Answers to Odd Exercises
826
h
= 6.0.
t
h
With t = 0.5, h = h(1.5) − h(1.0) = 2.5, so
= 5.0.
t
h
With t = 0.1, h = h(1.1) − h(1.0) = 0.42, so
= 4.2.
t
h
With t = 0.01, h = h(1.01) − h(1.0) = 0.0402, so
=
t
3. With t = 1.0, h = h(2.0) − h(1.0) = 6.0, so
4.02.
G
≈
5. With t = 1.0, G = G(1.0) − G(0.0) ≈ 6.389, so
t
6.389. With t = 0.5, G = G(0.5) − G(0.0) ≈ 1.718, so
G
≈ 3.436. With t = 0.1, G = G(0.1) − G(0.0) ≈
t
G
≈ 2.21. With t = 0.01, G = G(0.01) −
t
G
G(0.0) ≈ 0.0202, so
≈ 2.02.
t
0.221, so
7. Each secant line is f s (t) = 2 + 3t.
f (t)
8.5
8
7.5
7
6.5
6
5.5
5
4.5
1
1.2
1.4
1.6
1.8
2
t
9. The coordinates of the base point are (1, 2), so the secant
lines are as follows: with t = 1.0, h s (t) = 2 + 6(t − 1);
with t = 0.5, h s (t) = 2 + 5(t − 1); with t = 0.1, h s (t) =
2 + 4.2(t − 1); with t = 0.01, h s (t) = 2 + 4.02(t − 1).
h(t)
9
8
7
6
5
4
3
2
1
1
1.2
1.4
1.6
1.8
2
t
Section 2.1, page 143
f
= 3.0.
t
f
With t = 0.5, f = f(1.5) − f(1.0) = 1.5, so
= 3.0.
t
f
= 3.0.
With t = 0.1, f = f(1.1) − f(1.0) = 0.3, so
t
f
With t = 0.01, f = f(1.01) − f(1.0) = 0.03, so
=
t
1. With t = 1.0, f = f(2.0) − f(1.0) = 3.0, so
3.0.
10
9
8
7
6
5
4
3
2
1
0
G(t)
Chapter 2
11. The coordinates of the base point are (0, 1), so the secant lines
are as follows: with t = 1.0, G s (t) = 1 + 6.389t; with t =
0.5, G s (t) = 1 + 3.436t; with t = 0.1, G s (t) = 1 + 2.21t;
with t = 0.01, G s (t) = 1 + 2.02t.
0
0.2
0.4
0.6
t
0.8
1
Answers to Odd Exercises
13. The slope is 3, so the tangent line is f̂ (t) = 2 + 3t.
f.
827
3
15. It looks like the slopes are getting close to 4.0, so the tangent
line is ĥ(t) = 2 + 4(t − 1).
b(t)
2
17. It looks like the slopes are getting close to 2.0, so the tangent
line is Ĝ(t) = 1 + 2t.
1
19. The derivative of g(t), the slope of the tangent line.
0
21. a. b(0) = 1.0, b(1.0) = 1.5, b(2.0) = 2.25.
0
0.5
1
1.5
2
b(t)
t
2.4
2.2
2
1.8
1.6
1.4
1.2
1
0.8
25. a. The slope is (5 · 1.02 − 0.0)/1.0 = 5.0.
b. The slope is (5 · 0.12 − 0.0)/0.1 = 0.5.
c. The slope is (5 · 0.012 − 0.0)/0.01 = 0.05.
d. The slope is (5 · 0.0012 − 0.0)/0.001 = 0.005.
0
1
e. The slope gets close to 0.
2
t
f.
3
b. b = 1.5 − 1.0 = 0.5, so b/t = 0.5.
b(t)
2
Secant line connecting t = 0 and t = 1
2.4
1
2.2
2
0
b(t)
1.8
0
0.5
1
1.6
1.4
b 0.5
1.2
2
27. During the first hour, 3.0 bacteria/h. During the first half-hour,
2.485 bacteria/h. During the second half-hour, 3.515 bacteria/h. Changes faster during the second half-hour.
t 1.0
1
0.8
1.5
t
0
0.5
1
1.5
2
t
Secant line connecting t = 1 and t = 2
b(t)
c. b = 2.25 − 1.5 = 0.75, so b/t = 0.75.
4
3
6.5
6
5.5
5
4.5
4
3.5
3
2.5
2
0
0.5
1
b(t)
t
2
t 0.75
0
29. During the first hour, −0.79 bacteria/h. During the first halfhour, −0.88 bacteria/h. During the second half-hour, −0,69
bacteria/h. Changes more rapidly during the first half-hour.
t 1.0
1
0
0.5
1
1.5
2
2.5
3
23. a. b = 1.51.0 − 1.0 = 0.5, and b/t = 0.5.
b. b = 1.50.1 − 1.0 = 0.0413, and b/t = 0.414.
c. b = 1.50.01 − 1.0 = 0.00406, and b/t = 0.406.
d. b = 1.50.001 − 1.0 = 0.000405, and b/t = 0.405.
e. The limit looks like 0.405.
b(t)
t
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0
0.5
t
1
Answers to Odd Exercises
31.
Average Rate
of Change,
c.
Per Capita Rate
of Change
∆b
db
≈
∆t
dt
db
dt
b(t )
t
b(t)
Change,
∆b
0.0
1.0
—
—
—
1.0
2.0
1.0
1.0
0.5
2.0
4.0
2.0
2.0
0.5
3.0
8.0
4.0
4.0
0.5
Rate of change divided by height
Rate of change divided by height
828
db
= 0.5b(t)
dt
33.
Change,
∆b
dt
b(t )
—
—
0.0100 1.0070 0.00696
0.69556
0.69075
0.0200 1.0140 0.00700
0.70039
0.69075
0.0300 1.0210 0.00705
0.70526
0.69075
0.0400 1.0281 0.00710
0.71017
0.69075
0.0500 1.0353 0.00715
0.71511
0.69075
b(t)
0.0000 1.0000
—
0.09
0.08
0 1 2 3 4 5 6 7 8 9 10
Age
d. Both seem to jump around a bit, but the tree increases its
height by about 9% per year. The approximate differential
equation is
Per Capita Rate
Average Rate
of Change
of Change,
db
∆b
db
≈
∆t
dt
t
0.1
dh
= 0.09h
dt
37. Getting 5% interest corresponds to multiplying the principle
by 1.05, giving 1.05 · $1000 = $1050.00.
39.
5
≈ 0.417. We must multiply the principle by 1.00417 twelve
12
times, for (1.00417)12 · $1000 ≈ $1051.16, for an extra $1.16.
41. Breaking into intervals of length t, we get
1
lim (1 + 0.05t) t $1000
t→0
db
= 0.6907b(t)
dt
It looks like the limit is $1051.27.
35. a. Using the previous measurement as the point, we find
Age Rate of Change (m/year) Rate Divided by Height
1
1.07
0.0957
2
1.22
0.0984
3
1.34
0.0975
4
1.27
0.0846
5
1.60
0.0963
6
1.66
0.0909
7
1.90
0.0942
8
1.84
0.0836
9
2.44
0.0998
10
2.40
0.0894
b.
Average rate of change
2.5
43. With yearly compounding, you’d get 2.0 · $1000 = $2000.00.
100
With monthly compounding, you’d get
≈ 8.33 for each
12
of 12 months, giving 1.083312 · $1000 ≈ $2613.00. With daily
100
≈ 0.274 for each of 365 days,
compounding, you’d get
365
giving 1.00274365 · $1000 ≈ $2714.60. The compound interest is more important because the principal changes much
more during the year.
Section 2.2, page 153
1. At x = 0.1, value is 2.594; at x = 0.01 it is 2.705; at x = 0.001
it is 2.717; and at x = 0.0001 it is 2.718. The limit seems to
be e.
(1 x)1/x
5
Average rate of change
4
3
2
2
1
1.5
1
1
0 1 2 3 4 5 6 7 8 9 10
Age
0
1
x
3. At x = 0.1, value is 0.499; at x = 0.01 it is 0.005; at x = 0.001
it is 0.0005. The limit seems to be 0.
Answers to Odd Exercises
15. f 3 (x) < 0.1 if x < 0.316, f 3 (x) < 0.01 if x < 0.1. This approaches 0 quickly because small values of the input produce
tiny values of the output. Graphically, the value of the function
becomes small even when x is not that small.
[1-cos(x)]/x
1
1
829
0
1
x
1
0.8
1
x2
0.6
0.4
5. 0.1 ln(0.1) = −0.230, 0.01 ln(0.01) = −0.046, 0.001 ln(0.001)
= −0.006. It seems to approach 0.
0.2
0
1
0
0.2
0.4
0.6
0.8
1
x ln(x)
x
0
0.5
1
1.5
2
x
1
17. g1 (x) > 10 if x < 0.01. g1 (x) > 100 if x < 0.0001. This approaches infinity slowly because tiny values of the input are
required to produce large values of the output. Graphically,
the value of the function becomes large only when x is very
small.
1
7. Denoting the function by h(x), we find that h(0.9) = −2.558,
h(0.99) = −4.652, h(0.999) = −6.915, h(0.9999) = −9.211.
This seems to go to negative infinity.
1/√x
0.8
0
0.6
0.7
0.8
0.9
1
x
0.6
0.4
log(1 x)/x
0.2
0
5
0
0.4
0.6
0.8
1
x
10
9. This will be 5 times the previous limit, or 5e = 13.59, using
Theorem 2.2c.
19. g3 (x) > 10 if x < 0.316. g3 (x) > 100 if x < 0.1. This
proaches infinity quickly because small values of the
put produce huge values of the output. Graphically,
value of the function becomes large even when x is not
small.
200
13. f 1 (x) < 0.1 if x < 0.01, f 1 (x) < 0.01 if x < 0.0001. This approaches 0 slowly because tiny values of the input are required to produce small values of the output. Graphically, the
value of the function does not become small until x is very
small.
150
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1/x 2
11. This will be the product of the two earlier limits, or e · 0 = 0,
using Theorem 2.2b.
√x
0.2
apinthe
too
100
50
0
0
0.2
0.4
0.6
0.8
1
x
21. Both the left- and the right-hand limits are 1.
23. The left-hand limit is about 0.1, and the right-hand limit is
about 0.02.
0
0.2
0.4
0.6
x
0.8
1
25. f = 5x, so the average rate of change is 5 (unless x = 0).
The limit is also 5.
Answers to Odd Exercises
830
35. The average rate of change between t = 0 and t = t is
5x 7
15
b(t) − b(0) et − 1
=
t
t
f (x)
Plugging in small values of t, I found that the average rate
of change is 1.01 when t = 0.02.
10
f
37. a. This would cost only $5.
x
1
0
b. This would cost $50.
1
x
c. This would cost $500.
27. f = 5x 2 , so the average rate of change is 5x (unless
x = 0). The secant becomes flatter and flatter as x becomes
smaller, so the limit must be 0.
b. The difference in temperature is 0.11◦ C. This would cost
about $10 to detect.
c. The difference in temperature is 0.00077◦ C. This would
cost over $1000 to detect.
f (x)
5
41. a. The pressure is 11 atmospheres, and measuring it would
cost $121.
4
b. The pressure is 101 atmospheres, and measuring it would
cost $10,201.
3
2
c. The pressure is 501 atmospheres, and measuring it would
cost $251,001.
1
–1
39. a. The difference in temperature is 6.25◦ C. It looks like a $1
device would be good enough.
0
1
x
Section 2.3, page 163
1. This is a linear function and is continuous everywhere.
29. f = 5(1 + x)2 + 7(1 + x) − 12, so the average rate of
change is 17 + 5x (unless x = 0). The limit is 17.
f(x)
40
35
30
25
20
15
10
5
0
–5
3. This is constructed as the quotient of the continuous exponential function and a continuous linear function. It is guaranteed
to be continuous everywhere that the denominator is not equal
to 0. The only potential trouble point is x = −1.
5. This is a composition of the natural log with a linear function
divided by a polynomial. The theorems guarantee continuity
except at z = 1, where we are taking the natural log of 0. At
z = 0, where the denominator is 0, the logarithm is not defined.
7. This is a composition of the continuous
cosine function
π
with a continuous linear function x −
and is continuous
2
everywhere.
1
2
x
31. a. limT →0 V(T ) = 1.0 cm3 .
b. V(2.0) = 5.0.
9. This is the quotient of the constant 1 by the function (1 − w)4 ,
which is a polynomial. This is guaranteed to be continuous
except where the denominator is 0, or when w = 1.
11. l(5) = 31, l(5.1) = 31.5, l(5.01) = 31.05, l(4.9) = 30.5,
l(4.99) = 30.95.
c. V(1.0) = 2.0.
13. f(0) = e0 /(0 + 1) = 1. f(0.1) = 1.0047, f(0.01) = 1.00005,
f(−0.1) = 1.0054, and f(−0.01) = 1.00005.
d. We need to solve V(T ) = 1.01. But 1 + T 2 = 1.01 if T 2 =
0.01, or T = 0.1 K.
15. g(2) = 0. g(2.1) = 0.022, g(2.01) = 0.0024,
−0.029, g(1.99) = −0.0025.
33. The average rate of change between t = 0 and t = t is
b(t) − b(0) t + t 2 − 0
=
= 1 + t
t
t
So t would have to be less than 0.01 for the average
rate of change to be within 1% of the instantaneous rate of
change.
g(1.9) =
17. g(0) cannot be computed. In fact, g(z) does not make sense
for z ≤ 1.
19. r (1) cannot be computed. r (1.1) = 1.0 × 104 , r (1.01) =
1.0 × 108 , r (0.9) = 1.0 × 104 , r (0.99) = 1.0 × 108 . This limit
is infinity.
21. f(x) = 2.1 if x = 0.1 and f(x) = 1.9 if x = −0.1, so −0.1 ≤
x ≤ 0.1.
Answers to Odd Exercises
√
23. f(x) = 1.1 if x = 1.1 = 1.049 and f(x) = 0.9 if x = 0.949,
so 0.949 ≤ x ≤ 1.049.
43. We need that 2V = V ∗ at V = V0 = 50. In this case, V ∗ = 100.
200
25. Any value of x ≥ 0 works, so x must be within 1 away from
1.
27. a.
831
150
f (V)
S(x)
100
1
50
0.5
0
–1
– 0.5
0.5
1
x
0
50
V
100
– 0.5
45. a.
Pitch without hysteresis
–1
1000
800
if x ≤ −0.1
if 0.1 < x < 0.1
if x ≥ 0.1
Pitch
b. The slope on the central part is 10.


 −1
S(x) = −1 + 10(x + 0.1)


1
600
400
200
c. The input would have to be between −0.01 and 0.01.
0
4.8 < M < 5.2
29.
1
0
2
3
Position
4.8 < 2.0V < 5.2
2.4 < V < 2.6
Pitch with hysteresis
1000
800
0.95 < F < 1.05
600
31.
Pitch
V must be within 0.1 cm3 of 2.5 cm3 for M to be within 0.2 g
of 5.0 g.
0.95 < r 4 < 1.05
0.987 < r < 1.012
On way back
400
On way forward
200
The radius must be within about 1% of 1.0 to guarantee a flow
within 5%.
0
33. Between 4.5 × 108 and 5.5 × 108 . Tolerance is 5.0 × 107 .
0
1
2
3
Position
35. Between 8.789 × 105 and 10.742 × 105 . Tolerance is
9.76 × 104 .
37. Between 0.96 and 1.04 g/L. Tolerance is 0.04 g/L.
b. The second seems more likely. The pitch is usually different in different directions.
39. Between 491.52 and 532.48 g/L. Tolerance is 20.48 g/L.
c. Both sound awful.
41. Denote the function by f . Then
80
f(V ) =
2V
Section 2.4, page 171
if V < 50
1. x 2 + 2xx + x 2 .
if V ≥ 50
3. 9x 2 + 12xx + 4x 2 .
5.
200
2
150
f (V)
y
derivative 0 at points
marked with
not differentiable
(corner)
100
1
not continuous
not differentiable
50
0
0
50
V
100
0
0
1
2
x
832
Answers to Odd Exercises
dM
= 0.5. This function is increasing.
dx
dg
= −3. This function is decreasing.
9. g (y) = −3,
dy
7. M (x) = 0.5,
y
19.
4
3
11. f(1) = 3, f(1 + x) = 3 − 2x − x 2 , f = −2x −
f
x 2 , so the slope of the secant is
= −2 − x if x =
0.
1
Taking the limit of this constant function by substituting
x = 0, we find that the slope of the tangent is f (1) =
−2.
–1
x
13. f(x + x) = 4 − x 2 − 2xx − x 2 , f = −2xx − x 2 ,
f
so the slope of the secant is
= −2x − x if x =
0. Takx
ing the limit, we find that the derivative is f (x) = −2x, or
derivative
2
function
1
21.
y
5
4
3
15. The derivative is f (x) = −2x, so the critical point occurs at
x = 0. For x < 0, the derivative is positive, whereas for x > 0,
it is negative. The function thus switches from increasing to
decreasing at x = 0.
1
derivative
0
1
–1
f (x) decreasing
– 1 –0.5
2
1
2
x
3
23. At x = 0, the slope of any secant with x > 0 is 1, whereas the
slope of any secant with x < 0 is −1. The derivative wants
to be both 1 and −1, which is impossible. From the graph,
these slopes correspond to two possible tangent lines.
f (x)
critical point
–2
function
2
f (x) increasing
x
1
–2
df
= −2x.
dx
3.5
3
2.5
2
1.5
1
0.5
2
Derivative of 冷x冷
1
0.5
x
1
0.5
0
0.5
1
x
0.5
derivative positive
and f (x) increasing
–2
f (x)
4
3
2
1
–1
critical point of f(x)
1
–1
–2
–3
–4
x
2
25. The slope of any secant with x > 0 is 0. With x < 0, the
slope of the secant is −1/x, which has a limit of negative infinity. There is a sort of half-tangent line with equation
Ĥ (x) = 1 for positive x, but there is no candidate tangent
line at all for negative x.
Heaviside function
derivative negative
and f (x) decreasing
17.
1
Maximum
derivative
4
1
0
Position
27.
Minimum
derivative
2
0
derivative
negative
derivative
positive
1
derivative 0
2
Time
3
Rate of change of temperature
–2
3
–1
4
2
0
–2
–4
–6
–8
0
1
x
2
warming up
1
2
3
4
5
Time
cooling down
Answers to Odd Exercises
Rate of change of temperature
29.
40
30
20
10
0
–10
–20
–30
–40
13. x 3 + 3x 2 + 3x + 1.
warming up
1
15. 8x 3 + 12x 2 + 6x + 1.
3
2
cooling
down
4
5
Time
17. f (x) = −2 + 2x, which is negative for x < 1, zero at x = 1,
and positive for x > 1.
cooling
down
Position
f (x)
31.
capture
hiker
bear
33.
capture
bear
h(x)
Position
hiker
Time
9.78
35. The time solves M(t) = 0, or 100 =
t 2 , or t 2 = 20.44
2
or t = 4.52 s. The speed is the derivative of the position,
or M (t) = at = 9.78t. The speed when it hits the ground is
44.2 m/s.
22.88 2
t , or t = 2.96 s. The
37. The time solves M(t) = 0, or 100 =
2
speed is the derivative of the position, or M (t) = at = 22.88t.
The speed when it hits the ground is 67.64 m/s.
1. 5x 4 .
derivative negative
and f(x) decreasing
0
1
x
2
2
derivative positive
1.5
and f(x) increasing
1
0.5
0
1
2
–0.5
–1
–1.5
–2 derivative
negative and
f(x) decreasing
5. ex e−1 .
21. The constant 2 is the derivative of a linear function with slope
2. One such function is 2x.
23. x 15 .
25. x −1 .
1 1 −1
xe .
e
f (x) =
d(3x 2 ) d(3x) d1
+
+
dx
dx
dx
=3
d(x 2 )
dx
+3
dx
dx
= 3 · 2x + 3 · 1 = 6x + 3
g (z) =
d(3z 3 ) d(2z 2 )
+
dz
dz
=3
x
29. A(r ) = 2πr . The units are centimeters. The derivative is equal
to the perimeter of the circle, corresponding to the area of the
little ring around the circle.
3. 0.2x −0.8 .
11.
derivative positive
and f(x) increasing
27. A(t) = 525t 2 , with units of people per year. New cases went
up quickly.
Section 2.5, page 184
9.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
19. h (x) = 3x 2 − 3, which is negative for x < 1, zero at x = 1, and
positive for x > 1.
Time
7.
833
d(z 3 )
dz 2
+2
dz
dz
= 3 · 3z 2 + 2 · 2z = 9z 2 + 4z
sum rule
r
constant product rule
r
power rule
sum rule
constant product rule
power rule
Answers to Odd Exercises
834
7. Set u(x) = 1 + x and v(x) = 2 + x. Then u (x) = 1 and v (x) =
2. By the quotient rule,
Position
31.
Speed
Front car
(2 + x) · 1 − (1 + x) · 2
1
=
(2 + x)2
(2 + x)2
f (x) =
Car being towed
Time
Time
33. Subtract the 10 mph for the passenger from the 80 mph for the
train to get 70 mph.
35. Relative to passenger: 20 mph; relative to train: 10 mph; relative to ground: 90 mph.
9. g (z) =
−2z(−1 + 3z + z 3 )
.
(1 + 2z 3 )2
11. Set u(x) = 1 + x and v(x) = (2 + x)(3 + x). Then u (x) = 1
and v (x) = (2 + x) + (3 + x) = 5 + 2x, by the product rule.
Applying the quotient rule gives
37. a. P(t) = 2 − 2t + 2t 2 .
b. a (t) = 2t and b(t) = −2 + 2t. The units are millions of
bacteria per hour.
P(t)
e.
14
12
10
8
6
4
2
0
P(t)
a(t)
13. f = 0.2, g = −0.3, and ( f g) = −1.76. This is equal
to g(x0 ) f + f(x0 )g + f g = −0.2 − 1.5 − 0.06. With
x = 0.01, f = 0.02, g = −0.03, and ( f g) = −0.1706.
This is equal to g(x0 ) f + f(x0 )g + f g = −0.02 −
0.15 − 0.0006. The last term is now much smaller than the
rest.
15. Would get that
dx · x2
= 1 · 2x
dx
b(t)
0
1
2
which is wrong.
3
t
d f(x) · f(x)
= f (x) f(x) + f (x) f(x) = 2 f (x) f(x) > 0
dx
17.
39. The speed is M (t) = 10 − at = 10 − 9.78t. The critical point
is then when t = 1.02 s. Plugging in M(1.02) = 105.11. It
hits the ground when M(t) = 0, or at t = 5.66. The speed is
M (5.66) = 45.34 m/s.
41. The speed is M (t) = 10 − at = 10 − 22.88t. The critical point
is then when t = 0.44 s. Plugging in M(0.44) = 102.18. It
hits the ground when M(t) = 0, or at t = 3.42. The speed is
M (3.42) = 68.38 m/s.
43. S(t) = 3t 2 works. S(1) = 3 and S(2) = 12. That seems pretty
fast.
dS
S(t)
45. S(t) = t 6 because
= 6t 5 = 6
. S(1) = 1 and S(2) = 64.
dt
t
That is quite fast, because the derivative gets bigger the larger
the organism is.
Section 2.6, page 192
1. f (x) = 2 · (−3x + 2) + (−3) · (2x + 3) = −12x − 5.
because f (x) > 0 and f(x) > 0. Therefore, f(x)2 is increasing.
19. The power rule works for n = 1 because x 1 = x is a linear
function with derivative 1 = 1x 0 .
21.
dx3
dx
dx2
= x2 ·
+
· x = x 2 + 2x 2 = 3x 2 . It worked.
dx
dx
dx
23. We denote the total mass by T(t).
a. T(t) = P(t)W(T ) = (2.0 × 106 + 2.0 × 104 t)(80 − 0.5t).
b. T (t) = 6.0 × 105 − 2.0 × 104 t.
c. T (t) = 0 when 6.0 × 105 − 2.0 × 104 t = 0 or when t =
30 yr. At time 30, the population is 2.6 × 106 , the
weight per person is 65 kg, and the total weight is 1.69 ×
108 kg.
d.
200
3. r (y) = 5(y − 1) + 2y(5y − 3).
2
d(x + 2)
dh
=
(2x + 3)(−3x + 2)
dx
dx
+ (x + 2)
d(2x + 3)(−3x + 2)
dx
= 1 · (2x + 3)(−3x + 2) + (x + 2) · (−12x − 5)
= (2x + 3)(−3x + 2) − (x + 2)(12x + 5)
160
T(t)
5.
1 − 2x − x 2
(2 + x)2 (3 + x)2
=
c. P (t) = −2 + 4t, which is equal to a (t) + b(t) = 2t − 2 +
2t.
d. The population of a is always growing, the population of
b shrinks before it grows, and the whole population also
shrinks before it grows.
(2 + x)(3 + x) · 1 − (1 + x)(5 + 2x)
(2 + x)2 (3 + x)2
F (x) =
120
80
40
0
0
20
40
60
t
80
100
Answers to Odd Exercises
25. We denote the total mass by T(t).
835
4
a. T(t) = P(t)W(T ) = (2.0 × 10 + 1.0 × 10 t )(80 − 0.5t).
6
3 2
3
S(N)
b. T (t) = 1.0 × 106 + 1.6 × 105 t − 1.5 × 103 t 2 .
c. T (t) = 0 at t ≈ 6.67 and t = 100.0. At time 6.67, the population is 2.04 × 106 , the weight per person is 76.67 kg, and
the total weight is 1.56 × 108 kg. At time 100, the population is 1.2 × 107 , the weight per person is 30.0 kg, and the
total weight is 3.6 × 108 kg.
2
1
0
0
2
4
6
8
10
N
d.
400
33. This simplifies to
T(t)
350
300
35. a. ρ(t) =
250
b. ρ (t) =
200
150
0
20
40
60
80
1 + t2
.
1+t
t 2 + 2t − 1
.
1 + t2
c. This is positive when t 2 + 2t − 1 > 0. This occurs for t
larger than the solution of t 2 + 2t − 1 = 0, which can be
found with the quadratic formula to be 0.414. After that,
the density increases.
100
t
Above-ground mass
(t)
27. Total mass above ground is (3.0t + 20)(1.2 − 0.01t). The
derivative is −0.06t + 3.4. This switches from positive to negative at t ≈ 56.7, and thus the above-ground mass increases
during the entire first 30 days.
100
90
80
4.5
4
3.5
3
2.5
2
1.5
1
0.5
70
60
0
10
20
30
Day
29. S(1) = 0.92, S(5) = 3.0, S(10) = 2.0. S (N ) = 1 − 0.16N .
This bird does best by laying about 6 eggs.
3
2
1
4
5
37. The final population is the product of the per capita reproduc
b
tion and the initial population bt , or bt+1 = 2.0 1 − t bt .
1000
b
The derivative of f(b) = 2.0b 1 −
is f (b) = 2.0 −
1000
4.0b
, which is positive for b < 500 and negative for b > 500.
1000
4
bt + 1
3
S(N)
0
t
50
40
30
20
2.4
.
(1.2 p + 2.0(1 − p))2
2
1
0
0
2
4
6
8
500
450
400
350
300
150
100
150
100
50
0
10
0
200
400
600
800
1000
bt
N
31. S(1) = 0.67, S(5) = 1.43, S(10) = 1.67. S (N ) =
1
(1 + 0.5N )2
which is always positive. This bird does best by laying as many
eggs as it can.
39. a. h 3 (0) = 0, h 3 (1) = 0.5, h 3 (2) ≈ 0.89.
b. h 3(x) =
3x 2
. h (0) = 0 h 3(1) = 0.75, h 3(2) ≈ 0.148.
(1 + x 3 )2 3
Answers to Odd Exercises
836
c.
9. The point of inflection x = 0 has a negative third derivative
because the second derivative changes from positive to negative values and is therefore decreasing.
1
h(x)
0.8
0.6
11. s (x) = −1 + 2x − 3x 2 + 4x 3 . s (x) = 2 − 6x + 12x 2 .
slope .75
0.4
13. h (y) = 10y 9 − 9y 8 . h (y) = 90y 8 − 72y 7 .
0.2
0
2
1
0
15. F (z) = (1 + z)(2 + z) + z(2 + z) + z(1 + z). F (z) = (1 +
z) + (2 + z) + z + (2 + z) + z + (1 + z) = 6z + 6.
17. f (x) =
x
2x − 2(3 + x)
−3
−3
= 2 . f (x) = 3 .
4x 2
2x
x
19. f (x) = −3x −4 < 0, so the function is decreasing, f (x) =
12x −5 > 0, so the function is always concave up.
d. This is kind of a mushy response.
Section 2.7, page 201
1.
10
y
A
1
8
B, E
6
4
F
–2
–1
1
x
2
2
C, D
0
–1
0
1
2
3
4
x
A
3.
21. h (x) = −3x 2 + 12x − 11, which has solutions at 2.577 and
1.422. h (x) = −6x + 12, so the function is concave up for
x < 2 and concave down for x > 2.
y
4
A, E
A, D
–4
–3
–2
A, D
1
–1
5
2
A
2
A
3
4
A
0
x
h(x)
A
B
A
3
–2
1
2
3
4
5
x
5
–3
A
–4
5.
23. f (x) = 6x 2 . The first derivative is always positive. f (x) =
12x. The second derivative is positive when x > 0 and negative when x < 0, producing a point of inflection at x = 0.
f (x)
Positive increasing derivative
10
f(x)
100
derivative
50
x
Negative increasing derivative
7.
–5
5
x
50
100
f (x)
function
x
25. f (x) = 20x − 50. The first derivative is positive when x > 2.5
and negative when x < 2.5. f (x) = 20. The second derivative
is always positive.
Answers to Odd Exercises
f(x)
490
100
480
–5
5
derivative
500
200
470
p(t)
function
b.
x
837
460
100
450
200
440
430
0
0.5
1
1.5
t
2
2.5
3
27. It is 0, because the degree will have been reduced to 0.
c. The tower was 500.0 m high, and the object was thrown
downward at 20.0 m/s. The acceleration of gravity on Pluto
is tiny, so the object is falling only slightly faster after 3 s
than it was when it was thrown.
29. Positive.
31. The derivative is f (x) = −3x −4 and the second derivative is
f (x) = 12x −5 .
a. Because f(1) = 1 and f (1) = −3, the tangent line is
fˆ(x) = 1 − 3(x − 1).
ˆˆ = 1 − 3(x − 1) +
b. Because f (1) = 12, the quadratic f(x)
6(x − 1) matches both the first derivative and the second
derivative at x = 1.
2
39. The second derivative is 160000 − 3000t, which is positive for t ≤ 53.33 and negative thereafter. This matches a
graph that switches from being concave up to being concave
down.
41.
14
12
10
f(x)
Position
8
7
6
5
4
3
2
1
0
1
f (x)
c.
37. The second derivative is −20000, matching the graph, which
is always concave down.
tangent line
quadratic
x
1
1.5
8
4
b. The position at t = 3 is less than 0, so the object has already
hit the ground.
most rapid acceleration
fastest
2
0
33. a. Velocity is p (t) = −10.4t − 2.0. Acceleration is p (t) =
−10.4.
most rapid
deceleration
6
0
2
4
6
Time
8
10
12
b
is f (b) = 2.0 −
43. The derivative of f(b) = 2.0b 1 −
1000
4.0b
4.0
, and the second derivative is −
, which is always
1000
1000
negative. The graph is always concave down.
50
40
20
bt + 1
p(t)
30
10
0
10
t
0.5
1
1.5
2
2.5
c. The tower was 50.0 m high, and the object was thrown
downward at 2.0 m/s. The acceleration of gravity on Saturn is only slightly greater than that on Earth, probably
because of Saturn’s low density.
35. a. Velocity is p (t) = −0.65t − 20.0. Acceleration is p (t) =
−0.65.
500
450
400
350
300
150
100
150
100
50
0
0
200
400
600
800
1000
bt
45. Using the quotient rule, we find that the first derivative
is
h (x) =
1
(1 + x)2
Answers to Odd Exercises
838
The second derivative can also be found with the quotient rule
and product rule as
(1 + x)2 · 0 − 1 · ((1 + x) + (1 + x))
−2
=
4
(1 + x)
(1 + x)3
G(z)
h (x) =
1
h(x)
0.8
0.4
0
1
1
− , which is negative if x < 2 and positive if x > 2.
2
x
1
L (x) = 2 , which is always positive. This function is increasx
0
1
x
2
ing when x > 2 and is always concave up.
L(x)
Section 2.8, page 209
1. F (x) = 2x + 4e x . F (x) = 2 + 4e x .
3. f (x) = (x 2 + 2x)e x . f (x) = (x 2 + 4x + 2)e x .
e x (x − 1) e x (x 2 − 2x + 2)
. g (x) =
.
2
x
x3
−x
x −1
f (x) = x . f (x) = x .
e
e
4
−4
f (x) = 1 + . f (x) = 2 .
x
x
4 1
4
g (z) = ln(z) + 1 + . g (z) = − 2 .
z
z
z
ew
ew
ew
F (w) = ew ln(w) + . F (w) = ew ln(w) + 2 − 2 .
w
w
w
2
−2
ln(x 2 ) = 2 ln(x), so s (x) = . s (x) = 2 .
x
x
−z
−(1
+
e
)
.
F (z) =
1 + ez
5. g (x) =
11.
13.
15.
17.
19. f (x) = −xe x , which is positive if x < 0 and negative if x > 0.
f (x) = −(1 + x)e x , which is positive if x < −1 and negative if x > −1. This function is increasing when x < 0 and is
concave up when x < −1.
1
2
x
3
df
5x+h − 5x
= lim
d x h→0
h
25. a.
df
5 x 5h − 5 x
= lim
d x h→0
h
b.
5x (5h − 1)
h→0
h
h
5
−1
= 5x lim
h→0
h
= lim
c. For h = 1, (51 − 50 )/1 = 4.0. For h = 0.1, (50.1 − 50 )/
0.1 = 1.746. For h = 0.01, (50.01 − 50 )/0.01 = 1.622. For
h = 0.001, (50.001 − 50 )/0.001 = 1.611. For h = 0.0001,
(50.0001 − 50 )/0.0001 = 1.609.
d. e1.609 = 4.998, which is suspiciously close to 5.
29. h(x) = (x − 1)e x seems to work. h (x) = xe x , so the critical
point is at x = 0. h (x) = (x + 1)e x , so the point of inflection
is at x = −1.
1
0.8
0.6
31. a. The definition is
0.4
d ln(x)
ln(1 + h) − ln(1)
ln(1 + h)
= lim
= lim
h→0
h→0
dx
h
h
0.2
–1
0.5
0.48
0.46
0.44
0.42
0.4
0.38
0.36
0.34
0.32
0.3
27. h (x) = (x + 2)e x , h (x) = (x + 3)e x .
f (x)
–2
3
23. L (x) =
This is always negative, and thus the function is concave down.
9.
2
x
1
0.6
0.2
7.
2.8
2.7
2.6
2.5
2.4
2.3
2.2
2.1
2
1.9
1.8
0
1
x
e z (z − 2)
, which is negative if z < 2 and positive if
z3 z 2
e
(z − 4z + 6)
z > 2. G (z) =
, which is always positive. This
z4
21. G (z) =
function is increasing when z > 2 and is always concave
up.
b. The limit seems to be 1.0, meaning that the derivative is
1.0.
c. The derivative of ln(x) is
1
, which is 1.0 at x = 1.
x
33. The definition is
d ln(x)
ln(2 + h) − ln(2)
ln(1 + h/2)
= lim
= lim
h→0
h→0
dx
h
h
Answers to Odd Exercises
If we write x = h/2, then h = 2x and the limit can be written
lim
h→0
ln(1 + h/2)
ln(1 + x)
= lim
h
2x
x →0
=
1
ln(1 + x) 1
lim
=
2 x →0
x
2
where we used the limit in Exercise 31.
t
1−t
35. The total mass is M(t) = 10 1 −
10et =
et . M (t) =
2
2
5(1 − t)et , which is negative for t > 1, and M (t) = −5tet ,
which is always negative. The total mass is greater than the
initial value at t = 0. It reaches zero at t = 2.
M(t)
6
4
2
0
0.5
1
t
1.5
2
t
is
r (x) =
−6x(1 + 3x)
.
(1 + 2x)4
6(1 + 2x)3 (1 + 3x) − 6(1 + 3x)2 (1 + 2x)2
=
(1 + 2x)6
7. F(z) = g(h(z)) where h(z) = 1 +
2
and g(h) = h 3 . h (z) =
1+z
−2(1 + z)−2 , found using the chain rule. Then
F (z) = g (h(z))h (z) = 3h 2 · −2(1 + z)−2
= −6
8
6
4
2
M(t)
1. g(x) = f(h(x)) where h(x) = 1 + 3x and f(h) = h 2 . Then
tive
14
12
10
1+
2
1+z
2
(1 + z)2
9. Set f(x) = g(h(x)) where g(h) = eh and h(x) = −3x. Then
0
0.5
t
f (x) = g (h(x))h (x) = eh(x) · (−3) = −3e−3x
1
11. g (y) =
1 − 2x −x
√ e , which is positive for x < 1/2 and nega2 x
tive for x > 1/2. It also says that the curve is infinitely steep
4x 2 − 4x − 1 −x
e . Solving 4x 2 − 4x − 1 = 0
at x = 0. G (x) =
4x 3/2
√
1± 2
gives points of inflection at
. Only the larger one is
2
positive. Below this, the function is concave down, and above
it, it is concave up.
G(x)
Section 2.9, page 220
5. This is a quotient of two compositions. The numerator is (1 +
3x)2 with derivative 6(1 + 3x). The denominator is (1 + 2x)3
with derivative 6(1 + 2x)2 . By the quotient rule, the deriva-
37. The total mass is M(t) = 10(1√− t )e . M (t) = 10(1 − 2t −
t 2 )et , which is negative for t > 2 − 1, and M (t) = 10(−1 −
4t − t 2 )et , which is always √
negative. The total mass is greater
than the initial value at t = 2 − 1. It reaches zero at t = 1.
2
39. G (x) =
43. This equation says that the rate of change of the population
is equal to the negative of population size, meaning that it
shrinks faster the larger it is. The derivative of b(t) = e−t is
−e−t , which is indeed the negative of b(t). This solution is
decreasing.
3. f 1 (t) = h(g(t)) where g(t) = 1 + 3t and h(g) = g 30 . Then
g (t) = 3 and h (g) = 30g 29 . Therefore, f 1(t) = 3 · 30g 29 =
90(1 + t)29 .
8
0
41. This equation says that the rate of change of the population
increases exponentially over time. The derivative of b(t) = et
is et , so it works. This solution is increasing.
g (x) = f (h(x))h (x) = 2h · 3 = 6(1 + 3x)
14
12
10
0
839
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1
.
1+ y
13. G (x) = 16xe x .
2
15. L (x) =
1
.
x ln(x)
17. With the quotient rule, set u(x) = 1 and v(x) = 1 + e x . Then
H (x) =
=
u v
(x) =
v(x)u (x) − u(x)v (x)
v 2 (x)
(1 + e x ) · 0 − 1 · e x
ex
=−
x
2
(1 + e )
(1 + e x )2
With the chain rule, set H (x) = r ( p(x)) where p(x) = 1 + e x
−1
and r ( p) = 1/ p. Then p (x) = e x and r ( p) = 2 . By the chain
p
rule,
0
0.5
1
x
1.5
2
H (x) = r ( p) p (x) =
−1 x
−e x
·
e
=
p2
(1 + e x )2
Answers to Odd Exercises
840
19. With law 1 of logs, ln(3x) = ln(3) + ln(x). Then
9. As in the text, we can use the fact that tan(θ ) =
that the derivative is sec2 (θ ) = cos(θ )−2 . With the chain rule,
d ln(3x) d ln(3) d ln(x)
1 1
=
+
=0+ =
dx
dx
dx
x
x
the second derivative is
With the chain rule, set f(x) = 3x and l( f ) = ln( f ). Then
(l ◦ f )(x) = l ( f(x)) f (x) =
1
1
1
·3=
·3=
f
3x
x
2 tan(θ)sec(θ )2 .
dsec(θ) d cos (θ )−1
=
= sin (θ ) cos (θ)−2
dθ
dθ
21. F(x) = 1 + 4x + 4x 2 and F (x) = 4 + 8x. With the chain rule,
we think of F as the composition F(x) = g(h(x)) where
h(x) = 1 + 2x and g(h) = h 2 . Then h (x) = 2 and g (h) = 2h,
so F (x) = g (h)h (x) = 2h · 2 = 4(1 + 2x). Multiplying out,
we find that this matches the result found directly.
This can be rewritten as sec(θ ) tan(θ). We can find the second
derivative with the product rule as sec(θ) tan(θ )2 + sec(θ )3 .
13. The double angle formula is cos(2θ ) = cos(θ ) cos(θ ) −
sin(θ) sin(θ). Taking the derivative yields
23. F (x) = 3x 2 . We can write F(x) = e3 ln(x) , a composition
f(g(x)) where f(g) = e g and g(x) = 3 ln(x). Then f (g) = e g
and g (x) = 3/x, so
3
3
3
= e3 ln(x) = x 3 = 3x 2
x
x
x
y−1
25. The inverse is f −1 (y) =
with derivative ( f −1 )(y) =
3
1/3. Using the theorem, f (x) = 3 for any x, so f ( f −1 (y)) = 3
and ( f −1 )(y) = 1/3.
27. The inverse is h −1 (y) = (y − 2)1/3 with derivative (h −1 )(y) =
1
(y − 2)−2/3 . Using the theorem, h (x) = 3x 2 , so h (h −1 (y)) =
3
1
3
√
−1
+ 4+ y
1
29. The inverse is q −1 (y) =
. (q −1 )(y) = √
.
2
4
√ +y
3(y − 2)2/3 and (h −1 )(y) = (y − 2)−2/3 .
Using the theorem, q (x) = 1 + 2x. So q (q −1 (y)) =
1
and (q −1 )(y) = √
.
d
cos(2θ ) = −sin(θ ) cos(θ ) − sin(θ ) cos(θ )
dθ
−sin(θ ) cos(θ ) − sin(θ ) cos(θ )
= −4 sin(θ) cos(θ ) = −2 sin(2θ)
This is just what the chain rule gives.
15. The angle addition formula says that cos(θ + φ) =
cos(θ) cos(φ) − sin(θ ) sin(φ). Taking the derivative yields
d
cos(θ + φ) = −sin(θ) cos(φ) − cos(θ ) sin(φ)
dθ
= −sin(θ + φ)
This matches the result with the chain rule.
17. a (x) = 3 − sin(x). a(0) = 1, a (0) = 3, a(π/2) = 3π/2,
a (π/2) = 2, a(π) = 3π − 1, a (π ) = 3. The function is increasing, but with a slight slowing at around x = π.
4+ y
4+ y
n ln(x)
33. f (x) = √
.
−x
1 − x2
a(x)
31. x = e
n
.
35. B(M) = 0.5M, so B (M) = 0.5. M(W ) = 5W + 2, so
M (W ) = 5. Therefore, the derivative of the composition is
B (M)M (W ) = 2.5.
37. F (V ) = 0.4 and V (I ) = 10I , so the derivative of the composition is F (V )V (I ) = 4I .
39. Q (t) = −0.000122 · 6.0 × 1010 e−0.000122t = −7.32 ×
106 e−0.000122t .
db
41. The derivative is
= 300e3t = 3 · 100e3t = 3b(t). This soludt
tion is increasing.
3. h (x) = cos2 (θ) − sin2 (θ ).
0
/2
3 /2
2
19. c(z) = (cos(z) − sin(z))/e z . c(0) = 0, c(0) = 1, c(π/2) =
e−π/2 , c(π/2) = −e−π/2 , c(π) = 0, c(π ) = −e−π . The function zips up to a maximum at around x = π/2, dips down to
zero at x = π , becomes negative, and then returns to zero at
x = 2π .
0.35
db
0.3
0.25
0.2
c(z)
1. f (x) = 2x sin(x) + x 2 cos(x).
20
18
16
14
12
10
8
6
4
2
0
x
= 6e2t = 1 + 2(3e2t − 0.5) = 1 + 2b(t).
43. The derivative is
dt
This solution is increasing.
Section 2.10, page 230
2 sin(θ)
, which can be written
cos(θ)3
11. The first derivative is found with the chain rule.
The two answers check.
F (x) = f (g)g (x) = e g ·
sin(θ)
to find
cos(θ)
0.15
0.1
0.05
5. F (z) = −2 sin(2z − 1).
0
7. f (x) = − sin(x)ecos(x) .
–0.05
3 /2
/2
2
z
Answers to Odd Exercises
21. s (t) = 0.2e0.2t cos(t) − e0.2t sin(t). s(0) = 1, s (0) = 0.2,
s(π/2) = 0, s (π/2) = −1.369, s(π ) = −1.874, s (π ) =
−0.375. This oscillation expands exponentially.
35.
0
10
20
30
40
t
20
60
23. The derivative of sin(x) is cos(x). Therefore,
1
cos(sin−1 (x))
Using the identity, cos(x) = 1 − sin2 (x), so
cos(sin−1 (x)) = 1 − sin2 (sin−1 (x)) = 1 − x 2
(sin−1 )(x) =
Therefore, the derivative is (sin−1 )(x) = √
But
1
1 − x2
.
−2x − 2
1
sec2 (tan−1 (x))
−1
sec (tan (x)) = 1 + tan (tan (x)) = 1 + x
2
Therefore, the derivative is (tan−1 )(x) =
dx
= −2e−t cos(t) − 2(−e−t cos(t) − e−t sin(t))
dt
d2x
= 2e−t sin(t) = 2
dt
This is indeed a solution.
Using the identity,
−1
d2x
= 2e−t sin(t)
dt 2
dx
= −e−t cos(t) − e−t sin(t),
dt
2
41. This population is growing, but per capita production
oscillates.
43.
1
.
1 + x2
10
8
27. The derivative is s (t) = cos(t).
29. The fourth derivative of each piece repeats.
8.0π
x − 1.0
8.0π
31. f (x) = −
sin(2π
). f (0) = −
sin(−2π/5) >
5.0
5.0
5.0
0, which is consistent with the fact that this oscillation is increasing at x = 0.
Population
(tan−1 )(x) =
6
4
2
0
0
5
7
f(x)
2πt
T
dt
25. The derivative of tan(x) is sec (x). Therefore,
10
Time
15
20
45. b(t) = (0.1 + 0.1A cos(0.8t))b(t) = 0.1(1 + A cos(0.8t))
b(t).
3
2
4
6
8
10
x
47. a. t = π , June 21, 16.5 h.
b. t = π/2, March 21, 12.0 h.
5.0π
z − 3.0
5.0π
33. h (z) = −
sin(2π
). h (0) = −
sin(−3π/2) <
2.0
4.0
2.0
0, which is consistent with the fact that the curve begins by
decreasing.
6
h(z)
cos
39. This says that acceleration has two pieces: one proportional
to the negative of displacement (the −2x term) and one prodx
portional to the negative of velocity (the 2
term).
2
–4
2
37. The derivative of the sum is
2π
2π(t − 16)
−0.2 ·
sin
− 0.3 · 2π sin(2π(t − 0.583))
28
28
40
1
0
2π
T
2π 2 0.1
=
= 0.1
T
1.0
√
This has solution T = 0.12π . This weaker spring oscillates
more slowly with a larger period.
20
0
–1
p (t) = −
40
s(t)
2π
2π t
sin
,
T
T
We thus need
60
2
p (t) = −
841
c. L (t) = 4.5 sin(t) equal to 0 at t = 0 and t = π, when days
are shortest or longest.
d. The day length changes most rapidly at t = π/2 and t =
3π/2, or on March 21 and September 21. This should be
easiest for the plant to detect.
Supplementary Problems, page 233
2
4
6
8
10
z
1. We can substitute because the function is continuous. The limit
is 1.
3. This increases to infinity.
Answers to Odd Exercises
842
5. This increases to infinity.
c. The population is 1490 after 7 yr and 1640 after 8 yr. The
approximate growth rate is 150 during this year.
7. F (y) = 4y + 10y.
3
9. H (c) =
d. The derivative is 20t.
2c(1 + c)
for c =
−1/2.
(1 + 2c)2
11. b(y) = −0.75y −1.75 for y ≥ 0.
2000
13. g (x) = 60x(4 + 5x 2 )5 .
15. s (t) = 3/t.
1500
N(t)
17. s (x) = −3e−3x+1 + 5/x.
19. f (t) = (cos(t) − sin(t))et . The point t = π/4 is a critical
point.
21. h (y) =
1000
500
2(y − 2)
. h(y) is increasing when y > 2.
(1 + y)4
2 −x 3
0
0
2
4
6
23. f (x) = −3x e , which is 0 at x = 0. f (x) = (9x −
3
6x)e−x , which is 0 at x = 0 and x ≈ 0.873.
f(x)
e. The per capita rate of growth is
3
10
20t
. It starts at
(1000 + 10t 2 )
0 and increases for a while. Eventually, however, it will
decrease.
2
29. a. F (t) = 3t 2 − 12t + 9 is the derivative, with units of factoids per week.
1
–1
8
t
4
0
1
x
2
3
1
2
25. h (y) = −sin(y) + , which is 0 at y =
π
+ 2nπ and y =
6
b. With the quadratic formula, we find that F (t) = 0 when
t = 1 or t = 3. Because F (t) is a parabola that goes up, it
must be negative between times 1 and 3. This is the time
when knowledge is being lost.
c.
tπ
+ 2nπ for any integer n. h (y) = −cos(y), which is 0 at
6 π
y = + nπ for any integer n.
2
300
F(t)
200
h(y)
4
100
3
0
2
0
1
2
3
4
–6 –5 –4 –3 –2 –1
–1
1 2 3 4 5 6
8
9 10 11
b. p0 would have to be between 0.243 and 0.428, or within
about 0.1.
27. a. The 1000 has units of individuals, and the 10 has units of
individuals/yr2 .
c. It would get smaller and smaller also.
33. a.
With increasing temperature
2000
100
approximate growth
rate is slope
of secant
1000
80
60
V
1500
N(t)
7
31. a. p0 = 0.333.
–3
40
20
500
0
0
6
y
–2
b.
5
t
1
0
2
4
6
t
8
10
0
20
40
60
T
80
100
Answers to Odd Exercises
With decreasing temperature
100
80
V
60
40
20
0
0
40
20
60
80
100
T
b. limT →50+ Vi (T ) = 100. limT →50− Vi (T ) = 50. limT →50+ Vd
(T ) = limT →50− Vd (T ) = 100.
c. limT →20+ Vi (T ) = limT →20− Vi (T ) = 20. limT →20+ Vd (T ) =
100. limT →20− Vd (T ) = 20.
35. a.
12
first derivative positive
second derivative
negative
Area open
10
8
6
point of inflection
4
2
0
b.
0
2
4
6
Time of night
8
10
8
10
8
10
Derivative of area open
4
3
2
1
0
2
4
6
–1
–2
c.
Secaond derivative of area open
Time of night
3
2
1
0
2
4
6
–1
–2
–3
Time of night
843