CHAPTER 5:
Derivatives of Exponential and
Trigonometric Functions
Review of Prerequisite Skills,
pp. 224–225
1
32
1
5
9
2
5
32 R 2
b. 325 5 Q "
5 22
54
1
2
c. 2723 5 3
2
Q "27 R
1
5 2
3
1
5
9
3 2
2 22
d. a b 5 a b
3
2
9
5
4
2. a. log5 625 5 4
1
b. log4
5 22
16
c. logx 3 5 3
d. log10 450 5 w
e. log3 z 5 8
f. loga T 5 b
3. a.
y
3
2
1
1. a. 322 5
–3 –2 –1 0
–1
–2
–3
0 5 log10 (x 1 2)
100 5 x 1 2
x 5 21
The x-intercept is (21, 0).
b.
10 y
8
6
4
2
–8 –6 –4 –2 0
–2
x
2 4 6 8
An exponential function is always positive, so there
is no x-intercept.
y
4. a. sin u 5
r
x
b. cos u 5
r
y
c. tan u 5
x
5. To convert to radian measure from degree
p
x
1 2 3
The x-intercept occurs where y 5 0.
Calculus and Vectors Solutions Manual
measure, multiply the degree measure by 180°.
p
5 2p
a. 360° 3
180°
p
p
5
b. 45° 3
180°
4
p
p
52
c. 290° 3
180°
2
p
p
5
d. 30° 3
180°
6
3p
p
5
e. 270° 3
180°
2
2p
p
52
f. 2120° 3
180°
3
p
5p
5
g. 225° 3
180°
4
5-1
11p
p
5
180°
6
6. For the unit circle, sine is associated with the
y-coordinate of the point where the terminal arm of
the angle meets the circle, and cosine is associated
with the x-coordinate.
a. sin u 5 b
b
b. tan u 5
a
c. cos u 5 a
p
d. sin a 2 ub 5 a
2
p
e. cos a 2 ub 5 b
2
f. sin (2u) 5 2b
7. a. The angle is in the second quadrant, so cosine
and tangent will be negative.
12
cos u 5 2
13
5
tan u 5 2
12
b. The angle is in the third quadrant, so sine will be
negative and tangent will be positive.
sin2 u 1 cos2 u 5 1
4
sin2 u 1 5 1
9
5
sin2 u 5
9
!5
sin u 5 2
3
sin u
tan u 5
cos u
!5
5
2
c. The angle is in the fourth quadrant, so cosine
will be positive and sine will be negative.
Because tan u 5 22, the point (1, 22) is on the
terminal arm of the angle. The reference triangle for
this angle has a hypotenuse of "22 1 12 or !5.
2
sin u 5 2
!5
1
cos u 5
!5
d. The sine is only equal to 1 for one angle between
h. 330° 3
p
p
50
2
cos
p
tan 2 is undefined
2p
8. a. The period is 2 or p. The amplitude is 1.
2p
b. The period is 1 or 4p. The amplitude is 2.
2
2p
c. The period is p or 2. The amplitude is 3.
2p
p
d. The period is 12 or 6 . The amplitude is 27.
e. The period is 2p. The amplitude is 5.
f. The period is 2p. Because of the absolute value
being taken, the amplitude is 32.
2p
9. a. The period is 2 or p. Graph the function from
x 5 0 to x 5 2p.
4
y
3
2
1
x
0
p
2
p
3p
2
2p
b. The period is 2p, so graph the function from
x 5 0 to x 5 4p.
3
y
2
1
0
–1
x
p
2
p
3p 2p 5p 3p 7p 4p
2
2
2
–2
–3
10. a. tan x 1 cot x 5 sec x csc x
LS 5 tan x 1 cot x
sin x
cos x
5
1
cos x
sin x
sin2 x 1 cos2 x
5
cos x 1 sin x
0 and p, so u 5 2 .
5-2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
1
cos x 1 sin x
RS 5 sec x 1 csc x
1
1
5
?
cos x sin x
1
5
cos x sin x
Therefore, tan x 1 cot x 5 sec x csc x.
sin x
b.
5 tan x 1 sec x
1 2 sin2 x
sin x
LS 5
1 2 sin2 x
sin x
5
cos2 x
RS 5 tan x sec x
sin x
1
5
?
cos x cos x
sin x
5
cos2 x
sin x
5 tan x sec x.
Therefore,
1 2 sin2 x
11. a. 3 sin x 5 sin x 1 1
2 sin x 5 1
1
sin x 5
2
p 5p
x5 ,
6 6
b. cos x 2 1 5 2cos x
2 cos x 5 1
1
cos x 5
2
p 5p
x5 ,
3 3
5
5.1 Derivatives of Exponential
Functions, y 5 e x, pp. 232–234
1. You can only use the power rule when the term
containing variables is in the base of the exponential
expression. In the case of y 5 e x, the exponent
contains a variable.
2. a. y 5 e 3x
dy
5 3e 3x
dx
b. s 5 e 3t25
ds
5 3e 3t25
dt
Calculus and Vectors Solutions Manual
c. y 5 2e 10t
dy
5 20e 10t
dt
d. y 5 e 23x
dy
5 23e 23x
dx
2
e. y 5 e 526x1x
dy
2
5 (26 1 2x)e 526x1x
dx
f. y 5 e "x
dy
1 "x
5
e
dx
2 !x
3
3. a. y 5 2e x
dy
3
5 2(3x 2 )e x
dx
3
5 6x 2e x
3x
dy
d(xe )
5
b.
dx
dx
5 (x)(3e 3x ) 1 (e 3x )(1)
5 3xe 3x 1 e 3x
5 e 3x (3x 1 1)
3
e 2x
x
3
3
23x 2e 2x (x) 2 e 2x
f r (x) 5
x2
d. f(x) 5 !xe x
1
f r (x) 5 "xe x 1 e x a
b
2 !x
2
e. h(t) 5 e t 1 3e 2t
2
hr (t) 5 2te t 2 3e 2t
e 2t
f. g(t) 5
1 1 e 2t
2e 2t (1 1 e 2t ) 2 2e 2t (e 2t )
g r(t) 5
(1 1 e 2t )2
2t
2e
5
(1 1 e 2t )2
1
4. a. f r (x) 5 (3e 3x 2 3e 23x )
3
5 e 3x 2 e 23x
f r (1) 5 e 3 2 e 23
c. f(x) 5
1
b. f(x) 5 e 2x 1 1
f r (x) 5 e 2x 1 1 a
1
b
(x 1 1)2
f r (0) 5 e 21 (1)
1
5
e
1
5-3
c. hr(z) 5 2z(1 1 e 2z ) 1 z 2 (2e 2z )
hr(21) 5 2(21)(1 1 e) 1 (21)2 (2e 1 )
5 22 2 2e 2 e
5 22 2 3e
2e x
5. a. y 5
1 1 ex
dy
(1 1 e x )2e x 2 2e x (e x )
5
dx
(1 1 e x )2
dy
2(2) 2 2(1)(1)
5
dx
22
1
5
2
When x 5 0,
At the point (1, e 21 ), the slope is e 21 (0) 5 0. The
equation of the tangent line at the point A is
1
y 2 e 21 5 0(x 2 1) or y 5 e .
8. The slope of the tangent line at any point on the
curve is
5 (2x 2 x 2 )(e 2x )
2x 2 x 2
5
.
ex
Horizontal lines have slope equal to 0.
dy
the slope of the tangent is 12.
The equation of the tangent is y 5 12x 1 1, since the
y-intercept was given as (0, 1).
b.
c. The answers agree very well; the calculator does
not show a slope of exactly 0.5, due to internal
rounding.
6. y 5 e 2x
dy
5 2e 2x
dx
When x 5 21,
dy
5 2e. And
dx
when x 5 21, y 5 e.
The equation of the tangent is y 2 e 5 2e(x 1 1)
or ex 1 y 5 0.
3
2
1
–3 –2 –1 0
–1
–2
–3
y
x
1 2 3
7. The slope of the tangent line at any point is
given by
dy
5 (1)(e 2x ) 1 x(2e 2x )
dx
5 e 2x (1 2 x).
5-4
dy
5 2xe 2x 1 x 2 (e 2x )
dx
We solve 5 0
dx
x(2 2 x)
5 0.
ex
Since e x . 0 for all x, the solutions are x 5 0 and
x 5 2. The points on the curve at which the tangents
4
are horizontal are (0, 0) and (2, e2).
5 x
x
9. If y 5 (e5 1 e 25 ), then
2
5 1 x
1 x
yr 5 a e5 2 e25 b, and
2 5
5
5 1 x
1 x
ys 5 a e 5 1 e 25 b
2 25
25
1 5 5x
x
5
c (e 1 e 25 )d
25 2
1
5 y.
25
10. a. y 5 e 23x
dy
5 23e 23x
dx
d 2y
5 9e 23x
dx 2
d 3y
5 227e 23x
dx 3
d ny
b. n 5 (21)n (3n )e 23x
dx
dy
d(23e x )
5
11. a.
dx
dx
5 23e x
d 2y
5 23e x
dx 2
d(xe 2x )
dy
5
b.
dx
dx
5 (x)(2e 2x ) 1 (e 2x )(1)
5 2xe 2x 1 e 2x
5 e 2x (2x 1 1)
Chapter 5: Derivatives of Exponential and Trigonometric Functions
d 2y
5 e2x (2) 1 (2x 1 1)(2e2x )
dx2
5 4xe2x 1 4e2x
d(e x (4 2 x))
dy
c.
5
dx
dx
x
5 (e )(21) 1 (4 2 x)(e x )
5 2e x 1 4e x 2 xe x
5 3e x 2 xe x
5 e x (3 2 x)
d 2y
5 e x (21) 1 (3 2 x)(e x )
dx2
5 2e x 2 xe x
5 e x (2 2 x)
12. a. When t 5 0, N 5 1000330 1 e 04 5 31 000.
dN
1 2t
100 230t
b.
5 1000 c 0 2 e 30 d 5 2
e
dt
30
3
dN
100 223
c. When t 5 20h,
52
e 8 217 bacteria> h.
dt
3
t
d. Since e 230 . 0 for all t, there is no solution to
dN
dt 5 0.
Hence, the maximum number of bacteria in the
culture occurs at an endpoint of the interval of
domain.
5
When t 5 50, N 5 1000330 1 e 23 4 8 30 189.
The largest number of bacteria in the culture is
31 000 at time t 5 0.
e. The number of bacteria is constantly decreasing
as time passes.
ds
1
1 t
5 160a 2 e 24 b
13. a. v 5
dt
4
4
t
5 40(1 2 e24 )
dv
1 t
t
5 40a e 24 b 5 10e 24
b. a 5
dt
4
t
t
v
From a., v 5 40(1 2 e 24 ), which gives e 24 5 1 2 40.
Thus, a 5 10a1 2
1
v
b 5 10 2 v.
40
4
c. vT 5 lim v
tS`
24t
vT 5 lim 40(1 2 e )
tS`
1
5 40 lim a1 2 4t b
tS`
e
1
t 5 0
tS` e4
The terminal velocity of the skydiver is 40 m> s.
5 40(1), since lim
d. 95% of the terminal velocity is
95
(40) 5 38 m> s.
100
To determine when this velocity occurs, we solve
t
40(1 2 e 24 ) 5 38
38
t
1 2 e 24 5
40
1
t
e 24 5
20
t
e4 5 20
t
and 5 ln 20,
4
which gives t 5 4 ln 20 8 12 s.
The skydiver’s velocity is 38 m> s, 12 s after jumping.
The distance she has fallen at this time is
S 5 160(ln 20 2 1 1 e 220 )
1
5 160 aln 20 2 1 1 b
20
8 327.3 m.
x
14. a. i. Let f(x) 5 ( 1 1 1x) . Then,
x
f(x)
1
10
2
2.5937
100
2.7048
1000
2.7169
10 000
2.7181
x
So, from the table one can see that lim (1 1 1x) 5 e.
1
x
xS`
ii. Let f(x) 5 (1 1 x) .
x
f(x)
20.1
2.8680
20.01
2.7320
20.001
2.7196
20.0001
2.7184
?
?
0.0001
2.7181
0.001
2.7169
0.01
2.7048
0.1
2.5937
1
So, from the table one can see that lim (1 1 x)x 5 e.
xS0
That is, the limit approaches the value of
e 5 2.718 281 828c
b. The limits have the same value because as
1
x S `, x S 0.
Calculus and Vectors Solutions Manual
5-5
15. a. The given limit can be rewritten as
eh 2 1
e 01h 2 e 0
5 lim
lim
hS0
h
hS0
h
This expression is the limit definition of the derivative
at x 5 0 for f(x) 5 e x.
e 01h 2 e 0
d
cf r(0) 5 lim
hS0
h
dex
Since f r(x) 5 dx 5 e x, the value of the given limit
is e 0 5 1.
e 21h 2 e 2
b. Again, lim
is the derivative of e x at
hS0
h
x 5 2.
e 21h 2 e 2
5 e2.
Thus, lim
hS0
h
dy
d 2y
5 Ame ex and 2 5 Am 2e ex.
16. For y 5 Ae ex,
dt
dt
Substituting in the differential equation gives
Am 2e ex 1 Ame ex 2 6Ae ex 5 0
Ae ex (m 2 1 m 2 6) 5 0.
ex
Since Ae 2 0, m 2 1 m 2 6 5 0
(m 1 3)(m 2 2) 5 0
m 5 23 or m 5 2.
d 1 x
d
sinh x 5
c (e 2 e 2x ) d
17. a.
dx
dx 2
1
5 (e x 1 e 2x )
2
5 cosh x
1 x
d
cosh x 5 (e 2 e 2x )
b.
dx
2
5 sinh x
sinh x
,
c. Since tanh x 5
cosh x
d
tanh x
dx
d
d
(
sinh x) (cosh x) 2 (sinh x)(dx cosh x)
dx
5
(cosh x)2
5
1
x
2 (e
5-6
1
4
S (e
18. a. Four terms:
1
1
1
e511
1
1
5 2.666 666
1!
2!
3!
Five terms:
1
1
1
1
e511
1
1
1
5 2.708 333
1!
2!
3!
4!
Six terms:
1
1
1
1
1
e511
1
1
1
1
5 2.716 666
1!
2!
3!
4!
5!
Seven terms:
1
1
1
1
1
1
1
1
1
1
1
5 2.718 055
e511
1!
2!
3!
4!
5!
6!
b. The expression for e in part a. is a special case of
x1
1. a.
b.
c.
121e
) 2 (e 2 2 1 e
(cosh x)2
d(23x )
dy
5
dx
dx
5 3(23x ) ln 2
dy
d(3.1x 1 x 3 )
5
dx
dx
5 ln 3.1(3.1)x 1 3x 2
d(103t25 )
ds
5
dt
dt
5 3(103t25 ) ln 10
2
2 e )( 12) (e x 2 e 2x )
2x
x4
d(10526n1n )
dw
5
d.
dn
dn
2
5 (26 1 2n)(10526n1n )ln 10
2x
22x
x3
5.2 Derivatives of the General
Exponential Function, y 5 b x, p. 240
(cosh x)2
2x
x2
e x 5 1 1 1! 1 2! 1 3! 1 4! 1 c. in that it is the
case when x 5 1. Then e x 5 e 1 5 e is in fact
e 1 5 e 5 1 1 1!1 1 2!1 1 3!1 1 4!1 1 5!1 1 c. The
value of x is 1.
2
(cosh x)
1
x
2 (e
1
4 (4)
(cosh x)2
1
5
(cosh x)2
e.
2
2
5
1 e 2x )( 12) (cosh x)2 (e x 1 e 2x )
5
22x
)T
f.
d(3x 12 )
dy
5
dx
dx
2
5 2x(3x 12 )ln 3
d(400(2)x13 )
dy
5
dx
dx
5 400(2)x13 ln 2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
d(x 5 3 (5)x )
dy
5
dx
dx
5 (x 5 )((5)x (ln 5)) 1 ((5)x )(5x 4 )
5 5x 3(x 5 3 ln 5) 1 5x 44
2
d(x(3)x )
dy
5
b.
dx
dx
2
2
5 (x)(2x(3)x ln 3) 1 (3)x (1)
2
5 (3)x 3(2x 2 ln 3) 1 14
c. v 5 (2t )(t 21 )
d((2t )(t 21 ))
dv
5
dt
dx
5 (2t )(21t 22 ) 1 (t 21 )(2t ln 2)
2t
2t ln 2
52 21
t
t
x
2
3
d. f(x) 5 2
x
x
x
1
ln 3(32 )(x 2 ) 2 2x(32 )
f r(x) 5 2
x4
x
x
x ln 3(32 ) 2 4(32 )
5
x4
x
32 3x ln 3 2 44
5
x3
2
3t25
3. f(t) 5 10
? e 2t
2
2
fr (t) 5 (103t25 )(4te 2t ) 1 (e 2t )(3(10)3t25 ln 10)
2
5 103t25e 2t (4t 1 3 ln 10)
Now, set f r (t) 5 0.
2
So, f r (t) 5 0 5 103t25e 2t (4t 1 3ln 10)
2
So 103t25e 2t 5 0 and 4t 1 3 ln 10 5 0.
The first equation never equals zero because solving
would force one to take the natural log of both
sides, but ln 0 is undefined. So the first equation
does not produce any values for which fr (t) 5 0.
The second equation does give one value.
4t 1 3 ln 10 5 0
4t 5 23 ln 10
3 ln 10
t52
4
4. When x 5 3, the function y 5 f(x) evaluated at
3 is f(3) 5 3(23 ) 5 3(8) 5 24. Also,
d(3(2)x )
dy
5
dx
dx
5 3(2x )ln 2
So, at x 5 3,
dy
5 3(23 )(ln 2) 5 24(ln 2) 8 16.64
dx
Therefore, y 2 24 5 16.64(x 2 3)
y 2 24 5 16.64x 2 49.92
216.64x 1 y 1 25.92 5 0
2. a.
Calculus and Vectors Solutions Manual
d(10x )
dy
5
dx
dx
5 10x ln 10
So, at x 5 1,
dy
5 101 ln 10 5 10(ln 10) 8 23.03
dx
Therefore, y 2 10 5 23.03(x 2 1)
y 2 10 5 23.03x 2 23.03
223.03x 1 y 1 13.03 5 0
6. a. The half-life of the substance is the time
required for half of the substance to decay. That is, it
is when 50% of the substance is left, so P(t) 5 50.
50 5 100(1.2)2t
1
5 (1.2)2t
2
1
1
5
2
(1.2)t
t
(1.2) 5 2
t(ln 1.2) 5 ln 2
ln 1.2
t5
ln 2
t 8 3.80 years
Therefore, the half-life of the substance is about
3.80 years.
b. The problem asks for the rate of change when
t 8 3.80 years.
Pr (t) 5 2100(1.2)2t (ln 1.2)
Pr (3.80) 5 2100(1.2)2(3.80) (ln 1.2)
8 29.12
So, the substance is decaying at a rate of about
29.12 percent> year at the time 3.80 years where the
half-life is reached.
7. P 5 0.5(109 )e 0.200 15t
dP
5 0.5(109 )(0.200 15)e 0.200 15t
a.
dt
5.
dP
In 1968, t 5 1 and dt 5 0.5(109 )(0.200 15)e 0.200 15 8
0.122 25 3 109 dollars> annum
In 1978, t 5 11 and
dP
5 0.5(109 )(0.200 15)e 1130.200 15
dt
8 0.904 67 3 109 dollars> annum.
In 1978, the rate of increase of debt payments
was $904 670 000> annum compared to
$122 250 000> annum in 1968. As a ratio,
Rate in 1978
7.4
5 .
Rate in 1968
1
The rate of increase for 1978 is
7.4 times larger than that for 1968.
5-7
b. In 1988, t 5 21 and
dP
5 0.5(109 )(0.200 15)e 2130.200 15
dt
8 6.694 69 3 109 dollars> annum
In 1998, t 5 31 and
dP
5 0.5(109 )(0.200 15)e 3130.200 15
dt
8 49.541 69 3 109 dollars> annum
Rate in 1998
7.4
From the graph, one can notice that the values of v(t)
quickly rise in the range of about 0 # t # 15. The
slope for these values is positive and steep. Then as the
graph nears t 5 20 the steepness of the slope decreases
and seems to get very close to 0. One can reason that
the car quickly accelerates for the first 20 units of time.
Then, it seems to maintain a constant acceleration
for the rest of the time. To verify this, one could differentiate and look at values where vr (t) is increasing.
As a ratio, Rate in 1988 5 1 . The rate of increase
for 1998 is 7.4 times larger than that for 1988.
c. Answers may vary. For example, data from the
past are not necessarily good indicators of what will
happen in the future. Interest rates change, borrowing
may decrease, principal may be paid off early.
8. When x 5 0, the function y 5 f(x) evaluated at 0
2
is f(0) 5 220 5 20 5 1. Also,
2
d(22x )
dy
5
dx
dx
2
5 22x(22x )ln 2
So, at x 5 0,
dy
2
5 22(0)(220 )ln 2 5 0
dx
Therefore, y 2 1 5 0(x 2 0)
So, y 2 1 5 0 or y 5 1.
2
y
5.3 Optimization Problems Involving
Exponential Functions, pp. 245–247
1. a.
The maximum value is about 0.3849. The
minimum value is 0.
b.
1
–8 –6 –4 –2 0
x
2 4 6 8
–1
–2
9.
v(t)
120
100
80
60
40
20
0
5-8
t
0
20 40 60 80 100 120
The maximum value is about 10.043. The
minimum value is about 25961.916.
2. a. f(x) 5 e 2x 2 e 23x on 0 # x # 10
f r(x) 5 2e 2x 1 3e 23x
Let f r (x) 5 0, therefore e 2x 1 3e 23x 5 0.
Let e 2x 5 w, when 2w 1 3w 3 5 0.
w(21 1 3w 2 ) 5 0.
Therefore, w 5 0 or w 2 5 13
1
.
w56
"3
But w $ 0, w 5 1
1
"3
.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
1
1
When w 5 "3, e 2x 5 "3,
2x ln e 5 ln 1 2 ln "3
ln "3 2 ln 1
1
5 ln "3
8 0.55.
x5
f(0) 5 e 0 2 e 0
50
f(0.55) 8 0.3849
f(10) 5 e 210 2 e 230 8 0.000 05
Absolute maximum is about 0.3849 and absolute
minimum is 0.
m(x) 5 (x 1 2)e 22x on 24 # x # 4
mr(x) 5 e 22x 1 (22)(x 1 2)e 22x
Let mr(x) 5 0.
e 22x 2 0, therefore, 1 1 (22)(x 1 2) 5 0
23
x5
2
5 21.5.
m(24) 5 22e 8 8 25961
m(21.5) 5 0.5e 3 8 10
m(4) 5 6e 28 8 0.0002
The maximum value is about 10 and the minimum
value is about 25961.
b. The graphing approach seems to be easier to use for
the functions. It is quicker and it gives the graphs of
the functions in a good viewing rectangle. The only
problem may come in the second function, m(x),
because for x , 1.5 the function quickly approaches
values in the negative thousands.
20
3. a. P(t) 5
1 1 3e 20.02t
20
P(0) 5
1 1 3e 20.02(0)
20
5
1 1 3e 0
20
5
4
55
So, the population at the start of the study when
t 5 0 is 500 squirrels.
b. The question asks for lim P(t).
tS`
As t approaches `, e 20.02t 5
20
20.02t
tS`
tS` 1 1 3e
20
5
1 1 3(0)
5 20.
Therefore, the largest population of squirrels that
the forest can sustain is 2000 squirrels.
c. A point of inflection can only occur when
Ps (t) 5 0 and concavity changes around the point.
20
P(t) 5
1 1 3e 20.02t
P(t) 5 20(1 1 3e 20.02t )21
So, lim P(t) 5 lim
1
e0.02t
approaches 0.
Calculus and Vectors Solutions Manual
Pr (t) 5 20(2 (1 1 3e 20.02t )22 (20.06e 20.02t ))
5 (1.2e 20.02t )(1 1 3e 20.02t )22
Ps (t) 5 3(1.2e20.02t )(22(1 1 3e20.02t )23 (20.06e 20.02t )4
1 (1 1 3e 20.02t )22 (20.024e 20.02t )
0.144e 20.04t
0.024e 20.02t
2
(1 1 3e 20.02t )3
(1 1 3e 20.02t )2
0.144e 20.04t
0.024e 20.02t
Ps (0) when
2
50
(1 1 3e 20.02t )3
(1 1 3e 20.02t )2
Solving for t after setting the second derivative
equal to 0 is very tedious. Use a graphing calculator
to determine the value of t for which the second
derivative is 0, 54.9. Evaluate P(54.9). The point of
inflection is (54.9, 10).
d.
P(t)
25
5
20
15
10
5
0
t
0
50 100 150 200 250 300
e. P grows exponentially until the point of inflection,
then the growth rate decreases and the curve becomes
concave down.
4. a. P(x) 5 106 31 1 (x 2 1)e20.001x4, 0 # x # 2000
Using the Algorithm for Extreme Values, we have
P(0) 5 106 31 2 14 5 0
P(2000) 5 106 31 1 1999e 224 8 271.5 3 106.
Now,
Pr (x) 5 106 3(1)e 20.001x 1 (x 2 1)(20.001)e 20.001x4
5 106e 20.001x (1 2 0.001x 1 0.001)
5-9
7. C 5 0.015 3 109e 0.075 33t, 0 # t # 100
a.
C(t)
Capital investment from U.S. sources
($100 million)
Since e 20.001x . 2 for all x,
Pr (x) 5 0 when 1.001 2 0.001x 5 0
1.001
x5
5 1001.
0.001
P(1001) 5 106 31 1 1000e 21.0014 8 368.5 3 106
The maximum monthly profit will be 368.5 3 106
dollars when 1001 items are produced and sold.
b. The domain for P(x) becomes 0 # x # 500.
P(500) 5 106 31 1 499e 20.54 5 303.7 3 106
Since there are no critical values in the domain, the
maximum occurs at an endpoint. The maximum
monthly profit when 500 items are produced and
sold is 303.7 3 106 dollars.
5. R(x) 5 40x 2e 20.4x 1 30, 0 # x # 8
We use the Algorithm for Extreme Values:
Rr (x) 5 80xe 20.4x 1 40x 2 (20.4)e 20.4x
5 40xe 20.4x (2 2 0.4x)
Since e 20.4x . 0 for all x, Rr(x) 5 0 when
x 5 0 or 2 2 0.4x 5 0
x 5 5.
R(0) 5 30
R(5) 8 165.3
R(8) 8 134.4
The maximum monthly revenue of 165.3 thousand
dollars is achieved when 500 units are produced and
sold.
6. P(t) 5 100(e 2t 2 e 24t ), 0 # t # 3
Pr(t) 5 100(2e 2t 1 4e 24t )
5 100e 2t (21 1 4e 23t )
Since e 2t . 0 for all t, Pr(t) 5 0 when
4e 23t 5 1
1
e 23t 5
4
23t 5 ln (0.25)
2ln (0.25)
t5
3
5 0.462.
P(0) 5 0
P(0.462) 8 47.2
P(3) 8 4.98
The highest percentage of people spreading the
rumour is 47.2% and occurs at the 0.462 h point.
16
14
12
10
8
6
4
2
0
t
20 40 60 80 100
Years since 1867
dC
5 0.015 3 109 3 0.075 33e 0.075 33t
dt
In 1947, t 5 80 and the growth rate was
dC
5 0.468 05 3 109 dollars> year.
dt
In 1967, t 5 100 and the growth rate was
dC
5 2.1115 3 109 dollars> year.
dt
The ratio of growth rates of 1967 to that of 1947 is
2.1115 3 109
4.511
.
9 5
0.468 05 3 10
1
The growth rate of capital investment grew from
468 million dollars per year in 1947 to 2.112 billion
dollars per year in 1967.
c. In 1967, the growth rate of investment as a
percentage of the amount invested is
2.1115 3 109
3 100 5 7.5%.
28.0305 3 109
d. In 1977, t 5 110
C 5 59.537 3 109 dollars
dC
5 4.4849 3 109 dollars> year.
dt
e. Statistics Canada data shows the actual amount of
U.S. investment in 1977 was 62.5 3 109 dollars.
The error in the model is 3.5%.
f. In 2007, t 5 140.
The expected investment and growth rates are
b.
C 5 570.490 3 109 dollars and dC 5 42.975 3 109
dt
dollars> year.
5-10
Chapter 5: Derivatives of Exponential and Trigonometric Functions
t
1 210t
te b
10
1
30 2 t
30 2 t
1 0.6a2e 2 5 1 (30 2 t)e 2 20 b
20
30 2 t
210t
5 0.05e (10 2 t) 1 0.03e 2 5
8. a. The growth function is N 5 25.
t
Er(t) 5 0.5ae 210 2
t
3
The number killed is given by K 5 e .
After 60 minutes, N 5 212.
Let T be the number of minutes after 60 minutes.
The population of the colony at any time, T after
the first 60 minutes is
P5N2k
60 1 T
T
5 2 5 2 e3 , T $ 0
dP
1
1 T
60 1 T
5 2 5 a b ln 2 2 e 3
dt
5
3
ln 2
1 T
12 1 T
52 5 a
b 2 e3
5
3
ln 2
1 T
T
5 212 ? 2 5 a
b 2 e3
5
3
dP
ln
2
1 T
T
5 0 when 212
2 5 5 e 3 or
dt
5
3
ln 2 12 T5
T
3
? 2 2 5 e3 .
5
We take the natural logarithm of both sides:
ln 2
T
T
ln a3.212
b 1 ln 2 5
5
5
3
1
ln 2
b
7.4404 5 T a 2
3
5
7.4404
5 38.2 min.
T5
0.1947
At T 5 0, P 5 212 5 4096.
At T 5 38.2, P 5 478 158.
(220 1 30 2 t)
t
5 ( 0.05e 210 1 0.03e 2
t
Calculus and Vectors Solutions Manual
30 2 t
20
) (10 2 t)
Er (t) 5 0 when 10 2 t 5 0
t 5 10 (The first factor is always a positive number.)
3
E(0) 5 5 1 5.4 1 18e 22 5 14.42
E(10) 5 16.65
E(30) 5 11.15
For maximum study effectiveness, 10 h of study
should be assigned to the firs exam and 20 h of
study for the second exam.
10. Use the algorithm for finding extreme values.
First, find the derivative f r(x). Then, find any
critical points by setting fr (x) 5 0 and solving for x.
Also, find the values of x for which f r(x) is
undefined. Together these are the critical values.
Now, evaluate f(x) for the critical values and the
endpoints 2 and 22. The highest value will be the
absolute maximum on the interval and the lowest
value will be the absolute minimum on the interval.
11. a. f r (x) 5 (x 2 )(e x ) 1 (e x )(2x)
For T . 38.2, dP is always negative.
dt
The maximum number of bacteria in the colony
occurs 38.2 min after the drug was introduced.
At this time the population numbers 478 158.
60 1 T
T
b. P 5 0 when 2 5 5 e 3
T
60 1 T
ln 2 5
5
3
ln 2
1
b
12 ln 2 5 Ta 2
3
5
T 5 42.72
The colony will be obliterated 42.72 minutes after
the drug was introduced.
9. Let t be the number of minutes assigned to study
for the first exam and 30 2 t minutes assigned to
study for the second exam. The measure of study
effectiveness for the two exams is given by
E(t) 5 E1 (t) 1 E2 (30 2 t), 0 # t # 30
5 0.5( 10 1 te 210 ) 1 0.6( 9 1 (30 2 t)e 2
30 2 t
5
)
5 e x (x 2 1 2x)
The function is increasing when f r (x) . 0 and
decreasing when f r (x) , 0. First, find the critical
values of f r (x). Solve e x 5 0 and (x 2 1 2x) 5 0
ex is never equal to zero.
x 2 1 2x 5 0
x(x 1 2) 5 0.
So, the critical values are 0 and 22.
Interval
e x (x 2 1 2x)
x , 22
1
22 , x , 0
2
0,x
1
So, f(x) is increasing on the intervals (2 `, 22)
and (0, ` ).
Also, f(x) is decreasing on the interval (22, 0).
b. At x 5 0, f r (x) switches from decreasing on the
left of zero to increasing on the right of zero. So,
x 5 0 is a minimum. Since it is the only critical
point that is a minimum, it is the x-coordinate of the
5-11
absolute minimum value of f(x). The absolute
minimum value is f(0) 5 0.
12. a. yr 5 e x
Setting e x 5 0 yields no solutions for x. ex is a
function that is always increasing. So, there is no
maximum or minimum value for y 5 e x 1 2.
y
8
6
4
2
x
–8 –6 –4 –2 0
–2
2 4 6 8
ex (x 1 1)
x , 21
2
x . 21
1
x,2
1
2
2
x.2
1
2
1
So y is decreasing on the left of x 5 2 12 and
increasing on the right of x 5 2 12. So x 5 2 12 is the
x-coordinate of the minimum of y. The minimum
value is
1
1
2a2 b (e2(22 ) )
2
5 2e21
8 20.37. There is no maximum value.
x
2
d. yr 5 (3x)(2e 2x ) 1 (e 2x )(3) 1 1
5 3e 2x (1 2 x) 1 1
Solve 3e 2x (1 2 x) 1 1 5 0.
This gives no real solutions. By looking at the graph
of y 5 f(x), one can see that the function is always
increasing. So, there is no maximum or minimum
value for y 5 3xe 2x 1 x.
8
y
y
4
–8
–8 –6 –4 –2 0
–2
y
–8 –6 –4 –2 0
–2
So y is decreasing on the left of x 5 21 and
increasing on the right of x 5 21. So x 5 21 is the
x-coordinate of the minimum of y. The minimum
value is 2e21 1 3 8 2.63. There is no maximum
value.
8
6
4
2
2e 2x (2x 1 1)
8
6
4
2
b. yr 5 (x)(e x ) 1 (e x )(1)
5 e x (x 1 1)
Solve e x 5 0 and (x 1 1) 5 0
ex is never equal to zero.
x1150
x 5 21.
So there is one critical point: x 5 21.
Interval
Interval
x
2 4 6 8
–4
0
x
4
8
–4
–8
2x
2x
c. yr 5 (2x)(2e ) 1 (e )(2)
5 2e2x (2x 1 1)
Solve 2e2x 5 0 and (2x 1 1) 5 0
2e 2x is never equal to zero.
2x 1 1 5 0
1
x52
2
So there is one critical point: x 5 2 12.
5-12
2
2
13. Pr (x) 5 (x)(2xe 20.5x ) 1 (e 20.5x )(1)
2
5 e 20.5x (2x 2 1 1)
2
Solve e 20.5x 5 0 and (1 2 x 2 ) 5 0.
20.5x 2
e
gives no critical points.
1 2 x2 5 0
(1 2 x)(1 1 x) 5 0
Chapter 5: Derivatives of Exponential and Trigonometric Functions
So x 5 1 and x 5 21 are the critical points.
So P(x) is decreasing on the left of x 5 21 and on
for t ,
2
Interval
e20.5x (2x2 1 1)
x , 21
2
21 , x , 1
1
d
120
100
80
60
40
20
t
0
2
4
6
8
10
12
b. The speed is increasing when dr(t) . 0 and the
speed is decreasing when dr(t) , 0.
dr(t) 5 (200t)(222t )(ln 2) 1 (22t )(200)
5 200(2)2t (2t ln 2 1 1)
Solve 200(2)2t 5 0 and 2t ln 2 1 1 5 0.
200(2)2t gives no critical points.
2t ln 2 1 1 5 0
1
t5
8 1.44
ln 2
So t 5
1
ln 2
Interval
is the critical point.
1
ln2
1
t.
1
ln2
2
1
ln 2
and decreasing when t .
Calculus and Vectors Solutions Manual
2In1 2
1
.
ln 2
8 106.15 degrees> s
d. The door seems to be closed for t . 10 s.
15. The solution starts in a similar way to that of 9.
The effectiveness function is
t
25 2 t
E(t) 5 0.5( 10 1 te 210) 1 0.6( 9 1 (25 2 t)e 2 20 ) .
The derivative simplifies to
t
25 2 t
Er(t) 5 0.05e 210 (10 2 t) 1 0.03e 2 20 (5 2 t).
This expression is very difficult to solve analytically.
By calculation on a graphing calculator, we can
determine the maximum effectiveness occurs when
t 5 8.16 hours.
aL
16. P 5
a 1 (L 2 a)e 2kLt
a. We are given a 5 100, L 5 10 000, k 5 0.0001.
106
104
P5
2t 5
100 1 9900e
1 1 99e 2t
4
2t 21
5 10 (1 1 99e )
P
10
8
6
4
2
0
0
2
4
(dP)
So the speed of the closing door is increasing when
0,t,
since dr(t) , 0
6 8
Days
10
12
t
b. We need to determine when the derivative of the
200(2)2t (2t ln2 1 1)
t,
and dr(t) . 0 for t .
(ln12) 5 200 (ln12) (2)
Number of cells (thousands)
the right of x 5 1 and it is increasing between
x 5 21 and x 5 1. So x 5 21 is the x-coordinate of
the minimum of P(x). Also, x 5 1 is the x-coordinate
of the maximum of P(x). The minimum value is
2
P(21) 5 (21)(e 20.5(21) ) 5 2e 20.5 8 20.61.
The maximum value is
2
P(1) 5 (1)(e 20.5(1) ) 5 e 20.5 8 0.61.
14. a.
d(t)
0
1
ln 2
1
ln 2
The maximum speed is
2
1,x
c. There is a maximum at t 5
1
.
ln 2
d 2P
growth rate dt is zero, i.e., when dt 2 5 0.
dP
2104 (299e 2t )
990 000e 2t
5
5
dt
(1 1 99e 2t )2
(1 1 99e 2t )2
d 2P
2990 000e 2t (1 1 99e 2t )2 2 990 000e 2t
2 5
dt
(1 1 99e 2t )4
(2)(1 1 99e 2t )(299e 2t )
3
(1 1 99e 2t )4
2990 000e2t (1 1 99e 2t ) 1 198(990 000)e 22t
5
(1 1 99e 2t )3
5-13
2
dP
2 5 0
dt
990 000e 2t (21 2 99e 2t 1 198e 2t ) 5 0
99e 2t 5 1
e t 5 99
t 5 ln 99
8 4.6
After 4.6 days, the rate of change of the growth rate
is zero. At this time the population numbers 5012.
c. When t 5 3, dt 5 (1 1 99e 23 )2 8 1402 cells> day.
dP
990 000e 23
When t 5 8, dt 5 (1 1 99e 28 )2 8 311 cells> day.
The rate of growth is slowing down as the colony is
getting closer to its limiting value.
dP
990 000e 28
Mid-Chapter Review, pp. 248–249
d(5e 23x )
dy
5
dx
dx
5 (5e 23x )(23x)r
5 (5e 23x )(23)
5 215e 23x
1
d( 7e7 x)
dy
b.
5
dx
dx
1 r
1
5 ( 7e7 x) a xb
7
1
1
5 ( 7e7 x) a b
7
1
5 e7 x
dy
c.
5 (x 3 )(e 22x )r 1 (x 3 )r (e 22x )
dx
5 (x 3 )((e 22x )(22x)r) 1 (3x 2 )(e 22x )
5 (x 3 )((e 22x ))(22) 1 3x 2e 22x
5 22x 3e 22x 1 3x 2e 22x
5 e 22x (22x 3 1 3x 2 )
dy
5 (x 2 1)2 (e x )r 1 ( (x 2 1)2 )r (e x )
d.
dx
5 (x 2 1)2 (e x ) 1 (2(x 2 1))(e x )
5 (x 2 2 2x 1 1)(e x ) 1 (2x 2 2)(e x )
5 (e x )(x 2 2 2x 1 1 1 2x 2 2)
5 (e x )(x 2 2 1)
dy
e.
5 2(x 2 e 2x )(x 2 e 2x )r
dx
5 2(x 2 e 2x )(1 2 (e 2x )(2x)r)
5 2(x 2 e 2x )(1 2 (e 2x )(21))
5 2(x 2 e 2x )(1 1 e 2x )
5 2(x 1 xe 2x 2 e 2x 2 e 2x1 2x )
5 2(x 1 xe 2x 2 e 2x 2 e 22x )
1. a.
5-14
dy
(e x 1 e 2x )(e x 2 e 2x )r
5
dx
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 e 2x )r
2
(e x 1 e 2x )2
(e x 1 e 2x )(e x 2 (e 2x )(2x)r)
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 (e 2x )(2x)r)
2
(e x 1 e 2x )2
x
2x
(e 1 e )(e x 2 (e 2x )(21))
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 (e 2x )(21))
2
(e x 1 e 2x )2
(e x 1 e 2x )(e x 1 e 2x )
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 2 e 2x )
2
(e x 1 e 2x )2
e 2x 1 e 0 1 e 0 1 e 22x
5
(e x 1 e 2x )2
(e 2x 2 e 0 2 e 0 1 e 22x )
2
(e x 1 e 2x )2
e 2x 1 e 0 1 e 0 1 e 22x 2 e 2x
5
(e x 1 e 2x )2
e 0 1 e 0 2 e 22x
1
(e x 1 e 2x )2
4
5 x
(e 1 e 2x )2
dP
2. a.
5 100e 25t (25t)r
dt
5 100e 25t (25)
5 2500e 25t
b. The time is needed for when the sample of the
substance is at half of the original amount. So, find
t when P 5 12.
f.
when
P 5 100e 25t
1
5 100e 25t
2
1
5 e 25t
200
1
5 25t
ln
200
1
ln 200
5t
25
Chapter 5: Derivatives of Exponential and Trigonometric Functions
dP
Now, the question asks for dt 5 Pr when
t5
Pra
1
ln 200
8 1.06
25
1
ln 200
b 5 22.5 (using a calculator)
25
dy
5 (2x)(e x )r 1 (e x )(2x)r
dx
5 (2x)(e x ) 1 (e x )(21)
5 2xe x 2 e x
At the point x 5 0,
dy
5 20e 0 2 e 0 5 21.
dx
At the point x 5 0,
y 5 2 2 0e 0 5 2
So, an equation of the tangent to the curve at the
point x 5 0 is
y 2 2 5 21(x 2 0)
y 2 2 5 2x
y 5 2x 1 2
x1y2250
4. a. yr 5 23(e x )r
5 23e x
ys 5 23e x
b. yr 5 (x)(e 2x )r 1 (e 2x )(x)r
5 (x)((e 2x ) 1 (2x)r) 1 (e 2x )(1)
5 (x)((e 2x )(2)) 1 e 2x
5 2xe 2x 1 e 2x
ys 5 (2x)(e 2x )r 1 (e 2x )(2x)r 1 e 2x (2x)r
5 (2x)((e 2x )(2x)r) 1 (e 2x )(2) 1 (e 2x )(2)
5 (2x)((e 2x )(2)) 1 2e 2x 1 2e 2x
5 4xe 2x 1 4e 2x
c. yr 5 (e x )(4 2 x)r 1 (4 2 x)(e x )r
5 (e x )(21) 1 (4 2 x)(e x )
5 2e x 1 4e x 2 xe x
5 3e x 2 xe x
ys 5 (3e x )r 2 3 (x)(e x )r 1 (e x )(x)r4
5 3e x 2 3xe x 1 (e x )(1)4
5 3e x 2 xe x 2 e x
5 2e x 2 xe x
dy
5 (82x15 )(ln 8)(2x 1 5)r
5. a.
dx
5 (82x15 )(ln 8)(2)
5 2(ln 8)(82x15 )
3.
dy
5 3.2((10)0.2x )(ln 10)(0.2x)r
dx
5 3.2((10)0.2x )(ln 10)(0.2)
5 0.64(ln 10)((10).2x )
c. fr (x) 5 (x 2 )(2x )r 1 (2x )(x 2 )r
5 (x 2 )(2x )(ln 2) 1 (2x )(2x)
5 (ln 2)(x 22x ) 1 2x2x
5 2x ((ln 2)(x 2 ) 1 2x)
d. Hr (x) 5 300((5)3x21 )(ln 5)(3x 2 1)r
5 300((5)3x21 )(ln 5)(3)
5 900(ln 5)(5)3x21
5 900(ln 5)(5)3x21
e. qr (x) 5 (1.9)x ? (ln 1.9) 1 1.9(x)1.921
5 (1.9)x ? (ln 1.9) 1 1.9(x)0.9
5 (ln 1.9)(1.9)x 1 1.9x 0.9
f. f r (x) 5 (x 2 2)2 (4x )r 1 (4x )( (x 2 2)2 )r
5 (x 2 2)2 (4x )(ln 4) 1 (4x )(2(x 2 2))
5 (ln 4)(4x )(x 2 2)2 1 (4x )(2x 2 4)
5 4x ((ln 4)(x 2 2)2 1 2x 2 4)
6. a. The initial number of rabbits in the forest is
given by the time t 5 0.
0
R(0) 5 500( 10 1 e 2 10)
5 500(10 1 1)
5 500(11)
5 5500
b.
dR
b. The rate of change is the derivative, dt .
t
R(t) 5 5000 1 500( e210)
dR
t r
t
5 0 1 500( e210) a2 b
dt
10
1
t
5 500( e210) a2 b
10
t
5 250( e210)
c. 1 year 5 12 months
dR
The question asks for dt 5 Rr when t 5 12.
12
Rr (12) 5 250( e210)
8 215.06
d. To find the maximum number of rabbits,
optimize the function.
t
Rr (t) 5 250( e210)
t
0 5 250( e210)
t
0 5 e210
Calculus and Vectors Solutions Manual
5-15
When solving, the natural log (ln) of both sides
must be taken, but (ln 0) does not exist. So there
are no solutions to the equation. The function is
therefore always decreasing. So, the largest number
of rabbits will exist at the earliest time in the interval
at time t 5 0. To check, compare R(0) and R(36).
R(0) 5 5500 and R(36) 8 5013. So, the largest
number of rabbits in the forest during the first
3 years is 5500.
e. 6000
4000
2000
0
0
10
20 30 40
The graph is constantly decreasing. The y-intercept
is at the point (0, 5500). Rabbit populations normally
grow exponentially, but this population is shrinking
exponentially. Perhaps a large number of rabbit
predators such as snakes recently began to appear in
the forest. A large number of predators would
quickly shrink the rabbit population.
7. The highest concentration of the drug can be
found by optimizing the given function.
C(t) 5 10e 22t 2 10e 23t
Cr(t) 5 (10e 22t )(22t)r 2 (10e 23t )(23t)r
5 (10e 22t )(22) 2 (10e 23t )(23)
5 220e 22t 1 30e 23t
Set the derivative of the function equal to zero and
find the critical points.
0 5 220e 22t 1 30e 23t
20e 22t 5 30e 23t
2 22t
e 5 e 23t
3
e 23t
2
5 22t
3
e
2
5 (e 23t )(e 2t )
3
2
5 e 23t12t
3
2
5 e 2t
3
5-16
ln
2
5 2t
3
2
2 aln b 5 t
3
Therefore, t 5 2 ( ln 23) 8 0.41 is the critical value.
Now, use the algorithm for finding extreme values.
C(0) 5 10(e 0 2 e 0 ) 5 0
2
Ca2 aln bb 8 1.48 (using a calculator)
3
C(5) 5 0.0005
So, the function has a maximum when
t 5 2 ( ln 23) 8 0.41. Therefore, during the first five
hours, the highest concentration occurs at about
0.41 hours.
8. y 5 ce kx
yr 5 cke kx
The original function is increasing when its derivative is positive and decreasing when its derivative is
negative.
e kx . 0 for all k, xPR.
So, the original function represents growth when
ck . 0, meaning that c and k must have the same
sign. The original function represents decay when c
and k have opposite signs.
9. a. A(t) 5 5000e 0.02t
5 5000e 0.02(0)
5 5000
The initial population is 5000.
b. at t 5 7
A(7) 5 5000e 0.02(7) 5 5751
After a week, the population is 5751.
c. at t 5 30
A(30) 5 5000e 0.02(30) 5 9111
After 30 days, the population is 9111.
10. a. P(5) 5 760e 20.125(5)
8 406.80 mm Hg
b. P(7) 5 760e 20.125(7)
8 316.82 mm Hg
c. P(9) 5 760e 20.125(9)
8 246.74 mm Hg
11. A 5 100e 20.3x
Ar 5 100e 20.3x (20.3)
5 230e 20.3x
When 50% of the substance is gone, y 5 50
50 5 100e 20.3x
0.5 5 e 20.3x
ln (0.5) 5 ln e 20.3x
ln (0.5) 5 20.3x ln e
Chapter 5: Derivatives of Exponential and Trigonometric Functions
ln 0.5
5 20.3x
ln e
ln 0.5
5x
2
0.3 ln e
x 5 2.31
Ar 5 230e 20.3x
Ar(2.31) 5 230e 20.3(2.31)
Ar 8 215
When 50% of the substance is gone, the rate of
decay is 15% per year.
12. f(x) 5 xe x
f r(x) 5 xe x 1 (1)e x
5 e x (x 1 1)
x
So e . 0
x11.0
x . 21
This means that the function is increasing when
x . 21.
2
13. y 5 52x
When x 5 1,
1
y5
5
2
yr 5 52x (22x) ln 5
2
yr 5 2 ln 5
5
2
1
5y 2 5 2 ln 5(x 2 1)
5
5
5y 2 1 5 22 ln 5(x 2 1)
5y 2 1 5 (22 ln 5)x 1 2 ln 5
(2 ln 5)x 1 5y 5 2 ln 5 1 1
y
4
2
–4
–2
0
x
2
4
–2
–4
14. a. A 5 P(1 1 i)t
A(t) 5 1000(1 1 0.06)t
5 1000(1.06)t
b. Ar(t) 5 1000(1.06)t (1) ln (1.06)
5 1000(1.06)t ln 1.06
c. Ar (2) 5 1000(1.06)2 ln 1.06
5 $65.47
Calculus and Vectors Solutions Manual
Ar (5) 5 1000(1.06)5 ln 1.06
5 $77.98
Ar (10) 5 1000(1.06)10 ln 1.06
5 $104.35
d. No, the rate is not constant.
Ar (2)
5 ln 1.06
e.
A(2)
Ar (5)
5 ln 1.06
A(5)
Ar (10)
5 ln 1.06
A(10)
f. All the ratios are equivalent (they equal ln 1.06,
which is about 0.058 27), which means that
Ar (t)
A(t)
is
constant.
15. y 5 cex
yr 5 c(ex ) 1 (0)ex
5 cex
y 5 yr 5 ce x
5.4 The Derivatives of y 5 sin x and
y 5 cos x, pp. 256–257
d(2x)
dy
5 (cos 2x) ?
dx
dx
5 2 cos 2x
d(3x)
dy
b.
5 22 (sin 3x) ?
dx
dx
5 26 sin 3x
d(x 3 2 2x 1 4)
dy
c.
5 (cos (x 3 2 2x 1 4)) ?
dx
dx
2
3
5 (3x 2 2)(cos (x 2 2x 1 4))
d(24x)
dy
d.
5 22 sin (24x) ?
dx
dx
5 8 sin (24x)
d(3x)
d(4x)
dy
e.
5 cos (3x) ?
1 sin (4x) ?
dx
dx
dx
5 3 cos (3x) 1 4 sin (4x)
dy
f.
5 2x (ln 2) 1 2 cos x 1 2 sin x
dx
d(e x )
dy
g.
5 cos (e x ) ?
dx
dx
5 e xcos (e x )
d(3x 1 2p)
dy
h.
5 3 cos (3x 1 2p) ?
dx
dx
5 9 cos (3x 1 2p)
1. a.
5-17
dy
5 2x 2 sin x 1 0
dx
5 2x 2 sin x
d( 1x)
1
dy
j.
5 cos a b ?
x
dx
dx
1
1
5 2 2 cos a b
x
x
dy
2. a.
5 (2 sin x)(2sin x) 1 (cos x)(2 cos x)
dx
5 22 sin2 x 1 2 cos2 x
5 2(cos2 x 2 sin2 x)
5 2 cos (2x)
b. y 5 (x 21 )(cos 2x)
dy
5 (x 21 )(22 sin 2x) 1 (cos 2x)(2x 22 )
dx
2 sin 2x
cos 2x
52
2
x
x2
d(sin 2x)
dy
c.
5 2sin (sin 2x) ?
dx
dx
5 2sin (sin 2x) ? 2 cos 2x
d. y 5 (sin x)(1 1 cos x)21
dy
5 (sin x)(2 (1 1 cos x)22 ? (2sin x)
dx
1 (1 1 cos x)21 (cos x)
2sin2 x
cos x
5
2 1
2 (1 1 cos x)
1 1 cos x
sin2 x
cos x(1 1 cos x)
5
1
(1 1 cos x)2
(1 1 cos x)2
2
2
sin x 1 cos x 1 cos x
5
(1 1 cos x)2
1 1 cos x
5
(1 1 cos x)2
1
5
1 1 cos x
dy
e.
5 (e x )(2sin x 1 cos x) 1 (cos x 1 sin x)(e x )
dx
5 e x (2sin x 1 cos x 1 cos x 1 sin x)
5 e x (2 cos x)
i.
dy
f.
5 (2x 3 )(cos x) 1 (sin x)(6x 2 )
dx
2 3(3x)(2sin x) 1 (cos x)(3)4
5 2x 3 cos x 1 6x 2 sin x 1 3x sin x 2 3 cos x
p
p
p
3
3. a. When x 5 , f(x) 5 f a b 5 sin a b 5 # .
3
3
3
2
fr(x) 5 cos x
5-18
p
p
f r a b 5 cos
3
3
1
5
2
p
So an equation for the tangent at the point x 5 is
3
1
p
#3
5 ax 2 b
2
2
3
p
2y 2 #3 5 x 2
3
p
2x 1 2y 1 a 2 #3b 5 0
3
b. When x 5 0, f(x) 5 f(0) 5 0 1 sin (0) 5 0.
f r (x) 5 1 1 cos x
f r (0) 5 1 1 cos (0)
5111
52
So an equation for the tangent at the point x 5 0 is
y 2 0 5 2(x 2 0)
y 5 2x
22x 1 y 5 0
p
p
p
c. When x 5 , f(x) 5 f a b 5 cos a4 ? b
4
4
4
5 cos (p)
5 21
d(4x)
f r (x) 5 2sin (4x) ?
dx
5 24 sin (4x)
p
p
f ra b 5 24 sin a4 ? b
4
4
5 24 sin (p)
50
p
So an equation for the tangent at the point x 5 is
y2
4
y 2 (21) 5 0ax 2
p
b
4
y1150
y 5 21
d. f(x) 5 sin 2x 1 cos x, x 5
p
p
2
The point of contact is ( 2 , 0). The slope of the
tangent line at any point is fr (x) 5 2 cos 2x 2 sin x.
p
At ( 2 , 0), the slope of the tangent line is
p
2 cos p2sin 2 5 23.
p
An equation of the tangent line is y 5 23 x 2 2 .
(
)
Chapter 5: Derivatives of Exponential and Trigonometric Functions
e. f(x) 5 cos a2x 1
5 2 sin ( !t) ? cos ( !t) ?
p
p
b, x 5
3
4
The point of tangency is a , 2
p
4
!3
b. The
2
slope of the
p
tangent line at any point is f r (x) 5 22 sin (2x 1 3 ).
At a , 2
p
4
22 sin a
!3
b,
2
the slope of the tangent line is
5p
b 5 21.
6
An equation of the tangent line is
y1
"3
p
52 x24 .
2
(
)
p
p
p
p
, f(x) 5 f a b 5 2 sin a b cos a b
2
2
2
2
5 2(1)(0)
50
f r(x) 5 (2 sin x)(2sin x) 1 (cos x)(2 cos x)
5 22 sin2 x 1 2 cos2 x
5 2(cos2 x 2 sin2 x)
5 2 cos (2x)
p
p
f ra b 5 2 cos a2 ? b
2
2
5 2 cos p
5 22
p
So an equation for the tangent when x 5 is
f. When x 5
2
p
y 2 0 5 22ax 2 b
2
y 5 22x 1 p
2x 1 y 2 p 5 0
4. a. One could easily find f r (x) and gr (x) to see
that they both equal 2(sin x)(cos x). However, it
is easier to notice a fundamental trigonometric
identity. It is known that sin2 x 1 cos2 x 5 1. So,
sin2 x 5 1 2 cos2 x.
Therefore, one can notice that f(x) is in fact equal
to g(x). So, because f(x) 5 g(x), f r (x) 5 gr (x).
b. f r(x) and gr(x) are negatives of each other.
That is, f r (x) 5 2(sin x)(cos x) while
gr (x) 5 22(sin x) (cos x).
5. a. v(t) 5 (sin ( !t))2
d(sin ( !t))
dt
d( !t)
5 2 sin ( !t) ? cos ( !t) ?
dt
1 212
5 2 sin ( !t) ? cos ( !t) ? t
2
vr(t) 5 2 sin ( !t) ?
Calculus and Vectors Solutions Manual
5
sin ( !t) cos ( !t)
!t
1
2 !t
1
b. v(t) 5 (1 1 cos t 1 sin2 t)2
1
1
vr (t) 5 (1 1 cos t 1 sin2 t)22
2
d(1 1 cos t 1 (sin t)2 )
3
dt
d(sin t)
2sin t 1 2(sin t) ?
dt
5
2# 1 1 cos t 1 sin2 t
2sin t 1 2(sin t)(cos t)
5
2# 1 1 cos t 1 sin2 t
c. h(x) 5 sin x sin 2x sin 3x
So, treat sin x sin 2x as one function, say f(x) and
treat sin 3x as another function, say g(x).
Then, the product rule may be used with the
chain rule:
hr (x) 5 f(x)gr (x) 1 g(x)f r (x)
5 (sin x sin 2x)(3 cos 3x)
1(sin 3x)3 (sin x)(2 cos 2x)
1 (sin 2x)(cos x)4
5 3 sin x sin 2x cos 3x
1 2 sin x sin 3x cos 2x
1 sin 2x sin 3x cos x
d(x 2 1 (cos x)2 )
d. mr (x) 5 3(x 2 1 cos2 x)2 ?
dx
5 3(x2 1 cos2 x)2 ? (2x 1 2(cos x) (2sin x))
5 3(x 2 1 cos2 x)2 ? (2x 2 2 sin x cos x)
6. By the algorithm for finding extreme values, the
maximum and minimum values occur at points on
the graph where f r (x) 5 0, or at an endpoint of the
interval.
dy
5 2sin x 1 cos x
a.
dx
Set
dy
50
dx
and solve for x to find any critical points.
cos x 2 sin x 5 0
cos x 5 sin x
sin x
15
cos x
1 5 tan x
p 5p
x5 ,
4 4
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
5-19
x
0
p
4
5p
4
2p
f(x) 5 cos x 1 sin x
1
"2
2"2
1
Set
So, the absolute maximum value on the interval is
# 2 when x 5
p
4
and the absolute minimum value
on the interval is 2 # 2 when x 5
5p
.
4
y
2
1
x
0
p
–1
2p
–2
b.
dy
5 1 2 2 sin x
dx
Set
dy
50
dx
x
2p
f(x) 5 x 1 2 cos x
2p 2 2
2
p
6
p
6
p
p
1 #3
1 #3
6
6
8 25.14 8 1.21
8 2.26
2
p
p22
8 1.14
So, the absolute maximum value on the interval is
p
2.26 when x 5 and the absolute minimum value
6
on the interval is 25.14 when x 5 2p.
y
4
–p
2
0
–4
–8
5-20
and solve for x to find any critical points.
cos x 1 sin x 5 0
sin x 5 2cos x
sin x
5 21
cos x
tan x 5 21
3p 7p
,
x5
4 4
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
x
0
3p
4
7p
4
2
f(x) 5 sin x 2 cos x
21
"2
2"2
21
# 2 when x 5 4 and the absolute minimum value
3p
1 2 2 sin x 5 0
1 5 2 sin x
1
5 sin x
2
p 5p
x5 ,
6 6
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
–p
dy
50
dx
So, the absolute maximum value on the interval is
and solve for x to find any critical points.
8
dy
5 cos x 1 sin x
dx
c.
x
p
2
p
on the interval is 2 # 2 when x 5
7p
.
4
y
2
1
x
0
p
–1
2p
–2
d.
dy
5 3 cos x 2 4 sin x
dx
Set
dy
50
dx
and solve for x to find any critical points.
3 cos x 2 4 sin x 5 0
3 cos x 5 4 sin x
sin x
3
5
cos x
4
3
5 tan x
4
3
tan21 a b 5 tan21 (tan x)
4
Using a calculator, x 8 0.6435.
This is a critical value, but there is also one more in
the interval 0 # x # 2p. The period of tan x is p,
so adding p to the one solution will give another
solution in the interval.
x 5 0.6435 1 p 8 3.7851
Chapter 5: Derivatives of Exponential and Trigonometric Functions
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
0
x
f(x) 5 3 sin x 1 4 cos x
0.64
4
3.79
y
8
4
x
p
–4
2p
–8
7. a. The particle will change direction when the
velocity, sr (t), changes from positive to negative.
sr(t) 5 16 cos 2t
Set sr (t) 5 0 and solve for t to find any critical points.
0 5 16 cos 2t
0 5 cos 2t
p 3p
,
5 2t
2 2
p 3p
,
5t
4 4
Also, there is no given interval so it will be beneficial
to locate all solutions.
p
3p
1 pk for some positive
Therefore, t 5 1 pk,
4
4
integer k constitutes all solutions.
One can create a table and notice that on each side
of any value of t, the function is increasing on one
side and decreasing on the other. So, each t value is
either a maximum or a minimum.
t
p
4
3p
4
5p
4
7p
4
s(t) 5 8 sin 2t
8
28
8
28
The table continues in this pattern for all critical
values t. So, the particle changes direction at all
critical values. That is, it changes direction for
t5
p
3p
1 pk,
1 pk
4
4
for positive integers k.
b. From the table or a graph, one can see that
the particle’s maximum velocity is 8 at the time
p
t 5 1 pk.
4
Calculus and Vectors Solutions Manual
f(x)
1
x
0
4
So, the absolute maximum value on the interval is 5
when x 8 0.64 and the absolute minimum value on
the interval is 25 when x 8 3.79.
0
2
2p
25
5
8. a.
p
–1
2p
–2
b. The tangent to the curve f(x) is horizontal at the
point(s) where f r (x) is zero.
f r (x) 5 2sin x 1 cos x
Set f r (x) 5 0 and solve for x to find any critical
points.
cos x 2 sin x 5 0
cos x 5 sin x
sin x
15
cos x
1 5 tan x
x5
p
4
(Note: The solution x 5
5p
4
is not in the
interval 0 # x # p so it is not included.) When
p
4
x 5 , f(x) 5 f
( 4 ) 5 # 2.
p
So, the coordinates of the point where the tangent to
p
the curve of f(x) is horizontal is ( , # 2).
4
1
5 (sin x)21
9. csc x 5
sin x
1
sec x 5
5 (cos x)21
cos x
Now, the power rule can be used to compute the
derivates of csc x and sec x.
d(sin x)
((sin x)21 )r 5 2 (sin x)22 ?
dx
5 2 (sin x)22 ? cos x
cos x
52
(sin x)2
d(sin x)
((sin x)21 )r 5 2 (sin x)22 ?
dx
cos x
1
?
52
sin x sin x
5 2csc x cot x
d(cos x)
((cos x)21 )r 5 2 (cos x)22 ?
dx
5 2 (cos x)22 ? (2sin x)
sin x
5
(cos x)2
1
sin x
5
?
cos x cos x
5 sec x tan x
5-21
function is 1, so the maximum distance from the
origin is 4(1) or 4.
12.
dy
5 22 sin 2x
dx
p 1
At the point , ,
10.
( 6 2)
h
dy
p
5 22 sin a2 ? b
dx
6
p
5 22 sin a b
3
5 22a
cos u 5
x
5x
1
and sin u 5
area of a trapezoid is
(p 1)
Therefore, at the point 6 , 2 , the slope of the
tangent to the curve y 5 cos 2x is 2 #3.
11. a. The particle will change direction when the
velocity, sr(t) changes from positive to negative.
sr(t) 5 16 cos 4t
Set sr (t) 5 0 and solve for t to find any critical points.
0 5 16 cos 4t
0 5 cos 4t
p 3p
,
5 4t
2 2
p 3p
,
5t
8 8
Also, there is no given interval so it will be beneficial
to locate all solutions.
p
3p
1 pk,
1 pk
8
8
for some positive
integer k constitutes all solutions.
One can create a table and notice that on each side
of any value of t, the function is increasing on one
side and decreasing on the other. So, each t value is
either a maximum or a minimum.
t
p
8
3p
8
5p
8
7p
8
s(t) 5 4 sin 4t
4
24
4
24
The table continues in this pattern for all critical
values t. So, the particle changes direction at all
critical values. That is, it changes direction for
p
3p
1 pk,
1 pk
4
4
for positive integers k.
b. From the table or a graph, one can see that the
particle’s maximum velocity is 4 at the time
p
t 5 1 pk.
4
c. At t 5 0, s 5 0, so the minimum distance from
the origin is 0. The maximum value of the sine
5-22
h
h
5 h.
1
Therefore, x 5 cos u and h 5 sin u.
The irrigation channel forms a trapezoid and the
5 2#3
t5
1m
1m
u
x
Label the base of a triangle x and the height h. So
!3
b
2
Therefore, t 5
1m
u
x
(b1 1 b2 )h
2
where b1 and b2 are
the bottom and top bases of the trapezoid and h is
the height.
b1 5 1
b2 5 x 1 1 1 x 5 cos u 1 1 1 cos u 5 2 cos u 1 1
h 5 sin u
Therefore, the area equation is given by
(2 cos u 1 1 1 1) sin u
A5
2
(2 cos u 1 2) sin u
5
2
2 cos u sin u 1 2 sin u
5
2
5 sin u cos u 1 sin u
To maximize the cross-sectional area, differentiate:
Ar 5 (sin u)(2sin u) 1 (cos u)(cos u) 1 cos u
5 2sin2 u 1 cos2 u 1 cos u
Using the trig identity sin2 u 1 cos2 u 5 1, use the
fact that sin2 u 5 1 2 cos2 u.
Ar 5 2 (1 2 cos2 u) 1 cos2 u 1 cos u
5 21 1 cos2 u 1 cos2 u 1 cos u
5 2 cos2 u 1 cos u21
Set Ar 5 0 to find the critical points.
0 5 2 cos2 u 1 cos u 2 1
0 5 (2 cos u21)(cos u 1 1)
Solve the two expressions for u.
2 cos u 5 1
1
cos u 5
2
p
u5
3
Also, cos u 5 21
u5p
(Note: The question only seeks an answer around
p
0 # u # . So, there is no need to find all solutions
2
by adding kp for all integer values of k.)
The area, A, when u 5 p is 0 so that answer is
disregarded for this problem.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
p
!3
2!3
1
4
4
3 !3
5
4
p
The area is maximized by the angle u 5 .
14. First find ys.
y 5 A cos kt 1 B sin kt
yr 5 2kA sin kt 1 kB cos kt
ys 5 2k 2A cos kt 2 k 2B sin kt
So, ys 1 k 2y
5 2k 2A cos kt 2 k 2B sin kt
1 k 2 (A cos kt 1 B sin kt)
5 2k 2A cos kt 2 k 2B sin kt 1 k 2A cos kt
1 k 2B sin kt
50
Therefore, ys 1 k 2y 5 0.
13. Let O be the centre of the circle with line
segments drawn and labeled, as shown.
5.5 The Derivative of y 5 tan x, p. 260
When u 5 ,
3
p
p
p
A 5 sin cos 1 sin
3
3
3
!3 1
!3
5a
? b1
2
2
2
5
3
A
1. a.
uu
R
R
x 2u
y
C
B
In ^OCB, /COB 5 2u.
Thus,
y
5 sin 2u
R
and
x
5 cos 2u,
R
so y 5 R sin 2u and x 5 R cos 2u.
The area A of ^ ABD is
1
A 5 0 DB 0 0 AC 0
2
5 y(R 1 x)
5 R sin 2u(R 1 R cos 2u)
5 R 2 (sin 2u 1 sin 2u cos 2u), where 0 , 2u , p
dA
5 R 2 (2 cos 2u 1 2 cos 2u cos 2u
du
1 sin 2u(22 sin 2u)).
We solve
dA
5 0:
du
2 cos2 2u 2 2 sin2 2u 1 2 cos 2u 5 0
2 cos2 2u 1 cos 2u 2 1 5 0
(2 cos 2u21)(cos 2u 1 1) 5 0
1
cos 2u 5 or cos 2u 5 21
2
p
2u 5 or 2u 5 p (not in domain).
3
As 2u S 0, A S 0 and as 2u S p, A S 0. The
maximum area of the triangle is
p
3
d
dy
5 2 sec2 x 2 sec 2xa 2xb
dx
dx
5 2 sec2 x 2 2 sec 2x
d
dy
c.
5 2 tan (x 3 )a tan (x 3 )b
dx
dx
d
dy
5 2 tan (x 3 )a tan (x 3 )b
dx
dx
d
5 2 tan (x 3 ) sec2 (x 3 )a x 3 b
dx
5 6x 2 tan (x 3 ) sec2 (x 3 )
b.
O
D
d
dy
5 sec2 3xa 3xb
dx
dx
5 3 sec2 3x
p
6
3 !3 2
R
4
when 2u 5 , i.e., u 5 .
Calculus and Vectors Solutions Manual
(d )
2x tan px 2 x 2 sec2 px px
dy
dx
d.
5
dx
tan2 px
2x tan px 2 px 2 sec2 px
5
tan2 px
x(2 tan px 2 px sec2 px)
5
tan2 px
d
d
dy
e.
5 sec2 (x 2 )a x 2 b 2 2 tan xa b (tan x)
dx
dx
dx
5 2x sec2 (x 2 ) 2 2 tan x sec2 x
dy
d
f.
5 tan 5x(3 cos 5x)a 5xb
dx
dx
d
1 3 sin 5x sec2 5xa 5xb
dx
5 15 tan 5x cos 5x 1 15 sin 5x sec2 5x
5 15 (tan 5x cos 5x 1 sin 5x sec2 5x)
2. a. The general equation for the line tangent to
the function f(x) at the point (a, b) is
y 2 b 5 fr (x)(x 2 a).
f(x) 5 tan x
fr (x) 5 sec2 x
5-23
p
fa b 5 0
4
p
f ra b 5 2
4
The equation for the line tangent to the function
p
(
p
)
f(x) at x 5 4 is y 5 2 x 2 4 .
b. The general equation for the line tangent to the
function f(x) at the point (a, b) is
y 2 b 5 f r (x)(x 2 a).
f(x) 5 6 tan x 2 tan 2x
d
f r(x) 5 6 sec2 x 2 sec2 2xa 2xb
dx
f r(x) 5 6 sec2 x 2 2 sec2 2x
f(0) 5 0
f r(0) 5 22
The equation for the line tangent to the function
f(x) at x 5 0 is y 5 22x.
d
dy
5 sec2 x(sin x)a sin xb
3. a.
dx
dx
5 cos x sec2 (sin x)
d
dy
5 22 3tan (x 2 2 1)4 23 a tan (x 2 2 1)b
b.
dx
dx
5 22 3tan (x 2 2 1)4 23 sec2 (x 2 2 1)
d
3 a (x 2 2 1)b
dx
5 24x 3tan (x 2 2 1)4 23 sec2 (x 2 2 1)
d
dy
5 2 tan (cos x)a tan (cos x)b
c.
dx
dx
d
5 2 tan (cos x) sec2 (cos x)a cos xb
dx
5 22 tan (cos x) sec2 (cos x) sin x
dy
d
5 2 (tan x 1 cos x)a tan x 1 cos xb
d.
dx
dx
5 2 (tan x 1 cos x)(sec2 x 2 sin x)
dy
d
5 tan x (3 sin2 x)a sin xb 1 sin3 x sec2 x
e.
dx
dx
2
5 3 tan x sin x cos x 1 sin3 x sec2 x
5 sin2 x (3 tan x cos x 1 sin x sec2 x)
dy
d
5 e tan "x a tan "xb
f.
dx
dx
d
5 e tan "x (sec2 "x)a "xb
dx
1
5
e tan "x sec2 "x
2"x
5-24
dy
5 tan x cos x 1 sin x sec2 x
dx
sin x
1
5
? cos x 1 sin x ?
cos x
cos2 x
sin x
5 sin x 1
cos2 x
2
cos3 x
dy
5
cos
x
1
dx 2
cos4 x
4. a.
)
(d
sin x(2 cos x) dx cos x
2
cos4 x
cos3 x 1 2 sin2 x cos x
5 cos x 1
cos4 x
1
2 sin2 x
5 cos x 1
1
cos x
cos3 x
2 sin2 x
5 cos x 1 sec x 1
cos3 x
d
dy
5 2 tan xa tan xb
b.
dx
dx
5 2 tan x sec2 x
2 sin x
1
5
?
cos x cos2 x
2 sin x
5
cos3 x
(d
)
2 cos4 x 2 6 sin x cos2 x dxcos x
d 2y
5
dx 2
cos6 x
4
2 cos x 1 6 sin2 x cos2 x
5
cos6 x
6 sin2 x
1
2
1
?
5
2
2
cos x
cos x cos2 x
2
5 2 sec x 1 6 tan2 x sec2 x
5 2 sec2 x(1 1 3 tan2 x)
5. The slope of f(x) 5 sin x tan x equals zero when
the derivative equals zero.
f(x) 5 sin x tan x
f r (x) 5 sin x(sec2x) 1 tan x(cos x)
sin x
5 sin x(sec2 x) 1
(cos x)
cos x
5 sin x(sec2 x) 1 sin x
5 sin x(sec2 x 1 1)
2
sec x 1 1 is always positive, so the derivative is 0
only when sin x 5 0. So, fr (x) equals 0 when x 5 0,
x 5 p, and x 5 2p. The solutions can be verified
by examining the graph of the derivative function
shown below.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
y
4
3
2
1
0
–1
–2
–3
–4
cos2 x 1 sin x 1 sin2 x
cos2 x
1 1 sin x
5
cos2 x
The denominator is never negative. 1 1 sin x . 0
5
f'(x)
x
␲
p
p
for 2 2 , x , 2 , since sin x reaches its minimum
2␲
p
of 21 at x 5 2 . Since the derivative of the original
function is always positive in the specified interval,
the function is always increasing in that interval.
p
p
8. When x 5 , y 5 2 tan ( 4 )
4
52
yr 5 2 sec2 x
p
p 2
When x 5 , yr 5 2 asec b
4
4
6. The local maximum point occurs when the
derivative equals zero.
dy
5 2 2 sec2 x
dx
2 2 sec2 x 5 0
sec2 x 5 2
sec x 5 6"2
p
x56
4
dy
p
5 0 when x 5 6 4 , so the local maximum
dx
p
point occurs when x 5 6 4 . Solve for y
when
5 2( #2)2
54
p
So an equation for the tangent at the point x 5 is
4
p
y 2 2 5 4ax 2 b
4
y 2 2 5 4x 2 p
24x 1 y 2 (2 2 p) 5 0
sin x
9. Write tan x 5 cos x and use the quotient rule to
derive the derivative of the tangent function.
10. y 5 cot x
1
y5
tan x
tan x(0) 2 (1) sec2 x
dy
5
dx
tan2 x
2
2sec x
5
tan2 x
p
x 5 4.
p
p
y 5 2a b 2 tan a b
4
4
p
y5 21
2
y 5 0.57
p
4
Solve for y when x 5 2 .
p
p
y 5 2a2 b 2 tan a2 b
4
4
p
y52 11
2
y 5 20.57
p
The local maximum occurs at the point 4 , 0.57 .
7. y 5 sec x 1 tan x
1
sin x
5
1
cos x
cos x
1 1 sin x
5
cos x
cos2 x 2 (1 1 sin x)(2sin x)
dy
5
dx
cos2 x
2
cos x 2 (2sin x 2 sin2 x)
5
cos2 x
(
Calculus and Vectors Solutions Manual
)
21
cos 2 x
5 sin2 x
cos 2 x
21
sin2 x
5 2csc2 x
11. Using the fact from question 10 that the
derivative of cot x is 2csc2 x,
f r (x) 5 24 csc2 x
5 24 (csc x)2
d(csc x)
f s (x) 5 28 (csc x) ?
dx
5 28 (csc x) ? (2csc x cot x)
5 8 csc2 x cot x
5
5-25
Review Exercise, pp. 263–265
1. a. yr 5 0 2 e x
5 2e x
b. yr 5 2 1 3e x
d(2x 1 3)
c. yr 5 e 2x13 ?
dx
5 2e 2x13
d(23x 2 1 5x)
2
d. yr 5 e 23x 15x ?
dx
2
5 (26x 1 5)e 23x 15x
e. yr 5 (x)(e x ) 1 (e x )(1)
5 e x (x 1 1)
(e t 1 1)(e t ) 2 (e t 2 1)(e t )
f. sr 5
(e t 1 1)2
2t
t
e 1 e 2 (e 2t 2 e t )
5
(e t 1 1)2
t
2e
5 t
(e 1 1)2
dy
5 10x ln 10
2. a.
dx
d(3x 2 )
dy
2
5 43x ? ln 4 ?
b.
dx
dx
3x 2
5 6x(4 )ln 4
dy
5 (5x)(5x ln 5) 1 (5x )(5)
c.
dx
5 5 ? 5x (x ln 5 1 1)
dy
5 (x 4 )(2x ln 2) 1 (2x )(4x 3 )
d.
dx
5 x 3 ? 2x (x ln 2 1 4)
e. y 5 (4x)(42x )
dy
5 (4x)(242x ln 4) 1 (42x )(4)
dx
5 4 ? 42x (2x ln 4 1 1)
4 2 4x ln 4
5
4x
x
f. y 5 (5# )(x 21 )
dy
d( #x)
5 (5#x )(2x 22 ) 1 (x 21 )a5#x ? ln 5 ?
b
dx
dx
1
1
5 (5#x )a2 2 b 1 (x 21 )a5#x ? ln 5 ?
b
x
2!x
1
ln 5
b
5 5#x a2 2 1
x
2x!x
d(2x)
d(2x)
dy
5 3 cos (2x) ?
1 4 sin (2x) ?
3. a.
dx
dx
dx
5 6 cos (2x) 1 8 sin (2x)
5-26
d(3x)
dy
5 sec2 (3x) ?
dx
dx
5 3 sec2 (3x)
c. y 5 (2 2 cos x)21
dy
d(2 2 cos x)
5 2 (2 2 cos x)22 ?
dx
dx
sin x
52
(2 2 cos x)2
d(2x)
dy
5 (x)asec2 (2x) ?
b 1 (tan (2x))(1)
d.
dx
dx
5 2x sec2 (2x) 1 tan 2x
d(3x)
dy
5 (sin 2x)ae 3x ?
b
e.
dx
dx
d(2x)
b
1 (e 3x )acos 2x ?
dx
5 3e 3x sin 2x 1 2e 3x cos 2x
5 e 3x (3 sin 2x 1 2 cos 2x)
f. y 5 (cos (2x))2
d(cos (2x))
dy
5 2 (cos (2x)) ?
dx
dx
d(2x)
5 2(cos (2x)) ? 2sin (2x) ?
dx
5 24 cos (2x) sin (2x)
b.
4. a. f(x) 5 e x ? x 21
f r (x) 5 (e x )(2x 22 ) 1 (x 21 )(e x )
1
1
5 e x a2 2 1 b
x
x
2
2x
1
x
b
5 ex a
x3
Now, set f r (x) 5 0 and solve for x.
2x 1 x 2
0 5 ex a
b
x3
x2 2 x
5 0.
Solve e x 5 0 and
x3
x
e is never zero.
x2 2 x
50
x3
x2 2 x 5 0
x(x 2 1) 5 0
So x 5 0 or x 5 1.
(Note, however, that x cannot be zero because this
would cause division by zero in the original function.)
So x 5 1.
b. The function has a horizontal tangent at (1, e).
Chapter 5: Derivatives of Exponential and Trigonometric Functions
d(22x)
b 1 (e 22x )(1)
dx
5 22xe 22x 1 e 22x
5 e 22x (22x 1 1)
1
1
1
f ra b 5 e 22 2 a22 ? 1 1b
2
2
5 e 21 (21 1 1)
50
b. This means that the slope of the tangent to f(x)
at the point with x-coordinate 21 is 0.
6. a. yr 5 (x)(e x ) 1 (e x )(1) 2 e x
5 xe x
ys 5 (x)(e x ) 1 (e x )(1)
5 xe x 1 e x
5 e x (x 1 1)
b. yr 5 (x)(10e 10x ) 1 (e 10x )(1)
5 10xe 10x 1 e 10x
ys 5 (10x)(10e 10x ) 1 (e 10x )(10) 1 10e 10x
5 100xe 10x 1 10e 10x 1 10e 10x
5 100xe 10x 1 20e 10x
5 20e 10x (5x 1 1)
e 2x 2 1
7. y 5 2x
e 11
dy
2e 2x (e 2x 1 1) 2 (e 2x 2 1)(2e 2x )
5
dx
(e 2x 1 1)2
4x
2x
2e 1 2e 2 2e 4x 1 2e 2x
5
(e 2x 1 1)2
4e 2x
5 2x
(e 1 1)2
5. a. f r (x) 5 (x)ae 22x ?
Now, 1 2 y 2 5 1 2
4x
5
e 4x 2 2e 2x 1 1
(e
2x
2x
2
1 1)
4x
The equation of the tangent line is
y 1 ln 2 1 2 5 3(x 1 ln 2) or
3x 2 y 1 2 ln 2 2 2 5 0.
p
2
9. When x 5 ,
p
p
p
p
p
y 5 f(x) 5 f a b 5 sina b 5 (1) 5
2
2
2
2
2
yr 5 fr (x) 5 (x)(cos x) 1 (sin x)(1)
5 x cos x 1 sin x
p
p
p
p
f r a b 5 cos 1 sin
2
2
2
2
p
5 (0) 1 1
2
51
p
So an equation for the tangent at the point x 5 is
2
p
p
y 2 5 1ax 2 b
2
2
p
p
y2 5x2
2
2
y5x
2x 1 y 5 0
10. If s(t) 5
(
Calculus and Vectors Solutions Manual
)(
p
)
(
)
p
p
(
sin 4 )(22 sin 2 ? 4 )
2
(3 1 cos 2 ? p4 )
2
5
To find the point(s) on the curve where the tangent
has slope 3, we solve:
1 1 e 2x 5 3
e 2x 5 2
2x 5 ln 2
x 5 2ln 2.
The point of contact of the tangent is
(2ln 2,2ln 2 2 2).
p
3 1 cos 2 ? 4 cos 4
p
sr a b 5
4
p 2
3 1 cos 2 ? 4
e 1 2e 1 1 2 e 1 2e 2 1
(e 2x 1 1)2
dy
2x
dx 5 1 1 e .
is the function describing
an object’s position at time t, then v(t) 5 sr (t) is
the function describing the object’s velocity at
time t. So
v(t) 5 sr (t)
(3 1 cos 2t)(cos t) 2 (sin t)(22 sin 2t)
5
(3 1 cos 2t)2
2x
4e 2x
dy
2x
2 5
(3 1 1)
dx
8. The slope of the required tangent line is 3.
The slope at any point on the curve is given by
sin t
3 1 cos 2t
(3 1 cos p2 )("22) 2 ("22)(22 sin p2 )
5
(3 1 cos p2 )2
"2
(3 1 0)("2
2 ) 2 ( 2 )(22 ? 1)
5
(3 1 0)2
1 #2
9
3#2 1 2#2 1
5
?
2
9
5#2
5
18
5
3"2
2
5-27
So, the object’s velocity at time t 5
5#2
8 0.3928
18
p
4
Since e 2t . 0 for all t, c1r(t) 5 0 when t 5 1.
Since c1r(t) . 0 for 0 # t , 1, and c1r(t) , 0 for all
is
metres per unit of time.
1
11. a. The question asks for the time t when
Nr(t) 5 0.
t
N(t) 5 60 000 1 2000te220
1 t
t
Nr(t) 5 0 1 (2000t)a2 e 220 b 1 (e 220 )(2000)
20
t
t
5 2100te 220 1 2000e 220
t
5 100e 220 (2t 1 20)
Set Nr (t) 5 0 and solve for t.
t
0 5 100e 220 (2t 1 20)
t
100e 220 is never equal to zero.
2t 1 20 5 0
20 5 t
Therefore, the rate of change of the number of
bacteria is equal to zero when time t 5 20.
b. The question asks for
dM
5 Mr(t)
dt
when t 5 10.
That is, it asks for Mr (10).
1
M(t) 5 (N 1 1000)3
d(N 1 1000)
1
2
Mr (t) 5 (N 1 1000)23 ?
3
dt
dN
1
5
2 ?
3(N 1 1000)3 dt
From part a.,
t
dN
5 Nr(t) 5 100e 220 (2t 1 20)
dt
220t
and
N(t) 5 60 000 1 2000te
1
100e220 (2t 1 20)
So Mr (t) 5
2
3(N 1 1000)3
First calculate N(10).
10
N(10) 5 60 000 1 2000(10)e220
1
5 60 000 1 20 000e 22
8 72 131
10
100e 220 (210 1 20)
So Mr (10) 5
2
3(N(10) 1 1000)3
1
100e 22 (10)
5
2
3(72 131 1 1000)3
606.53
8
5246.33
8 0.1156
So, after 10 days, about 0.1156 mice are infected
per day. Essentially, almost 0 mice are infected per
day when t 5 10.
12. a. c1 (t) 5 te 2t; c1 (0) 5 0
c1r (t) 5 e 2t 2 te 2t
5 e 2t (1 2 t)
5-28
t . 1, c1 (t) has a maximum value of e 8 0.368
at t 5 1 h.
c2 (t) 5 t 2e 2t; c2 (0) 5 0
c2r(t) 5 2te 2t 2 t 2e 2t
5 te 2t (2 2 t)
c2r(t) 5 0 when t 5 0 or t 5 2.
Since c2r(t) . 0 for 0 , t , 2 and c2r(t) , 0 for all
4
t . 2, c2 (t) has a maximum value of e 2 8 0.541 at
t 5 2 h. The larger concentration occurs for
medicine c2.
b. c1 (0.5) 5 0.303
c2 (0.5) 5 0.152
In the first half-hour, the concentration of c1
increases from 0 to 0.303, and that of c2 increases
from 0 to 0.152. Thus, c1 has the larger concentration
over this interval.
13. a. y 5 (2 1 3e 2x )3
yr 5 3(2 1 3e 2x )2 30 1 3e 2x (21)4
5 3(2 1 3e 2x )2 (23e 2x )
5 29e 2x (2 1 3e 2x )2
b. y 5 x e
yr 5 ex e21
x
c. y 5 e e
x
yr 5 e e (e x )(1)
x
5 e x1e
d. y 5 (1 2 e 5x )5
yr 5 5(1 2 e 5x )4 30 2 e 5x (5)4
5 225e 5x (1 2 e 5x )4
14. a. y 5 5x
yr 5 5x ln 5
b. y 5 (0.47)x
yr 5 (0.47)x ln (0.47)
c. y 5 (52)2x
yr 5 (52)2x (2) ln 52
5 2(52)2x ln 52
d. y 5 5(2)x
yr 5 5(2)x ln 2
e. y 5 4e x
yr 5 4e x (1) ln e
5 4e x
f. y 5 22(10)3x
yr 5 22(3)103x ln 10
5 26(10)3x ln 10
d(2x )
15. a. yr 5 cos 2x ?
dx
5 2x ln 2 cos 2x
Chapter 5: Derivatives of Exponential and Trigonometric Functions
b. yr 5 (x 2 )(cos x) 1 (sin x)(2x)
5 x 2 cos x 1 2x sin x
d( p 2 x)
p
c. yr 5 cos a 2 xb ? 2
2
dx
p
5 2cos a 2 xb
2
d. yr 5 (cos x)(cos x) 1 (sin x)(2sin x)
5 cos2 x 2 sin2 x
e. y 5 (cos x)2
d(cos x)
yr 5 2(cos x) ?
dx
5 22 cos x sin x
f. y 5 cos x (sin x)2
yr 5 (cos x)(2 (sin x)(cos x)) 1 (sin x)2 (2sin x)
5 2 sin x cos2 x 2 sin3 x
16. Compute
dy
dx
when x 5
p
2
to find the slope of the
p
dv
5 210 cos 2t 1
dt
4
p
5 220 cos a2t 1 b.
4
The maximum values of the displacement,
velocity, and acceleration are 5, 10, and 20,
respectively.
(
and a 5
(
p
19. Let the base angle be u, 0 , u , , and let the
2
sides of the triangle have lengths x and y, as shown.
Let the perimeter of the triangle be P cm.
12
y
u
x
x
5 cos u
12
y
5 sin u
12
line at the given point.
yr 5 2sin x
p
So, at the point x 5 , yr 5 f r(x) is
Now,
p
p
f ra b 5 2sin a b 5 21.
2
2
Therefore, an equation of the line tangent to the
curve at the given point is
p
y 2 0 5 21ax 2 b
2
p
y 5 2x 1
2
p
x1y2 50
2
17. The velocity of the object at any
dP
5 212 sin u 1 12 cos u.
du
ds
.
dt
Thus, v 5 8 (cos (10pt))(10p)
5 80p cos (10pt).
The acceleration at any time t is a 5
dv
d 2s
5 2.
dt
dt
Hence, a 5 80p(2sin (10pt))(10p) 5
2800p2 sin (10pt).
d 2s
Now, dt 2 1 100p2s
5 2800p2 sin (10pt)
1 100p2 (8 sin (10pt)) 5 0.
p
18. Since s 5 5 cos 2t 1 ,
(
v5
4
)
p
ds
5 5 a 2sin a 2t 1 bb
4
dt
p
5 210 sin a2t 1 b
4
Calculus and Vectors Solutions Manual
and
so x 5 12 cos u and y 5 12 sin u.
Therefore, P 5 12 1 12 cos u 1 12 sin u and
2
time t is v 5
))
For critical values, 212 sin u 1 12 cos u 5 0
sin u 5 cos u
tan u 5 1
p
p
u 5 , since 0 , u , .
4
2
12
p
12
When u 5 , P 5 12 1 !2 1 !2
4
24
5 12 1
!2
5 12 1 12#2.
As u S 0 1 , cos u S 1, sin u S 0, and
P S 12 1 12 1 0 5 24.
p
As u S , cos u S 0, sin u S 1 and
2
P S 12 1 0 1 12 5 24.
Therefore, the maximum value of the perimeter is
12 1 12!2 cm, and occurs when the other two
p
angles are each rad, or 45°.
4
20. Let l be the length of the ladder, u be the angle
between the foot of the ladder and the ground, and x
be the distance of the foot of the ladder from the
fence, as shown.
Thus,
x11
5 cos u
l
1.5
and x 5 tan u
1.5
x 1 1 5 l cos u where x 5 tan u.
5-29
dl
Solving du 5 0 yields:
l
0.8 sin3 u 2 cos3 u 5 0
tan3 u 5 1.25
3
tan u 5 !
1.25
tan u 8 1.077
u 8 0.822.
wall
1.5
u
0.8
1
x
1.5
1 1 5 l cos u
tan u
Replacing x,
1.5
1
p
1
,0,u,
sin u
cos u
2
dl
1.5 cos u
sin u
52
1
du
sin2 u
cos2 u
3
21.5 cos u 1 sin3 u
.
5
sin2 u cos2 u
dl
5 0 yields:
Solving
du
sin3 u 2 1.5 cos3 u 5 0
tan3 u 5 1.5
3
tan u 5 !
1.5
u 8 0.46365.
The length of the ladder corresponding to this value
p2
of u is l 8 4.5 m. As u S 0 1 and 2 , l increases
without bound. Therefore, the shortest ladder that
goes over the fence and reaches the wall has a
length of 4.5 m.
21. The longest pole that can fit around the corner is
determined by the minimum value of x 1 y. Thus,
we need to find the minimum value of l 5 x 1 y.
l5
0.8
x
u
1
0.8
1
From the diagram, y 5 sin u and x 5 cos u.
1
0.8
p
l 5 cos u 1 sin u, 0 # u # 2 :
dl
1 sin u
0.8 cos u
5
2
2
du
cos u
sin2 u
3
0.8 sin u 2 cos3 u
5
.
cos2 u sin2 u
5-30
A
6
3
B
C
x
a
u
b
D
9
3
From the diagram, tan a 5 x and tan b 5 x.
Hence,
9
3
x2x
tan u 5
27
1 1 x2
9x 2 3x
x 2 1 27
6x
5 2
.
x 1 27
We differentiate implicitly with respect to x:
du
6(x 2 1 27) 2 6x(2x)
sec2 u
5
dx
(x 2 1 27)2
du
162 2 6x 2
5
dx
sec2 u(x 2 1 27)2
5
y
u
Thus,
1
Now, l 5 cos (0.822) 1 sin (0.822) 8 2.5.
When u 5 0, the longest possible pole would have a
p
length of 0.8 m. When u 5 2 , the longest possible
pole would have a length of 1 m. Therefore, the
longest pole that can be carried horizontally around
the corner is one of length 2.5 m.
22. We want to find the value of x that maximizes u.
Let /ADC 5 a and /BDC 5 b.
Thus, u 5 a 2 b:
tan u 5 tan (a 2 b)
tan a 2 tan b
5
.
1 1 tan a tan b
du
Solving dx 5 0 yields:
162 2 6x 2 5 0
x 2 5 27
x 5 3#3.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
23. a. f (x) 5 4 (sin (x 2 2))2
f r (x) 5 8 sin (x 2 2) cos (x 2 2)
f s (x) 5 (8 sin (x 2 2))(2sin (x 2 2))
1 (cos (x 2 2))(8 cos (x 2 2))
5 28 sin2 (x 2 2) 1 8 cos2 (x 2 2)
b. f(x) 5 (2 cos x)(sec x)2
f r(x) 5 (2 cos x)(2 sec x ? sec x tan x)
1 (sec x)2 ( 2 2 sin x)
5 (4 cos x)(sec2 x tan x) 2 2 sin x (sec x)2
Using the product rule multiple times,
f s (x) 5 (4 cos x) S sec2 x ? sec2 x
1 tan x (2 sec x ? sec x tan x) T
1 (sec2 x tan x)(24 sin x)
1 (22 sin x)(2 sec x ? sec x tan x)
1 (sec x)2 (22 cos x)
5 4 cos x sec4 x 1 8 cos x tan2 x sec2 x
2 4 sin x tan x sec2 x 2 4 sin x tan x sec2 x
2 2 cos x sec2 x
5 4 cos x sec4 x 1 8 cos x tan2 x sec2 x
2 8 sin x tan x sec2 x 2 2 cos x sec2 x
Chapter 5 Test, p. 266
2
1. a. y 5 e 22x
dy
2
5 24xe 22x
dx
2
b. y 5 3x 13x
dy
2
5 3x 13x ? ln 3 ? (2x 1 3)
dx
e 3x 1 e 23x
c. y 5
2
1 3x
dy
5 33e 2 3e 23x4
dx
2
3
5 3e 3x 2 e 23x4
2
d. y 5 2 sin x 2 3 cos 5x
dy
5 2 cos x 2 3(2sin 5x)(5)
dx
5 2 cos x 1 15 sin 5x
e. y 5 sin3 (x 2 )
dy
5 3 sin2 (x 2 )(cos (x 2 )(2x))
dx
5 6x sin2 (x 2 ) cos (x 2 )
f. y 5 tan "1 2 x
dy
1
1
5 sec2 "1 2 x a 3
b (21)
dx
2
"1 2 x
sec2 "1 2 x
52
2"1 2 x
Calculus and Vectors Solutions Manual
2. The given line is 26x 1 y 5 2 or y 5 6x 1 2, so
the slope is 6.
y 5 2e 3x
dy
5 2e 3x (3)
dx
5 6e 3x
In order for the tangent line to be parallel to the
given line, the derivative has to equal 6 at the
tangent point.
6e 3x 5 6
e 3x 5 1
x50
When x 5 0, y 5 2.
The equation of the tangent line is y 2 2 5 6(x 2 0)
or 26x 1 y 5 2. The tangent line is the given line.
3. y 5 e x 1 sin x
dy
5 e x 1 cos x
dx
dy
When x 5 0, dx 5 1 1 1 or 2, so the slope of the
tangent line at (0, 1) is 2.
The equation of the tangent line at (0, 1) is
y 2 1 5 2(x 2 0) or 22x 1 y 5 1.
4. v(t) 5 10e2kt
a. a(t) 5 vr (t) 5 210ke 2kt
5 2k(10e 2kt )
5 2kv(t)
Thus, the acceleration is a constant multiple of the
velocity. As the velocity of the particle decreases,
the acceleration increases by a factor of k.
b. At time t 5 0, v 5 10 cm> s.
c. When v 5 5, we have 10e 2kt 5 5
1
e 2kt 5
2
1
2kt 5 ln a b 5 2ln 2
2
ln 2
t5
.
k
After
ln 2
k
s have elapsed, the velocity of the particle
is 5 cm> s. The acceleration of the particle is 25k at
this time.
5. a. f(x) 5 (cos x)2
d(cos x)
f r (x) 5 2 (cos x) ?
dx
5 2 (cos x) ? (2sin x)
5 22 sin x cos x
f s (x) 5 (22 sin x)(2sin x) 1 (cos x)(22 cos x)
5 2 sin2 x 2 2 cos2 x
5 2 (sin2 x 2 cos2 x)
5-31
b. f(x) 5 cos x cot x
f r(x) 5 (cos x)(2csc2 x) 1 (cot x)(2sin x)
cos x
1
2
? sin x
5 2cos x ?
sin2 x
sin x
1
cos x
?
2 cos x
52
sin x sin x
5 2cot x csc x 2 cos x
f s (x) 5
(2cot x)(2csc x cot x) 1 (csc x)(csc2 x) 1 sin x
5 csc x cot2 x 1 csc3 x 1 sin x
6. f(x) 5 (sin x)2
To find the absolute extreme values, first find the
derivative, set it equal to zero, and solve for x.
d(sin x)
f r(x) 5 2 (sin x) ?
dx
5 2 sin x cos x
5 sin 2x
Now set f r (x) 5 0 and solve for x.
0 5 sin 2x
2x 5 0, p, 2p
p
x 5 0, , p in the interval 0 # x # p.
2
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
x
0
p
2
p
f(x) 5 (sin 2 x)
0
1
0
So, the absolute maximum value on the interval is 1
p
when x 5 2 and the absolute minimum value on
the interval is 0 when x 5 0 and x 5 p.
7. y 5 f(x) 5 5x
Find the derivative, f r(x), and evaluate the
derivative at x 5 2 to find the slope of the tangent
when x 5 2.
f r(x) 5 5x ln 5
f r(2) 5 52 ln 5
5 25 ln 5
8 40.24
8. y 5 xe x 1 3e x
To find the maximum and minimum values, first
find the derivative, set it equal to zero, and solve
for x.
yr 5 (x)(e x ) 1 (e x )(1) 1 3e x
5 xe x 1 e x 1 3e x
5 xe x 1 4e x
5 e x (x 1 4)
Now set yr 5 0 and solve for x.
0 5 e x (x 1 4)
5-32
e x is never equal to zero.
(x 1 4) 5 0
So x 5 24.
Therefore, the critical value is 24.
Interval
ex (x 1 4)
x , 24
2
1
24 , x
So f(x) is decreasing on the left of x 5 24 and
increasing on the right of x 5 24. Therefore, the
function has a minimum value at a24, 2 e4 b . There
is no maximum value.
9. f(x) 5 2 cos x 2 sin 2x
So, f(x) 5 2 cos x 2 2 sin x cos x.
a. f r (x) 5 22 sin x 2 (2 sin x)(2sin x)
2 (cos x)(2 cos x)
5 22 sin x 1 2 sin2 x 2 2 cos2 x
Set f r (x) 5 0 to solve for the critical values.
22 sin x 1 2 sin2 x 2 2 cos2 x 5 0
22 sin x 1 2 sin2 x 2 2(1 2 sin2 x) 5 0
22 sin x 1 2 sin2 x 2 2 1 2 sin2 x 5 0
4 sin2 x 2 2 sin x 2 2 5 0
(2 sin x 1 1)(2 sin x 2 2) 5 0
2 sin x 1 1 5 0 and 2 sin x 2 2 5 0
So, sinx 5 2 12.
p
5p
In the given interval, this occurs when x 5 2 6 , 2 6 .
Also, sin x 5 1.
p
In the given interval, this occurs when x 5 2 .
Therefore, on the given interval, the critical
p
5p p
numbers for f(x) are x 5 2 6 , 2 6 , 2 .
b. To determine the intervals where f(x) is increasing
and where f(x) is decreasing, find the slope of f(x)
in the intervals between the endpoints and the critical
numbers. To do this, it helps to make a table.
1
x
slope of f(x)
5p
6
p
,x,2
6
p
,x,
2
2p # x , 2
5p
6
p
2
6
p
,x#p
2
2
2
1
2
2
So, f(x) is increasing on the interval
5p
p
2 6 , x , 2 6 and f(x) is decreasing on the
5p
p
intervals 2p # x , 2 6 and 2 6 , x , p.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
c. From the table in part b., it can be seen that there is
p
a local maximum at the point where x 5 2 6 and
5p
there is a local minimum at the point where x 5 2 6 .
d.
y
4
3
2
1
–p
–p
2
0
–1
–2
–3
–4
p
2
p x
Cumulative Review of Calculus
1. a. f(x) 5 3x2 1 4x 2 5
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 15
5 lim
hS0
h
3(2 1 h)2 1 4(2 1 h) 2 5 2 15
5 lim
hS0
h
2
12 1 12h 1 3h 1 8 1 4h 2 20
5 lim
hS0
h
2
3h 1 16h
5 lim
hS0
h
5 lim 3h 1 16
hS0
5 16
2
b. f(x) 5
x21
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 2
5 lim
hS0
h
5 lim
hS0
5 lim
2
22
21h21
h
2
2(1 1 h)
2 11h
11h
h
2 2 2(1 1 h)
5 lim
hS0
h(1 1 h)
22h
5 lim
hS0 h(1 1 h)
hS0
Calculus and Vectors Solutions Manual
22
hS0 1 1 h
5 22
c. f(x) 5 !x 1 3
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(6 1 h) 2 3
5 lim
hS0
h
!h 1 9 2 3
5 lim
hS0
h
( !h 1 9 2 3)( !h 1 9 1 3)
5 lim
hS0
h( !h 1 9 1 3)
h1929
5 lim
hS0 h( !h 1 9 1 3)
h
5 lim
hS0 h( !h 1 9 1 3)
1
5 lim
hS0 ( !h 1 9 1 3)
1
5
6
d. f(x) 5 25x
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
25(11h) 2 32
5 lim
hS0
h
25 ? 25h 2 32
5 lim
hS0
h
32(25h 2 1)
5 lim
hS0
h
5(25h 2 1)
5 32 lim
hS0
5h
(25h 2 1)
5 160 lim
hS0
5h
5 160 ln 2
change in distance
2. a. average velocity 5
change in time
s(t2 ) 2 s(t1 )
5
t2 2 t1
32(4)2 1 3(4) 1 14 2 3(2(1)2 1 3(1) 1 1)4
5
421
45 2 6
5
3
5 13 m> s
b. instantaneous velocity 5 slope of the tangent
s(a 1 h) 2 s(a)
m 5 lim
hS0
h
5 lim
5-33
s(3 1 h) 2 s(3)
hS0
h
2(3 1 h)2 1 3(3 1 h) 1 1
5 lim c
hS0
h
(2(3)2 1 3(3) 1 1)
d
2
h
18 1 12h 1 2h 2 1 9 1 3h 1 1 2 28
5 lim
hS0
h
15h 1 2h 2
5 lim
hS0
h
5 lim (15 1 2h)
5 lim
hS0
5 15 m> s
f(a 1 h) 2 f(a)
h
3
(4 1 h) 2 64
f(4 1 h) 2 f(4)
5 lim
lim
hS0
h
hS0
h
3
(4 1 h) 2 64 5 f(4 1 h) 2 f(4)
Therefore, f(x) 5 x 3.
4. a. Average rate of change in distance with respect
to time is average velocity, so
s(t2 ) 2 s(t1 )
average velocity 5
t2 2 t1
s(3) 2 s(1)
5
321
4.9(3)2 2 4.9(1)
5
321
5 19.6 m> s
b. Instantaneous rate of change in distance with
respect to time 5 slope of the tangent.
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 f(2)
5 lim
hS0
h
4.9(2 1 h)2 2 4.9(2)2
5 lim
hS0
h
19.6 1 19.6h 1 4.9h 2 2 19.6
5 lim
hS0
h
19.6h 1 4.9h 2
5 lim
hS0
h
5 lim 19.6 1 4.9h
m 5 lim
3.
hS0
hS0
5 19.6 m> s
c. First, we need to determine t for the given
distance:
146.9 5 4.9t 2
29.98 5 t 2
5.475 5 t
5-34
Now use the slope of the tangent to determine the
instantaneous velocity for t 5 5.475:
f(5.475 1 h) 2 f(5.475)
m 5 lim
hS0
h
4.9(5.475 1 h)2 2 4.9(5.475)2
5 lim
hS0
h
146.9 1 53.655h 1 4.9h 2 2 146.9
5 lim
hS0
h
53.655h 1 4.9h 2
5 lim
hS0
h
5 lim 353.655 1 4.9h4
hS0
5 53.655 m> s
5. a. Average rate of population change
p(t2 ) 2 p(t1 )
5
t2 2 t1
2(8)2 1 3(8) 1 1 2 (2(0) 1 3(0) 1 1)
5
820
128 1 24 1 1 2 1
5
820
5 19 thousand fish> year
b. Instantaneous rate of population change
p(t 1 h) 2 p(t)
5 lim
hS0
h
p(5 1 h) 2 p(5)
5 lim
hS0
h
2(5 1 h)2 1 3(5 1 h) 1 1
5 lim c
hS0
h
(2(5)2 1 3(5) 1 1)
d
2
h
50 1 20h 1 2h 2 1 15 1 3h 1 1 2 66
5 lim
hS0
h
2h 2 1 23h
5 lim
hS0
h
5 lim 2h 1 23
5 23 thousand fish> year
6. a. i. f(2) 5 3
ii. lim2 f(x) 5 1
hS0
xS2
iii. lim1 f(x) 5 3
xS2
iv. lim f(x) 5 2
xS6
b. No, lim f(x) does not exist. In order for the limit
xS4
to exist, lim2 f(x) and lim1 f(x) must exist and they
xS4
xS4
must be equal. In this case, lim2 f(x) 5 `, but
xS4
lim1 f(x) 5 2 `, so lim f(x) does not exist.
xS4
xS4
Chapter 5: Derivatives of Exponential and Trigonometric Functions
7. f(x) is discontinuous at x 5 2. lim2 f(x) 5 5, but
xS2
lim f(x) 5 3.
xS2 1
2(0)2 1 1
2x 2 1 1
5
xS0 x 2 5
025
1
52
5
x23
b. lim
xS3 !x 1 6 2 3
(x 2 3)( !x 1 6 1 3)
5 lim
xS3 ( !x 1 6 2 3)( !x 1 6 1 3)
(x 2 3)( !x 1 6 1 3)
5 lim
xS3
x1629
(x 2 3)( !x 1 6 1 3)
5 lim
xS3
x23
5 lim !x 1 6 1 3
8. a. lim
xS3
56
c. lim
xS23
5 lim
1
1
13
x
x13
x13
3x
x13
x13
5 lim
xS23 3x(x 1 3)
1
5 lim
xS23 3x
1
52
9
x2 2 4
d. lim 2
xS2 x 2 x 2 2
(x 1 2)(x 2 2)
5 lim
xS2 (x 1 1)(x 2 2)
x12
5 lim
xS2 x 1 1
4
5
3
x22
e. lim 3
xS2 x 2 8
x22
5 lim
2
xS2 (x 2 2)(x 1 2x 1 4)
1
5 lim 2
xS2 x 1 2x 1 4
1
5
12
xS23
Calculus and Vectors Solutions Manual
!x 1 4 2 !4 2 x
x
( !x 1 4 2 !4 2 x)( !x 1 4 1 !4 2 x)
5 lim
xS0
x( !x 1 4 1 !4 2 x)
x 1 4 2 (4 2 x)
5 lim
xS0 x( !x 1 4 1 !4 2 x)
2x
5 lim
xS0 x( !x 1 4 1 !4 2 x)
2
5 lim
xS0 ( !x 1 4 1 !4 2 x)
1
5
2
9. a. f(x) 5 3x2 1 x 1 1
f(x 1 h) 2 f(x)
f r (x) 5 lim
hS0
h
3(x 1 h)2 1 (x 1 h) 1 1
5 lim c
hS0
h
(3x 2 1 x 1 1)
d
2
h
3x 2 1 6hx 1 6h 2 1 x 1 h
5 lim c
hS0
h
2
1 2 3x 2 x 2 1
d
1
h
6hx 1 6h 2 1 h
5 lim
hS0
h
5 lim 6x 1 6h 1 1
f. lim
xS0
hS0
5 6x 1 1
1
b. f(x) 5
x
f(x 1 h) 2 f(x)
f r (x) 5 lim
hS0
h
1
1
2x
x
1
h
5 lim
h
x 2 (x 1 h)
5 lim
hS0 h(x)(x 1 h)
2h
5 lim
hS0 h(x)(x 1 h)
21
5 lim
hS0 x(x 1 h)
1
52 2
x
10. a. To determine the derivative, use the power rule:
y 5 x 3 2 4x 2 1 5x 1 2
dy
5 3x 2 2 8x 1 5
dx
hS0
5-35
b. To determine the derivative, use the chain rule:
y 5 "2x 3 1 1
dy
1
5
(6x 2 )
dx
2"2x 3 1 1
3x 2
5
"2x 3 1 1
c. To determine the derivative, use the quotient rule:
2x
y5
x13
dy
2(x 1 3) 2 2x
5
dx
(x 1 3)2
6
5
(x 1 3)2
d. To determine the derivative, use the product rule:
y 5 (x 2 1 3)2 (4x 5 1 5x 1 1)
dy
5 2(x 2 1 3)(2x)(4x 5 1 5x 1 1)
dx
1 (x 2 1 3)2 (20x 4 1 5)
5 4x(x 2 1 3)(4x 5 1 5x 1 1)
1 (x 2 1 3)2 (20x 4 1 5)
e. To determine the derivative, use the quotient rule:
(4x 2 1 1)5
y5
(3x 2 2)3
dy
5(4x 2 1 1)4 (8x)(3x 2 2)3
5
dx
(3x 2 2)6
3(3x 2 2)2 (3)(4x 2 1 1)5
2
(3x 2 2)6
5 (4x 2 1 1)4 (3x 2 2)2
40x(3x 2 2) 2 9(4x 2 1 1)
3
(3x 2 2)6
(4x 2 1 1)4 (120x 2 2 80x 2 36x 2 2 9)
5
(3x 2 2)4
(4x 2 1 1)4 (84x 2 2 80x 2 9)
5
(3x 2 2)4
f. y 5 3x 2 1 (2x 1 1)34 5
Use the chain rule
dy
5 53x 2 1 (2x 1 1)34 4 32x 1 6(2x 1 1)24
dx
11. To determine the equation of the tangent line,
we need to determine its slope at the point (1, 2).
To do this, determine the derivative of y and
evaluate for x 5 1:
18
y5
(x 1 2)2
5 18(x 1 2)22
5-36
dy
5 236(x 1 2)23
dx
236
5
(x 1 2)3
236
m5
(x 1 2)3
24
236
5
5
27
3
Since we have a given point and we know the slope,
use point-slope form to write the equation of the
tangent line:
24
(x 2 1)
y225
3
3y 2 6 5 24x 1 4
4x 1 3y 2 10 5 0
12. The intersection point of the two curves
occurs when
x 2 1 9x 1 9 5 3x
x 2 1 6x 1 9 5 0
(x 1 3)2 5 0
x 5 23.
At a point x, the slope of the line tangent to the
curve y 5 x 2 1 9x 1 9 is given by
d 2
dy
5
(x 1 9x 1 9)
dx
dx
5 2x 1 9.
At x 5 23, this slope is 2(23) 1 9 5 3.
d
13. a. pr (t) 5 (2t2 1 6t 1 1100)
dt
5 4t 1 6
b. 1990 is 10 years after 1980, so the rate of change
of population in 1990 corresponds to the value
pr (10) 5 4(10) 1 6
5 46 people per year.
c. The rate of change of the population will be 110
people per year when
4t 1 6 5 110
t 5 26.
This corresponds to 26 years after 1980, which is
the year 2006.
d
14. a. f r (x) 5 (x 5 2 5x 3 1 x 1 12)
dx
5 5x 4 2 15x 2 1 1
d
(5x 4 2 15x 2 1 1)
f s (x) 5
dx
5 20x 3 2 30x
Chapter 5: Derivatives of Exponential and Trigonometric Functions
b. f(x) can be rewritten as f(x) 5 22x 22
d
f r(x) 5
(22x 22 )
dx
5 4x 23
4
5 3
x
d
f s (x) 5
(4x 23 )
dx
5 212x 24
12
52 4
x
1
c. f(x) can be rewritten as f(x) 5 4x 22
d
1
f r(x) 5
(4x 2 2 )
dx
3
5 22x22
2
52
"x 3
d
3
f s (x) 5
(22x 22 )
dx
5
5 3x 22
3
5
"x 5
d. f(x) can be rewritten as f(x) 5 x 4 2 x 24
d 4
f r(x) 5
(x 2 x 24 )
dx
5 4x 3 1 4x 25
4
5 4x 3 1 5
x
d
f s (x) 5
(4x 3 1 4x 25 )
dx
5 12x 2 2 20x 26
20
5 12x 2 2 6
x
15. Extreme values of a function on an interval will
only occur at the endpoints of the interval or at a
critical point of the function.
d
a. f r (x) 5 (1 1 (x 1 3)2 )
dx
5 2(x 1 3)
The only place where f r(x) 5 0 is at x 5 23, but
that point is outside of the interval in question. The
extreme values therefore occur at the endpoints of
the interval:
f(22) 5 1 1 (22 1 3)2 5 2
f(6) 5 1 1 (6 1 3)2 5 82
The maximum value is 82, and the minimum
value is 6
Calculus and Vectors Solutions Manual
1
b. f(x) can be rewritten as f(x) 5 x 1 x 22
d
1
f r (x) 5
(x 1 x 22 )
dx
1 3
5 1 1 2 x 22
2
1
512
2"x 3
On this interval, x $ 1, so the fraction on the right
is always less than or equal to 12. This means that
f r (x) . 0 on this interval and so the extreme values
occur at the endpoints.
1
52
f(1) 5 1 1
!1
1
1
59
f(9) 5 9 1
!9
3
The maximum value is 9 13, and the minimum
value is 2.
d
ex
b
c. f r (x) 5 a
dx 1 1 e x
(1 1 e x )(e x ) 2 (e x )(e x )
(1 1 e x )2
x
e
5
(1 1 e x )2
x
Since e is never equal to zero, f r (x) is never zero,
and so the extreme values occur at the endpoints of
the interval.
e0
1
f(0) 5
0 5
11e
2
e4
f(4) 5
1 1 e4
e4
The maximum value is 1 1 e 4, and the minimum
value is 12.
d
d. f r (x) 5 (2 sin (4x) 1 3)
dx
5 8 cos (4x)
p
Cosine is 0 when its argument is a multiple of 2
5
3p
or 2 .
p
3p
1 2kp or 4x 5
1 2kp
2
2
p
p
p
3p
x5 1 k
1 k
x5
8
2
8
2
4x 5
p 3p 5p 7p
Since xP30, p4, x 5 8 , 8 , 8 , 8 .
Also test the function at the endpoints of the interval.
f(0) 5 2 sin 0 1 3 5 3
p
p
f a b 5 2 sin 1 3 5 5
8
2
5-37
3p
3p
b 5 2 sin
1351
8
2
5p
5p
1355
f a b 5 2 sin
8
2
7p
7p
f a b 5 2 sin
1351
8
2
f(p) 5 2 sin (4p) 1 3 5 3
The maximum value is 5, and the minimum
value is 1.
16. a. The velocity of the particle is given by
v(t) 5 sr (t)
d
5 (3t 3 2 40.5t 2 1 162t)
dt
5 9t 2 2 81t 1 162.
The acceleration is
a(t) 5 vr (t)
d
5 (9t 2 2 81t 1 162)
dt
5 18t 2 81
b. The object is stationary when v(t) 5 0:
9t 2 2 81t 1 162 5 0
9(t 2 6)(t 2 3) 5 0
t 5 6 or t 5 3
The object is advancing when v(t) . 0 and retreating
when v(t) , 0. Since v(t) is the product of two
linear factors, its sign can be determined using the
signs of the factors:
fa
t-values
t23
t26
v(t)
Object
0,t,3
,0
,0
.0
Advancing
3,t,6
.0
,0
,0
Retreating
.0
Advancing
6,t,8
.0
.0
c. The velocity of the object is unchanging when the
acceleration is 0; that is, when
a(t) 5 18t 2 81 5 0
t 5 4.5
d. The object is decelerating when a(t) , 0, which
occurs when
18t 2 81 , 0
0 # t , 4.5
e. The object is accelerating when a(t) . 0, which
occurs when
18t 2 81 . 0
4.5 , t # 8
5-38
17.
w
l
Let the length and width of the field be l and w, as
shown. The total amount of fencing used is then
2l 1 5w. Since there is 750 m of fencing available,
this gives
2l 1 5w 5 750
5
l 5 375 2 w
2
The total area of the pens is
A 5 lw
5
5 375w 2 w 2
2
The maximum value of this area can be found by
expressing A as a function of w and examining its
derivative to determine critical points.
A(w) 5 375w 2 52w 2, which is defined for 0 # w
and 0 # l. Since l 5 375 2 52w, 0 # l gives the
restriction w # 150. The maximum area is therefore
the maximum value of the function A(w) on the
interval 0 # w # 150.
5
d
a375w 2 w 2 b
Ar (w) 5
dw
2
5 375 2 5w
Setting Ar (w) 5 0 shows that w 5 75 is the only
critical point of the function. The only values of
interest are therefore:
5
A(0) 5 375(0) 2 (0)2 5 0
2
5
A(75) 5 375(75) 2 (75)2 5 14 062.5
2
5
A(150) 5 375(150) 2 (150)2 5 0
2
The maximum area is 14 062.5 m2
18.
r
h
Let the height and radius of the can be h and r, as
shown. The total volume of the can is then pr 2h.
The volume of the can is also give at 500 mL, so
pr 2h 5 500
500
h5
pr 2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
The total surface area of the can is
A 5 2prh 1 2pr 2
1000
1 2pr 2
5
r
The minimum value of this surface area can be
found by expressing A as a function of r and
examining its derivative to determine critical points.
1000
A(r) 5
1 2pr 2, which is defined for 0 , r and
r
500
0 , h. Since h 5 pr 2 , 0 , h gives no additional
restriction on r. The maximum area is therefore the
maximum value of the function A(r) on the interval
0 , r.
d 1000
Ar(r) 5 a
1 2pr 2 b
r
dr
1000
5 2 2 1 4pr
r
The critical points of A(r) can be found by setting
Ar(r) 5 0:
2
1000
1 4pr 5 0
r2
4pr 3 5 1000
3
r 5 # 1000 8 4.3 cm
4p
So r 5 4.3 cm is the only critical point of the
function. This gives the value
500
h5
8 8.6 cm.
p(4.3)2
19.
r
20
0.02p
r 8 6.8
Using the max min algorithm:
C(1) 5 20.03, C(6.8) 5 4.39, C(36) 5 41.27.
The dimensions for the cheapest container are a
radius of 6.8 cm and a height of 27.5 cm.
20. a. Let the length, width, and depth be l, w, and
d, respectively. Then, the given information is that
l 5 x, w 5 x, and
l 1 w 1 d 5 140. Substituting gives
2x 1 d 5 140
d 5 140 2 2x
b. The volume of the box is V 5 lwh. Substituting
in the values from part a. gives
V 5 (x)(x)(140 2 2x)
5 140x 2 2 2x 3
In order for the dimensions of the box to make sense,
the inequalities l $ 0, w $ 0, and h $ 0 must be
satisfied. The first two give x $ 0, the third requires
x # 70. The maximum volume is therefore the
maximum value of V(x) 5 140x 2 2 2x 3 on the
interval 0 # x # 70, which can be found by
determining the critical points of the derivative Vr(x).
d
Vr (x) 5
(140x 2 2 2x 3 )
dx
5 280x 2 6x 2
5 2x(140 2 3x)
Setting Vr (x) 5 0 shows that x 5 0 and
r3 5
140
h
Let the radius be r and the height h.
Minimize the cost:
C 5 2pr 2 (0.005) 1 2prh(0.0025)
V 5 pr2 h 5 4000
4000
h5
pr 2
4000
b (0.0025)
C(r) 5 2pr 2 (0.005) 1 2pr a
pr 2
20
5 0.01pr 2 1 , 1 # r # 36
r
20
C r(r) 5 0.02pr 2 2 .
r
For a maximum or minimum value, let C r (r) 5 0.
0.02pr 2 2
20
r2
50
Calculus and Vectors Solutions Manual
x 5 3 8 46.7 are the critical points of the function.
The maximum value therefore occurs at one of these
points or at one of the endpoints of the interval:
V(0) 5 140(0)2 2 2(0)3 5 0
V(46.7) 5 140(46.7)2 2 2(46.7)3 5 101 629.5
V(0) 5 140(70)2 2 2(70)3 5 0
So the maximum volume is 101 629.5 cm3, from a
box with length and width 46.7 cm and depth
140 2 2(46.7) 5 46.6 cm.
21. The revenue function is
R(x) 5 x(50 2 x 2 )
5 50x 2 x 3. Its maximum for x $ 0 can be
found by examining its derivative to determine
critical points.
d
Rr (x) 5
(50x 2 x 3 )
dx
5 50 2 3x 2
The critical points can be found by setting Rr (x) 5 0:
50 2 3x 2 5 0
50
x5 6
8 64.1
Å3
5-39
Only the positive root is of interest since the number
of MP3 players sold must be positive. The number
must also be an integer, so both x 5 4 and x 5 5
must be tested to see which is larger.
R(4) 5 50(4) 2 43 5 136
R(4) 5 50(5) 2 53 5 125
So the maximum possible revenue is $136, coming
from a sale of 4 MP3 players.
22. Let x be the fare, and p(x) be the number of
passengers per year. The given information shows
that p is a linear function of x such that an increase
of 10 in x results in a decrease of 1000 in p. This
means that the slope of the line described by p(x) is
21000
10 5 2100. Using the initial point given,
p(x) 5 2100(x 2 50) 1 10 000
5 2100x 1 15 000
The revenue function can now be written:
R(x) 5 xp(x)
5 x(2100x 1 15 000)
5 15 000x 2 100x 2
Its maximum for x $ 0 can be found by examining
its derivative to determine critical points.
d
(15 000x 2 100x 2 )
Rr(x) 5
dx
5 15 000 2 200x
Setting Rr (x) 5 0 shows that x 5 75 is the only
critical point of the function. The problem states
that only $10 increases in fare are possible, however,
so the two nearest must be tried to determine the
maximum possible revenue:
R(70) 5 15 000(70) 2 100(70)2 5 560 000
R(80) 5 15 000(80) 2 100(80)2 5 560 000
So the maximum possible revenue is $560 000,
which can be achieved by a fare of either $70 or $80.
23. Let the number of $30 price reductions be n.
The resulting number of tourists will be 80 1 n
where 0 # n # 70. The price per tourist will be
5000 2 30n dollars. The revenue to the travel
agency will be (5000 2 30n)(80 1 n) dollars. The
cost to the agency will be 250 000 1 300(80 1 n)
dollars.
Profit 5 Revenue 2 Cost
P(n) 5 (5000 2 30n)(80 1 n)
2 250 000 2 300(80 1 n), 0 # n # 70
P r (n) 5 230(80 1 n) 1 (5000 2 30n)(1) 2 300
5 2300 2 60n
1
P r(n) 5 0 when n 5 38
3
Since n must be an integer, we now evaluate P(n)
for n 5 0, 38, 39, and 70. (Since P(n) is a quadratic
5-40
function whose graph opens downward with vertex
at 38 13 , we know P(38) . P(39).)
P(0) 5 126 000
P(38) 5 (3860)(118) 2 250 000 2 300(118)
5 170 080
P(39) 5 (3830)(119) 2 250 000 2 300(119)
5 170 070
P(70) 5 (2900)(150) 2 250 000 2 300(150)
5 140 000
The price per person should be lowered by $1140
(38 decrements of $30) to realize a maximum profit
of $170 080.
dy
d
5
(25x 2 1 20x 1 2)
24. a.
dx
dx
5 210x 1 20
dy
Setting dx 5 0 shows that x 5 2 is the only critical
number of the function.
b.
x
x,2
x52
x.2
y9
1
0
2
Graph
Inc.
Local Max
Dec.
dy
d
5
(6x 2 1 16x 2 40)
dx
dx
5 12x 1 16
dy
Setting dx 5 0 shows that x 5 2 43 is the only
critical number of the function.
x
x,2
4
3
x52
4
3
x.2
y9
2
0
1
Graph
Dec.
Local Min
Inc.
4
3
dy
d
5
(2x 3 2 24x)
dx
dx
5 6x 2 2 24
dy
The critical numbers are found by setting dx 5 0:
6x 2 2 24 5 0
6x 2 5 24
x 5 62
c.
x
x , 22
x 5 22
22 , x , 2
x52
x.2
y9
1
0
2
0
1
Graph
Inc.
Local Max
Dec.
Local Min
Inc.
d.
dy
d
x
5
a
b
dx
dx x 2 2
(x 2 2)(1) 2 x(1)
5
(x 2 2)2
22
5
(x 2 2)2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
This derivative is never equal to zero, so the
function has no critical numbers. Since the
numerator is always negative and the denominator
is never negative, the derivative is always negative.
This means that the function is decreasing
everywhere it is defined, that is, x 2 2.
25. a. This function is discontinuous when
x2 2 9 5 0
x 5 63. The numerator is non-zero at these
points, so these are the equations of the vertical
asymptotes.
To check for a horizontal asymptote:
8
8
lim 2
5 lim
xS` x 2 9
xS` 2
9
x a1 2 2 b
x
lim (8)
xS`
5
9
lim x 2 a1 2 2 b
xS`
x
lim (8)
xS`
5
9
lim (x)2 3 lim a1 2 2 b
xS`
xS`
x
1
8
5 lim 2 3
xS` x
120
50
8
Similarly, lim x 2 2 9 5 0, so y 5 0 is a horizontal
xS2`
asymptote of the function.
There is no oblique asymptote because the degree
of the numerator does not exceed the degree of the
denominator by 1.
Local extrema can be found by examining the
derivative to determine critical points:
(x 2 2 9)(0) 2 (8)(2x)
yr 5
(x 2 2 9)2
216x
5 2
(x 2 9)2
Setting yr 5 0 shows that x 5 0 is the only critical
point of the function.
x
x,0
x50
x.0
y9
1
0
1
Graph
Inc.
Local Max
Dec.
So (0, 2 89 ) is a local maximum.
b. This function is discontinuous when
x2 2 1 5 0
x 5 61. The numerator is non-zero at these
points, so these are the equations of the vertical
asymptotes.
Calculus and Vectors Solutions Manual
To check for a horizontal asymptote:
4x 3
x 3 (4)
lim 2
5 lim
xS` x 2 1
xS` 2
1
x 12 2
x
x(4)
5 lim
1
xS`
1 2 x2
(
)
lim (x(4))
5
xS`
(
1
)
lim 1 2 x 2
lim (x) 3 lim (4)
xS`
5
xS`
xS`
(
1
)
lim 1 2 x 2
xS`
4
5 lim (x) 3
xS`
120
5`
4x 3
Similarly, lim x 2 2 1 5 lim (x) 5 2`, so this
xS 2`
xS 2`
function has no horizontal asymptote.
To check for an oblique asymptote:
4x
2
3
2
q
x 2 1 4x 1 0x 1 0x 1 0
4x3 1 0x2 2 4x
0 1 0 1 4x 1 0
So y can be written in the form
4x
y 5 4x 1 2
. Since
x 21
4x
x(4)
lim 2
5 lim
xS` x 2 1
xS` 2
1
x 12 2
x
(
5 lim
xS`
5
)
4
(
1
x 1 2 x2
lim (4)
)
xS`
((
1
lim x 1 2 x 2
xS`
))
lim (4)
5
xS`
(
1
lim (x) 3 lim 1 2 x 2
xS`
xS`
)
1
4
5 lim a b 3
x
120
5 0,
4x
and similarly lim x 2 2 1 5 0, the line y 5 4x is an
xS 2`
asymptote to the function y.
Local extrema can be found by examining the
derivative to determine critical points:
5-41
(x 2 2 1)(12x 2 ) 2 (4x 3 )(2x)
(x 2 2 1)2
12x 4 2 12x 2 2 8x 4
5
(x 2 2 1)2
4x 4 2 12x 2
5
(x 2 2 1)2
Setting yr 5 0:
4x 4 2 12x 2 5 0
x 2 (x 2 2 3) 5 0
so x 5 0, x 5 6 !3 are the critical points of the
function
(2!3, 26 !3) is a local maximum, ( !3, 6 !3) is
a local minimum, and (0, 0) is neither.
yr 5
x
x , 2 !3
x 5 2 !3
2 !3 , x , 0
x50
y9
1
0
2
0
Graph
Inc.
Local Max
Dec.
Horiz.
x
0 , x , !3
x 5 !3
x . !3
y9
2
0
2
Graph
Dec.
Local Min
Inc.
26. a. This function is continuous everywhere, so it
has no vertical asymptotes. To check for a horizontal
asymptote:
lim (4x 3 1 6x 2 2 24x 2 2)
xS`
xS`
5`
Similarly,
lim (4x 3 1 6x 2 2 24x 2 2) 5 lim (x 3 ) 5 2`,
xS 2`
so this function has no horizontal asymptote.
The y-intercept can be found by letting x 5 0,
which gives y 5 4(0)3 1 6(0)2 2 24(0) 2 2
5 22
The derivative is of the function is
d
(4x 3 1 6x 2 2 24x 2 2)
yr 5
dx
5 12x 2 1 12x 2 24
5 12(x 1 2)(x 2 1), and the second derivative is
d
(12x 2 1 12x 2 24)
ys 5
dx
5-42
x
x , 22
x 5 22
22 , x
y9
1
0
2
2
Graph
Inc.
Local Max
Dec.
Dec.
y0
2
2
2
0
Concavity
Down
Down
Down
Infl.
x51
x.1
0
1
1
,x,1
2
2
2
x
y9
Graph
Dec.
Local Min
Inc.
y0
1
1
1
Concavity
Up
Up
Up
x52
1
2
y
30
20
10
x
–3 –2 –1 0
–10
1 2
y = 4x3 + 6x2 – 24x – 2
24
6
2
5 lim x a4 1 2 2 2 3 b
x
xS`
x
x
6
2
24
5 lim (x 3 ) 3 lim a4 1 2 2 2 3 b
x
xS`
xS`
x
x
5 lim (x 3 ) 3 (4 1 0 2 0 2 0)
3
xS2`
5 24x 1 12
Letting f r (x) 5 0 shows that x 5 22 and x 5 1 are
critical points of the function. Letting ys 5 0 shows
that x 5 2 12 is an inflection point of the function.
b. This function is discontinuous when
x2 2 4 5 0
(x 1 2)(x 2 2) 5 0
x 5 2 or x 5 22. The numerator is non-zero at
these points, so the function has vertical asymptotes
at both of them. The behaviour of the function near
these asymptotes is:
3x
x12
x22
y
2
,0
,0
,0
,0
1
x-values
x S 22
lim y
xS`
2`
,0
.0
,0
.0
1`
x S 22
.0
.0
,0
,0
2`
x S 21
.0
.0
.0
.0
1`
x S 22
To check for a horizontal asymptote:
3x
x(3)
lim 2
5 lim
xS` x 2 4
xS` 2
4
x 1 2 x2
(
5 lim
xS`
)
3
(
4
x 1 2 x2
)
Chapter 5: Derivatives of Exponential and Trigonometric Functions
lim (3)
5
xS`
((
4
lim x 1 2 2
x
xS`
6
4
2
))
lim (3)
5
xS`
(
4
lim (x) 3 lim 1 2 x 2
xS`
xS`
1
3
5 lim 3
x
xS`
120
50
–6 –4 –2 0
–2
–4
–6
)
Similarly,
so y 5 0 is a horizontal
asymptote of the function.
This function has y 5 0 when x 5 0, so the origin is
both the x- and y-intercept.
The derivative is
(x 2 2 4)(3) 2 (3x)(2x)
yr 5
(x 2 2 4)2
23x 2 2 12
5
, and the second derivative is
(x 2 2 4)2
(x 2 2 4)2 (26x)
ys 5
(x 2 2 4)4
(23x 2 2 12)(2(x 2 2 4)(2x))
2
(x 2 2 4)4
3
26x 1 24x 1 12x 3 1 48x
5
(x 2 2 4)3
6x 3 1 72x
5
(x 2 2 4)3
The critical points of the function can be found by
letting yr 5 0, so
23x 2 2 12 5 0
x 2 1 4 5 0. This has no real solutions, so the
function y has no critical points.
The inflection points can be found by letting
ys 5 0, so
6x 3 1 72x 5 0
6x(x 2 1 12) 5 0
The only real solution to this equation is x 5 0, so
that is the only possible inflection point.
x , 22 22 , x , 0
x50
0,x,2
x.2
y9
2
2
2
2
2
Graph
Dec.
Dec.
Dec.
Dec.
Dec.
y0
2
1
0
2
1
Concavity
Down
Up
Infl.
Down
Up
Calculus and Vectors Solutions Manual
3x
y = ——–—
x2 – 4
x
2 4 6
d
((24)e 5x11 )
dx
d
5 (24)e 5x11 3
(5x 1 1)
dx
5 (220)e 5x11
d
b. f r (x) 5 (xe 3x )
dx
d
(3x) 1 (1)e 3x
5 xe 3x 3
dx
5 e 3x (3x 1 1)
d
c. yr 5 (63x28 )
dx
d
5 (ln 6)63x28 3
(3x 2 8)
dx
5 (3 ln 6)63x28
d
d. yr 5 (e sin x )
dx
d
5 e sin x 3
(sin x)
dx
5 (cos x)e sin x
28. The slope of the tangent line at x 5 1 can be
27. a. f r (x) 5
3x
lim 2
5 0,
xS` x 2 4
x
y
dy
found by evaluating the derivative dx for x 5 1:
dy
d 2x21
5
(e
)
dx
dx
d
5 e 2x21 3
(2x 2 1)
dx
2x21
5 2e
Substituting x 5 1 shows that the slope is 2e. The
value of the original function at x 5 1 is e, so the
equation of the tangent line at x 5 1 is
y 5 2e(x 2 1) 1 e.
29. a. The maximum of the function modelling the
number of bacteria infected can be found by
examining its derivative.
d
t
N r (t) 5 ((15t)e 2 5 )
dt
d
t
t
t
5 15te 25 3 a2 b 1 (15)e 25
dt
5
t
5 e25 (15 2 3t)
5-43
Setting Nr(t) 5 0 shows that t 5 5 is the only
critical point of the function (since the exponential
function is never zero). The maximum number of
infected bacteria therefore occurs after 5 days.
5
b. N(5) 5 (15(5))e 25
5 27 bacteria
d
dy
5
(2 sin x 2 3 cos 5x)
30. a.
dx
dx
d
(5x)
5 2 cos x 2 3(2sin 5x) 3
dx
5 2 cos x 1 15 sin 5x
d
dy
5
(sin 2x 1 1)4
b.
dx
dx
d
(sin 2x 1 1)
5 4(sin 2x 1 1)3 3
dx
d
5 4(sin 2x 1 1)3 3 (cos 2x) 3
(2x)
dx
5 8 cos 2x(sin 2x 1 1)3
1
c. y can be rewritten as y 5 (x 2 1 sin 3x)2 . Then,
d
dy
1
5
(x 2 1 sin 3x)2
dx
dx
1 2
d
1
5 (x 1 sin 3x)22 3
(x 2 1 sin 3x)
2
dx
1
1
5 (x 2 1 sin 3x)22
2
d
3 a2x 1 cos 3x 3
(3x)b
dx
2x 1 3 cos 3x
5
2 !x 2 1 sin 3x
dy
d
sin x
5
a
b
d.
dx
dx cos x 1 2
(cos x 1 2)(cos x) 2 (sin x)(2sin x)
5
(cos x 1 2)2
cos2 x 1 sin2 x 1 2 cos x
5
(cos x 1 2)2
1 1 2 cos x
5
(cos x 1 2)2
d
dy
5
(tan x 2 2 tan2 x)
e.
dx
dx
d
d
sec2 x 2 3
(x 2 )
5
dx
dx
d
(tan x)
2 2 tan x 3
dx
5 2x sec2 x 2 2 2 tan x sec2 x
5-44
f.
d
dy
5
(sin (cos x 2 ))
dx
dx
d
(cos x 2 )
5 cos (cos x 2 ) 3
dx
5 cos (cos x 2 ) 3 (2sin x 2 ) 3
d
(x 2 )
dx
5 22x sin x 2 cos(cos x 2 )
31.
l2
u
100
l1
250
u
As shown in the diagram, let u be the angle between
the ladder and the ground, and let the total length
of the ladder be l 5 l1 1 l2, where l1 is the length
from the ground to the top corner of the shed and
l2is the length from the corner of the shed to the
wall.
250
100
sin u 5
cos u 5
l1
l2
l1 5 250 csc u
l2 5 100 sec u
l 5 250 csc u 1 100 sec u
dl
5 2250 csc u cot u 1 100 sec u tan u
du
100 sin u
250 cos u
1
52
sin2 u
cos2 u
dl
To determine the minimum, solve du 5 0.
100 sin u
250 cos u
5
2
sin u
cos2 u
3
250 cos u 5 100 sin3 u
2.5 5 tan3 u
3
tan u 5 "2.5
u 8 0.94
At u 5 0.94, l 5 250 csc 0.94 1 100 sec 0.94
8 479 cm
The shortest ladder is about 4.8 m long.
32. The longest rod that can fit around the corner is
determined by the minimum value of x 1 y. So,
determine the minimum value of l 5 x 1 y.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
dl
y
Solving du 5 0 yields:
3
u
x
u
3
3
3
From the diagram, sin u 5 y and cos u 5 x. So,
l5
3
3
1
,
cos u
sin u
p
for 0 # u # 2 .
3 sin u
3 cos u
dl
5
2
du
cos2 u
sin2 u
3
3 sin u 2 3 cos3 u
5
cos2 u sin2 u
Calculus and Vectors Solutions Manual
3 sin3 u 2 3 cos3 u 5 0
tan3 u 5 1
tan u 5 1
p
u5
4
3
3
So l 5
p 1
p
cos 4
sin 4
5 3 !2 1 3 !2
5 6 !2
p
When u 5 0 or u 5 2 , the longest possible rod
would have a length of 3 m. Therefore the longest
rod that can be carried horizontally around the
corner is one of length 6 !2, or about 8.5 m.
5-45
CHAPTER 5:
Derivatives of Exponential and
Trigonometric Functions
Review of Prerequisite Skills,
pp. 224–225
1
32
1
5
9
2
5
32 R 2
b. 325 5 Q "
5 22
54
1
2
c. 2723 5 3
2
Q "27 R
1
5 2
3
1
5
9
3 2
2 22
d. a b 5 a b
3
2
9
5
4
2. a. log5 625 5 4
1
b. log4
5 22
16
c. logx 3 5 3
d. log10 450 5 w
e. log3 z 5 8
f. loga T 5 b
3. a.
y
3
2
1
1. a. 322 5
–3 –2 –1 0
–1
–2
–3
0 5 log10 (x 1 2)
100 5 x 1 2
x 5 21
The x-intercept is (21, 0).
b.
10 y
8
6
4
2
–8 –6 –4 –2 0
–2
x
2 4 6 8
An exponential function is always positive, so there
is no x-intercept.
y
4. a. sin u 5
r
x
b. cos u 5
r
y
c. tan u 5
x
5. To convert to radian measure from degree
p
x
1 2 3
The x-intercept occurs where y 5 0.
Calculus and Vectors Solutions Manual
measure, multiply the degree measure by 180°.
p
5 2p
a. 360° 3
180°
p
p
5
b. 45° 3
180°
4
p
p
52
c. 290° 3
180°
2
p
p
5
d. 30° 3
180°
6
3p
p
5
e. 270° 3
180°
2
2p
p
52
f. 2120° 3
180°
3
p
5p
5
g. 225° 3
180°
4
5-1
11p
p
5
180°
6
6. For the unit circle, sine is associated with the
y-coordinate of the point where the terminal arm of
the angle meets the circle, and cosine is associated
with the x-coordinate.
a. sin u 5 b
b
b. tan u 5
a
c. cos u 5 a
p
d. sin a 2 ub 5 a
2
p
e. cos a 2 ub 5 b
2
f. sin (2u) 5 2b
7. a. The angle is in the second quadrant, so cosine
and tangent will be negative.
12
cos u 5 2
13
5
tan u 5 2
12
b. The angle is in the third quadrant, so sine will be
negative and tangent will be positive.
sin2 u 1 cos2 u 5 1
4
sin2 u 1 5 1
9
5
sin2 u 5
9
!5
sin u 5 2
3
sin u
tan u 5
cos u
!5
5
2
c. The angle is in the fourth quadrant, so cosine
will be positive and sine will be negative.
Because tan u 5 22, the point (1, 22) is on the
terminal arm of the angle. The reference triangle for
this angle has a hypotenuse of "22 1 12 or !5.
2
sin u 5 2
!5
1
cos u 5
!5
d. The sine is only equal to 1 for one angle between
h. 330° 3
p
p
50
2
cos
p
tan 2 is undefined
2p
8. a. The period is 2 or p. The amplitude is 1.
2p
b. The period is 1 or 4p. The amplitude is 2.
2
2p
c. The period is p or 2. The amplitude is 3.
2p
p
d. The period is 12 or 6 . The amplitude is 27.
e. The period is 2p. The amplitude is 5.
f. The period is 2p. Because of the absolute value
being taken, the amplitude is 32.
2p
9. a. The period is 2 or p. Graph the function from
x 5 0 to x 5 2p.
4
y
3
2
1
x
0
p
2
p
3p
2
2p
b. The period is 2p, so graph the function from
x 5 0 to x 5 4p.
3
y
2
1
0
–1
x
p
2
p
3p 2p 5p 3p 7p 4p
2
2
2
–2
–3
10. a. tan x 1 cot x 5 sec x csc x
LS 5 tan x 1 cot x
sin x
cos x
5
1
cos x
sin x
sin2 x 1 cos2 x
5
cos x 1 sin x
0 and p, so u 5 2 .
5-2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
1
cos x 1 sin x
RS 5 sec x 1 csc x
1
1
5
?
cos x sin x
1
5
cos x sin x
Therefore, tan x 1 cot x 5 sec x csc x.
sin x
b.
5 tan x 1 sec x
1 2 sin2 x
sin x
LS 5
1 2 sin2 x
sin x
5
cos2 x
RS 5 tan x sec x
sin x
1
5
?
cos x cos x
sin x
5
cos2 x
sin x
5 tan x sec x.
Therefore,
1 2 sin2 x
11. a. 3 sin x 5 sin x 1 1
2 sin x 5 1
1
sin x 5
2
p 5p
x5 ,
6 6
b. cos x 2 1 5 2cos x
2 cos x 5 1
1
cos x 5
2
p 5p
x5 ,
3 3
5
5.1 Derivatives of Exponential
Functions, y 5 e x, pp. 232–234
1. You can only use the power rule when the term
containing variables is in the base of the exponential
expression. In the case of y 5 e x, the exponent
contains a variable.
2. a. y 5 e 3x
dy
5 3e 3x
dx
b. s 5 e 3t25
ds
5 3e 3t25
dt
Calculus and Vectors Solutions Manual
c. y 5 2e 10t
dy
5 20e 10t
dt
d. y 5 e 23x
dy
5 23e 23x
dx
2
e. y 5 e 526x1x
dy
2
5 (26 1 2x)e 526x1x
dx
f. y 5 e "x
dy
1 "x
5
e
dx
2 !x
3
3. a. y 5 2e x
dy
3
5 2(3x 2 )e x
dx
3
5 6x 2e x
3x
dy
d(xe )
5
b.
dx
dx
5 (x)(3e 3x ) 1 (e 3x )(1)
5 3xe 3x 1 e 3x
5 e 3x (3x 1 1)
3
e 2x
x
3
3
23x 2e 2x (x) 2 e 2x
f r (x) 5
x2
d. f(x) 5 !xe x
1
f r (x) 5 "xe x 1 e x a
b
2 !x
2
e. h(t) 5 e t 1 3e 2t
2
hr (t) 5 2te t 2 3e 2t
e 2t
f. g(t) 5
1 1 e 2t
2e 2t (1 1 e 2t ) 2 2e 2t (e 2t )
g r(t) 5
(1 1 e 2t )2
2t
2e
5
(1 1 e 2t )2
1
4. a. f r (x) 5 (3e 3x 2 3e 23x )
3
5 e 3x 2 e 23x
f r (1) 5 e 3 2 e 23
c. f(x) 5
1
b. f(x) 5 e 2x 1 1
f r (x) 5 e 2x 1 1 a
1
b
(x 1 1)2
f r (0) 5 e 21 (1)
1
5
e
1
5-3
1
cos x 1 sin x
RS 5 sec x 1 csc x
1
1
5
?
cos x sin x
1
5
cos x sin x
Therefore, tan x 1 cot x 5 sec x csc x.
sin x
b.
5 tan x 1 sec x
1 2 sin2 x
sin x
LS 5
1 2 sin2 x
sin x
5
cos2 x
RS 5 tan x sec x
sin x
1
5
?
cos x cos x
sin x
5
cos2 x
sin x
5 tan x sec x.
Therefore,
1 2 sin2 x
11. a. 3 sin x 5 sin x 1 1
2 sin x 5 1
1
sin x 5
2
p 5p
x5 ,
6 6
b. cos x 2 1 5 2cos x
2 cos x 5 1
1
cos x 5
2
p 5p
x5 ,
3 3
5
5.1 Derivatives of Exponential
Functions, y 5 e x, pp. 232–234
1. You can only use the power rule when the term
containing variables is in the base of the exponential
expression. In the case of y 5 e x, the exponent
contains a variable.
2. a. y 5 e 3x
dy
5 3e 3x
dx
b. s 5 e 3t25
ds
5 3e 3t25
dt
Calculus and Vectors Solutions Manual
c. y 5 2e 10t
dy
5 20e 10t
dt
d. y 5 e 23x
dy
5 23e 23x
dx
2
e. y 5 e 526x1x
dy
2
5 (26 1 2x)e 526x1x
dx
f. y 5 e "x
dy
1 "x
5
e
dx
2 !x
3
3. a. y 5 2e x
dy
3
5 2(3x 2 )e x
dx
3
5 6x 2e x
3x
dy
d(xe )
5
b.
dx
dx
5 (x)(3e 3x ) 1 (e 3x )(1)
5 3xe 3x 1 e 3x
5 e 3x (3x 1 1)
3
e 2x
x
3
3
23x 2e 2x (x) 2 e 2x
f r (x) 5
x2
d. f(x) 5 !xe x
1
f r (x) 5 "xe x 1 e x a
b
2 !x
2
e. h(t) 5 e t 1 3e 2t
2
hr (t) 5 2te t 2 3e 2t
e 2t
f. g(t) 5
1 1 e 2t
2e 2t (1 1 e 2t ) 2 2e 2t (e 2t )
g r(t) 5
(1 1 e 2t )2
2t
2e
5
(1 1 e 2t )2
1
4. a. f r (x) 5 (3e 3x 2 3e 23x )
3
5 e 3x 2 e 23x
f r (1) 5 e 3 2 e 23
c. f(x) 5
1
b. f(x) 5 e 2x 1 1
f r (x) 5 e 2x 1 1 a
1
b
(x 1 1)2
f r (0) 5 e 21 (1)
1
5
e
1
5-3
c. hr(z) 5 2z(1 1 e 2z ) 1 z 2 (2e 2z )
hr(21) 5 2(21)(1 1 e) 1 (21)2 (2e 1 )
5 22 2 2e 2 e
5 22 2 3e
2e x
5. a. y 5
1 1 ex
dy
(1 1 e x )2e x 2 2e x (e x )
5
dx
(1 1 e x )2
dy
2(2) 2 2(1)(1)
5
dx
22
1
5
2
When x 5 0,
At the point (1, e 21 ), the slope is e 21 (0) 5 0. The
equation of the tangent line at the point A is
1
y 2 e 21 5 0(x 2 1) or y 5 e .
8. The slope of the tangent line at any point on the
curve is
5 (2x 2 x 2 )(e 2x )
2x 2 x 2
5
.
ex
Horizontal lines have slope equal to 0.
dy
the slope of the tangent is 12.
The equation of the tangent is y 5 12x 1 1, since the
y-intercept was given as (0, 1).
b.
c. The answers agree very well; the calculator does
not show a slope of exactly 0.5, due to internal
rounding.
6. y 5 e 2x
dy
5 2e 2x
dx
When x 5 21,
dy
5 2e. And
dx
when x 5 21, y 5 e.
The equation of the tangent is y 2 e 5 2e(x 1 1)
or ex 1 y 5 0.
3
2
1
–3 –2 –1 0
–1
–2
–3
y
x
1 2 3
7. The slope of the tangent line at any point is
given by
dy
5 (1)(e 2x ) 1 x(2e 2x )
dx
5 e 2x (1 2 x).
5-4
dy
5 2xe 2x 1 x 2 (e 2x )
dx
We solve 5 0
dx
x(2 2 x)
5 0.
ex
Since e x . 0 for all x, the solutions are x 5 0 and
x 5 2. The points on the curve at which the tangents
4
are horizontal are (0, 0) and (2, e2).
5 x
x
9. If y 5 (e5 1 e 25 ), then
2
5 1 x
1 x
yr 5 a e5 2 e25 b, and
2 5
5
5 1 x
1 x
ys 5 a e 5 1 e 25 b
2 25
25
1 5 5x
x
5
c (e 1 e 25 )d
25 2
1
5 y.
25
10. a. y 5 e 23x
dy
5 23e 23x
dx
d 2y
5 9e 23x
dx 2
d 3y
5 227e 23x
dx 3
d ny
b. n 5 (21)n (3n )e 23x
dx
dy
d(23e x )
5
11. a.
dx
dx
5 23e x
d 2y
5 23e x
dx 2
d(xe 2x )
dy
5
b.
dx
dx
5 (x)(2e 2x ) 1 (e 2x )(1)
5 2xe 2x 1 e 2x
5 e 2x (2x 1 1)
Chapter 5: Derivatives of Exponential and Trigonometric Functions
d 2y
5 e2x (2) 1 (2x 1 1)(2e2x )
dx2
5 4xe2x 1 4e2x
d(e x (4 2 x))
dy
c.
5
dx
dx
x
5 (e )(21) 1 (4 2 x)(e x )
5 2e x 1 4e x 2 xe x
5 3e x 2 xe x
5 e x (3 2 x)
d 2y
5 e x (21) 1 (3 2 x)(e x )
dx2
5 2e x 2 xe x
5 e x (2 2 x)
12. a. When t 5 0, N 5 1000330 1 e 04 5 31 000.
dN
1 2t
100 230t
b.
5 1000 c 0 2 e 30 d 5 2
e
dt
30
3
dN
100 223
c. When t 5 20h,
52
e 8 217 bacteria> h.
dt
3
t
d. Since e 230 . 0 for all t, there is no solution to
dN
dt 5 0.
Hence, the maximum number of bacteria in the
culture occurs at an endpoint of the interval of
domain.
5
When t 5 50, N 5 1000330 1 e 23 4 8 30 189.
The largest number of bacteria in the culture is
31 000 at time t 5 0.
e. The number of bacteria is constantly decreasing
as time passes.
ds
1
1 t
5 160a 2 e 24 b
13. a. v 5
dt
4
4
t
5 40(1 2 e24 )
dv
1 t
t
5 40a e 24 b 5 10e 24
b. a 5
dt
4
t
t
v
From a., v 5 40(1 2 e 24 ), which gives e 24 5 1 2 40.
Thus, a 5 10a1 2
1
v
b 5 10 2 v.
40
4
c. vT 5 lim v
tS`
24t
vT 5 lim 40(1 2 e )
tS`
1
5 40 lim a1 2 4t b
tS`
e
1
t 5 0
tS` e4
The terminal velocity of the skydiver is 40 m> s.
5 40(1), since lim
d. 95% of the terminal velocity is
95
(40) 5 38 m> s.
100
To determine when this velocity occurs, we solve
t
40(1 2 e 24 ) 5 38
38
t
1 2 e 24 5
40
1
t
e 24 5
20
t
e4 5 20
t
and 5 ln 20,
4
which gives t 5 4 ln 20 8 12 s.
The skydiver’s velocity is 38 m> s, 12 s after jumping.
The distance she has fallen at this time is
S 5 160(ln 20 2 1 1 e 220 )
1
5 160 aln 20 2 1 1 b
20
8 327.3 m.
x
14. a. i. Let f(x) 5 ( 1 1 1x) . Then,
x
f(x)
1
10
2
2.5937
100
2.7048
1000
2.7169
10 000
2.7181
x
So, from the table one can see that lim (1 1 1x) 5 e.
1
x
xS`
ii. Let f(x) 5 (1 1 x) .
x
f(x)
20.1
2.8680
20.01
2.7320
20.001
2.7196
20.0001
2.7184
?
?
0.0001
2.7181
0.001
2.7169
0.01
2.7048
0.1
2.5937
1
So, from the table one can see that lim (1 1 x)x 5 e.
xS0
That is, the limit approaches the value of
e 5 2.718 281 828c
b. The limits have the same value because as
1
x S `, x S 0.
Calculus and Vectors Solutions Manual
5-5
15. a. The given limit can be rewritten as
eh 2 1
e 01h 2 e 0
5 lim
lim
hS0
h
hS0
h
This expression is the limit definition of the derivative
at x 5 0 for f(x) 5 e x.
e 01h 2 e 0
d
cf r(0) 5 lim
hS0
h
dex
Since f r(x) 5 dx 5 e x, the value of the given limit
is e 0 5 1.
e 21h 2 e 2
b. Again, lim
is the derivative of e x at
hS0
h
x 5 2.
e 21h 2 e 2
5 e2.
Thus, lim
hS0
h
dy
d 2y
5 Ame ex and 2 5 Am 2e ex.
16. For y 5 Ae ex,
dt
dt
Substituting in the differential equation gives
Am 2e ex 1 Ame ex 2 6Ae ex 5 0
Ae ex (m 2 1 m 2 6) 5 0.
ex
Since Ae 2 0, m 2 1 m 2 6 5 0
(m 1 3)(m 2 2) 5 0
m 5 23 or m 5 2.
d 1 x
d
sinh x 5
c (e 2 e 2x ) d
17. a.
dx
dx 2
1
5 (e x 1 e 2x )
2
5 cosh x
1 x
d
cosh x 5 (e 2 e 2x )
b.
dx
2
5 sinh x
sinh x
,
c. Since tanh x 5
cosh x
d
tanh x
dx
d
d
(
sinh x) (cosh x) 2 (sinh x)(dx cosh x)
dx
5
(cosh x)2
5
1
x
2 (e
5-6
1
4
S (e
18. a. Four terms:
1
1
1
e511
1
1
5 2.666 666
1!
2!
3!
Five terms:
1
1
1
1
e511
1
1
1
5 2.708 333
1!
2!
3!
4!
Six terms:
1
1
1
1
1
e511
1
1
1
1
5 2.716 666
1!
2!
3!
4!
5!
Seven terms:
1
1
1
1
1
1
1
1
1
1
1
5 2.718 055
e511
1!
2!
3!
4!
5!
6!
b. The expression for e in part a. is a special case of
x1
1. a.
b.
c.
121e
) 2 (e 2 2 1 e
(cosh x)2
d(23x )
dy
5
dx
dx
5 3(23x ) ln 2
dy
d(3.1x 1 x 3 )
5
dx
dx
5 ln 3.1(3.1)x 1 3x 2
d(103t25 )
ds
5
dt
dt
5 3(103t25 ) ln 10
2
2 e )( 12) (e x 2 e 2x )
2x
x4
d(10526n1n )
dw
5
d.
dn
dn
2
5 (26 1 2n)(10526n1n )ln 10
2x
22x
x3
5.2 Derivatives of the General
Exponential Function, y 5 b x, p. 240
(cosh x)2
2x
x2
e x 5 1 1 1! 1 2! 1 3! 1 4! 1 c. in that it is the
case when x 5 1. Then e x 5 e 1 5 e is in fact
e 1 5 e 5 1 1 1!1 1 2!1 1 3!1 1 4!1 1 5!1 1 c. The
value of x is 1.
2
(cosh x)
1
x
2 (e
1
4 (4)
(cosh x)2
1
5
(cosh x)2
e.
2
2
5
1 e 2x )( 12) (cosh x)2 (e x 1 e 2x )
5
22x
)T
f.
d(3x 12 )
dy
5
dx
dx
2
5 2x(3x 12 )ln 3
d(400(2)x13 )
dy
5
dx
dx
5 400(2)x13 ln 2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
15. a. The given limit can be rewritten as
eh 2 1
e 01h 2 e 0
5 lim
lim
hS0
h
hS0
h
This expression is the limit definition of the derivative
at x 5 0 for f(x) 5 e x.
e 01h 2 e 0
d
cf r(0) 5 lim
hS0
h
dex
Since f r(x) 5 dx 5 e x, the value of the given limit
is e 0 5 1.
e 21h 2 e 2
b. Again, lim
is the derivative of e x at
hS0
h
x 5 2.
e 21h 2 e 2
5 e2.
Thus, lim
hS0
h
dy
d 2y
5 Ame ex and 2 5 Am 2e ex.
16. For y 5 Ae ex,
dt
dt
Substituting in the differential equation gives
Am 2e ex 1 Ame ex 2 6Ae ex 5 0
Ae ex (m 2 1 m 2 6) 5 0.
ex
Since Ae 2 0, m 2 1 m 2 6 5 0
(m 1 3)(m 2 2) 5 0
m 5 23 or m 5 2.
d 1 x
d
sinh x 5
c (e 2 e 2x ) d
17. a.
dx
dx 2
1
5 (e x 1 e 2x )
2
5 cosh x
1 x
d
cosh x 5 (e 2 e 2x )
b.
dx
2
5 sinh x
sinh x
,
c. Since tanh x 5
cosh x
d
tanh x
dx
d
d
(
sinh x) (cosh x) 2 (sinh x)(dx cosh x)
dx
5
(cosh x)2
5
1
x
2 (e
5-6
1
4
S (e
18. a. Four terms:
1
1
1
e511
1
1
5 2.666 666
1!
2!
3!
Five terms:
1
1
1
1
e511
1
1
1
5 2.708 333
1!
2!
3!
4!
Six terms:
1
1
1
1
1
e511
1
1
1
1
5 2.716 666
1!
2!
3!
4!
5!
Seven terms:
1
1
1
1
1
1
1
1
1
1
1
5 2.718 055
e511
1!
2!
3!
4!
5!
6!
b. The expression for e in part a. is a special case of
x1
1. a.
b.
c.
121e
) 2 (e 2 2 1 e
(cosh x)2
d(23x )
dy
5
dx
dx
5 3(23x ) ln 2
dy
d(3.1x 1 x 3 )
5
dx
dx
5 ln 3.1(3.1)x 1 3x 2
d(103t25 )
ds
5
dt
dt
5 3(103t25 ) ln 10
2
2 e )( 12) (e x 2 e 2x )
2x
x4
d(10526n1n )
dw
5
d.
dn
dn
2
5 (26 1 2n)(10526n1n )ln 10
2x
22x
x3
5.2 Derivatives of the General
Exponential Function, y 5 b x, p. 240
(cosh x)2
2x
x2
e x 5 1 1 1! 1 2! 1 3! 1 4! 1 c. in that it is the
case when x 5 1. Then e x 5 e 1 5 e is in fact
e 1 5 e 5 1 1 1!1 1 2!1 1 3!1 1 4!1 1 5!1 1 c. The
value of x is 1.
2
(cosh x)
1
x
2 (e
1
4 (4)
(cosh x)2
1
5
(cosh x)2
e.
2
2
5
1 e 2x )( 12) (cosh x)2 (e x 1 e 2x )
5
22x
)T
f.
d(3x 12 )
dy
5
dx
dx
2
5 2x(3x 12 )ln 3
d(400(2)x13 )
dy
5
dx
dx
5 400(2)x13 ln 2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
d(x 5 3 (5)x )
dy
5
dx
dx
5 (x 5 )((5)x (ln 5)) 1 ((5)x )(5x 4 )
5 5x 3(x 5 3 ln 5) 1 5x 44
2
d(x(3)x )
dy
5
b.
dx
dx
2
2
5 (x)(2x(3)x ln 3) 1 (3)x (1)
2
5 (3)x 3(2x 2 ln 3) 1 14
c. v 5 (2t )(t 21 )
d((2t )(t 21 ))
dv
5
dt
dx
5 (2t )(21t 22 ) 1 (t 21 )(2t ln 2)
2t
2t ln 2
52 21
t
t
x
2
3
d. f(x) 5 2
x
x
x
1
ln 3(32 )(x 2 ) 2 2x(32 )
f r(x) 5 2
x4
x
x
x ln 3(32 ) 2 4(32 )
5
x4
x
32 3x ln 3 2 44
5
x3
2
3t25
3. f(t) 5 10
? e 2t
2
2
fr (t) 5 (103t25 )(4te 2t ) 1 (e 2t )(3(10)3t25 ln 10)
2
5 103t25e 2t (4t 1 3 ln 10)
Now, set f r (t) 5 0.
2
So, f r (t) 5 0 5 103t25e 2t (4t 1 3ln 10)
2
So 103t25e 2t 5 0 and 4t 1 3 ln 10 5 0.
The first equation never equals zero because solving
would force one to take the natural log of both
sides, but ln 0 is undefined. So the first equation
does not produce any values for which fr (t) 5 0.
The second equation does give one value.
4t 1 3 ln 10 5 0
4t 5 23 ln 10
3 ln 10
t52
4
4. When x 5 3, the function y 5 f(x) evaluated at
3 is f(3) 5 3(23 ) 5 3(8) 5 24. Also,
d(3(2)x )
dy
5
dx
dx
5 3(2x )ln 2
So, at x 5 3,
dy
5 3(23 )(ln 2) 5 24(ln 2) 8 16.64
dx
Therefore, y 2 24 5 16.64(x 2 3)
y 2 24 5 16.64x 2 49.92
216.64x 1 y 1 25.92 5 0
2. a.
Calculus and Vectors Solutions Manual
d(10x )
dy
5
dx
dx
5 10x ln 10
So, at x 5 1,
dy
5 101 ln 10 5 10(ln 10) 8 23.03
dx
Therefore, y 2 10 5 23.03(x 2 1)
y 2 10 5 23.03x 2 23.03
223.03x 1 y 1 13.03 5 0
6. a. The half-life of the substance is the time
required for half of the substance to decay. That is, it
is when 50% of the substance is left, so P(t) 5 50.
50 5 100(1.2)2t
1
5 (1.2)2t
2
1
1
5
2
(1.2)t
t
(1.2) 5 2
t(ln 1.2) 5 ln 2
ln 1.2
t5
ln 2
t 8 3.80 years
Therefore, the half-life of the substance is about
3.80 years.
b. The problem asks for the rate of change when
t 8 3.80 years.
Pr (t) 5 2100(1.2)2t (ln 1.2)
Pr (3.80) 5 2100(1.2)2(3.80) (ln 1.2)
8 29.12
So, the substance is decaying at a rate of about
29.12 percent> year at the time 3.80 years where the
half-life is reached.
7. P 5 0.5(109 )e 0.200 15t
dP
5 0.5(109 )(0.200 15)e 0.200 15t
a.
dt
5.
dP
In 1968, t 5 1 and dt 5 0.5(109 )(0.200 15)e 0.200 15 8
0.122 25 3 109 dollars> annum
In 1978, t 5 11 and
dP
5 0.5(109 )(0.200 15)e 1130.200 15
dt
8 0.904 67 3 109 dollars> annum.
In 1978, the rate of increase of debt payments
was $904 670 000> annum compared to
$122 250 000> annum in 1968. As a ratio,
Rate in 1978
7.4
5 .
Rate in 1968
1
The rate of increase for 1978 is
7.4 times larger than that for 1968.
5-7
b. In 1988, t 5 21 and
dP
5 0.5(109 )(0.200 15)e 2130.200 15
dt
8 6.694 69 3 109 dollars> annum
In 1998, t 5 31 and
dP
5 0.5(109 )(0.200 15)e 3130.200 15
dt
8 49.541 69 3 109 dollars> annum
Rate in 1998
7.4
From the graph, one can notice that the values of v(t)
quickly rise in the range of about 0 # t # 15. The
slope for these values is positive and steep. Then as the
graph nears t 5 20 the steepness of the slope decreases
and seems to get very close to 0. One can reason that
the car quickly accelerates for the first 20 units of time.
Then, it seems to maintain a constant acceleration
for the rest of the time. To verify this, one could differentiate and look at values where vr (t) is increasing.
As a ratio, Rate in 1988 5 1 . The rate of increase
for 1998 is 7.4 times larger than that for 1988.
c. Answers may vary. For example, data from the
past are not necessarily good indicators of what will
happen in the future. Interest rates change, borrowing
may decrease, principal may be paid off early.
8. When x 5 0, the function y 5 f(x) evaluated at 0
2
is f(0) 5 220 5 20 5 1. Also,
2
d(22x )
dy
5
dx
dx
2
5 22x(22x )ln 2
So, at x 5 0,
dy
2
5 22(0)(220 )ln 2 5 0
dx
Therefore, y 2 1 5 0(x 2 0)
So, y 2 1 5 0 or y 5 1.
2
y
5.3 Optimization Problems Involving
Exponential Functions, pp. 245–247
1. a.
The maximum value is about 0.3849. The
minimum value is 0.
b.
1
–8 –6 –4 –2 0
x
2 4 6 8
–1
–2
9.
v(t)
120
100
80
60
40
20
0
5-8
t
0
20 40 60 80 100 120
The maximum value is about 10.043. The
minimum value is about 25961.916.
2. a. f(x) 5 e 2x 2 e 23x on 0 # x # 10
f r(x) 5 2e 2x 1 3e 23x
Let f r (x) 5 0, therefore e 2x 1 3e 23x 5 0.
Let e 2x 5 w, when 2w 1 3w 3 5 0.
w(21 1 3w 2 ) 5 0.
Therefore, w 5 0 or w 2 5 13
1
.
w56
"3
But w $ 0, w 5 1
1
"3
.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
b. In 1988, t 5 21 and
dP
5 0.5(109 )(0.200 15)e 2130.200 15
dt
8 6.694 69 3 109 dollars> annum
In 1998, t 5 31 and
dP
5 0.5(109 )(0.200 15)e 3130.200 15
dt
8 49.541 69 3 109 dollars> annum
Rate in 1998
7.4
From the graph, one can notice that the values of v(t)
quickly rise in the range of about 0 # t # 15. The
slope for these values is positive and steep. Then as the
graph nears t 5 20 the steepness of the slope decreases
and seems to get very close to 0. One can reason that
the car quickly accelerates for the first 20 units of time.
Then, it seems to maintain a constant acceleration
for the rest of the time. To verify this, one could differentiate and look at values where vr (t) is increasing.
As a ratio, Rate in 1988 5 1 . The rate of increase
for 1998 is 7.4 times larger than that for 1988.
c. Answers may vary. For example, data from the
past are not necessarily good indicators of what will
happen in the future. Interest rates change, borrowing
may decrease, principal may be paid off early.
8. When x 5 0, the function y 5 f(x) evaluated at 0
2
is f(0) 5 220 5 20 5 1. Also,
2
d(22x )
dy
5
dx
dx
2
5 22x(22x )ln 2
So, at x 5 0,
dy
2
5 22(0)(220 )ln 2 5 0
dx
Therefore, y 2 1 5 0(x 2 0)
So, y 2 1 5 0 or y 5 1.
2
y
5.3 Optimization Problems Involving
Exponential Functions, pp. 245–247
1. a.
The maximum value is about 0.3849. The
minimum value is 0.
b.
1
–8 –6 –4 –2 0
x
2 4 6 8
–1
–2
9.
v(t)
120
100
80
60
40
20
0
5-8
t
0
20 40 60 80 100 120
The maximum value is about 10.043. The
minimum value is about 25961.916.
2. a. f(x) 5 e 2x 2 e 23x on 0 # x # 10
f r(x) 5 2e 2x 1 3e 23x
Let f r (x) 5 0, therefore e 2x 1 3e 23x 5 0.
Let e 2x 5 w, when 2w 1 3w 3 5 0.
w(21 1 3w 2 ) 5 0.
Therefore, w 5 0 or w 2 5 13
1
.
w56
"3
But w $ 0, w 5 1
1
"3
.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
1
1
When w 5 "3, e 2x 5 "3,
2x ln e 5 ln 1 2 ln "3
ln "3 2 ln 1
1
5 ln "3
8 0.55.
x5
f(0) 5 e 0 2 e 0
50
f(0.55) 8 0.3849
f(10) 5 e 210 2 e 230 8 0.000 05
Absolute maximum is about 0.3849 and absolute
minimum is 0.
m(x) 5 (x 1 2)e 22x on 24 # x # 4
mr(x) 5 e 22x 1 (22)(x 1 2)e 22x
Let mr(x) 5 0.
e 22x 2 0, therefore, 1 1 (22)(x 1 2) 5 0
23
x5
2
5 21.5.
m(24) 5 22e 8 8 25961
m(21.5) 5 0.5e 3 8 10
m(4) 5 6e 28 8 0.0002
The maximum value is about 10 and the minimum
value is about 25961.
b. The graphing approach seems to be easier to use for
the functions. It is quicker and it gives the graphs of
the functions in a good viewing rectangle. The only
problem may come in the second function, m(x),
because for x , 1.5 the function quickly approaches
values in the negative thousands.
20
3. a. P(t) 5
1 1 3e 20.02t
20
P(0) 5
1 1 3e 20.02(0)
20
5
1 1 3e 0
20
5
4
55
So, the population at the start of the study when
t 5 0 is 500 squirrels.
b. The question asks for lim P(t).
tS`
As t approaches `, e 20.02t 5
20
20.02t
tS`
tS` 1 1 3e
20
5
1 1 3(0)
5 20.
Therefore, the largest population of squirrels that
the forest can sustain is 2000 squirrels.
c. A point of inflection can only occur when
Ps (t) 5 0 and concavity changes around the point.
20
P(t) 5
1 1 3e 20.02t
P(t) 5 20(1 1 3e 20.02t )21
So, lim P(t) 5 lim
1
e0.02t
approaches 0.
Calculus and Vectors Solutions Manual
Pr (t) 5 20(2 (1 1 3e 20.02t )22 (20.06e 20.02t ))
5 (1.2e 20.02t )(1 1 3e 20.02t )22
Ps (t) 5 3(1.2e20.02t )(22(1 1 3e20.02t )23 (20.06e 20.02t )4
1 (1 1 3e 20.02t )22 (20.024e 20.02t )
0.144e 20.04t
0.024e 20.02t
2
(1 1 3e 20.02t )3
(1 1 3e 20.02t )2
0.144e 20.04t
0.024e 20.02t
Ps (0) when
2
50
(1 1 3e 20.02t )3
(1 1 3e 20.02t )2
Solving for t after setting the second derivative
equal to 0 is very tedious. Use a graphing calculator
to determine the value of t for which the second
derivative is 0, 54.9. Evaluate P(54.9). The point of
inflection is (54.9, 10).
d.
P(t)
25
5
20
15
10
5
0
t
0
50 100 150 200 250 300
e. P grows exponentially until the point of inflection,
then the growth rate decreases and the curve becomes
concave down.
4. a. P(x) 5 106 31 1 (x 2 1)e20.001x4, 0 # x # 2000
Using the Algorithm for Extreme Values, we have
P(0) 5 106 31 2 14 5 0
P(2000) 5 106 31 1 1999e 224 8 271.5 3 106.
Now,
Pr (x) 5 106 3(1)e 20.001x 1 (x 2 1)(20.001)e 20.001x4
5 106e 20.001x (1 2 0.001x 1 0.001)
5-9
7. C 5 0.015 3 109e 0.075 33t, 0 # t # 100
a.
C(t)
Capital investment from U.S. sources
($100 million)
Since e 20.001x . 2 for all x,
Pr (x) 5 0 when 1.001 2 0.001x 5 0
1.001
x5
5 1001.
0.001
P(1001) 5 106 31 1 1000e 21.0014 8 368.5 3 106
The maximum monthly profit will be 368.5 3 106
dollars when 1001 items are produced and sold.
b. The domain for P(x) becomes 0 # x # 500.
P(500) 5 106 31 1 499e 20.54 5 303.7 3 106
Since there are no critical values in the domain, the
maximum occurs at an endpoint. The maximum
monthly profit when 500 items are produced and
sold is 303.7 3 106 dollars.
5. R(x) 5 40x 2e 20.4x 1 30, 0 # x # 8
We use the Algorithm for Extreme Values:
Rr (x) 5 80xe 20.4x 1 40x 2 (20.4)e 20.4x
5 40xe 20.4x (2 2 0.4x)
Since e 20.4x . 0 for all x, Rr(x) 5 0 when
x 5 0 or 2 2 0.4x 5 0
x 5 5.
R(0) 5 30
R(5) 8 165.3
R(8) 8 134.4
The maximum monthly revenue of 165.3 thousand
dollars is achieved when 500 units are produced and
sold.
6. P(t) 5 100(e 2t 2 e 24t ), 0 # t # 3
Pr(t) 5 100(2e 2t 1 4e 24t )
5 100e 2t (21 1 4e 23t )
Since e 2t . 0 for all t, Pr(t) 5 0 when
4e 23t 5 1
1
e 23t 5
4
23t 5 ln (0.25)
2ln (0.25)
t5
3
5 0.462.
P(0) 5 0
P(0.462) 8 47.2
P(3) 8 4.98
The highest percentage of people spreading the
rumour is 47.2% and occurs at the 0.462 h point.
16
14
12
10
8
6
4
2
0
t
20 40 60 80 100
Years since 1867
dC
5 0.015 3 109 3 0.075 33e 0.075 33t
dt
In 1947, t 5 80 and the growth rate was
dC
5 0.468 05 3 109 dollars> year.
dt
In 1967, t 5 100 and the growth rate was
dC
5 2.1115 3 109 dollars> year.
dt
The ratio of growth rates of 1967 to that of 1947 is
2.1115 3 109
4.511
.
9 5
0.468 05 3 10
1
The growth rate of capital investment grew from
468 million dollars per year in 1947 to 2.112 billion
dollars per year in 1967.
c. In 1967, the growth rate of investment as a
percentage of the amount invested is
2.1115 3 109
3 100 5 7.5%.
28.0305 3 109
d. In 1977, t 5 110
C 5 59.537 3 109 dollars
dC
5 4.4849 3 109 dollars> year.
dt
e. Statistics Canada data shows the actual amount of
U.S. investment in 1977 was 62.5 3 109 dollars.
The error in the model is 3.5%.
f. In 2007, t 5 140.
The expected investment and growth rates are
b.
C 5 570.490 3 109 dollars and dC 5 42.975 3 109
dt
dollars> year.
5-10
Chapter 5: Derivatives of Exponential and Trigonometric Functions
t
1 210t
te b
10
1
30 2 t
30 2 t
1 0.6a2e 2 5 1 (30 2 t)e 2 20 b
20
30 2 t
210t
5 0.05e (10 2 t) 1 0.03e 2 5
8. a. The growth function is N 5 25.
t
Er(t) 5 0.5ae 210 2
t
3
The number killed is given by K 5 e .
After 60 minutes, N 5 212.
Let T be the number of minutes after 60 minutes.
The population of the colony at any time, T after
the first 60 minutes is
P5N2k
60 1 T
T
5 2 5 2 e3 , T $ 0
dP
1
1 T
60 1 T
5 2 5 a b ln 2 2 e 3
dt
5
3
ln 2
1 T
12 1 T
52 5 a
b 2 e3
5
3
ln 2
1 T
T
5 212 ? 2 5 a
b 2 e3
5
3
dP
ln
2
1 T
T
5 0 when 212
2 5 5 e 3 or
dt
5
3
ln 2 12 T5
T
3
? 2 2 5 e3 .
5
We take the natural logarithm of both sides:
ln 2
T
T
ln a3.212
b 1 ln 2 5
5
5
3
1
ln 2
b
7.4404 5 T a 2
3
5
7.4404
5 38.2 min.
T5
0.1947
At T 5 0, P 5 212 5 4096.
At T 5 38.2, P 5 478 158.
(220 1 30 2 t)
t
5 ( 0.05e 210 1 0.03e 2
t
Calculus and Vectors Solutions Manual
30 2 t
20
) (10 2 t)
Er (t) 5 0 when 10 2 t 5 0
t 5 10 (The first factor is always a positive number.)
3
E(0) 5 5 1 5.4 1 18e 22 5 14.42
E(10) 5 16.65
E(30) 5 11.15
For maximum study effectiveness, 10 h of study
should be assigned to the firs exam and 20 h of
study for the second exam.
10. Use the algorithm for finding extreme values.
First, find the derivative f r(x). Then, find any
critical points by setting fr (x) 5 0 and solving for x.
Also, find the values of x for which f r(x) is
undefined. Together these are the critical values.
Now, evaluate f(x) for the critical values and the
endpoints 2 and 22. The highest value will be the
absolute maximum on the interval and the lowest
value will be the absolute minimum on the interval.
11. a. f r (x) 5 (x 2 )(e x ) 1 (e x )(2x)
For T . 38.2, dP is always negative.
dt
The maximum number of bacteria in the colony
occurs 38.2 min after the drug was introduced.
At this time the population numbers 478 158.
60 1 T
T
b. P 5 0 when 2 5 5 e 3
T
60 1 T
ln 2 5
5
3
ln 2
1
b
12 ln 2 5 Ta 2
3
5
T 5 42.72
The colony will be obliterated 42.72 minutes after
the drug was introduced.
9. Let t be the number of minutes assigned to study
for the first exam and 30 2 t minutes assigned to
study for the second exam. The measure of study
effectiveness for the two exams is given by
E(t) 5 E1 (t) 1 E2 (30 2 t), 0 # t # 30
5 0.5( 10 1 te 210 ) 1 0.6( 9 1 (30 2 t)e 2
30 2 t
5
)
5 e x (x 2 1 2x)
The function is increasing when f r (x) . 0 and
decreasing when f r (x) , 0. First, find the critical
values of f r (x). Solve e x 5 0 and (x 2 1 2x) 5 0
ex is never equal to zero.
x 2 1 2x 5 0
x(x 1 2) 5 0.
So, the critical values are 0 and 22.
Interval
e x (x 2 1 2x)
x , 22
1
22 , x , 0
2
0,x
1
So, f(x) is increasing on the intervals (2 `, 22)
and (0, ` ).
Also, f(x) is decreasing on the interval (22, 0).
b. At x 5 0, f r (x) switches from decreasing on the
left of zero to increasing on the right of zero. So,
x 5 0 is a minimum. Since it is the only critical
point that is a minimum, it is the x-coordinate of the
5-11
absolute minimum value of f(x). The absolute
minimum value is f(0) 5 0.
12. a. yr 5 e x
Setting e x 5 0 yields no solutions for x. ex is a
function that is always increasing. So, there is no
maximum or minimum value for y 5 e x 1 2.
y
8
6
4
2
x
–8 –6 –4 –2 0
–2
2 4 6 8
ex (x 1 1)
x , 21
2
x . 21
1
x,2
1
2
2
x.2
1
2
1
So y is decreasing on the left of x 5 2 12 and
increasing on the right of x 5 2 12. So x 5 2 12 is the
x-coordinate of the minimum of y. The minimum
value is
1
1
2a2 b (e2(22 ) )
2
5 2e21
8 20.37. There is no maximum value.
x
2
d. yr 5 (3x)(2e 2x ) 1 (e 2x )(3) 1 1
5 3e 2x (1 2 x) 1 1
Solve 3e 2x (1 2 x) 1 1 5 0.
This gives no real solutions. By looking at the graph
of y 5 f(x), one can see that the function is always
increasing. So, there is no maximum or minimum
value for y 5 3xe 2x 1 x.
8
y
y
4
–8
–8 –6 –4 –2 0
–2
y
–8 –6 –4 –2 0
–2
So y is decreasing on the left of x 5 21 and
increasing on the right of x 5 21. So x 5 21 is the
x-coordinate of the minimum of y. The minimum
value is 2e21 1 3 8 2.63. There is no maximum
value.
8
6
4
2
2e 2x (2x 1 1)
8
6
4
2
b. yr 5 (x)(e x ) 1 (e x )(1)
5 e x (x 1 1)
Solve e x 5 0 and (x 1 1) 5 0
ex is never equal to zero.
x1150
x 5 21.
So there is one critical point: x 5 21.
Interval
Interval
x
2 4 6 8
–4
0
x
4
8
–4
–8
2x
2x
c. yr 5 (2x)(2e ) 1 (e )(2)
5 2e2x (2x 1 1)
Solve 2e2x 5 0 and (2x 1 1) 5 0
2e 2x is never equal to zero.
2x 1 1 5 0
1
x52
2
So there is one critical point: x 5 2 12.
5-12
2
2
13. Pr (x) 5 (x)(2xe 20.5x ) 1 (e 20.5x )(1)
2
5 e 20.5x (2x 2 1 1)
2
Solve e 20.5x 5 0 and (1 2 x 2 ) 5 0.
20.5x 2
e
gives no critical points.
1 2 x2 5 0
(1 2 x)(1 1 x) 5 0
Chapter 5: Derivatives of Exponential and Trigonometric Functions
So x 5 1 and x 5 21 are the critical points.
So P(x) is decreasing on the left of x 5 21 and on
for t ,
2
Interval
e20.5x (2x2 1 1)
x , 21
2
21 , x , 1
1
d
120
100
80
60
40
20
t
0
2
4
6
8
10
12
b. The speed is increasing when dr(t) . 0 and the
speed is decreasing when dr(t) , 0.
dr(t) 5 (200t)(222t )(ln 2) 1 (22t )(200)
5 200(2)2t (2t ln 2 1 1)
Solve 200(2)2t 5 0 and 2t ln 2 1 1 5 0.
200(2)2t gives no critical points.
2t ln 2 1 1 5 0
1
t5
8 1.44
ln 2
So t 5
1
ln 2
Interval
is the critical point.
1
ln2
1
t.
1
ln2
2
1
ln 2
and decreasing when t .
Calculus and Vectors Solutions Manual
2In1 2
1
.
ln 2
8 106.15 degrees> s
d. The door seems to be closed for t . 10 s.
15. The solution starts in a similar way to that of 9.
The effectiveness function is
t
25 2 t
E(t) 5 0.5( 10 1 te 210) 1 0.6( 9 1 (25 2 t)e 2 20 ) .
The derivative simplifies to
t
25 2 t
Er(t) 5 0.05e 210 (10 2 t) 1 0.03e 2 20 (5 2 t).
This expression is very difficult to solve analytically.
By calculation on a graphing calculator, we can
determine the maximum effectiveness occurs when
t 5 8.16 hours.
aL
16. P 5
a 1 (L 2 a)e 2kLt
a. We are given a 5 100, L 5 10 000, k 5 0.0001.
106
104
P5
2t 5
100 1 9900e
1 1 99e 2t
4
2t 21
5 10 (1 1 99e )
P
10
8
6
4
2
0
0
2
4
(dP)
So the speed of the closing door is increasing when
0,t,
since dr(t) , 0
6 8
Days
10
12
t
b. We need to determine when the derivative of the
200(2)2t (2t ln2 1 1)
t,
and dr(t) . 0 for t .
(ln12) 5 200 (ln12) (2)
Number of cells (thousands)
the right of x 5 1 and it is increasing between
x 5 21 and x 5 1. So x 5 21 is the x-coordinate of
the minimum of P(x). Also, x 5 1 is the x-coordinate
of the maximum of P(x). The minimum value is
2
P(21) 5 (21)(e 20.5(21) ) 5 2e 20.5 8 20.61.
The maximum value is
2
P(1) 5 (1)(e 20.5(1) ) 5 e 20.5 8 0.61.
14. a.
d(t)
0
1
ln 2
1
ln 2
The maximum speed is
2
1,x
c. There is a maximum at t 5
1
.
ln 2
d 2P
growth rate dt is zero, i.e., when dt 2 5 0.
dP
2104 (299e 2t )
990 000e 2t
5
5
dt
(1 1 99e 2t )2
(1 1 99e 2t )2
d 2P
2990 000e 2t (1 1 99e 2t )2 2 990 000e 2t
2 5
dt
(1 1 99e 2t )4
(2)(1 1 99e 2t )(299e 2t )
3
(1 1 99e 2t )4
2990 000e2t (1 1 99e 2t ) 1 198(990 000)e 22t
5
(1 1 99e 2t )3
5-13
2
dP
2 5 0
dt
990 000e 2t (21 2 99e 2t 1 198e 2t ) 5 0
99e 2t 5 1
e t 5 99
t 5 ln 99
8 4.6
After 4.6 days, the rate of change of the growth rate
is zero. At this time the population numbers 5012.
c. When t 5 3, dt 5 (1 1 99e 23 )2 8 1402 cells> day.
dP
990 000e 23
When t 5 8, dt 5 (1 1 99e 28 )2 8 311 cells> day.
The rate of growth is slowing down as the colony is
getting closer to its limiting value.
dP
990 000e 28
Mid-Chapter Review, pp. 248–249
d(5e 23x )
dy
5
dx
dx
5 (5e 23x )(23x)r
5 (5e 23x )(23)
5 215e 23x
1
d( 7e7 x)
dy
b.
5
dx
dx
1 r
1
5 ( 7e7 x) a xb
7
1
1
5 ( 7e7 x) a b
7
1
5 e7 x
dy
c.
5 (x 3 )(e 22x )r 1 (x 3 )r (e 22x )
dx
5 (x 3 )((e 22x )(22x)r) 1 (3x 2 )(e 22x )
5 (x 3 )((e 22x ))(22) 1 3x 2e 22x
5 22x 3e 22x 1 3x 2e 22x
5 e 22x (22x 3 1 3x 2 )
dy
5 (x 2 1)2 (e x )r 1 ( (x 2 1)2 )r (e x )
d.
dx
5 (x 2 1)2 (e x ) 1 (2(x 2 1))(e x )
5 (x 2 2 2x 1 1)(e x ) 1 (2x 2 2)(e x )
5 (e x )(x 2 2 2x 1 1 1 2x 2 2)
5 (e x )(x 2 2 1)
dy
e.
5 2(x 2 e 2x )(x 2 e 2x )r
dx
5 2(x 2 e 2x )(1 2 (e 2x )(2x)r)
5 2(x 2 e 2x )(1 2 (e 2x )(21))
5 2(x 2 e 2x )(1 1 e 2x )
5 2(x 1 xe 2x 2 e 2x 2 e 2x1 2x )
5 2(x 1 xe 2x 2 e 2x 2 e 22x )
1. a.
5-14
dy
(e x 1 e 2x )(e x 2 e 2x )r
5
dx
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 e 2x )r
2
(e x 1 e 2x )2
(e x 1 e 2x )(e x 2 (e 2x )(2x)r)
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 (e 2x )(2x)r)
2
(e x 1 e 2x )2
x
2x
(e 1 e )(e x 2 (e 2x )(21))
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 (e 2x )(21))
2
(e x 1 e 2x )2
(e x 1 e 2x )(e x 1 e 2x )
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 2 e 2x )
2
(e x 1 e 2x )2
e 2x 1 e 0 1 e 0 1 e 22x
5
(e x 1 e 2x )2
(e 2x 2 e 0 2 e 0 1 e 22x )
2
(e x 1 e 2x )2
e 2x 1 e 0 1 e 0 1 e 22x 2 e 2x
5
(e x 1 e 2x )2
e 0 1 e 0 2 e 22x
1
(e x 1 e 2x )2
4
5 x
(e 1 e 2x )2
dP
2. a.
5 100e 25t (25t)r
dt
5 100e 25t (25)
5 2500e 25t
b. The time is needed for when the sample of the
substance is at half of the original amount. So, find
t when P 5 12.
f.
when
P 5 100e 25t
1
5 100e 25t
2
1
5 e 25t
200
1
5 25t
ln
200
1
ln 200
5t
25
Chapter 5: Derivatives of Exponential and Trigonometric Functions
2
dP
2 5 0
dt
990 000e 2t (21 2 99e 2t 1 198e 2t ) 5 0
99e 2t 5 1
e t 5 99
t 5 ln 99
8 4.6
After 4.6 days, the rate of change of the growth rate
is zero. At this time the population numbers 5012.
c. When t 5 3, dt 5 (1 1 99e 23 )2 8 1402 cells> day.
dP
990 000e 23
When t 5 8, dt 5 (1 1 99e 28 )2 8 311 cells> day.
The rate of growth is slowing down as the colony is
getting closer to its limiting value.
dP
990 000e 28
Mid-Chapter Review, pp. 248–249
d(5e 23x )
dy
5
dx
dx
5 (5e 23x )(23x)r
5 (5e 23x )(23)
5 215e 23x
1
d( 7e7 x)
dy
b.
5
dx
dx
1 r
1
5 ( 7e7 x) a xb
7
1
1
5 ( 7e7 x) a b
7
1
5 e7 x
dy
c.
5 (x 3 )(e 22x )r 1 (x 3 )r (e 22x )
dx
5 (x 3 )((e 22x )(22x)r) 1 (3x 2 )(e 22x )
5 (x 3 )((e 22x ))(22) 1 3x 2e 22x
5 22x 3e 22x 1 3x 2e 22x
5 e 22x (22x 3 1 3x 2 )
dy
5 (x 2 1)2 (e x )r 1 ( (x 2 1)2 )r (e x )
d.
dx
5 (x 2 1)2 (e x ) 1 (2(x 2 1))(e x )
5 (x 2 2 2x 1 1)(e x ) 1 (2x 2 2)(e x )
5 (e x )(x 2 2 2x 1 1 1 2x 2 2)
5 (e x )(x 2 2 1)
dy
e.
5 2(x 2 e 2x )(x 2 e 2x )r
dx
5 2(x 2 e 2x )(1 2 (e 2x )(2x)r)
5 2(x 2 e 2x )(1 2 (e 2x )(21))
5 2(x 2 e 2x )(1 1 e 2x )
5 2(x 1 xe 2x 2 e 2x 2 e 2x1 2x )
5 2(x 1 xe 2x 2 e 2x 2 e 22x )
1. a.
5-14
dy
(e x 1 e 2x )(e x 2 e 2x )r
5
dx
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 e 2x )r
2
(e x 1 e 2x )2
(e x 1 e 2x )(e x 2 (e 2x )(2x)r)
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 (e 2x )(2x)r)
2
(e x 1 e 2x )2
x
2x
(e 1 e )(e x 2 (e 2x )(21))
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 1 (e 2x )(21))
2
(e x 1 e 2x )2
(e x 1 e 2x )(e x 1 e 2x )
5
(e x 1 e 2x )2
(e x 2 e 2x )(e x 2 e 2x )
2
(e x 1 e 2x )2
e 2x 1 e 0 1 e 0 1 e 22x
5
(e x 1 e 2x )2
(e 2x 2 e 0 2 e 0 1 e 22x )
2
(e x 1 e 2x )2
e 2x 1 e 0 1 e 0 1 e 22x 2 e 2x
5
(e x 1 e 2x )2
e 0 1 e 0 2 e 22x
1
(e x 1 e 2x )2
4
5 x
(e 1 e 2x )2
dP
2. a.
5 100e 25t (25t)r
dt
5 100e 25t (25)
5 2500e 25t
b. The time is needed for when the sample of the
substance is at half of the original amount. So, find
t when P 5 12.
f.
when
P 5 100e 25t
1
5 100e 25t
2
1
5 e 25t
200
1
5 25t
ln
200
1
ln 200
5t
25
Chapter 5: Derivatives of Exponential and Trigonometric Functions
dP
Now, the question asks for dt 5 Pr when
t5
Pra
1
ln 200
8 1.06
25
1
ln 200
b 5 22.5 (using a calculator)
25
dy
5 (2x)(e x )r 1 (e x )(2x)r
dx
5 (2x)(e x ) 1 (e x )(21)
5 2xe x 2 e x
At the point x 5 0,
dy
5 20e 0 2 e 0 5 21.
dx
At the point x 5 0,
y 5 2 2 0e 0 5 2
So, an equation of the tangent to the curve at the
point x 5 0 is
y 2 2 5 21(x 2 0)
y 2 2 5 2x
y 5 2x 1 2
x1y2250
4. a. yr 5 23(e x )r
5 23e x
ys 5 23e x
b. yr 5 (x)(e 2x )r 1 (e 2x )(x)r
5 (x)((e 2x ) 1 (2x)r) 1 (e 2x )(1)
5 (x)((e 2x )(2)) 1 e 2x
5 2xe 2x 1 e 2x
ys 5 (2x)(e 2x )r 1 (e 2x )(2x)r 1 e 2x (2x)r
5 (2x)((e 2x )(2x)r) 1 (e 2x )(2) 1 (e 2x )(2)
5 (2x)((e 2x )(2)) 1 2e 2x 1 2e 2x
5 4xe 2x 1 4e 2x
c. yr 5 (e x )(4 2 x)r 1 (4 2 x)(e x )r
5 (e x )(21) 1 (4 2 x)(e x )
5 2e x 1 4e x 2 xe x
5 3e x 2 xe x
ys 5 (3e x )r 2 3 (x)(e x )r 1 (e x )(x)r4
5 3e x 2 3xe x 1 (e x )(1)4
5 3e x 2 xe x 2 e x
5 2e x 2 xe x
dy
5 (82x15 )(ln 8)(2x 1 5)r
5. a.
dx
5 (82x15 )(ln 8)(2)
5 2(ln 8)(82x15 )
3.
dy
5 3.2((10)0.2x )(ln 10)(0.2x)r
dx
5 3.2((10)0.2x )(ln 10)(0.2)
5 0.64(ln 10)((10).2x )
c. fr (x) 5 (x 2 )(2x )r 1 (2x )(x 2 )r
5 (x 2 )(2x )(ln 2) 1 (2x )(2x)
5 (ln 2)(x 22x ) 1 2x2x
5 2x ((ln 2)(x 2 ) 1 2x)
d. Hr (x) 5 300((5)3x21 )(ln 5)(3x 2 1)r
5 300((5)3x21 )(ln 5)(3)
5 900(ln 5)(5)3x21
5 900(ln 5)(5)3x21
e. qr (x) 5 (1.9)x ? (ln 1.9) 1 1.9(x)1.921
5 (1.9)x ? (ln 1.9) 1 1.9(x)0.9
5 (ln 1.9)(1.9)x 1 1.9x 0.9
f. f r (x) 5 (x 2 2)2 (4x )r 1 (4x )( (x 2 2)2 )r
5 (x 2 2)2 (4x )(ln 4) 1 (4x )(2(x 2 2))
5 (ln 4)(4x )(x 2 2)2 1 (4x )(2x 2 4)
5 4x ((ln 4)(x 2 2)2 1 2x 2 4)
6. a. The initial number of rabbits in the forest is
given by the time t 5 0.
0
R(0) 5 500( 10 1 e 2 10)
5 500(10 1 1)
5 500(11)
5 5500
b.
dR
b. The rate of change is the derivative, dt .
t
R(t) 5 5000 1 500( e210)
dR
t r
t
5 0 1 500( e210) a2 b
dt
10
1
t
5 500( e210) a2 b
10
t
5 250( e210)
c. 1 year 5 12 months
dR
The question asks for dt 5 Rr when t 5 12.
12
Rr (12) 5 250( e210)
8 215.06
d. To find the maximum number of rabbits,
optimize the function.
t
Rr (t) 5 250( e210)
t
0 5 250( e210)
t
0 5 e210
Calculus and Vectors Solutions Manual
5-15
When solving, the natural log (ln) of both sides
must be taken, but (ln 0) does not exist. So there
are no solutions to the equation. The function is
therefore always decreasing. So, the largest number
of rabbits will exist at the earliest time in the interval
at time t 5 0. To check, compare R(0) and R(36).
R(0) 5 5500 and R(36) 8 5013. So, the largest
number of rabbits in the forest during the first
3 years is 5500.
e. 6000
4000
2000
0
0
10
20 30 40
The graph is constantly decreasing. The y-intercept
is at the point (0, 5500). Rabbit populations normally
grow exponentially, but this population is shrinking
exponentially. Perhaps a large number of rabbit
predators such as snakes recently began to appear in
the forest. A large number of predators would
quickly shrink the rabbit population.
7. The highest concentration of the drug can be
found by optimizing the given function.
C(t) 5 10e 22t 2 10e 23t
Cr(t) 5 (10e 22t )(22t)r 2 (10e 23t )(23t)r
5 (10e 22t )(22) 2 (10e 23t )(23)
5 220e 22t 1 30e 23t
Set the derivative of the function equal to zero and
find the critical points.
0 5 220e 22t 1 30e 23t
20e 22t 5 30e 23t
2 22t
e 5 e 23t
3
e 23t
2
5 22t
3
e
2
5 (e 23t )(e 2t )
3
2
5 e 23t12t
3
2
5 e 2t
3
5-16
ln
2
5 2t
3
2
2 aln b 5 t
3
Therefore, t 5 2 ( ln 23) 8 0.41 is the critical value.
Now, use the algorithm for finding extreme values.
C(0) 5 10(e 0 2 e 0 ) 5 0
2
Ca2 aln bb 8 1.48 (using a calculator)
3
C(5) 5 0.0005
So, the function has a maximum when
t 5 2 ( ln 23) 8 0.41. Therefore, during the first five
hours, the highest concentration occurs at about
0.41 hours.
8. y 5 ce kx
yr 5 cke kx
The original function is increasing when its derivative is positive and decreasing when its derivative is
negative.
e kx . 0 for all k, xPR.
So, the original function represents growth when
ck . 0, meaning that c and k must have the same
sign. The original function represents decay when c
and k have opposite signs.
9. a. A(t) 5 5000e 0.02t
5 5000e 0.02(0)
5 5000
The initial population is 5000.
b. at t 5 7
A(7) 5 5000e 0.02(7) 5 5751
After a week, the population is 5751.
c. at t 5 30
A(30) 5 5000e 0.02(30) 5 9111
After 30 days, the population is 9111.
10. a. P(5) 5 760e 20.125(5)
8 406.80 mm Hg
b. P(7) 5 760e 20.125(7)
8 316.82 mm Hg
c. P(9) 5 760e 20.125(9)
8 246.74 mm Hg
11. A 5 100e 20.3x
Ar 5 100e 20.3x (20.3)
5 230e 20.3x
When 50% of the substance is gone, y 5 50
50 5 100e 20.3x
0.5 5 e 20.3x
ln (0.5) 5 ln e 20.3x
ln (0.5) 5 20.3x ln e
Chapter 5: Derivatives of Exponential and Trigonometric Functions
ln 0.5
5 20.3x
ln e
ln 0.5
5x
2
0.3 ln e
x 5 2.31
Ar 5 230e 20.3x
Ar(2.31) 5 230e 20.3(2.31)
Ar 8 215
When 50% of the substance is gone, the rate of
decay is 15% per year.
12. f(x) 5 xe x
f r(x) 5 xe x 1 (1)e x
5 e x (x 1 1)
x
So e . 0
x11.0
x . 21
This means that the function is increasing when
x . 21.
2
13. y 5 52x
When x 5 1,
1
y5
5
2
yr 5 52x (22x) ln 5
2
yr 5 2 ln 5
5
2
1
5y 2 5 2 ln 5(x 2 1)
5
5
5y 2 1 5 22 ln 5(x 2 1)
5y 2 1 5 (22 ln 5)x 1 2 ln 5
(2 ln 5)x 1 5y 5 2 ln 5 1 1
y
4
2
–4
–2
0
x
2
4
–2
–4
14. a. A 5 P(1 1 i)t
A(t) 5 1000(1 1 0.06)t
5 1000(1.06)t
b. Ar(t) 5 1000(1.06)t (1) ln (1.06)
5 1000(1.06)t ln 1.06
c. Ar (2) 5 1000(1.06)2 ln 1.06
5 $65.47
Calculus and Vectors Solutions Manual
Ar (5) 5 1000(1.06)5 ln 1.06
5 $77.98
Ar (10) 5 1000(1.06)10 ln 1.06
5 $104.35
d. No, the rate is not constant.
Ar (2)
5 ln 1.06
e.
A(2)
Ar (5)
5 ln 1.06
A(5)
Ar (10)
5 ln 1.06
A(10)
f. All the ratios are equivalent (they equal ln 1.06,
which is about 0.058 27), which means that
Ar (t)
A(t)
is
constant.
15. y 5 cex
yr 5 c(ex ) 1 (0)ex
5 cex
y 5 yr 5 ce x
5.4 The Derivatives of y 5 sin x and
y 5 cos x, pp. 256–257
d(2x)
dy
5 (cos 2x) ?
dx
dx
5 2 cos 2x
d(3x)
dy
b.
5 22 (sin 3x) ?
dx
dx
5 26 sin 3x
d(x 3 2 2x 1 4)
dy
c.
5 (cos (x 3 2 2x 1 4)) ?
dx
dx
2
3
5 (3x 2 2)(cos (x 2 2x 1 4))
d(24x)
dy
d.
5 22 sin (24x) ?
dx
dx
5 8 sin (24x)
d(3x)
d(4x)
dy
e.
5 cos (3x) ?
1 sin (4x) ?
dx
dx
dx
5 3 cos (3x) 1 4 sin (4x)
dy
f.
5 2x (ln 2) 1 2 cos x 1 2 sin x
dx
d(e x )
dy
g.
5 cos (e x ) ?
dx
dx
5 e xcos (e x )
d(3x 1 2p)
dy
h.
5 3 cos (3x 1 2p) ?
dx
dx
5 9 cos (3x 1 2p)
1. a.
5-17
ln 0.5
5 20.3x
ln e
ln 0.5
5x
2
0.3 ln e
x 5 2.31
Ar 5 230e 20.3x
Ar(2.31) 5 230e 20.3(2.31)
Ar 8 215
When 50% of the substance is gone, the rate of
decay is 15% per year.
12. f(x) 5 xe x
f r(x) 5 xe x 1 (1)e x
5 e x (x 1 1)
x
So e . 0
x11.0
x . 21
This means that the function is increasing when
x . 21.
2
13. y 5 52x
When x 5 1,
1
y5
5
2
yr 5 52x (22x) ln 5
2
yr 5 2 ln 5
5
2
1
5y 2 5 2 ln 5(x 2 1)
5
5
5y 2 1 5 22 ln 5(x 2 1)
5y 2 1 5 (22 ln 5)x 1 2 ln 5
(2 ln 5)x 1 5y 5 2 ln 5 1 1
y
4
2
–4
–2
0
x
2
4
–2
–4
14. a. A 5 P(1 1 i)t
A(t) 5 1000(1 1 0.06)t
5 1000(1.06)t
b. Ar(t) 5 1000(1.06)t (1) ln (1.06)
5 1000(1.06)t ln 1.06
c. Ar (2) 5 1000(1.06)2 ln 1.06
5 $65.47
Calculus and Vectors Solutions Manual
Ar (5) 5 1000(1.06)5 ln 1.06
5 $77.98
Ar (10) 5 1000(1.06)10 ln 1.06
5 $104.35
d. No, the rate is not constant.
Ar (2)
5 ln 1.06
e.
A(2)
Ar (5)
5 ln 1.06
A(5)
Ar (10)
5 ln 1.06
A(10)
f. All the ratios are equivalent (they equal ln 1.06,
which is about 0.058 27), which means that
Ar (t)
A(t)
is
constant.
15. y 5 cex
yr 5 c(ex ) 1 (0)ex
5 cex
y 5 yr 5 ce x
5.4 The Derivatives of y 5 sin x and
y 5 cos x, pp. 256–257
d(2x)
dy
5 (cos 2x) ?
dx
dx
5 2 cos 2x
d(3x)
dy
b.
5 22 (sin 3x) ?
dx
dx
5 26 sin 3x
d(x 3 2 2x 1 4)
dy
c.
5 (cos (x 3 2 2x 1 4)) ?
dx
dx
2
3
5 (3x 2 2)(cos (x 2 2x 1 4))
d(24x)
dy
d.
5 22 sin (24x) ?
dx
dx
5 8 sin (24x)
d(3x)
d(4x)
dy
e.
5 cos (3x) ?
1 sin (4x) ?
dx
dx
dx
5 3 cos (3x) 1 4 sin (4x)
dy
f.
5 2x (ln 2) 1 2 cos x 1 2 sin x
dx
d(e x )
dy
g.
5 cos (e x ) ?
dx
dx
5 e xcos (e x )
d(3x 1 2p)
dy
h.
5 3 cos (3x 1 2p) ?
dx
dx
5 9 cos (3x 1 2p)
1. a.
5-17
dy
5 2x 2 sin x 1 0
dx
5 2x 2 sin x
d( 1x)
1
dy
j.
5 cos a b ?
x
dx
dx
1
1
5 2 2 cos a b
x
x
dy
2. a.
5 (2 sin x)(2sin x) 1 (cos x)(2 cos x)
dx
5 22 sin2 x 1 2 cos2 x
5 2(cos2 x 2 sin2 x)
5 2 cos (2x)
b. y 5 (x 21 )(cos 2x)
dy
5 (x 21 )(22 sin 2x) 1 (cos 2x)(2x 22 )
dx
2 sin 2x
cos 2x
52
2
x
x2
d(sin 2x)
dy
c.
5 2sin (sin 2x) ?
dx
dx
5 2sin (sin 2x) ? 2 cos 2x
d. y 5 (sin x)(1 1 cos x)21
dy
5 (sin x)(2 (1 1 cos x)22 ? (2sin x)
dx
1 (1 1 cos x)21 (cos x)
2sin2 x
cos x
5
2 1
2 (1 1 cos x)
1 1 cos x
sin2 x
cos x(1 1 cos x)
5
1
(1 1 cos x)2
(1 1 cos x)2
2
2
sin x 1 cos x 1 cos x
5
(1 1 cos x)2
1 1 cos x
5
(1 1 cos x)2
1
5
1 1 cos x
dy
e.
5 (e x )(2sin x 1 cos x) 1 (cos x 1 sin x)(e x )
dx
5 e x (2sin x 1 cos x 1 cos x 1 sin x)
5 e x (2 cos x)
i.
dy
f.
5 (2x 3 )(cos x) 1 (sin x)(6x 2 )
dx
2 3(3x)(2sin x) 1 (cos x)(3)4
5 2x 3 cos x 1 6x 2 sin x 1 3x sin x 2 3 cos x
p
p
p
3
3. a. When x 5 , f(x) 5 f a b 5 sin a b 5 # .
3
3
3
2
fr(x) 5 cos x
5-18
p
p
f r a b 5 cos
3
3
1
5
2
p
So an equation for the tangent at the point x 5 is
3
1
p
#3
5 ax 2 b
2
2
3
p
2y 2 #3 5 x 2
3
p
2x 1 2y 1 a 2 #3b 5 0
3
b. When x 5 0, f(x) 5 f(0) 5 0 1 sin (0) 5 0.
f r (x) 5 1 1 cos x
f r (0) 5 1 1 cos (0)
5111
52
So an equation for the tangent at the point x 5 0 is
y 2 0 5 2(x 2 0)
y 5 2x
22x 1 y 5 0
p
p
p
c. When x 5 , f(x) 5 f a b 5 cos a4 ? b
4
4
4
5 cos (p)
5 21
d(4x)
f r (x) 5 2sin (4x) ?
dx
5 24 sin (4x)
p
p
f ra b 5 24 sin a4 ? b
4
4
5 24 sin (p)
50
p
So an equation for the tangent at the point x 5 is
y2
4
y 2 (21) 5 0ax 2
p
b
4
y1150
y 5 21
d. f(x) 5 sin 2x 1 cos x, x 5
p
p
2
The point of contact is ( 2 , 0). The slope of the
tangent line at any point is fr (x) 5 2 cos 2x 2 sin x.
p
At ( 2 , 0), the slope of the tangent line is
p
2 cos p2sin 2 5 23.
p
An equation of the tangent line is y 5 23 x 2 2 .
(
)
Chapter 5: Derivatives of Exponential and Trigonometric Functions
e. f(x) 5 cos a2x 1
5 2 sin ( !t) ? cos ( !t) ?
p
p
b, x 5
3
4
The point of tangency is a , 2
p
4
!3
b. The
2
slope of the
p
tangent line at any point is f r (x) 5 22 sin (2x 1 3 ).
At a , 2
p
4
22 sin a
!3
b,
2
the slope of the tangent line is
5p
b 5 21.
6
An equation of the tangent line is
y1
"3
p
52 x24 .
2
(
)
p
p
p
p
, f(x) 5 f a b 5 2 sin a b cos a b
2
2
2
2
5 2(1)(0)
50
f r(x) 5 (2 sin x)(2sin x) 1 (cos x)(2 cos x)
5 22 sin2 x 1 2 cos2 x
5 2(cos2 x 2 sin2 x)
5 2 cos (2x)
p
p
f ra b 5 2 cos a2 ? b
2
2
5 2 cos p
5 22
p
So an equation for the tangent when x 5 is
f. When x 5
2
p
y 2 0 5 22ax 2 b
2
y 5 22x 1 p
2x 1 y 2 p 5 0
4. a. One could easily find f r (x) and gr (x) to see
that they both equal 2(sin x)(cos x). However, it
is easier to notice a fundamental trigonometric
identity. It is known that sin2 x 1 cos2 x 5 1. So,
sin2 x 5 1 2 cos2 x.
Therefore, one can notice that f(x) is in fact equal
to g(x). So, because f(x) 5 g(x), f r (x) 5 gr (x).
b. f r(x) and gr(x) are negatives of each other.
That is, f r (x) 5 2(sin x)(cos x) while
gr (x) 5 22(sin x) (cos x).
5. a. v(t) 5 (sin ( !t))2
d(sin ( !t))
dt
d( !t)
5 2 sin ( !t) ? cos ( !t) ?
dt
1 212
5 2 sin ( !t) ? cos ( !t) ? t
2
vr(t) 5 2 sin ( !t) ?
Calculus and Vectors Solutions Manual
5
sin ( !t) cos ( !t)
!t
1
2 !t
1
b. v(t) 5 (1 1 cos t 1 sin2 t)2
1
1
vr (t) 5 (1 1 cos t 1 sin2 t)22
2
d(1 1 cos t 1 (sin t)2 )
3
dt
d(sin t)
2sin t 1 2(sin t) ?
dt
5
2# 1 1 cos t 1 sin2 t
2sin t 1 2(sin t)(cos t)
5
2# 1 1 cos t 1 sin2 t
c. h(x) 5 sin x sin 2x sin 3x
So, treat sin x sin 2x as one function, say f(x) and
treat sin 3x as another function, say g(x).
Then, the product rule may be used with the
chain rule:
hr (x) 5 f(x)gr (x) 1 g(x)f r (x)
5 (sin x sin 2x)(3 cos 3x)
1(sin 3x)3 (sin x)(2 cos 2x)
1 (sin 2x)(cos x)4
5 3 sin x sin 2x cos 3x
1 2 sin x sin 3x cos 2x
1 sin 2x sin 3x cos x
d(x 2 1 (cos x)2 )
d. mr (x) 5 3(x 2 1 cos2 x)2 ?
dx
5 3(x2 1 cos2 x)2 ? (2x 1 2(cos x) (2sin x))
5 3(x 2 1 cos2 x)2 ? (2x 2 2 sin x cos x)
6. By the algorithm for finding extreme values, the
maximum and minimum values occur at points on
the graph where f r (x) 5 0, or at an endpoint of the
interval.
dy
5 2sin x 1 cos x
a.
dx
Set
dy
50
dx
and solve for x to find any critical points.
cos x 2 sin x 5 0
cos x 5 sin x
sin x
15
cos x
1 5 tan x
p 5p
x5 ,
4 4
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
5-19
x
0
p
4
5p
4
2p
f(x) 5 cos x 1 sin x
1
"2
2"2
1
Set
So, the absolute maximum value on the interval is
# 2 when x 5
p
4
and the absolute minimum value
on the interval is 2 # 2 when x 5
5p
.
4
y
2
1
x
0
p
–1
2p
–2
b.
dy
5 1 2 2 sin x
dx
Set
dy
50
dx
x
2p
f(x) 5 x 1 2 cos x
2p 2 2
2
p
6
p
6
p
p
1 #3
1 #3
6
6
8 25.14 8 1.21
8 2.26
2
p
p22
8 1.14
So, the absolute maximum value on the interval is
p
2.26 when x 5 and the absolute minimum value
6
on the interval is 25.14 when x 5 2p.
y
4
–p
2
0
–4
–8
5-20
and solve for x to find any critical points.
cos x 1 sin x 5 0
sin x 5 2cos x
sin x
5 21
cos x
tan x 5 21
3p 7p
,
x5
4 4
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
x
0
3p
4
7p
4
2
f(x) 5 sin x 2 cos x
21
"2
2"2
21
# 2 when x 5 4 and the absolute minimum value
3p
1 2 2 sin x 5 0
1 5 2 sin x
1
5 sin x
2
p 5p
x5 ,
6 6
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
–p
dy
50
dx
So, the absolute maximum value on the interval is
and solve for x to find any critical points.
8
dy
5 cos x 1 sin x
dx
c.
x
p
2
p
on the interval is 2 # 2 when x 5
7p
.
4
y
2
1
x
0
p
–1
2p
–2
d.
dy
5 3 cos x 2 4 sin x
dx
Set
dy
50
dx
and solve for x to find any critical points.
3 cos x 2 4 sin x 5 0
3 cos x 5 4 sin x
sin x
3
5
cos x
4
3
5 tan x
4
3
tan21 a b 5 tan21 (tan x)
4
Using a calculator, x 8 0.6435.
This is a critical value, but there is also one more in
the interval 0 # x # 2p. The period of tan x is p,
so adding p to the one solution will give another
solution in the interval.
x 5 0.6435 1 p 8 3.7851
Chapter 5: Derivatives of Exponential and Trigonometric Functions
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
0
x
f(x) 5 3 sin x 1 4 cos x
0.64
4
3.79
y
8
4
x
p
–4
2p
–8
7. a. The particle will change direction when the
velocity, sr (t), changes from positive to negative.
sr(t) 5 16 cos 2t
Set sr (t) 5 0 and solve for t to find any critical points.
0 5 16 cos 2t
0 5 cos 2t
p 3p
,
5 2t
2 2
p 3p
,
5t
4 4
Also, there is no given interval so it will be beneficial
to locate all solutions.
p
3p
1 pk for some positive
Therefore, t 5 1 pk,
4
4
integer k constitutes all solutions.
One can create a table and notice that on each side
of any value of t, the function is increasing on one
side and decreasing on the other. So, each t value is
either a maximum or a minimum.
t
p
4
3p
4
5p
4
7p
4
s(t) 5 8 sin 2t
8
28
8
28
The table continues in this pattern for all critical
values t. So, the particle changes direction at all
critical values. That is, it changes direction for
t5
p
3p
1 pk,
1 pk
4
4
for positive integers k.
b. From the table or a graph, one can see that
the particle’s maximum velocity is 8 at the time
p
t 5 1 pk.
4
Calculus and Vectors Solutions Manual
f(x)
1
x
0
4
So, the absolute maximum value on the interval is 5
when x 8 0.64 and the absolute minimum value on
the interval is 25 when x 8 3.79.
0
2
2p
25
5
8. a.
p
–1
2p
–2
b. The tangent to the curve f(x) is horizontal at the
point(s) where f r (x) is zero.
f r (x) 5 2sin x 1 cos x
Set f r (x) 5 0 and solve for x to find any critical
points.
cos x 2 sin x 5 0
cos x 5 sin x
sin x
15
cos x
1 5 tan x
x5
p
4
(Note: The solution x 5
5p
4
is not in the
interval 0 # x # p so it is not included.) When
p
4
x 5 , f(x) 5 f
( 4 ) 5 # 2.
p
So, the coordinates of the point where the tangent to
p
the curve of f(x) is horizontal is ( , # 2).
4
1
5 (sin x)21
9. csc x 5
sin x
1
sec x 5
5 (cos x)21
cos x
Now, the power rule can be used to compute the
derivates of csc x and sec x.
d(sin x)
((sin x)21 )r 5 2 (sin x)22 ?
dx
5 2 (sin x)22 ? cos x
cos x
52
(sin x)2
d(sin x)
((sin x)21 )r 5 2 (sin x)22 ?
dx
cos x
1
?
52
sin x sin x
5 2csc x cot x
d(cos x)
((cos x)21 )r 5 2 (cos x)22 ?
dx
5 2 (cos x)22 ? (2sin x)
sin x
5
(cos x)2
1
sin x
5
?
cos x cos x
5 sec x tan x
5-21
function is 1, so the maximum distance from the
origin is 4(1) or 4.
12.
dy
5 22 sin 2x
dx
p 1
At the point , ,
10.
( 6 2)
h
dy
p
5 22 sin a2 ? b
dx
6
p
5 22 sin a b
3
5 22a
cos u 5
x
5x
1
and sin u 5
area of a trapezoid is
(p 1)
Therefore, at the point 6 , 2 , the slope of the
tangent to the curve y 5 cos 2x is 2 #3.
11. a. The particle will change direction when the
velocity, sr(t) changes from positive to negative.
sr(t) 5 16 cos 4t
Set sr (t) 5 0 and solve for t to find any critical points.
0 5 16 cos 4t
0 5 cos 4t
p 3p
,
5 4t
2 2
p 3p
,
5t
8 8
Also, there is no given interval so it will be beneficial
to locate all solutions.
p
3p
1 pk,
1 pk
8
8
for some positive
integer k constitutes all solutions.
One can create a table and notice that on each side
of any value of t, the function is increasing on one
side and decreasing on the other. So, each t value is
either a maximum or a minimum.
t
p
8
3p
8
5p
8
7p
8
s(t) 5 4 sin 4t
4
24
4
24
The table continues in this pattern for all critical
values t. So, the particle changes direction at all
critical values. That is, it changes direction for
p
3p
1 pk,
1 pk
4
4
for positive integers k.
b. From the table or a graph, one can see that the
particle’s maximum velocity is 4 at the time
p
t 5 1 pk.
4
c. At t 5 0, s 5 0, so the minimum distance from
the origin is 0. The maximum value of the sine
5-22
h
h
5 h.
1
Therefore, x 5 cos u and h 5 sin u.
The irrigation channel forms a trapezoid and the
5 2#3
t5
1m
1m
u
x
Label the base of a triangle x and the height h. So
!3
b
2
Therefore, t 5
1m
u
x
(b1 1 b2 )h
2
where b1 and b2 are
the bottom and top bases of the trapezoid and h is
the height.
b1 5 1
b2 5 x 1 1 1 x 5 cos u 1 1 1 cos u 5 2 cos u 1 1
h 5 sin u
Therefore, the area equation is given by
(2 cos u 1 1 1 1) sin u
A5
2
(2 cos u 1 2) sin u
5
2
2 cos u sin u 1 2 sin u
5
2
5 sin u cos u 1 sin u
To maximize the cross-sectional area, differentiate:
Ar 5 (sin u)(2sin u) 1 (cos u)(cos u) 1 cos u
5 2sin2 u 1 cos2 u 1 cos u
Using the trig identity sin2 u 1 cos2 u 5 1, use the
fact that sin2 u 5 1 2 cos2 u.
Ar 5 2 (1 2 cos2 u) 1 cos2 u 1 cos u
5 21 1 cos2 u 1 cos2 u 1 cos u
5 2 cos2 u 1 cos u21
Set Ar 5 0 to find the critical points.
0 5 2 cos2 u 1 cos u 2 1
0 5 (2 cos u21)(cos u 1 1)
Solve the two expressions for u.
2 cos u 5 1
1
cos u 5
2
p
u5
3
Also, cos u 5 21
u5p
(Note: The question only seeks an answer around
p
0 # u # . So, there is no need to find all solutions
2
by adding kp for all integer values of k.)
The area, A, when u 5 p is 0 so that answer is
disregarded for this problem.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
p
!3
2!3
1
4
4
3 !3
5
4
p
The area is maximized by the angle u 5 .
14. First find ys.
y 5 A cos kt 1 B sin kt
yr 5 2kA sin kt 1 kB cos kt
ys 5 2k 2A cos kt 2 k 2B sin kt
So, ys 1 k 2y
5 2k 2A cos kt 2 k 2B sin kt
1 k 2 (A cos kt 1 B sin kt)
5 2k 2A cos kt 2 k 2B sin kt 1 k 2A cos kt
1 k 2B sin kt
50
Therefore, ys 1 k 2y 5 0.
13. Let O be the centre of the circle with line
segments drawn and labeled, as shown.
5.5 The Derivative of y 5 tan x, p. 260
When u 5 ,
3
p
p
p
A 5 sin cos 1 sin
3
3
3
!3 1
!3
5a
? b1
2
2
2
5
3
A
1. a.
uu
R
R
x 2u
y
C
B
In ^OCB, /COB 5 2u.
Thus,
y
5 sin 2u
R
and
x
5 cos 2u,
R
so y 5 R sin 2u and x 5 R cos 2u.
The area A of ^ ABD is
1
A 5 0 DB 0 0 AC 0
2
5 y(R 1 x)
5 R sin 2u(R 1 R cos 2u)
5 R 2 (sin 2u 1 sin 2u cos 2u), where 0 , 2u , p
dA
5 R 2 (2 cos 2u 1 2 cos 2u cos 2u
du
1 sin 2u(22 sin 2u)).
We solve
dA
5 0:
du
2 cos2 2u 2 2 sin2 2u 1 2 cos 2u 5 0
2 cos2 2u 1 cos 2u 2 1 5 0
(2 cos 2u21)(cos 2u 1 1) 5 0
1
cos 2u 5 or cos 2u 5 21
2
p
2u 5 or 2u 5 p (not in domain).
3
As 2u S 0, A S 0 and as 2u S p, A S 0. The
maximum area of the triangle is
p
3
d
dy
5 2 sec2 x 2 sec 2xa 2xb
dx
dx
5 2 sec2 x 2 2 sec 2x
d
dy
c.
5 2 tan (x 3 )a tan (x 3 )b
dx
dx
d
dy
5 2 tan (x 3 )a tan (x 3 )b
dx
dx
d
5 2 tan (x 3 ) sec2 (x 3 )a x 3 b
dx
5 6x 2 tan (x 3 ) sec2 (x 3 )
b.
O
D
d
dy
5 sec2 3xa 3xb
dx
dx
5 3 sec2 3x
p
6
3 !3 2
R
4
when 2u 5 , i.e., u 5 .
Calculus and Vectors Solutions Manual
(d )
2x tan px 2 x 2 sec2 px px
dy
dx
d.
5
dx
tan2 px
2x tan px 2 px 2 sec2 px
5
tan2 px
x(2 tan px 2 px sec2 px)
5
tan2 px
d
d
dy
e.
5 sec2 (x 2 )a x 2 b 2 2 tan xa b (tan x)
dx
dx
dx
5 2x sec2 (x 2 ) 2 2 tan x sec2 x
dy
d
f.
5 tan 5x(3 cos 5x)a 5xb
dx
dx
d
1 3 sin 5x sec2 5xa 5xb
dx
5 15 tan 5x cos 5x 1 15 sin 5x sec2 5x
5 15 (tan 5x cos 5x 1 sin 5x sec2 5x)
2. a. The general equation for the line tangent to
the function f(x) at the point (a, b) is
y 2 b 5 fr (x)(x 2 a).
f(x) 5 tan x
fr (x) 5 sec2 x
5-23
p
!3
2!3
1
4
4
3 !3
5
4
p
The area is maximized by the angle u 5 .
14. First find ys.
y 5 A cos kt 1 B sin kt
yr 5 2kA sin kt 1 kB cos kt
ys 5 2k 2A cos kt 2 k 2B sin kt
So, ys 1 k 2y
5 2k 2A cos kt 2 k 2B sin kt
1 k 2 (A cos kt 1 B sin kt)
5 2k 2A cos kt 2 k 2B sin kt 1 k 2A cos kt
1 k 2B sin kt
50
Therefore, ys 1 k 2y 5 0.
13. Let O be the centre of the circle with line
segments drawn and labeled, as shown.
5.5 The Derivative of y 5 tan x, p. 260
When u 5 ,
3
p
p
p
A 5 sin cos 1 sin
3
3
3
!3 1
!3
5a
? b1
2
2
2
5
3
A
1. a.
uu
R
R
x 2u
y
C
B
In ^OCB, /COB 5 2u.
Thus,
y
5 sin 2u
R
and
x
5 cos 2u,
R
so y 5 R sin 2u and x 5 R cos 2u.
The area A of ^ ABD is
1
A 5 0 DB 0 0 AC 0
2
5 y(R 1 x)
5 R sin 2u(R 1 R cos 2u)
5 R 2 (sin 2u 1 sin 2u cos 2u), where 0 , 2u , p
dA
5 R 2 (2 cos 2u 1 2 cos 2u cos 2u
du
1 sin 2u(22 sin 2u)).
We solve
dA
5 0:
du
2 cos2 2u 2 2 sin2 2u 1 2 cos 2u 5 0
2 cos2 2u 1 cos 2u 2 1 5 0
(2 cos 2u21)(cos 2u 1 1) 5 0
1
cos 2u 5 or cos 2u 5 21
2
p
2u 5 or 2u 5 p (not in domain).
3
As 2u S 0, A S 0 and as 2u S p, A S 0. The
maximum area of the triangle is
p
3
d
dy
5 2 sec2 x 2 sec 2xa 2xb
dx
dx
5 2 sec2 x 2 2 sec 2x
d
dy
c.
5 2 tan (x 3 )a tan (x 3 )b
dx
dx
d
dy
5 2 tan (x 3 )a tan (x 3 )b
dx
dx
d
5 2 tan (x 3 ) sec2 (x 3 )a x 3 b
dx
5 6x 2 tan (x 3 ) sec2 (x 3 )
b.
O
D
d
dy
5 sec2 3xa 3xb
dx
dx
5 3 sec2 3x
p
6
3 !3 2
R
4
when 2u 5 , i.e., u 5 .
Calculus and Vectors Solutions Manual
(d )
2x tan px 2 x 2 sec2 px px
dy
dx
d.
5
dx
tan2 px
2x tan px 2 px 2 sec2 px
5
tan2 px
x(2 tan px 2 px sec2 px)
5
tan2 px
d
d
dy
e.
5 sec2 (x 2 )a x 2 b 2 2 tan xa b (tan x)
dx
dx
dx
5 2x sec2 (x 2 ) 2 2 tan x sec2 x
dy
d
f.
5 tan 5x(3 cos 5x)a 5xb
dx
dx
d
1 3 sin 5x sec2 5xa 5xb
dx
5 15 tan 5x cos 5x 1 15 sin 5x sec2 5x
5 15 (tan 5x cos 5x 1 sin 5x sec2 5x)
2. a. The general equation for the line tangent to
the function f(x) at the point (a, b) is
y 2 b 5 fr (x)(x 2 a).
f(x) 5 tan x
fr (x) 5 sec2 x
5-23
p
fa b 5 0
4
p
f ra b 5 2
4
The equation for the line tangent to the function
p
(
p
)
f(x) at x 5 4 is y 5 2 x 2 4 .
b. The general equation for the line tangent to the
function f(x) at the point (a, b) is
y 2 b 5 f r (x)(x 2 a).
f(x) 5 6 tan x 2 tan 2x
d
f r(x) 5 6 sec2 x 2 sec2 2xa 2xb
dx
f r(x) 5 6 sec2 x 2 2 sec2 2x
f(0) 5 0
f r(0) 5 22
The equation for the line tangent to the function
f(x) at x 5 0 is y 5 22x.
d
dy
5 sec2 x(sin x)a sin xb
3. a.
dx
dx
5 cos x sec2 (sin x)
d
dy
5 22 3tan (x 2 2 1)4 23 a tan (x 2 2 1)b
b.
dx
dx
5 22 3tan (x 2 2 1)4 23 sec2 (x 2 2 1)
d
3 a (x 2 2 1)b
dx
5 24x 3tan (x 2 2 1)4 23 sec2 (x 2 2 1)
d
dy
5 2 tan (cos x)a tan (cos x)b
c.
dx
dx
d
5 2 tan (cos x) sec2 (cos x)a cos xb
dx
5 22 tan (cos x) sec2 (cos x) sin x
dy
d
5 2 (tan x 1 cos x)a tan x 1 cos xb
d.
dx
dx
5 2 (tan x 1 cos x)(sec2 x 2 sin x)
dy
d
5 tan x (3 sin2 x)a sin xb 1 sin3 x sec2 x
e.
dx
dx
2
5 3 tan x sin x cos x 1 sin3 x sec2 x
5 sin2 x (3 tan x cos x 1 sin x sec2 x)
dy
d
5 e tan "x a tan "xb
f.
dx
dx
d
5 e tan "x (sec2 "x)a "xb
dx
1
5
e tan "x sec2 "x
2"x
5-24
dy
5 tan x cos x 1 sin x sec2 x
dx
sin x
1
5
? cos x 1 sin x ?
cos x
cos2 x
sin x
5 sin x 1
cos2 x
2
cos3 x
dy
5
cos
x
1
dx 2
cos4 x
4. a.
)
(d
sin x(2 cos x) dx cos x
2
cos4 x
cos3 x 1 2 sin2 x cos x
5 cos x 1
cos4 x
1
2 sin2 x
5 cos x 1
1
cos x
cos3 x
2 sin2 x
5 cos x 1 sec x 1
cos3 x
d
dy
5 2 tan xa tan xb
b.
dx
dx
5 2 tan x sec2 x
2 sin x
1
5
?
cos x cos2 x
2 sin x
5
cos3 x
(d
)
2 cos4 x 2 6 sin x cos2 x dxcos x
d 2y
5
dx 2
cos6 x
4
2 cos x 1 6 sin2 x cos2 x
5
cos6 x
6 sin2 x
1
2
1
?
5
2
2
cos x
cos x cos2 x
2
5 2 sec x 1 6 tan2 x sec2 x
5 2 sec2 x(1 1 3 tan2 x)
5. The slope of f(x) 5 sin x tan x equals zero when
the derivative equals zero.
f(x) 5 sin x tan x
f r (x) 5 sin x(sec2x) 1 tan x(cos x)
sin x
5 sin x(sec2 x) 1
(cos x)
cos x
5 sin x(sec2 x) 1 sin x
5 sin x(sec2 x 1 1)
2
sec x 1 1 is always positive, so the derivative is 0
only when sin x 5 0. So, fr (x) equals 0 when x 5 0,
x 5 p, and x 5 2p. The solutions can be verified
by examining the graph of the derivative function
shown below.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
y
4
3
2
1
0
–1
–2
–3
–4
cos2 x 1 sin x 1 sin2 x
cos2 x
1 1 sin x
5
cos2 x
The denominator is never negative. 1 1 sin x . 0
5
f'(x)
x
␲
p
p
for 2 2 , x , 2 , since sin x reaches its minimum
2␲
p
of 21 at x 5 2 . Since the derivative of the original
function is always positive in the specified interval,
the function is always increasing in that interval.
p
p
8. When x 5 , y 5 2 tan ( 4 )
4
52
yr 5 2 sec2 x
p
p 2
When x 5 , yr 5 2 asec b
4
4
6. The local maximum point occurs when the
derivative equals zero.
dy
5 2 2 sec2 x
dx
2 2 sec2 x 5 0
sec2 x 5 2
sec x 5 6"2
p
x56
4
dy
p
5 0 when x 5 6 4 , so the local maximum
dx
p
point occurs when x 5 6 4 . Solve for y
when
5 2( #2)2
54
p
So an equation for the tangent at the point x 5 is
4
p
y 2 2 5 4ax 2 b
4
y 2 2 5 4x 2 p
24x 1 y 2 (2 2 p) 5 0
sin x
9. Write tan x 5 cos x and use the quotient rule to
derive the derivative of the tangent function.
10. y 5 cot x
1
y5
tan x
tan x(0) 2 (1) sec2 x
dy
5
dx
tan2 x
2
2sec x
5
tan2 x
p
x 5 4.
p
p
y 5 2a b 2 tan a b
4
4
p
y5 21
2
y 5 0.57
p
4
Solve for y when x 5 2 .
p
p
y 5 2a2 b 2 tan a2 b
4
4
p
y52 11
2
y 5 20.57
p
The local maximum occurs at the point 4 , 0.57 .
7. y 5 sec x 1 tan x
1
sin x
5
1
cos x
cos x
1 1 sin x
5
cos x
cos2 x 2 (1 1 sin x)(2sin x)
dy
5
dx
cos2 x
2
cos x 2 (2sin x 2 sin2 x)
5
cos2 x
(
Calculus and Vectors Solutions Manual
)
21
cos 2 x
5 sin2 x
cos 2 x
21
sin2 x
5 2csc2 x
11. Using the fact from question 10 that the
derivative of cot x is 2csc2 x,
f r (x) 5 24 csc2 x
5 24 (csc x)2
d(csc x)
f s (x) 5 28 (csc x) ?
dx
5 28 (csc x) ? (2csc x cot x)
5 8 csc2 x cot x
5
5-25
Review Exercise, pp. 263–265
1. a. yr 5 0 2 e x
5 2e x
b. yr 5 2 1 3e x
d(2x 1 3)
c. yr 5 e 2x13 ?
dx
5 2e 2x13
d(23x 2 1 5x)
2
d. yr 5 e 23x 15x ?
dx
2
5 (26x 1 5)e 23x 15x
e. yr 5 (x)(e x ) 1 (e x )(1)
5 e x (x 1 1)
(e t 1 1)(e t ) 2 (e t 2 1)(e t )
f. sr 5
(e t 1 1)2
2t
t
e 1 e 2 (e 2t 2 e t )
5
(e t 1 1)2
t
2e
5 t
(e 1 1)2
dy
5 10x ln 10
2. a.
dx
d(3x 2 )
dy
2
5 43x ? ln 4 ?
b.
dx
dx
3x 2
5 6x(4 )ln 4
dy
5 (5x)(5x ln 5) 1 (5x )(5)
c.
dx
5 5 ? 5x (x ln 5 1 1)
dy
5 (x 4 )(2x ln 2) 1 (2x )(4x 3 )
d.
dx
5 x 3 ? 2x (x ln 2 1 4)
e. y 5 (4x)(42x )
dy
5 (4x)(242x ln 4) 1 (42x )(4)
dx
5 4 ? 42x (2x ln 4 1 1)
4 2 4x ln 4
5
4x
x
f. y 5 (5# )(x 21 )
dy
d( #x)
5 (5#x )(2x 22 ) 1 (x 21 )a5#x ? ln 5 ?
b
dx
dx
1
1
5 (5#x )a2 2 b 1 (x 21 )a5#x ? ln 5 ?
b
x
2!x
1
ln 5
b
5 5#x a2 2 1
x
2x!x
d(2x)
d(2x)
dy
5 3 cos (2x) ?
1 4 sin (2x) ?
3. a.
dx
dx
dx
5 6 cos (2x) 1 8 sin (2x)
5-26
d(3x)
dy
5 sec2 (3x) ?
dx
dx
5 3 sec2 (3x)
c. y 5 (2 2 cos x)21
dy
d(2 2 cos x)
5 2 (2 2 cos x)22 ?
dx
dx
sin x
52
(2 2 cos x)2
d(2x)
dy
5 (x)asec2 (2x) ?
b 1 (tan (2x))(1)
d.
dx
dx
5 2x sec2 (2x) 1 tan 2x
d(3x)
dy
5 (sin 2x)ae 3x ?
b
e.
dx
dx
d(2x)
b
1 (e 3x )acos 2x ?
dx
5 3e 3x sin 2x 1 2e 3x cos 2x
5 e 3x (3 sin 2x 1 2 cos 2x)
f. y 5 (cos (2x))2
d(cos (2x))
dy
5 2 (cos (2x)) ?
dx
dx
d(2x)
5 2(cos (2x)) ? 2sin (2x) ?
dx
5 24 cos (2x) sin (2x)
b.
4. a. f(x) 5 e x ? x 21
f r (x) 5 (e x )(2x 22 ) 1 (x 21 )(e x )
1
1
5 e x a2 2 1 b
x
x
2
2x
1
x
b
5 ex a
x3
Now, set f r (x) 5 0 and solve for x.
2x 1 x 2
0 5 ex a
b
x3
x2 2 x
5 0.
Solve e x 5 0 and
x3
x
e is never zero.
x2 2 x
50
x3
x2 2 x 5 0
x(x 2 1) 5 0
So x 5 0 or x 5 1.
(Note, however, that x cannot be zero because this
would cause division by zero in the original function.)
So x 5 1.
b. The function has a horizontal tangent at (1, e).
Chapter 5: Derivatives of Exponential and Trigonometric Functions
d(22x)
b 1 (e 22x )(1)
dx
5 22xe 22x 1 e 22x
5 e 22x (22x 1 1)
1
1
1
f ra b 5 e 22 2 a22 ? 1 1b
2
2
5 e 21 (21 1 1)
50
b. This means that the slope of the tangent to f(x)
at the point with x-coordinate 21 is 0.
6. a. yr 5 (x)(e x ) 1 (e x )(1) 2 e x
5 xe x
ys 5 (x)(e x ) 1 (e x )(1)
5 xe x 1 e x
5 e x (x 1 1)
b. yr 5 (x)(10e 10x ) 1 (e 10x )(1)
5 10xe 10x 1 e 10x
ys 5 (10x)(10e 10x ) 1 (e 10x )(10) 1 10e 10x
5 100xe 10x 1 10e 10x 1 10e 10x
5 100xe 10x 1 20e 10x
5 20e 10x (5x 1 1)
e 2x 2 1
7. y 5 2x
e 11
dy
2e 2x (e 2x 1 1) 2 (e 2x 2 1)(2e 2x )
5
dx
(e 2x 1 1)2
4x
2x
2e 1 2e 2 2e 4x 1 2e 2x
5
(e 2x 1 1)2
4e 2x
5 2x
(e 1 1)2
5. a. f r (x) 5 (x)ae 22x ?
Now, 1 2 y 2 5 1 2
4x
5
e 4x 2 2e 2x 1 1
(e
2x
2x
2
1 1)
4x
The equation of the tangent line is
y 1 ln 2 1 2 5 3(x 1 ln 2) or
3x 2 y 1 2 ln 2 2 2 5 0.
p
2
9. When x 5 ,
p
p
p
p
p
y 5 f(x) 5 f a b 5 sina b 5 (1) 5
2
2
2
2
2
yr 5 fr (x) 5 (x)(cos x) 1 (sin x)(1)
5 x cos x 1 sin x
p
p
p
p
f r a b 5 cos 1 sin
2
2
2
2
p
5 (0) 1 1
2
51
p
So an equation for the tangent at the point x 5 is
2
p
p
y 2 5 1ax 2 b
2
2
p
p
y2 5x2
2
2
y5x
2x 1 y 5 0
10. If s(t) 5
(
Calculus and Vectors Solutions Manual
)(
p
)
(
)
p
p
(
sin 4 )(22 sin 2 ? 4 )
2
(3 1 cos 2 ? p4 )
2
5
To find the point(s) on the curve where the tangent
has slope 3, we solve:
1 1 e 2x 5 3
e 2x 5 2
2x 5 ln 2
x 5 2ln 2.
The point of contact of the tangent is
(2ln 2,2ln 2 2 2).
p
3 1 cos 2 ? 4 cos 4
p
sr a b 5
4
p 2
3 1 cos 2 ? 4
e 1 2e 1 1 2 e 1 2e 2 1
(e 2x 1 1)2
dy
2x
dx 5 1 1 e .
is the function describing
an object’s position at time t, then v(t) 5 sr (t) is
the function describing the object’s velocity at
time t. So
v(t) 5 sr (t)
(3 1 cos 2t)(cos t) 2 (sin t)(22 sin 2t)
5
(3 1 cos 2t)2
2x
4e 2x
dy
2x
2 5
(3 1 1)
dx
8. The slope of the required tangent line is 3.
The slope at any point on the curve is given by
sin t
3 1 cos 2t
(3 1 cos p2 )("22) 2 ("22)(22 sin p2 )
5
(3 1 cos p2 )2
"2
(3 1 0)("2
2 ) 2 ( 2 )(22 ? 1)
5
(3 1 0)2
1 #2
9
3#2 1 2#2 1
5
?
2
9
5#2
5
18
5
3"2
2
5-27
So, the object’s velocity at time t 5
5#2
8 0.3928
18
p
4
Since e 2t . 0 for all t, c1r(t) 5 0 when t 5 1.
Since c1r(t) . 0 for 0 # t , 1, and c1r(t) , 0 for all
is
metres per unit of time.
1
11. a. The question asks for the time t when
Nr(t) 5 0.
t
N(t) 5 60 000 1 2000te220
1 t
t
Nr(t) 5 0 1 (2000t)a2 e 220 b 1 (e 220 )(2000)
20
t
t
5 2100te 220 1 2000e 220
t
5 100e 220 (2t 1 20)
Set Nr (t) 5 0 and solve for t.
t
0 5 100e 220 (2t 1 20)
t
100e 220 is never equal to zero.
2t 1 20 5 0
20 5 t
Therefore, the rate of change of the number of
bacteria is equal to zero when time t 5 20.
b. The question asks for
dM
5 Mr(t)
dt
when t 5 10.
That is, it asks for Mr (10).
1
M(t) 5 (N 1 1000)3
d(N 1 1000)
1
2
Mr (t) 5 (N 1 1000)23 ?
3
dt
dN
1
5
2 ?
3(N 1 1000)3 dt
From part a.,
t
dN
5 Nr(t) 5 100e 220 (2t 1 20)
dt
220t
and
N(t) 5 60 000 1 2000te
1
100e220 (2t 1 20)
So Mr (t) 5
2
3(N 1 1000)3
First calculate N(10).
10
N(10) 5 60 000 1 2000(10)e220
1
5 60 000 1 20 000e 22
8 72 131
10
100e 220 (210 1 20)
So Mr (10) 5
2
3(N(10) 1 1000)3
1
100e 22 (10)
5
2
3(72 131 1 1000)3
606.53
8
5246.33
8 0.1156
So, after 10 days, about 0.1156 mice are infected
per day. Essentially, almost 0 mice are infected per
day when t 5 10.
12. a. c1 (t) 5 te 2t; c1 (0) 5 0
c1r (t) 5 e 2t 2 te 2t
5 e 2t (1 2 t)
5-28
t . 1, c1 (t) has a maximum value of e 8 0.368
at t 5 1 h.
c2 (t) 5 t 2e 2t; c2 (0) 5 0
c2r(t) 5 2te 2t 2 t 2e 2t
5 te 2t (2 2 t)
c2r(t) 5 0 when t 5 0 or t 5 2.
Since c2r(t) . 0 for 0 , t , 2 and c2r(t) , 0 for all
4
t . 2, c2 (t) has a maximum value of e 2 8 0.541 at
t 5 2 h. The larger concentration occurs for
medicine c2.
b. c1 (0.5) 5 0.303
c2 (0.5) 5 0.152
In the first half-hour, the concentration of c1
increases from 0 to 0.303, and that of c2 increases
from 0 to 0.152. Thus, c1 has the larger concentration
over this interval.
13. a. y 5 (2 1 3e 2x )3
yr 5 3(2 1 3e 2x )2 30 1 3e 2x (21)4
5 3(2 1 3e 2x )2 (23e 2x )
5 29e 2x (2 1 3e 2x )2
b. y 5 x e
yr 5 ex e21
x
c. y 5 e e
x
yr 5 e e (e x )(1)
x
5 e x1e
d. y 5 (1 2 e 5x )5
yr 5 5(1 2 e 5x )4 30 2 e 5x (5)4
5 225e 5x (1 2 e 5x )4
14. a. y 5 5x
yr 5 5x ln 5
b. y 5 (0.47)x
yr 5 (0.47)x ln (0.47)
c. y 5 (52)2x
yr 5 (52)2x (2) ln 52
5 2(52)2x ln 52
d. y 5 5(2)x
yr 5 5(2)x ln 2
e. y 5 4e x
yr 5 4e x (1) ln e
5 4e x
f. y 5 22(10)3x
yr 5 22(3)103x ln 10
5 26(10)3x ln 10
d(2x )
15. a. yr 5 cos 2x ?
dx
5 2x ln 2 cos 2x
Chapter 5: Derivatives of Exponential and Trigonometric Functions
b. yr 5 (x 2 )(cos x) 1 (sin x)(2x)
5 x 2 cos x 1 2x sin x
d( p 2 x)
p
c. yr 5 cos a 2 xb ? 2
2
dx
p
5 2cos a 2 xb
2
d. yr 5 (cos x)(cos x) 1 (sin x)(2sin x)
5 cos2 x 2 sin2 x
e. y 5 (cos x)2
d(cos x)
yr 5 2(cos x) ?
dx
5 22 cos x sin x
f. y 5 cos x (sin x)2
yr 5 (cos x)(2 (sin x)(cos x)) 1 (sin x)2 (2sin x)
5 2 sin x cos2 x 2 sin3 x
16. Compute
dy
dx
when x 5
p
2
to find the slope of the
p
dv
5 210 cos 2t 1
dt
4
p
5 220 cos a2t 1 b.
4
The maximum values of the displacement,
velocity, and acceleration are 5, 10, and 20,
respectively.
(
and a 5
(
p
19. Let the base angle be u, 0 , u , , and let the
2
sides of the triangle have lengths x and y, as shown.
Let the perimeter of the triangle be P cm.
12
y
u
x
x
5 cos u
12
y
5 sin u
12
line at the given point.
yr 5 2sin x
p
So, at the point x 5 , yr 5 f r(x) is
Now,
p
p
f ra b 5 2sin a b 5 21.
2
2
Therefore, an equation of the line tangent to the
curve at the given point is
p
y 2 0 5 21ax 2 b
2
p
y 5 2x 1
2
p
x1y2 50
2
17. The velocity of the object at any
dP
5 212 sin u 1 12 cos u.
du
ds
.
dt
Thus, v 5 8 (cos (10pt))(10p)
5 80p cos (10pt).
The acceleration at any time t is a 5
dv
d 2s
5 2.
dt
dt
Hence, a 5 80p(2sin (10pt))(10p) 5
2800p2 sin (10pt).
d 2s
Now, dt 2 1 100p2s
5 2800p2 sin (10pt)
1 100p2 (8 sin (10pt)) 5 0.
p
18. Since s 5 5 cos 2t 1 ,
(
v5
4
)
p
ds
5 5 a 2sin a 2t 1 bb
4
dt
p
5 210 sin a2t 1 b
4
Calculus and Vectors Solutions Manual
and
so x 5 12 cos u and y 5 12 sin u.
Therefore, P 5 12 1 12 cos u 1 12 sin u and
2
time t is v 5
))
For critical values, 212 sin u 1 12 cos u 5 0
sin u 5 cos u
tan u 5 1
p
p
u 5 , since 0 , u , .
4
2
12
p
12
When u 5 , P 5 12 1 !2 1 !2
4
24
5 12 1
!2
5 12 1 12#2.
As u S 0 1 , cos u S 1, sin u S 0, and
P S 12 1 12 1 0 5 24.
p
As u S , cos u S 0, sin u S 1 and
2
P S 12 1 0 1 12 5 24.
Therefore, the maximum value of the perimeter is
12 1 12!2 cm, and occurs when the other two
p
angles are each rad, or 45°.
4
20. Let l be the length of the ladder, u be the angle
between the foot of the ladder and the ground, and x
be the distance of the foot of the ladder from the
fence, as shown.
Thus,
x11
5 cos u
l
1.5
and x 5 tan u
1.5
x 1 1 5 l cos u where x 5 tan u.
5-29
dl
Solving du 5 0 yields:
l
0.8 sin3 u 2 cos3 u 5 0
tan3 u 5 1.25
3
tan u 5 !
1.25
tan u 8 1.077
u 8 0.822.
wall
1.5
u
0.8
1
x
1.5
1 1 5 l cos u
tan u
Replacing x,
1.5
1
p
1
,0,u,
sin u
cos u
2
dl
1.5 cos u
sin u
52
1
du
sin2 u
cos2 u
3
21.5 cos u 1 sin3 u
.
5
sin2 u cos2 u
dl
5 0 yields:
Solving
du
sin3 u 2 1.5 cos3 u 5 0
tan3 u 5 1.5
3
tan u 5 !
1.5
u 8 0.46365.
The length of the ladder corresponding to this value
p2
of u is l 8 4.5 m. As u S 0 1 and 2 , l increases
without bound. Therefore, the shortest ladder that
goes over the fence and reaches the wall has a
length of 4.5 m.
21. The longest pole that can fit around the corner is
determined by the minimum value of x 1 y. Thus,
we need to find the minimum value of l 5 x 1 y.
l5
0.8
x
u
1
0.8
1
From the diagram, y 5 sin u and x 5 cos u.
1
0.8
p
l 5 cos u 1 sin u, 0 # u # 2 :
dl
1 sin u
0.8 cos u
5
2
2
du
cos u
sin2 u
3
0.8 sin u 2 cos3 u
5
.
cos2 u sin2 u
5-30
A
6
3
B
C
x
a
u
b
D
9
3
From the diagram, tan a 5 x and tan b 5 x.
Hence,
9
3
x2x
tan u 5
27
1 1 x2
9x 2 3x
x 2 1 27
6x
5 2
.
x 1 27
We differentiate implicitly with respect to x:
du
6(x 2 1 27) 2 6x(2x)
sec2 u
5
dx
(x 2 1 27)2
du
162 2 6x 2
5
dx
sec2 u(x 2 1 27)2
5
y
u
Thus,
1
Now, l 5 cos (0.822) 1 sin (0.822) 8 2.5.
When u 5 0, the longest possible pole would have a
p
length of 0.8 m. When u 5 2 , the longest possible
pole would have a length of 1 m. Therefore, the
longest pole that can be carried horizontally around
the corner is one of length 2.5 m.
22. We want to find the value of x that maximizes u.
Let /ADC 5 a and /BDC 5 b.
Thus, u 5 a 2 b:
tan u 5 tan (a 2 b)
tan a 2 tan b
5
.
1 1 tan a tan b
du
Solving dx 5 0 yields:
162 2 6x 2 5 0
x 2 5 27
x 5 3#3.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
23. a. f (x) 5 4 (sin (x 2 2))2
f r (x) 5 8 sin (x 2 2) cos (x 2 2)
f s (x) 5 (8 sin (x 2 2))(2sin (x 2 2))
1 (cos (x 2 2))(8 cos (x 2 2))
5 28 sin2 (x 2 2) 1 8 cos2 (x 2 2)
b. f(x) 5 (2 cos x)(sec x)2
f r(x) 5 (2 cos x)(2 sec x ? sec x tan x)
1 (sec x)2 ( 2 2 sin x)
5 (4 cos x)(sec2 x tan x) 2 2 sin x (sec x)2
Using the product rule multiple times,
f s (x) 5 (4 cos x) S sec2 x ? sec2 x
1 tan x (2 sec x ? sec x tan x) T
1 (sec2 x tan x)(24 sin x)
1 (22 sin x)(2 sec x ? sec x tan x)
1 (sec x)2 (22 cos x)
5 4 cos x sec4 x 1 8 cos x tan2 x sec2 x
2 4 sin x tan x sec2 x 2 4 sin x tan x sec2 x
2 2 cos x sec2 x
5 4 cos x sec4 x 1 8 cos x tan2 x sec2 x
2 8 sin x tan x sec2 x 2 2 cos x sec2 x
Chapter 5 Test, p. 266
2
1. a. y 5 e 22x
dy
2
5 24xe 22x
dx
2
b. y 5 3x 13x
dy
2
5 3x 13x ? ln 3 ? (2x 1 3)
dx
e 3x 1 e 23x
c. y 5
2
1 3x
dy
5 33e 2 3e 23x4
dx
2
3
5 3e 3x 2 e 23x4
2
d. y 5 2 sin x 2 3 cos 5x
dy
5 2 cos x 2 3(2sin 5x)(5)
dx
5 2 cos x 1 15 sin 5x
e. y 5 sin3 (x 2 )
dy
5 3 sin2 (x 2 )(cos (x 2 )(2x))
dx
5 6x sin2 (x 2 ) cos (x 2 )
f. y 5 tan "1 2 x
dy
1
1
5 sec2 "1 2 x a 3
b (21)
dx
2
"1 2 x
sec2 "1 2 x
52
2"1 2 x
Calculus and Vectors Solutions Manual
2. The given line is 26x 1 y 5 2 or y 5 6x 1 2, so
the slope is 6.
y 5 2e 3x
dy
5 2e 3x (3)
dx
5 6e 3x
In order for the tangent line to be parallel to the
given line, the derivative has to equal 6 at the
tangent point.
6e 3x 5 6
e 3x 5 1
x50
When x 5 0, y 5 2.
The equation of the tangent line is y 2 2 5 6(x 2 0)
or 26x 1 y 5 2. The tangent line is the given line.
3. y 5 e x 1 sin x
dy
5 e x 1 cos x
dx
dy
When x 5 0, dx 5 1 1 1 or 2, so the slope of the
tangent line at (0, 1) is 2.
The equation of the tangent line at (0, 1) is
y 2 1 5 2(x 2 0) or 22x 1 y 5 1.
4. v(t) 5 10e2kt
a. a(t) 5 vr (t) 5 210ke 2kt
5 2k(10e 2kt )
5 2kv(t)
Thus, the acceleration is a constant multiple of the
velocity. As the velocity of the particle decreases,
the acceleration increases by a factor of k.
b. At time t 5 0, v 5 10 cm> s.
c. When v 5 5, we have 10e 2kt 5 5
1
e 2kt 5
2
1
2kt 5 ln a b 5 2ln 2
2
ln 2
t5
.
k
After
ln 2
k
s have elapsed, the velocity of the particle
is 5 cm> s. The acceleration of the particle is 25k at
this time.
5. a. f(x) 5 (cos x)2
d(cos x)
f r (x) 5 2 (cos x) ?
dx
5 2 (cos x) ? (2sin x)
5 22 sin x cos x
f s (x) 5 (22 sin x)(2sin x) 1 (cos x)(22 cos x)
5 2 sin2 x 2 2 cos2 x
5 2 (sin2 x 2 cos2 x)
5-31
23. a. f (x) 5 4 (sin (x 2 2))2
f r (x) 5 8 sin (x 2 2) cos (x 2 2)
f s (x) 5 (8 sin (x 2 2))(2sin (x 2 2))
1 (cos (x 2 2))(8 cos (x 2 2))
5 28 sin2 (x 2 2) 1 8 cos2 (x 2 2)
b. f(x) 5 (2 cos x)(sec x)2
f r(x) 5 (2 cos x)(2 sec x ? sec x tan x)
1 (sec x)2 ( 2 2 sin x)
5 (4 cos x)(sec2 x tan x) 2 2 sin x (sec x)2
Using the product rule multiple times,
f s (x) 5 (4 cos x) S sec2 x ? sec2 x
1 tan x (2 sec x ? sec x tan x) T
1 (sec2 x tan x)(24 sin x)
1 (22 sin x)(2 sec x ? sec x tan x)
1 (sec x)2 (22 cos x)
5 4 cos x sec4 x 1 8 cos x tan2 x sec2 x
2 4 sin x tan x sec2 x 2 4 sin x tan x sec2 x
2 2 cos x sec2 x
5 4 cos x sec4 x 1 8 cos x tan2 x sec2 x
2 8 sin x tan x sec2 x 2 2 cos x sec2 x
Chapter 5 Test, p. 266
2
1. a. y 5 e 22x
dy
2
5 24xe 22x
dx
2
b. y 5 3x 13x
dy
2
5 3x 13x ? ln 3 ? (2x 1 3)
dx
e 3x 1 e 23x
c. y 5
2
1 3x
dy
5 33e 2 3e 23x4
dx
2
3
5 3e 3x 2 e 23x4
2
d. y 5 2 sin x 2 3 cos 5x
dy
5 2 cos x 2 3(2sin 5x)(5)
dx
5 2 cos x 1 15 sin 5x
e. y 5 sin3 (x 2 )
dy
5 3 sin2 (x 2 )(cos (x 2 )(2x))
dx
5 6x sin2 (x 2 ) cos (x 2 )
f. y 5 tan "1 2 x
dy
1
1
5 sec2 "1 2 x a 3
b (21)
dx
2
"1 2 x
sec2 "1 2 x
52
2"1 2 x
Calculus and Vectors Solutions Manual
2. The given line is 26x 1 y 5 2 or y 5 6x 1 2, so
the slope is 6.
y 5 2e 3x
dy
5 2e 3x (3)
dx
5 6e 3x
In order for the tangent line to be parallel to the
given line, the derivative has to equal 6 at the
tangent point.
6e 3x 5 6
e 3x 5 1
x50
When x 5 0, y 5 2.
The equation of the tangent line is y 2 2 5 6(x 2 0)
or 26x 1 y 5 2. The tangent line is the given line.
3. y 5 e x 1 sin x
dy
5 e x 1 cos x
dx
dy
When x 5 0, dx 5 1 1 1 or 2, so the slope of the
tangent line at (0, 1) is 2.
The equation of the tangent line at (0, 1) is
y 2 1 5 2(x 2 0) or 22x 1 y 5 1.
4. v(t) 5 10e2kt
a. a(t) 5 vr (t) 5 210ke 2kt
5 2k(10e 2kt )
5 2kv(t)
Thus, the acceleration is a constant multiple of the
velocity. As the velocity of the particle decreases,
the acceleration increases by a factor of k.
b. At time t 5 0, v 5 10 cm> s.
c. When v 5 5, we have 10e 2kt 5 5
1
e 2kt 5
2
1
2kt 5 ln a b 5 2ln 2
2
ln 2
t5
.
k
After
ln 2
k
s have elapsed, the velocity of the particle
is 5 cm> s. The acceleration of the particle is 25k at
this time.
5. a. f(x) 5 (cos x)2
d(cos x)
f r (x) 5 2 (cos x) ?
dx
5 2 (cos x) ? (2sin x)
5 22 sin x cos x
f s (x) 5 (22 sin x)(2sin x) 1 (cos x)(22 cos x)
5 2 sin2 x 2 2 cos2 x
5 2 (sin2 x 2 cos2 x)
5-31
b. f(x) 5 cos x cot x
f r(x) 5 (cos x)(2csc2 x) 1 (cot x)(2sin x)
cos x
1
2
? sin x
5 2cos x ?
sin2 x
sin x
1
cos x
?
2 cos x
52
sin x sin x
5 2cot x csc x 2 cos x
f s (x) 5
(2cot x)(2csc x cot x) 1 (csc x)(csc2 x) 1 sin x
5 csc x cot2 x 1 csc3 x 1 sin x
6. f(x) 5 (sin x)2
To find the absolute extreme values, first find the
derivative, set it equal to zero, and solve for x.
d(sin x)
f r(x) 5 2 (sin x) ?
dx
5 2 sin x cos x
5 sin 2x
Now set f r (x) 5 0 and solve for x.
0 5 sin 2x
2x 5 0, p, 2p
p
x 5 0, , p in the interval 0 # x # p.
2
Evaluate f(x) at the critical numbers, including the
endpoints of the interval.
x
0
p
2
p
f(x) 5 (sin 2 x)
0
1
0
So, the absolute maximum value on the interval is 1
p
when x 5 2 and the absolute minimum value on
the interval is 0 when x 5 0 and x 5 p.
7. y 5 f(x) 5 5x
Find the derivative, f r(x), and evaluate the
derivative at x 5 2 to find the slope of the tangent
when x 5 2.
f r(x) 5 5x ln 5
f r(2) 5 52 ln 5
5 25 ln 5
8 40.24
8. y 5 xe x 1 3e x
To find the maximum and minimum values, first
find the derivative, set it equal to zero, and solve
for x.
yr 5 (x)(e x ) 1 (e x )(1) 1 3e x
5 xe x 1 e x 1 3e x
5 xe x 1 4e x
5 e x (x 1 4)
Now set yr 5 0 and solve for x.
0 5 e x (x 1 4)
5-32
e x is never equal to zero.
(x 1 4) 5 0
So x 5 24.
Therefore, the critical value is 24.
Interval
ex (x 1 4)
x , 24
2
1
24 , x
So f(x) is decreasing on the left of x 5 24 and
increasing on the right of x 5 24. Therefore, the
function has a minimum value at a24, 2 e4 b . There
is no maximum value.
9. f(x) 5 2 cos x 2 sin 2x
So, f(x) 5 2 cos x 2 2 sin x cos x.
a. f r (x) 5 22 sin x 2 (2 sin x)(2sin x)
2 (cos x)(2 cos x)
5 22 sin x 1 2 sin2 x 2 2 cos2 x
Set f r (x) 5 0 to solve for the critical values.
22 sin x 1 2 sin2 x 2 2 cos2 x 5 0
22 sin x 1 2 sin2 x 2 2(1 2 sin2 x) 5 0
22 sin x 1 2 sin2 x 2 2 1 2 sin2 x 5 0
4 sin2 x 2 2 sin x 2 2 5 0
(2 sin x 1 1)(2 sin x 2 2) 5 0
2 sin x 1 1 5 0 and 2 sin x 2 2 5 0
So, sinx 5 2 12.
p
5p
In the given interval, this occurs when x 5 2 6 , 2 6 .
Also, sin x 5 1.
p
In the given interval, this occurs when x 5 2 .
Therefore, on the given interval, the critical
p
5p p
numbers for f(x) are x 5 2 6 , 2 6 , 2 .
b. To determine the intervals where f(x) is increasing
and where f(x) is decreasing, find the slope of f(x)
in the intervals between the endpoints and the critical
numbers. To do this, it helps to make a table.
1
x
slope of f(x)
5p
6
p
,x,2
6
p
,x,
2
2p # x , 2
5p
6
p
2
6
p
,x#p
2
2
2
1
2
2
So, f(x) is increasing on the interval
5p
p
2 6 , x , 2 6 and f(x) is decreasing on the
5p
p
intervals 2p # x , 2 6 and 2 6 , x , p.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
c. From the table in part b., it can be seen that there is
p
a local maximum at the point where x 5 2 6 and
5p
there is a local minimum at the point where x 5 2 6 .
d.
y
4
3
2
1
–p
–p
2
0
–1
–2
–3
–4
p
2
p x
Cumulative Review of Calculus
1. a. f(x) 5 3x2 1 4x 2 5
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 15
5 lim
hS0
h
3(2 1 h)2 1 4(2 1 h) 2 5 2 15
5 lim
hS0
h
2
12 1 12h 1 3h 1 8 1 4h 2 20
5 lim
hS0
h
2
3h 1 16h
5 lim
hS0
h
5 lim 3h 1 16
hS0
5 16
2
b. f(x) 5
x21
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 2
5 lim
hS0
h
5 lim
hS0
5 lim
2
22
21h21
h
2
2(1 1 h)
2 11h
11h
h
2 2 2(1 1 h)
5 lim
hS0
h(1 1 h)
22h
5 lim
hS0 h(1 1 h)
hS0
Calculus and Vectors Solutions Manual
22
hS0 1 1 h
5 22
c. f(x) 5 !x 1 3
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(6 1 h) 2 3
5 lim
hS0
h
!h 1 9 2 3
5 lim
hS0
h
( !h 1 9 2 3)( !h 1 9 1 3)
5 lim
hS0
h( !h 1 9 1 3)
h1929
5 lim
hS0 h( !h 1 9 1 3)
h
5 lim
hS0 h( !h 1 9 1 3)
1
5 lim
hS0 ( !h 1 9 1 3)
1
5
6
d. f(x) 5 25x
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
25(11h) 2 32
5 lim
hS0
h
25 ? 25h 2 32
5 lim
hS0
h
32(25h 2 1)
5 lim
hS0
h
5(25h 2 1)
5 32 lim
hS0
5h
(25h 2 1)
5 160 lim
hS0
5h
5 160 ln 2
change in distance
2. a. average velocity 5
change in time
s(t2 ) 2 s(t1 )
5
t2 2 t1
32(4)2 1 3(4) 1 14 2 3(2(1)2 1 3(1) 1 1)4
5
421
45 2 6
5
3
5 13 m> s
b. instantaneous velocity 5 slope of the tangent
s(a 1 h) 2 s(a)
m 5 lim
hS0
h
5 lim
5-33
c. From the table in part b., it can be seen that there is
p
a local maximum at the point where x 5 2 6 and
5p
there is a local minimum at the point where x 5 2 6 .
d.
y
4
3
2
1
–p
–p
2
0
–1
–2
–3
–4
p
2
p x
Cumulative Review of Calculus
1. a. f(x) 5 3x2 1 4x 2 5
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 15
5 lim
hS0
h
3(2 1 h)2 1 4(2 1 h) 2 5 2 15
5 lim
hS0
h
2
12 1 12h 1 3h 1 8 1 4h 2 20
5 lim
hS0
h
2
3h 1 16h
5 lim
hS0
h
5 lim 3h 1 16
hS0
5 16
2
b. f(x) 5
x21
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 2
5 lim
hS0
h
5 lim
hS0
5 lim
2
22
21h21
h
2
2(1 1 h)
2 11h
11h
h
2 2 2(1 1 h)
5 lim
hS0
h(1 1 h)
22h
5 lim
hS0 h(1 1 h)
hS0
Calculus and Vectors Solutions Manual
22
hS0 1 1 h
5 22
c. f(x) 5 !x 1 3
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(6 1 h) 2 3
5 lim
hS0
h
!h 1 9 2 3
5 lim
hS0
h
( !h 1 9 2 3)( !h 1 9 1 3)
5 lim
hS0
h( !h 1 9 1 3)
h1929
5 lim
hS0 h( !h 1 9 1 3)
h
5 lim
hS0 h( !h 1 9 1 3)
1
5 lim
hS0 ( !h 1 9 1 3)
1
5
6
d. f(x) 5 25x
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
25(11h) 2 32
5 lim
hS0
h
25 ? 25h 2 32
5 lim
hS0
h
32(25h 2 1)
5 lim
hS0
h
5(25h 2 1)
5 32 lim
hS0
5h
(25h 2 1)
5 160 lim
hS0
5h
5 160 ln 2
change in distance
2. a. average velocity 5
change in time
s(t2 ) 2 s(t1 )
5
t2 2 t1
32(4)2 1 3(4) 1 14 2 3(2(1)2 1 3(1) 1 1)4
5
421
45 2 6
5
3
5 13 m> s
b. instantaneous velocity 5 slope of the tangent
s(a 1 h) 2 s(a)
m 5 lim
hS0
h
5 lim
5-33
s(3 1 h) 2 s(3)
hS0
h
2(3 1 h)2 1 3(3 1 h) 1 1
5 lim c
hS0
h
(2(3)2 1 3(3) 1 1)
d
2
h
18 1 12h 1 2h 2 1 9 1 3h 1 1 2 28
5 lim
hS0
h
15h 1 2h 2
5 lim
hS0
h
5 lim (15 1 2h)
5 lim
hS0
5 15 m> s
f(a 1 h) 2 f(a)
h
3
(4 1 h) 2 64
f(4 1 h) 2 f(4)
5 lim
lim
hS0
h
hS0
h
3
(4 1 h) 2 64 5 f(4 1 h) 2 f(4)
Therefore, f(x) 5 x 3.
4. a. Average rate of change in distance with respect
to time is average velocity, so
s(t2 ) 2 s(t1 )
average velocity 5
t2 2 t1
s(3) 2 s(1)
5
321
4.9(3)2 2 4.9(1)
5
321
5 19.6 m> s
b. Instantaneous rate of change in distance with
respect to time 5 slope of the tangent.
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
f(2 1 h) 2 f(2)
5 lim
hS0
h
4.9(2 1 h)2 2 4.9(2)2
5 lim
hS0
h
19.6 1 19.6h 1 4.9h 2 2 19.6
5 lim
hS0
h
19.6h 1 4.9h 2
5 lim
hS0
h
5 lim 19.6 1 4.9h
m 5 lim
3.
hS0
hS0
5 19.6 m> s
c. First, we need to determine t for the given
distance:
146.9 5 4.9t 2
29.98 5 t 2
5.475 5 t
5-34
Now use the slope of the tangent to determine the
instantaneous velocity for t 5 5.475:
f(5.475 1 h) 2 f(5.475)
m 5 lim
hS0
h
4.9(5.475 1 h)2 2 4.9(5.475)2
5 lim
hS0
h
146.9 1 53.655h 1 4.9h 2 2 146.9
5 lim
hS0
h
53.655h 1 4.9h 2
5 lim
hS0
h
5 lim 353.655 1 4.9h4
hS0
5 53.655 m> s
5. a. Average rate of population change
p(t2 ) 2 p(t1 )
5
t2 2 t1
2(8)2 1 3(8) 1 1 2 (2(0) 1 3(0) 1 1)
5
820
128 1 24 1 1 2 1
5
820
5 19 thousand fish> year
b. Instantaneous rate of population change
p(t 1 h) 2 p(t)
5 lim
hS0
h
p(5 1 h) 2 p(5)
5 lim
hS0
h
2(5 1 h)2 1 3(5 1 h) 1 1
5 lim c
hS0
h
(2(5)2 1 3(5) 1 1)
d
2
h
50 1 20h 1 2h 2 1 15 1 3h 1 1 2 66
5 lim
hS0
h
2h 2 1 23h
5 lim
hS0
h
5 lim 2h 1 23
5 23 thousand fish> year
6. a. i. f(2) 5 3
ii. lim2 f(x) 5 1
hS0
xS2
iii. lim1 f(x) 5 3
xS2
iv. lim f(x) 5 2
xS6
b. No, lim f(x) does not exist. In order for the limit
xS4
to exist, lim2 f(x) and lim1 f(x) must exist and they
xS4
xS4
must be equal. In this case, lim2 f(x) 5 `, but
xS4
lim1 f(x) 5 2 `, so lim f(x) does not exist.
xS4
xS4
Chapter 5: Derivatives of Exponential and Trigonometric Functions
7. f(x) is discontinuous at x 5 2. lim2 f(x) 5 5, but
xS2
lim f(x) 5 3.
xS2 1
2(0)2 1 1
2x 2 1 1
5
xS0 x 2 5
025
1
52
5
x23
b. lim
xS3 !x 1 6 2 3
(x 2 3)( !x 1 6 1 3)
5 lim
xS3 ( !x 1 6 2 3)( !x 1 6 1 3)
(x 2 3)( !x 1 6 1 3)
5 lim
xS3
x1629
(x 2 3)( !x 1 6 1 3)
5 lim
xS3
x23
5 lim !x 1 6 1 3
8. a. lim
xS3
56
c. lim
xS23
5 lim
1
1
13
x
x13
x13
3x
x13
x13
5 lim
xS23 3x(x 1 3)
1
5 lim
xS23 3x
1
52
9
x2 2 4
d. lim 2
xS2 x 2 x 2 2
(x 1 2)(x 2 2)
5 lim
xS2 (x 1 1)(x 2 2)
x12
5 lim
xS2 x 1 1
4
5
3
x22
e. lim 3
xS2 x 2 8
x22
5 lim
2
xS2 (x 2 2)(x 1 2x 1 4)
1
5 lim 2
xS2 x 1 2x 1 4
1
5
12
xS23
Calculus and Vectors Solutions Manual
!x 1 4 2 !4 2 x
x
( !x 1 4 2 !4 2 x)( !x 1 4 1 !4 2 x)
5 lim
xS0
x( !x 1 4 1 !4 2 x)
x 1 4 2 (4 2 x)
5 lim
xS0 x( !x 1 4 1 !4 2 x)
2x
5 lim
xS0 x( !x 1 4 1 !4 2 x)
2
5 lim
xS0 ( !x 1 4 1 !4 2 x)
1
5
2
9. a. f(x) 5 3x2 1 x 1 1
f(x 1 h) 2 f(x)
f r (x) 5 lim
hS0
h
3(x 1 h)2 1 (x 1 h) 1 1
5 lim c
hS0
h
(3x 2 1 x 1 1)
d
2
h
3x 2 1 6hx 1 6h 2 1 x 1 h
5 lim c
hS0
h
2
1 2 3x 2 x 2 1
d
1
h
6hx 1 6h 2 1 h
5 lim
hS0
h
5 lim 6x 1 6h 1 1
f. lim
xS0
hS0
5 6x 1 1
1
b. f(x) 5
x
f(x 1 h) 2 f(x)
f r (x) 5 lim
hS0
h
1
1
2x
x
1
h
5 lim
h
x 2 (x 1 h)
5 lim
hS0 h(x)(x 1 h)
2h
5 lim
hS0 h(x)(x 1 h)
21
5 lim
hS0 x(x 1 h)
1
52 2
x
10. a. To determine the derivative, use the power rule:
y 5 x 3 2 4x 2 1 5x 1 2
dy
5 3x 2 2 8x 1 5
dx
hS0
5-35
b. To determine the derivative, use the chain rule:
y 5 "2x 3 1 1
dy
1
5
(6x 2 )
dx
2"2x 3 1 1
3x 2
5
"2x 3 1 1
c. To determine the derivative, use the quotient rule:
2x
y5
x13
dy
2(x 1 3) 2 2x
5
dx
(x 1 3)2
6
5
(x 1 3)2
d. To determine the derivative, use the product rule:
y 5 (x 2 1 3)2 (4x 5 1 5x 1 1)
dy
5 2(x 2 1 3)(2x)(4x 5 1 5x 1 1)
dx
1 (x 2 1 3)2 (20x 4 1 5)
5 4x(x 2 1 3)(4x 5 1 5x 1 1)
1 (x 2 1 3)2 (20x 4 1 5)
e. To determine the derivative, use the quotient rule:
(4x 2 1 1)5
y5
(3x 2 2)3
dy
5(4x 2 1 1)4 (8x)(3x 2 2)3
5
dx
(3x 2 2)6
3(3x 2 2)2 (3)(4x 2 1 1)5
2
(3x 2 2)6
5 (4x 2 1 1)4 (3x 2 2)2
40x(3x 2 2) 2 9(4x 2 1 1)
3
(3x 2 2)6
(4x 2 1 1)4 (120x 2 2 80x 2 36x 2 2 9)
5
(3x 2 2)4
(4x 2 1 1)4 (84x 2 2 80x 2 9)
5
(3x 2 2)4
f. y 5 3x 2 1 (2x 1 1)34 5
Use the chain rule
dy
5 53x 2 1 (2x 1 1)34 4 32x 1 6(2x 1 1)24
dx
11. To determine the equation of the tangent line,
we need to determine its slope at the point (1, 2).
To do this, determine the derivative of y and
evaluate for x 5 1:
18
y5
(x 1 2)2
5 18(x 1 2)22
5-36
dy
5 236(x 1 2)23
dx
236
5
(x 1 2)3
236
m5
(x 1 2)3
24
236
5
5
27
3
Since we have a given point and we know the slope,
use point-slope form to write the equation of the
tangent line:
24
(x 2 1)
y225
3
3y 2 6 5 24x 1 4
4x 1 3y 2 10 5 0
12. The intersection point of the two curves
occurs when
x 2 1 9x 1 9 5 3x
x 2 1 6x 1 9 5 0
(x 1 3)2 5 0
x 5 23.
At a point x, the slope of the line tangent to the
curve y 5 x 2 1 9x 1 9 is given by
d 2
dy
5
(x 1 9x 1 9)
dx
dx
5 2x 1 9.
At x 5 23, this slope is 2(23) 1 9 5 3.
d
13. a. pr (t) 5 (2t2 1 6t 1 1100)
dt
5 4t 1 6
b. 1990 is 10 years after 1980, so the rate of change
of population in 1990 corresponds to the value
pr (10) 5 4(10) 1 6
5 46 people per year.
c. The rate of change of the population will be 110
people per year when
4t 1 6 5 110
t 5 26.
This corresponds to 26 years after 1980, which is
the year 2006.
d
14. a. f r (x) 5 (x 5 2 5x 3 1 x 1 12)
dx
5 5x 4 2 15x 2 1 1
d
(5x 4 2 15x 2 1 1)
f s (x) 5
dx
5 20x 3 2 30x
Chapter 5: Derivatives of Exponential and Trigonometric Functions
b. f(x) can be rewritten as f(x) 5 22x 22
d
f r(x) 5
(22x 22 )
dx
5 4x 23
4
5 3
x
d
f s (x) 5
(4x 23 )
dx
5 212x 24
12
52 4
x
1
c. f(x) can be rewritten as f(x) 5 4x 22
d
1
f r(x) 5
(4x 2 2 )
dx
3
5 22x22
2
52
"x 3
d
3
f s (x) 5
(22x 22 )
dx
5
5 3x 22
3
5
"x 5
d. f(x) can be rewritten as f(x) 5 x 4 2 x 24
d 4
f r(x) 5
(x 2 x 24 )
dx
5 4x 3 1 4x 25
4
5 4x 3 1 5
x
d
f s (x) 5
(4x 3 1 4x 25 )
dx
5 12x 2 2 20x 26
20
5 12x 2 2 6
x
15. Extreme values of a function on an interval will
only occur at the endpoints of the interval or at a
critical point of the function.
d
a. f r (x) 5 (1 1 (x 1 3)2 )
dx
5 2(x 1 3)
The only place where f r(x) 5 0 is at x 5 23, but
that point is outside of the interval in question. The
extreme values therefore occur at the endpoints of
the interval:
f(22) 5 1 1 (22 1 3)2 5 2
f(6) 5 1 1 (6 1 3)2 5 82
The maximum value is 82, and the minimum
value is 6
Calculus and Vectors Solutions Manual
1
b. f(x) can be rewritten as f(x) 5 x 1 x 22
d
1
f r (x) 5
(x 1 x 22 )
dx
1 3
5 1 1 2 x 22
2
1
512
2"x 3
On this interval, x $ 1, so the fraction on the right
is always less than or equal to 12. This means that
f r (x) . 0 on this interval and so the extreme values
occur at the endpoints.
1
52
f(1) 5 1 1
!1
1
1
59
f(9) 5 9 1
!9
3
The maximum value is 9 13, and the minimum
value is 2.
d
ex
b
c. f r (x) 5 a
dx 1 1 e x
(1 1 e x )(e x ) 2 (e x )(e x )
(1 1 e x )2
x
e
5
(1 1 e x )2
x
Since e is never equal to zero, f r (x) is never zero,
and so the extreme values occur at the endpoints of
the interval.
e0
1
f(0) 5
0 5
11e
2
e4
f(4) 5
1 1 e4
e4
The maximum value is 1 1 e 4, and the minimum
value is 12.
d
d. f r (x) 5 (2 sin (4x) 1 3)
dx
5 8 cos (4x)
p
Cosine is 0 when its argument is a multiple of 2
5
3p
or 2 .
p
3p
1 2kp or 4x 5
1 2kp
2
2
p
p
p
3p
x5 1 k
1 k
x5
8
2
8
2
4x 5
p 3p 5p 7p
Since xP30, p4, x 5 8 , 8 , 8 , 8 .
Also test the function at the endpoints of the interval.
f(0) 5 2 sin 0 1 3 5 3
p
p
f a b 5 2 sin 1 3 5 5
8
2
5-37
3p
3p
b 5 2 sin
1351
8
2
5p
5p
1355
f a b 5 2 sin
8
2
7p
7p
f a b 5 2 sin
1351
8
2
f(p) 5 2 sin (4p) 1 3 5 3
The maximum value is 5, and the minimum
value is 1.
16. a. The velocity of the particle is given by
v(t) 5 sr (t)
d
5 (3t 3 2 40.5t 2 1 162t)
dt
5 9t 2 2 81t 1 162.
The acceleration is
a(t) 5 vr (t)
d
5 (9t 2 2 81t 1 162)
dt
5 18t 2 81
b. The object is stationary when v(t) 5 0:
9t 2 2 81t 1 162 5 0
9(t 2 6)(t 2 3) 5 0
t 5 6 or t 5 3
The object is advancing when v(t) . 0 and retreating
when v(t) , 0. Since v(t) is the product of two
linear factors, its sign can be determined using the
signs of the factors:
fa
t-values
t23
t26
v(t)
Object
0,t,3
,0
,0
.0
Advancing
3,t,6
.0
,0
,0
Retreating
.0
Advancing
6,t,8
.0
.0
c. The velocity of the object is unchanging when the
acceleration is 0; that is, when
a(t) 5 18t 2 81 5 0
t 5 4.5
d. The object is decelerating when a(t) , 0, which
occurs when
18t 2 81 , 0
0 # t , 4.5
e. The object is accelerating when a(t) . 0, which
occurs when
18t 2 81 . 0
4.5 , t # 8
5-38
17.
w
l
Let the length and width of the field be l and w, as
shown. The total amount of fencing used is then
2l 1 5w. Since there is 750 m of fencing available,
this gives
2l 1 5w 5 750
5
l 5 375 2 w
2
The total area of the pens is
A 5 lw
5
5 375w 2 w 2
2
The maximum value of this area can be found by
expressing A as a function of w and examining its
derivative to determine critical points.
A(w) 5 375w 2 52w 2, which is defined for 0 # w
and 0 # l. Since l 5 375 2 52w, 0 # l gives the
restriction w # 150. The maximum area is therefore
the maximum value of the function A(w) on the
interval 0 # w # 150.
5
d
a375w 2 w 2 b
Ar (w) 5
dw
2
5 375 2 5w
Setting Ar (w) 5 0 shows that w 5 75 is the only
critical point of the function. The only values of
interest are therefore:
5
A(0) 5 375(0) 2 (0)2 5 0
2
5
A(75) 5 375(75) 2 (75)2 5 14 062.5
2
5
A(150) 5 375(150) 2 (150)2 5 0
2
The maximum area is 14 062.5 m2
18.
r
h
Let the height and radius of the can be h and r, as
shown. The total volume of the can is then pr 2h.
The volume of the can is also give at 500 mL, so
pr 2h 5 500
500
h5
pr 2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
The total surface area of the can is
A 5 2prh 1 2pr 2
1000
1 2pr 2
5
r
The minimum value of this surface area can be
found by expressing A as a function of r and
examining its derivative to determine critical points.
1000
A(r) 5
1 2pr 2, which is defined for 0 , r and
r
500
0 , h. Since h 5 pr 2 , 0 , h gives no additional
restriction on r. The maximum area is therefore the
maximum value of the function A(r) on the interval
0 , r.
d 1000
Ar(r) 5 a
1 2pr 2 b
r
dr
1000
5 2 2 1 4pr
r
The critical points of A(r) can be found by setting
Ar(r) 5 0:
2
1000
1 4pr 5 0
r2
4pr 3 5 1000
3
r 5 # 1000 8 4.3 cm
4p
So r 5 4.3 cm is the only critical point of the
function. This gives the value
500
h5
8 8.6 cm.
p(4.3)2
19.
r
20
0.02p
r 8 6.8
Using the max min algorithm:
C(1) 5 20.03, C(6.8) 5 4.39, C(36) 5 41.27.
The dimensions for the cheapest container are a
radius of 6.8 cm and a height of 27.5 cm.
20. a. Let the length, width, and depth be l, w, and
d, respectively. Then, the given information is that
l 5 x, w 5 x, and
l 1 w 1 d 5 140. Substituting gives
2x 1 d 5 140
d 5 140 2 2x
b. The volume of the box is V 5 lwh. Substituting
in the values from part a. gives
V 5 (x)(x)(140 2 2x)
5 140x 2 2 2x 3
In order for the dimensions of the box to make sense,
the inequalities l $ 0, w $ 0, and h $ 0 must be
satisfied. The first two give x $ 0, the third requires
x # 70. The maximum volume is therefore the
maximum value of V(x) 5 140x 2 2 2x 3 on the
interval 0 # x # 70, which can be found by
determining the critical points of the derivative Vr(x).
d
Vr (x) 5
(140x 2 2 2x 3 )
dx
5 280x 2 6x 2
5 2x(140 2 3x)
Setting Vr (x) 5 0 shows that x 5 0 and
r3 5
140
h
Let the radius be r and the height h.
Minimize the cost:
C 5 2pr 2 (0.005) 1 2prh(0.0025)
V 5 pr2 h 5 4000
4000
h5
pr 2
4000
b (0.0025)
C(r) 5 2pr 2 (0.005) 1 2pr a
pr 2
20
5 0.01pr 2 1 , 1 # r # 36
r
20
C r(r) 5 0.02pr 2 2 .
r
For a maximum or minimum value, let C r (r) 5 0.
0.02pr 2 2
20
r2
50
Calculus and Vectors Solutions Manual
x 5 3 8 46.7 are the critical points of the function.
The maximum value therefore occurs at one of these
points or at one of the endpoints of the interval:
V(0) 5 140(0)2 2 2(0)3 5 0
V(46.7) 5 140(46.7)2 2 2(46.7)3 5 101 629.5
V(0) 5 140(70)2 2 2(70)3 5 0
So the maximum volume is 101 629.5 cm3, from a
box with length and width 46.7 cm and depth
140 2 2(46.7) 5 46.6 cm.
21. The revenue function is
R(x) 5 x(50 2 x 2 )
5 50x 2 x 3. Its maximum for x $ 0 can be
found by examining its derivative to determine
critical points.
d
Rr (x) 5
(50x 2 x 3 )
dx
5 50 2 3x 2
The critical points can be found by setting Rr (x) 5 0:
50 2 3x 2 5 0
50
x5 6
8 64.1
Å3
5-39
Only the positive root is of interest since the number
of MP3 players sold must be positive. The number
must also be an integer, so both x 5 4 and x 5 5
must be tested to see which is larger.
R(4) 5 50(4) 2 43 5 136
R(4) 5 50(5) 2 53 5 125
So the maximum possible revenue is $136, coming
from a sale of 4 MP3 players.
22. Let x be the fare, and p(x) be the number of
passengers per year. The given information shows
that p is a linear function of x such that an increase
of 10 in x results in a decrease of 1000 in p. This
means that the slope of the line described by p(x) is
21000
10 5 2100. Using the initial point given,
p(x) 5 2100(x 2 50) 1 10 000
5 2100x 1 15 000
The revenue function can now be written:
R(x) 5 xp(x)
5 x(2100x 1 15 000)
5 15 000x 2 100x 2
Its maximum for x $ 0 can be found by examining
its derivative to determine critical points.
d
(15 000x 2 100x 2 )
Rr(x) 5
dx
5 15 000 2 200x
Setting Rr (x) 5 0 shows that x 5 75 is the only
critical point of the function. The problem states
that only $10 increases in fare are possible, however,
so the two nearest must be tried to determine the
maximum possible revenue:
R(70) 5 15 000(70) 2 100(70)2 5 560 000
R(80) 5 15 000(80) 2 100(80)2 5 560 000
So the maximum possible revenue is $560 000,
which can be achieved by a fare of either $70 or $80.
23. Let the number of $30 price reductions be n.
The resulting number of tourists will be 80 1 n
where 0 # n # 70. The price per tourist will be
5000 2 30n dollars. The revenue to the travel
agency will be (5000 2 30n)(80 1 n) dollars. The
cost to the agency will be 250 000 1 300(80 1 n)
dollars.
Profit 5 Revenue 2 Cost
P(n) 5 (5000 2 30n)(80 1 n)
2 250 000 2 300(80 1 n), 0 # n # 70
P r (n) 5 230(80 1 n) 1 (5000 2 30n)(1) 2 300
5 2300 2 60n
1
P r(n) 5 0 when n 5 38
3
Since n must be an integer, we now evaluate P(n)
for n 5 0, 38, 39, and 70. (Since P(n) is a quadratic
5-40
function whose graph opens downward with vertex
at 38 13 , we know P(38) . P(39).)
P(0) 5 126 000
P(38) 5 (3860)(118) 2 250 000 2 300(118)
5 170 080
P(39) 5 (3830)(119) 2 250 000 2 300(119)
5 170 070
P(70) 5 (2900)(150) 2 250 000 2 300(150)
5 140 000
The price per person should be lowered by $1140
(38 decrements of $30) to realize a maximum profit
of $170 080.
dy
d
5
(25x 2 1 20x 1 2)
24. a.
dx
dx
5 210x 1 20
dy
Setting dx 5 0 shows that x 5 2 is the only critical
number of the function.
b.
x
x,2
x52
x.2
y9
1
0
2
Graph
Inc.
Local Max
Dec.
dy
d
5
(6x 2 1 16x 2 40)
dx
dx
5 12x 1 16
dy
Setting dx 5 0 shows that x 5 2 43 is the only
critical number of the function.
x
x,2
4
3
x52
4
3
x.2
y9
2
0
1
Graph
Dec.
Local Min
Inc.
4
3
dy
d
5
(2x 3 2 24x)
dx
dx
5 6x 2 2 24
dy
The critical numbers are found by setting dx 5 0:
6x 2 2 24 5 0
6x 2 5 24
x 5 62
c.
x
x , 22
x 5 22
22 , x , 2
x52
x.2
y9
1
0
2
0
1
Graph
Inc.
Local Max
Dec.
Local Min
Inc.
d.
dy
d
x
5
a
b
dx
dx x 2 2
(x 2 2)(1) 2 x(1)
5
(x 2 2)2
22
5
(x 2 2)2
Chapter 5: Derivatives of Exponential and Trigonometric Functions
This derivative is never equal to zero, so the
function has no critical numbers. Since the
numerator is always negative and the denominator
is never negative, the derivative is always negative.
This means that the function is decreasing
everywhere it is defined, that is, x 2 2.
25. a. This function is discontinuous when
x2 2 9 5 0
x 5 63. The numerator is non-zero at these
points, so these are the equations of the vertical
asymptotes.
To check for a horizontal asymptote:
8
8
lim 2
5 lim
xS` x 2 9
xS` 2
9
x a1 2 2 b
x
lim (8)
xS`
5
9
lim x 2 a1 2 2 b
xS`
x
lim (8)
xS`
5
9
lim (x)2 3 lim a1 2 2 b
xS`
xS`
x
1
8
5 lim 2 3
xS` x
120
50
8
Similarly, lim x 2 2 9 5 0, so y 5 0 is a horizontal
xS2`
asymptote of the function.
There is no oblique asymptote because the degree
of the numerator does not exceed the degree of the
denominator by 1.
Local extrema can be found by examining the
derivative to determine critical points:
(x 2 2 9)(0) 2 (8)(2x)
yr 5
(x 2 2 9)2
216x
5 2
(x 2 9)2
Setting yr 5 0 shows that x 5 0 is the only critical
point of the function.
x
x,0
x50
x.0
y9
1
0
1
Graph
Inc.
Local Max
Dec.
So (0, 2 89 ) is a local maximum.
b. This function is discontinuous when
x2 2 1 5 0
x 5 61. The numerator is non-zero at these
points, so these are the equations of the vertical
asymptotes.
Calculus and Vectors Solutions Manual
To check for a horizontal asymptote:
4x 3
x 3 (4)
lim 2
5 lim
xS` x 2 1
xS` 2
1
x 12 2
x
x(4)
5 lim
1
xS`
1 2 x2
(
)
lim (x(4))
5
xS`
(
1
)
lim 1 2 x 2
lim (x) 3 lim (4)
xS`
5
xS`
xS`
(
1
)
lim 1 2 x 2
xS`
4
5 lim (x) 3
xS`
120
5`
4x 3
Similarly, lim x 2 2 1 5 lim (x) 5 2`, so this
xS 2`
xS 2`
function has no horizontal asymptote.
To check for an oblique asymptote:
4x
2
3
2
q
x 2 1 4x 1 0x 1 0x 1 0
4x3 1 0x2 2 4x
0 1 0 1 4x 1 0
So y can be written in the form
4x
y 5 4x 1 2
. Since
x 21
4x
x(4)
lim 2
5 lim
xS` x 2 1
xS` 2
1
x 12 2
x
(
5 lim
xS`
5
)
4
(
1
x 1 2 x2
lim (4)
)
xS`
((
1
lim x 1 2 x 2
xS`
))
lim (4)
5
xS`
(
1
lim (x) 3 lim 1 2 x 2
xS`
xS`
)
1
4
5 lim a b 3
x
120
5 0,
4x
and similarly lim x 2 2 1 5 0, the line y 5 4x is an
xS 2`
asymptote to the function y.
Local extrema can be found by examining the
derivative to determine critical points:
5-41
(x 2 2 1)(12x 2 ) 2 (4x 3 )(2x)
(x 2 2 1)2
12x 4 2 12x 2 2 8x 4
5
(x 2 2 1)2
4x 4 2 12x 2
5
(x 2 2 1)2
Setting yr 5 0:
4x 4 2 12x 2 5 0
x 2 (x 2 2 3) 5 0
so x 5 0, x 5 6 !3 are the critical points of the
function
(2!3, 26 !3) is a local maximum, ( !3, 6 !3) is
a local minimum, and (0, 0) is neither.
yr 5
x
x , 2 !3
x 5 2 !3
2 !3 , x , 0
x50
y9
1
0
2
0
Graph
Inc.
Local Max
Dec.
Horiz.
x
0 , x , !3
x 5 !3
x . !3
y9
2
0
2
Graph
Dec.
Local Min
Inc.
26. a. This function is continuous everywhere, so it
has no vertical asymptotes. To check for a horizontal
asymptote:
lim (4x 3 1 6x 2 2 24x 2 2)
xS`
xS`
5`
Similarly,
lim (4x 3 1 6x 2 2 24x 2 2) 5 lim (x 3 ) 5 2`,
xS 2`
so this function has no horizontal asymptote.
The y-intercept can be found by letting x 5 0,
which gives y 5 4(0)3 1 6(0)2 2 24(0) 2 2
5 22
The derivative is of the function is
d
(4x 3 1 6x 2 2 24x 2 2)
yr 5
dx
5 12x 2 1 12x 2 24
5 12(x 1 2)(x 2 1), and the second derivative is
d
(12x 2 1 12x 2 24)
ys 5
dx
5-42
x
x , 22
x 5 22
22 , x
y9
1
0
2
2
Graph
Inc.
Local Max
Dec.
Dec.
y0
2
2
2
0
Concavity
Down
Down
Down
Infl.
x51
x.1
0
1
1
,x,1
2
2
2
x
y9
Graph
Dec.
Local Min
Inc.
y0
1
1
1
Concavity
Up
Up
Up
x52
1
2
y
30
20
10
x
–3 –2 –1 0
–10
1 2
y = 4x3 + 6x2 – 24x – 2
24
6
2
5 lim x a4 1 2 2 2 3 b
x
xS`
x
x
6
2
24
5 lim (x 3 ) 3 lim a4 1 2 2 2 3 b
x
xS`
xS`
x
x
5 lim (x 3 ) 3 (4 1 0 2 0 2 0)
3
xS2`
5 24x 1 12
Letting f r (x) 5 0 shows that x 5 22 and x 5 1 are
critical points of the function. Letting ys 5 0 shows
that x 5 2 12 is an inflection point of the function.
b. This function is discontinuous when
x2 2 4 5 0
(x 1 2)(x 2 2) 5 0
x 5 2 or x 5 22. The numerator is non-zero at
these points, so the function has vertical asymptotes
at both of them. The behaviour of the function near
these asymptotes is:
3x
x12
x22
y
2
,0
,0
,0
,0
1
x-values
x S 22
lim y
xS`
2`
,0
.0
,0
.0
1`
x S 22
.0
.0
,0
,0
2`
x S 21
.0
.0
.0
.0
1`
x S 22
To check for a horizontal asymptote:
3x
x(3)
lim 2
5 lim
xS` x 2 4
xS` 2
4
x 1 2 x2
(
5 lim
xS`
)
3
(
4
x 1 2 x2
)
Chapter 5: Derivatives of Exponential and Trigonometric Functions
lim (3)
5
xS`
((
4
lim x 1 2 2
x
xS`
6
4
2
))
lim (3)
5
xS`
(
4
lim (x) 3 lim 1 2 x 2
xS`
xS`
1
3
5 lim 3
x
xS`
120
50
–6 –4 –2 0
–2
–4
–6
)
Similarly,
so y 5 0 is a horizontal
asymptote of the function.
This function has y 5 0 when x 5 0, so the origin is
both the x- and y-intercept.
The derivative is
(x 2 2 4)(3) 2 (3x)(2x)
yr 5
(x 2 2 4)2
23x 2 2 12
5
, and the second derivative is
(x 2 2 4)2
(x 2 2 4)2 (26x)
ys 5
(x 2 2 4)4
(23x 2 2 12)(2(x 2 2 4)(2x))
2
(x 2 2 4)4
3
26x 1 24x 1 12x 3 1 48x
5
(x 2 2 4)3
6x 3 1 72x
5
(x 2 2 4)3
The critical points of the function can be found by
letting yr 5 0, so
23x 2 2 12 5 0
x 2 1 4 5 0. This has no real solutions, so the
function y has no critical points.
The inflection points can be found by letting
ys 5 0, so
6x 3 1 72x 5 0
6x(x 2 1 12) 5 0
The only real solution to this equation is x 5 0, so
that is the only possible inflection point.
x , 22 22 , x , 0
x50
0,x,2
x.2
y9
2
2
2
2
2
Graph
Dec.
Dec.
Dec.
Dec.
Dec.
y0
2
1
0
2
1
Concavity
Down
Up
Infl.
Down
Up
Calculus and Vectors Solutions Manual
3x
y = ——–—
x2 – 4
x
2 4 6
d
((24)e 5x11 )
dx
d
5 (24)e 5x11 3
(5x 1 1)
dx
5 (220)e 5x11
d
b. f r (x) 5 (xe 3x )
dx
d
(3x) 1 (1)e 3x
5 xe 3x 3
dx
5 e 3x (3x 1 1)
d
c. yr 5 (63x28 )
dx
d
5 (ln 6)63x28 3
(3x 2 8)
dx
5 (3 ln 6)63x28
d
d. yr 5 (e sin x )
dx
d
5 e sin x 3
(sin x)
dx
5 (cos x)e sin x
28. The slope of the tangent line at x 5 1 can be
27. a. f r (x) 5
3x
lim 2
5 0,
xS` x 2 4
x
y
dy
found by evaluating the derivative dx for x 5 1:
dy
d 2x21
5
(e
)
dx
dx
d
5 e 2x21 3
(2x 2 1)
dx
2x21
5 2e
Substituting x 5 1 shows that the slope is 2e. The
value of the original function at x 5 1 is e, so the
equation of the tangent line at x 5 1 is
y 5 2e(x 2 1) 1 e.
29. a. The maximum of the function modelling the
number of bacteria infected can be found by
examining its derivative.
d
t
N r (t) 5 ((15t)e 2 5 )
dt
d
t
t
t
5 15te 25 3 a2 b 1 (15)e 25
dt
5
t
5 e25 (15 2 3t)
5-43
Setting Nr(t) 5 0 shows that t 5 5 is the only
critical point of the function (since the exponential
function is never zero). The maximum number of
infected bacteria therefore occurs after 5 days.
5
b. N(5) 5 (15(5))e 25
5 27 bacteria
d
dy
5
(2 sin x 2 3 cos 5x)
30. a.
dx
dx
d
(5x)
5 2 cos x 2 3(2sin 5x) 3
dx
5 2 cos x 1 15 sin 5x
d
dy
5
(sin 2x 1 1)4
b.
dx
dx
d
(sin 2x 1 1)
5 4(sin 2x 1 1)3 3
dx
d
5 4(sin 2x 1 1)3 3 (cos 2x) 3
(2x)
dx
5 8 cos 2x(sin 2x 1 1)3
1
c. y can be rewritten as y 5 (x 2 1 sin 3x)2 . Then,
d
dy
1
5
(x 2 1 sin 3x)2
dx
dx
1 2
d
1
5 (x 1 sin 3x)22 3
(x 2 1 sin 3x)
2
dx
1
1
5 (x 2 1 sin 3x)22
2
d
3 a2x 1 cos 3x 3
(3x)b
dx
2x 1 3 cos 3x
5
2 !x 2 1 sin 3x
dy
d
sin x
5
a
b
d.
dx
dx cos x 1 2
(cos x 1 2)(cos x) 2 (sin x)(2sin x)
5
(cos x 1 2)2
cos2 x 1 sin2 x 1 2 cos x
5
(cos x 1 2)2
1 1 2 cos x
5
(cos x 1 2)2
d
dy
5
(tan x 2 2 tan2 x)
e.
dx
dx
d
d
sec2 x 2 3
(x 2 )
5
dx
dx
d
(tan x)
2 2 tan x 3
dx
5 2x sec2 x 2 2 2 tan x sec2 x
5-44
f.
d
dy
5
(sin (cos x 2 ))
dx
dx
d
(cos x 2 )
5 cos (cos x 2 ) 3
dx
5 cos (cos x 2 ) 3 (2sin x 2 ) 3
d
(x 2 )
dx
5 22x sin x 2 cos(cos x 2 )
31.
l2
u
100
l1
250
u
As shown in the diagram, let u be the angle between
the ladder and the ground, and let the total length
of the ladder be l 5 l1 1 l2, where l1 is the length
from the ground to the top corner of the shed and
l2is the length from the corner of the shed to the
wall.
250
100
sin u 5
cos u 5
l1
l2
l1 5 250 csc u
l2 5 100 sec u
l 5 250 csc u 1 100 sec u
dl
5 2250 csc u cot u 1 100 sec u tan u
du
100 sin u
250 cos u
1
52
sin2 u
cos2 u
dl
To determine the minimum, solve du 5 0.
100 sin u
250 cos u
5
2
sin u
cos2 u
3
250 cos u 5 100 sin3 u
2.5 5 tan3 u
3
tan u 5 "2.5
u 8 0.94
At u 5 0.94, l 5 250 csc 0.94 1 100 sec 0.94
8 479 cm
The shortest ladder is about 4.8 m long.
32. The longest rod that can fit around the corner is
determined by the minimum value of x 1 y. So,
determine the minimum value of l 5 x 1 y.
Chapter 5: Derivatives of Exponential and Trigonometric Functions
dl
y
Solving du 5 0 yields:
3
u
x
u
3
3
3
From the diagram, sin u 5 y and cos u 5 x. So,
l5
3
3
1
,
cos u
sin u
p
for 0 # u # 2 .
3 sin u
3 cos u
dl
5
2
du
cos2 u
sin2 u
3
3 sin u 2 3 cos3 u
5
cos2 u sin2 u
Calculus and Vectors Solutions Manual
3 sin3 u 2 3 cos3 u 5 0
tan3 u 5 1
tan u 5 1
p
u5
4
3
3
So l 5
p 1
p
cos 4
sin 4
5 3 !2 1 3 !2
5 6 !2
p
When u 5 0 or u 5 2 , the longest possible rod
would have a length of 3 m. Therefore the longest
rod that can be carried horizontally around the
corner is one of length 6 !2, or about 8.5 m.
5-45