08-035_06_AFSM_C06_001-034.qxd
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11:14 AM
Page 23
6.7 Rates of Change in Trigonometric
Functions, pp. 369–373
1. a) The average rate of change is zero in the
intervals of 0 , x , p, p , x , 2p.
b) The average rate of change is negative in the
p
p 3p
3p 5p
intervals of , x , ,
, x , 3p.
2
2 2
2. a) Two points where the instantaneous rate of
p
5p
change is zero are x 5 , x 5 .
4
4
b) Two points where the instantaneous rate of
p
5p
change is a negative value are x 5 , x 5 .
2
2
c) Two points where the instantaneous rate of
change is a positive value are x 5 0, x 5 2p.
3. Average rate of change for the interval
2 # x # 5:
020
0
` 5 ` ` 50
`
522
3
4. 4
3
2
p
3
p
2
p
6
2
3
20.27
5
0.5236
8 20.5157
d)
p
5p
#x#
2
4
Average rate of change 5
20.932 2 2.73
5p
p
2
4
2
23.662
2.356
8 21.554
5
5.
6
4
2
y
x
0
–2
–4
–6
p
2
p
3p
2
2p
p
3p
p
5p
intervals of 0 , x , , p , x , .
2
2
b) The average rate of change is negative in the
0
p
–1
2p
intervals of 0 , x , , p , x , .
4
4
c) The average rate of change is positive in the
5p
4
–2
intervals of
p
2
Average rate of change 5 p
2
20
0.73
1.57
8 0.465
5
p
p
#x#
6
2
Average rate of change 5
2.73 2 2.73
p
p
2
6
2
0
1.047
50
5
Advanced Functions Solutions Manual
p
p 5p
3p
,x, ,
,x, .
4
2 4
2
y
6.
2.73 2 2
b)
2.73 2 3
Average rate of change 5 p p
2
a) The average rate of change is zero in the
1
a) 0 # x #
p
p
#x#
3
2
5p
intervals of 2 2 , x , 2 , 2 , x , 2 .
c) The average rate of change is positive in the
p
c)
8
6
4
2
0
–2
–4
–6
x
p
2
p
3p
2
2p
a) Two points where the instantaneous rate of
change is zero are x 5 14, x 5 34.
b) Two points where the instantaneous rate of
change is a negative value are x 5 0, x 5 1.
c) Two points where the instantaneous rate of
change is a positive value are x 5 12, x 5 32.
6-23
08-035_06_AFSM_C06_001-034.qxd
7/22/08
7. a) y 5 6 cos (3x) 1 2 for
4:11 PM
Page 24
p
#x#p
4
p
For x 5 ,
4
6.25 2 6.25
p
2p
4
0
22.3562
50
8. The tip is at its minimum height at t 5 0.
Normally the sine function is 0 at 0, so the function
in this case is translated to the right by 14 of its
period. The propeller makes 200 revolutions per
1
second, so the period is 200
. The amplitude of the
function is the length of the propeller, which is
positive. Assume that it is 1 m for this exercise.
Then the function that describes the height of the
1
tip of the propeller is h(t) 5 sin (400p (t 2 800
)).
The graph of this function is shown below.
5
p
y 5 6 cos a3a bb 1 2
4
y 8 22.2426
For x 5 p,
y 5 6 cos (3(p)) 1 2
y 5 24
Average rate of change 5
22.2426 2 (24)
p
2p
4
1.7574
22.3562
8 20.7459
p
1
b) y 5 25 sin a xb 2 9 for # x # p
2
4
p
For x 5 ,
4
1 p
y 5 25 sin a a bb 2 9
2 4
y 8 210.9134
For x 5 p,
1
y 5 25 sin a pb 2 9
2
y 5 214
210.9134 2 (214)
Average rate of change 5
p
2p
5
4
3.0866
5
22.3562
8 21.310
p
1
c) y 5 cos (8x) 1 6 for # x # p
4
4
p
For x 5 ,
4
1
p
y 5 cos a8a bb 1 6
4
4
y 5 6.25
For x 5 p,
1
y 5 cos (8p) 1 6
4
y 5 6.25
6-24
Average rate of change 5
3.2
2.4
1.6
0.8
0
–0.8
–1.8
–2.4
–3.2
h(t)
t
0.001 0.002 0.003 0.004
1
8 0.0033
300
From the graph, it is clear that the instantaneous
1
rate of change at t 5 300
is negative.
9. a) The axis is at 20.2. So, the equation of the axis
is y 5 20.2 and the amplitude is 4.5.
2p
p
k5
5
24
12
p
R(t) 5 4.5 cos a tb 1 20.2
12
b) fastest: t 5 6 months, t 5 18 months,
t 5 30 months, t 5 42 months;
slowest: t 5 0 months, t 5 12 months,
t 5 24 months, t 5 36 months, t 5 48 months
c) For 5 # t # 7
t55
p
R(5) 5 4.5 cos a (5)b 1 20.2
12
R(5) 8 21.364
t57
t5
Chapter 6: Trigonometric Functions
R(7) 5 4.5 cos a
7/22/08
4:11 PM
Page 25
p
(7)b 1 20.2
12
R(7) 8 19.035
Use the points (5, 21.364) and (7, 19.035).
19.035 2 21.364
22.329
8
8 21.164
725
2
8 1.164 mice per owl> s
10. a)
y
4
3
2
1
x
0 4 8 12 16 20 24
–1
The instantaneous rate of change appears to be at its
greatest at 12 hours.
i) For 11 # t # 13
t 5 11
p
y 5 0.5 sin a (11)b 1 4
6
y 5 3.75
t 5 13
p
y 5 0.5 sin a (13)b 1 4
6
y 5 4.25
Use the points (11, 3.75) and (13, 4.25).
4.25 2 3.75
0.5
5
5 0.25 t> h
13 2 11
2
ii) For 11.5 # t # 12.5
t 5 11.5
p
y 5 0.5 sin a (11.5)b 1 4
6
y 8 3.8706
t 5 12.5
p
y 5 0.5 sin a (12.5)b 1 4
6
y 8 4.1294
Use the points (11.5, 3.8706) and (12.5, 4.1294).
4.1294 2 3.8706
8 0.2588 t> h
12.5 2 11.5
iii) For 11.75 # t # 12.25
t 5 11.75
p
y 5 0.5 sin a (11.75)b 1 4
6
y 8 3.9347
Advanced Functions Solutions Manual
t 5 12.25
p
y 5 0.5 sin a (12.25)b 1 4
6
y 8 4.0653
Use the points (11.75, 3.9347 and (12.25, 4.0653).
0.1306
4.0653 2 3.9347
5
5 0.2612 t> h
12.25 2 11.75
0.5
b) The estimate calculated in part iii) is the most
accurate. The smaller the interval, the more accurate
the estimate.
y
11. a)
8
Distance from rest
position (cm)
08-035_06_AFSM_C06_001-034.qxd
4
0
–4
x
0.2 0.4 0.6 0.8 1.0
–8
Time (s)
b) half of one cycle
27.2 2 7.2
c)
5 214.4 cm> s
120
d) The bob is moving the fastest when it passes
through its rest position. You can tell because the
images of the balls are farthest apart at this point.
e) The pendulum’s rest position is halfway between
the maximum and minimum values on the graph.
Therefore, at this point, the pendulum’s instantaneous rate of change is at its maximum.
p
12. h(t) 5 sin a tb
5
a) For 0 # t # 5,
p
h(0) 5 sin a (0)b 5 0
5
p
h(5) 5 sin a (5)b 5 0
5
020
50
520
b) For 5.5 # t # 6.5
t 5 5.5
p
h(5.5) 5 sin a (5.5)b
5
h(5.5) 5 20.309
p
h(6.5) 5 sin a (6.5)b
5
h(6.5) 8 20.809
6-25
08-035_06_AFSM_C06_001-034.qxd
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11:14 AM
Page 26
Use the points (5.5, 20.309) and (6.5, 20.809).
20.809 2 (20.309)
20.5
5
5 20.5 m> s
6.5 2 5.5
1
13. a)
u
0.20
0.15
0.10
0.05
t
0 1 2 3 4 5 6 7 8
–0.05
–0.10
–0.15
–0.20
b) When t 5 0, u 5 0. When t 5 1, u 5 0.2. The
average rate of change is 0.2 radians> s.
c) Answers may vary. For example, from the graph, it
appears that the instantaneous rate of at t 5 1.5 is
about 2 23 radians> s.
d) The pendulum speed seems to be the greatest for
t 5 0, 2, 4, 6, and 8.
14. Answers may vary. For example, for x 5 0, the
instantaneous rate of change of f(x) 5 sin x is
approximately 0.9003, while the instantaneous rate
of change of f(x) 5 3 sin x is approximately 2.7009.
p
4
(The interval 2 , x ,
p
4
was used.) Therefore, the
instantaneous rate of change of f(x) 5 3 sin x is at its
maximum three times more than the instantaneous
rate of change of f(x) 5 sin x. However, there are
points where the instantaneous rate of change is the
p
same for the two functions. For example, at x 5 , it
2
is 0 for both functions.
15. a) By examining the graph of f(x) 5 sin x, it
appears that the instantaneous rate of change at the
given values of x are 21, 0, 1, 0, and 21.
b)
y
4
2
0
– 3p –p– p
2
2
–4
–6
x
p p 3p
2
2
The function is f (x) 5 cos x. Based on this
information, the derivative of f (x) 5 sin x is
cos x.
6-26
16. a) By examining the graph of f (x) 5 cos x, it
appears that the instantaneous rate of change at the
given values of x are 0, 1, 0, 21, and 0.
b)
y
4
2
0
– 3p –p– p
2
2
–4
–6
x
p p 3p
2
2
The function is f (x) 5 2sin x. Based on this
information, the derivative of f (x) 5 cos x is
2sin x.
Chapter Review, pp. 376–377
1. Circumference 5 2pr 5 2p(16) 5 32p
33
x
5
32p
2p
(32p)(x) 5 (33)(2p)
(32p)(x) 5 66p
66p
x5
32p
33
x5
16
2. Circumference 5 2pr 5 2p(75) 5 150p
14p
x
5 15
150p
2p
(2p)(x) 5 (150p)a
14p
b
15
(2p)(x) 5 140p2
x 5 70p
3. a) 20° 5 20° 3 a
p radians
p
b 5 radians
180°
9
p radians
25p
b) 250° 5 250° 3 a
b5
radians
180°
18
p radians
8p
c) 160° 5 160° 3 a
b5
radians
180°
9
p radians
7p
d) 420° 5 420° 3 a
b5
radians
180°
3
p
4. a) radians;
4
p
180 °
radians 3 a
b 5 45°
4
p radians
Chapter 6: Trigonometric Functions