dx x

x /5]2

1

d

arctan(x 2 + ex ) = · [2x + ex ].

(f)

1 + (x 2 + ex )2

dx

7. Of course v(t) = p (t) = 12t ’ 5 so v(4) = 43 feet/second. The average

velocity from t = 2 to t = 8 is

p(8) ’ p(2) 364 ’ 34

vav = = = 55.

6 6

The derivative of the velocity function is (v ) (t) = 12. This derivative never

vanishes, so the extrema of the velocity function on the interval [5, 10] occur

at t = 5 and t = 10. Since v(5) = 55 and v(10) = 115, we see that the

maximum velocity on this time interval is 115 feet per second at t = 10.

280 Solutions to Exercises

8. (a) We know that

1 1

[f ’1 ] (1) = =.

3

f (0)

(b) We know that

1 1

[f ’1 ] (1) = =.

8

f (3)

(c) We know that

1 1

[f ’1 ] (1) = = 2.

f (2) π

(d) We know that

1 1

[f ’1 ] (1) = =.

40

f (1)

Chapter 3

1.

Fig. S3.1

2. Figure S3.2 shows a schematic of the imbedded cylinder. We see that the

volume of the imbedded cylinder, as a function of height h, is

V (h) = π · h · (3 ’ h/2)2 .

Then we solve

0 = V (h) = π · 9 ’ 6h + 3h2 /4 .

281

Chapter 3

Fig. S3.2

The roots of this equation are h = 2, 6. Of course height 6 gives a trivial

cylinder, as does height 0. We ¬nd that the solution of our problem is height

2, radius 2.

3. We know that

V = ·w·h

hence

dV d dw dh

= ·w·h+ · ·h+ ·w·

dt dt dt dt

= 1 · 60 · 15 + 100 · (’0.5) · 15 + 100 · 60 · 0.3

= 900 ’ 750 + 1800 = 1950 in/min.

4. We know that v0 = ’15. Therefore the position of the body is given by

p(t) = ’16t 2 ’ 15t + h0 .

Since

0 = p(5) = ’16 · 52 ’ 15 · 5 + h0 ,

we ¬nd that h0 = 475. The body has initial height 475 feet.

5. We know that

1

V= · π r 2 · h.

3

Therefore

1 dh 1

d dr

0= V = · π · r2 · + · π · 2r · · h.

3 3

dt dt dt

At the moment of the problem, dh/dt = 3, r = 5, h = 12/(5π ). Hence

dr 12

0 = π · 52 · 32 + π · (2 · 5) · ·

dt 5π

282 Solutions to Exercises

or

dr

0 = 225π + 24 ·.

dt

We conclude that dr/dt = ’75π/8 microns per minute.

6. Of course

10000 = V = π · r 2 · h.

We conclude that

10000

h= .

π · r2

We wish to minimize

A = (area of top) + (area of sides)

10000

= π · r 2 + 2π · r · h = π · r 2 + 2π · r · .

πr2

Thus the function to minimize is

20000

A(r) = π · r 2 + .

r

Thus

20000

0 = A (r) = 2π r ’ .

r2

We ¬nd therefore that

10000

r3 =

π

√

or r = 3 10000/π . Since the problem makes sense for 0√ r < ∞, and

<

since it clearly has no maximum, we conclude that r = 10000/π, h =

3

√3

10000/π .

7. We calculate that g (x) = sin x +x cos x and g (x) = 2 cos x ’x sin x. The

roots of these transcendental functions are best estimated with a calculator

or computer. Figure S3.7 gives an idea of where the extrema and in¬‚ection

points are located.

8. We know that v0 = ’5 and h0 = 400. Hence

p(t) = ’16t 2 ’ 5t + 400.

The body hits the ground when

0 = p(t) = ’16t 2 ’ 5t + 400.

Solving, we ¬nd that t ≈ 4.85 seconds.

283

Chapter 3

Fig. S3.7

9. We see that

x

h(x) =

x2 ’ 1

x2 + 1

h (x) = ’ 2

(x ’ 1)2

2x(x 2 + 3)

h (x) =

(x 2 ’ 1)3

We see that the function is unde¬ned at ±1, decreasing everywhere, and

has an in¬‚ection point only at 0. The sketch is shown in Fig. S3.9.

10. We know that

4π 3

V= r.

3

Therefore

4π

dV dr

= · 3r 2 .

3

dt dt

Using the values V = 36π, r = 3, dV /dt = ’2, we ¬nd that

dr

’2 = 4π · 32 ·

dt

hence

1

dr

=’ in. per sec.

18π

dt

284 Solutions to Exercises

Fig. S3.9

11. The acceleration due to gravity, near the surface of the earth, is about

’32 ft/sec2 regardless of the mass of the object being dropped. The two

stones will strike the ground at the same time.

12. He can drop a rock into the well and time how long it takes the rock to strike

the water. Then he can use the equation

p(t) = ’16t 2 + 0t + h0

to solve for the depth. If the well is very deep, then he will have to know

the speed of sound and compensate for how long it takes the splash to reach

his ears.

13. Refer to Fig. S3.13 to see the geometry of the situation.

Let (x, y) be the point where the rectangle touches the line. Then the area

of the rectangle is

A = x · y.

But of course 3x + 5y = 15 or y = 3 ’ (3/5)x. Hence

A = x · [3 ’ (3/5)x].

We may differentiate and set equal to zero to ¬nd that x = 5/2 and y = 3/2

is the solution to our problem.

14. Let s be a side of the base and let h be the height. The area of the base is s 2

and the same for the top. The area of each side is s · h. Thus the cost of the

base and top is

C1 = [s 2 + s 2 ] · 10 cents

285

Chapter 3

y

x

Fig. S3.13

while the cost of the sides is

C2 = 4(s · h) · 20 cents.

We ¬nd that the total cost is

C = C1 + C2 = 20s 2 + 80sh. (—)

But

100 = volume = s 2 · h

hence

h = 100/s 2 .

Substituting this last formula into (—) gives

8000

C(s) = 20s 2 + 80s · [100/s 2 ] = 20s 2 + .

s

We may calculate that

8000

0 = C (s) = 40s ’ .

s2

Solving for s gives the solution s ≈ 5.8479 and then h ≈ 2.9241.

15. We see that

x2 ’ 1

f (x) = 2

x +1

4x

f (x) = 2

(x + 1)2

’12x 2 + 4

f (x) =

(x 2 + 1)3

286 Solutions to Exercises

√

Thus there are a critical point at x = 0 and in¬‚ection points at x = ±1/ 3.

Figure S3.15 exhibits the complete graph.

Fig. S3.15

16. We see that the equation for the position of a falling body will now be

20 2

p(t) = ’ t + v0 t + h 0 .

2

It is given that v0 = 0 and h0 = 100. Hence

p(t) = ’10t 2 + 0t + 100.

The body hits the surface when

0 = p(t) = ’10t 2 + 100.

√

This occurs at time t = 10.

Chapter 4

F (x) = x 3 /3 + cos x + C

1. (a)

F (x) = e3x /3 + x 5 /5 ’ 2x + C

(b)

F (t) = t 3 + [ln t]2 /2

(c)

F (x) = ’ ln(cos x) ’ sin x ’ [cos 3x]/3 + C

(d)

F (x) = [sin 3x]/3 ’ [cos 4x]/4 + x + C

(e)

F (x) = esin x + C

(f)

’ cos x 2

x sin x dx = +C

2

2. (a)

2

287

Chapter 4

3 3

ln x 2 dx = ln2 x 2 + C

(b)

4

x

1

sin x · cos x dx = sin2 x + C

(c)

2

1

tan x · ln cos x dx = ’ ln2 cos x + C

(d)

2

sec2 x · etan x dx = etan x + C

(e)

12

(2x + 1) · (x 2 + x + 7)43 dx = (x + x + 7)44 + C

(f)

44

3. (a) We have

k 2

2 1

j j

x + x dx = lim 1+ + 1+ ·

2

k k k

k’∞

1 j =1

k

j2

2j j1

= lim 1+ + 2 +1+

k kk

k

k’∞

j =1

k

j2

2 3j

= lim + 2+ 3

k k k

k’∞

j =1

2 k2 + k 3 2k 3 + 3k 2 + k 1

= lim k· + · 2+ ·3

2 6

k k k

k’∞

3 3 1 1 1

= lim 2+ + ++ +2

2 2k 3 2k 6k

k’∞

31

=2+ +

23

23

= .

6

(b) We have

2

2j

’1 +

k

1 x2 2

k

’ dx = lim ’ ·

3 3 k

k’∞

’1 j =1

k

’2 4j 2

4j

= lim 1’ +2

3k k k

k’∞

j =1

288 Solutions to Exercises

k

8j 2

2 8j

= lim ’ + 2’ 3

3k 3k 3k

k’∞

j =1

’2 k 2 + k 8 2k 3 + 3k 2 + k ’8

= lim k · + · 2+ ·3

3k 2 6

3k 3k

k’∞

2 4 16

=’ + ’

3 3 18

2

=’ .

9

3

3 x3

x ’ 4x + 7 dx = ’ x + 7x

2 3 4

4. (a)

3

1 1

33 13

= ’ 34 + 7 · 3 ’ ’ 14 + 7 · 1

3 3

172

= (9 ’ 81 + 21) ’ (1/3 ’ 1 + 7) = ’ .

Y

3

6

2

sin2 x