
2
SA  4  4  x 

SA  4 16  8x  x 2

 dx
dSA
dx 
 4  8
 2x

dt
dt 
 dt

dSA
5 
 4 8  2 2   

dt
 72 
dSA
10

dt
3
So, at this time, the outer surface area of the ice is decreasing at a rate of


10 in2
.
3 min
45.
In the figure below at the right, x represents Frodo's distance from the door and
s represents the length of Frodo's shadow. At the specific time in question,
dx
m
ds
5
x = 16 and
. Similar triangles can be used to determine
.
dt
sec
dt
BC 1,2 Semester Review Solutions p. 24
12
1.25

x s
s
12s  1.25x  1.25s
10.75s  1.25x
5
s 
x
43
ds
5 dx

dt 43 dt
ds 25

dt 43
12 m
1.25 m
s m
x m
So, at this time, the length of Frodo's shadow is decreasing at a rate of
approximately 0.58 meters per second.
46.
1
:
2
(a)
Since the width of the interval is 2, the 4 rectangles or trapezoids have width
(b)
 3

 1

1 0
 e  1   e 2  1   e 1  1   e 2  1    2.923




2 




  5
  7

  3
1  1
Midpoint :   e 4  1    e 4  1    e 4  1    e 4  1    4.323
 
 
 

2  
 
 
 


 3

 1

1 1
Trapezoid sum :   e 0  1  2  e 2  1   2 e 1  1  2  e 2  1   e 2  1   4.521





2 2 





The left-hand Riemann sum underestimates the definite integral since the graph of
Left-hand :





46.
25
or
43





f  x   e x  1 is increasing on the interval [0, 2]. The midpoint Riemann sum
underestimates and the trapezoidal approximations overestimates the definite integral
since the graph of f  x   e x  1 is concave up on the interval [0, 2].
46.
46.
(c)
(d)
2


0
40
j 1
47
2

 

e x  1 dx  e x  x   e 2  2  e 0  0  4.389


0
2j

2  40
 e  1  or

40 


40

j 1
j

1  20
e  1 

20 


Since the width of the entire interval is 120 seconds and you want 6 equal subintervals,
the width of each of these subintervals will be 20 seconds. In order to generate a
Riemann sum that provides the largest possible upper estimate for the length of the
BC 1,2 Semester Review Solutions p. 25
road, we must choose the greatest speed attained on each of the 6 equal subintervals.
To do this, our Riemann sum must be as follows.
20  44  35  44  35  44  44   4920
Therefore, the largest possible upper estimate for the length of the road is 4920 ft.
48.
Using n rectangles, the width of each rectangle is
edge at 1 
3k
. So we add up width times height for all n rectangles, and let n go to
n
n
infinity, we get : lim
n 
n
49.
lim
n 

i 1

k 1

2
3k 


 1
3 
3k 
 n
1


e


n 
n 
2
 


7  3  6i   5   4  lim 2
 
 n n  3
n 


n

i 1
.
2
 


7  3  6i   5   6 .
 
 n
n 


The latter sun can be represented by the integral
n
lim
n 

i 1
3
, and the kth rectangle has its right
n
2
 


6
i
7  3    5   4  lim
 
 n n 
n 


n

i 1
2
3

3
7x 2  5 dx OR


3
2
  2 


3
4
i
7     2 
 4 .

5

 2 
 n
n 


Here, the latter sum can be represented by the integral
Another possibility would be the integral
50.
(a)
(b)


2
 63 2

 x  5 dx OR

2  4
1
7( 3  6x )2  5 4 dx .


0
9
4
F  5   3  3   2   4
F  5   
y
9
4
Since F   f , F  2   1 and
F  2   1.
(c)
F has exactly one zero on [–6, 5],
which occurs at x = –2.
(d)
F is increasing when F   f is
positive, which occurs on the interval [–5,
3].
3
2
BC 1,2 Semester Review Solutions p. 26
3
3
x
2
(e)
F will have a local maximum when F   f changes from positive to negative,
which occurs at x = 3.
F will have inflection points when F   f changes from increasing to decreasing,
or vice versa, which occurs at x = –2, x = 0, and x = 1.
F is concave down when F   f is decreasing, which occurs on the
intervals [–2, 0] and [1, 5].
(f)
(g)
51.
These are both FTC problems:
FTC immediately gives sin10  x  .
(a)
b 
3x 1

4
t  1 dt 
ln x

3x 1
0
4
t  1 dt 
4
 3 3x  1   1 
1
x

ln  x 
0
ln x 
using the generalized form of the FTC:
52.
(a)
t 4  1 dt
4
d 
dx 

1

g x 
a

f t  dt   f g  x  g   x  .



Since y  2t and y  sin t  are both odd
y
functions, y  2t  sin t  is an odd function.
Thus,
52.
(b)
52.
(c)
5


5
2
5
9
1
2  x dx    2  2  
2
2

0
4
9
2
16  y 2 dy represents the area of a quarter-circle of radius 4.
Thus,
(a)
2
x
The graph of y  16  x 2 is a semicircle of radius 4, which means the integral

53.
2

2t  sin t  dt  0 .


0
4
16  y 2 dy 

2
 4
4
 4
Displacement is simply the integral of velocity:
BC 1,2 Semester Review Solutions p. 27
4

0
53.
(b)
4
t 3 3t 2 
64
8
t  3t dt 

 24   .
 
3
2 
3
3
0
2

Distance traveled is the integral of speed, which is the absolute value of velocity:

4
0
t 2  3t dt 

3
0


3t  t 2 dt 
3

4
3
t
2

 3t dt
4
3t 2 t 3 
t 3 3t 2 

  


2
3 
3
2 
0
3
 27
  64
 
27 

 9  
 24    9 

2 
 2
  3
 

9 8  9
   
2 3  2
19
3
Average value of a function over an interval is the integral of the function over the
interval

54.
1
divided by the width of interval. This produces the integral
2

2
0
x sin  x  dx , which
can
be evaluated using integration by parts with u  x  u   1 and v   sin  x   v   cos  x  .
1
2

2
0
2
x cos  x  
1
 
x sin  x  dx 
2
2

0

2
0
cos  x  dx
2
x cos  x   sin  x  


2

0

 2 cos 2   sin 2    0 cos  0   sin  0 
2
2
 1
2

BC 1,2 Semester Review Solutions p. 28
55.
(a)
The area enclosed by these
graphs is above the x-axis on the
interval (–5, 0) and below it on the
interval (0, 2). Thus, the positive
area enclosed by the graphs can
be determined by calculating the
following difference of integrals:

0
5
x  x  2  x  5  dx 

0

5
3

2
0
x  x  2  x  5  dx

2
x  3x  10x dx 
0
2

0

x 3  3x 2  10x dx
2
x4

x4


 x 3  5x 2    
 x 3  5x 2  
 4




  5  4
0


 625
   16

  0  0  0   
 125  125      8  20    0  0  0  
 4
   4



375

8
4
407

4
BC 1,2 Semester Review Solutions p. 29