## Chapter 1 Limits, Derivatives, Integrals, and Integrals

Chapter 1—Limits, Derivatives, Integrals, and Integrals Problem Set 1-1 1. a. 95 cm b. From 5 to 5.1: average rate ≈ 26.34 cm/s From 5 to 5.01: averag...
Author: Karin Hodge
Chapter 1—Limits, Derivatives, Integrals, and Integrals Problem Set 1-1 1. a. 95 cm b. From 5 to 5.1: average rate ≈ 26.34 cm/s From 5 to 5.01: average rate ≈ 27.12 cm/s From 5 to 5.001: average rate ≈ 27.20 cm/s So the instantaneous rate of change of d at t = 5 is about 27.20 cm/s. c. Instantaneous rate would involve division by zero. d. For t = 1.5 to 1.501, rate ≈ −31.42 cm/s. The pendulum is approaching the wall: The rate of change is negative, so the distance is decreasing. e. The instantaneous rate of change is the limit of the average rates as the time interval approaches zero. It is called the derivative. f. Before t = 0, the pendulum was not yet moving. For large values of t, the pendulum’s motion will die out because of friction. 2. a. x = 5: y = 305, price is \$3.05 x = 10: y = 520, price is \$5.20 x = 20: y = 1280, price is \$12.80 b. x = 5.1, rate ≈ 46.822 ¢/ft x = 5.01, rate ≈ 46.9820… ¢/ft x = 5.001, rate ≈ 46.9982… ¢/ft c. 47 ¢/ft. It is called the derivative. d. x = 10: 44 ¢/ft. x = 20: 128 ¢/ft e. The 20-ft board costs more per foot than the 10-ft board. The reason is that longer boards require taller trees, which are harder to find.

Problem Set 1-2 Q1. Power function, or polynomial function Q2. f (2) = 8 Q3. Exponential function Q4. g (2) = 9 Q5. h (x )

3. a. Decreasing fast 4. a. Decreasing slowly 5. a. Increasing fast c. Decreasing slowly 6. a. Decreasing fast c. Increasing fast 7. a. Increasing slowly c. Increasing slowly 8. a. Decreasing fast c. Decreasing fast 9. a. Increasing fast

b. b. b. d. b. d. b.

Decreasing slowly Increasing slowly Increasing slowly Increasing fast Increasing slowly Decreasing fast Increasing slowly

b. Decreasing fast b. Neither increasing nor decreasing d. Increasing slowly b. Decreasing fast d. Neither increasing nor decreasing

c. Increasing fast 10. a. Decreasing slowly c. Decreasing fast 11. a. T(x) (°C) 100

50

x (s) 100

200

x = 40: rate ≈ 1.1°/s x = 100: rate = 0°/s x = 140: rate ≈ −0.8°/s b. • Between 0 and 80 s the water is warming up, but at a decreasing rate. • Between 80 and 120 s the water is boiling, thus staying at a constant temperature. • Beyond 120 s the water is cooling down, rapidly at first, then more slowly. 12. a. v ( x ) (ft/s)

1

x 1

Q6. Q8. Q10. 1. 2.

h (5) = 25 Q7. y=x Q9. Derivative a. Increasing slowly a. Increasing fast

Calculus Solutions Manual © 2005 Key Curriculum Press

y = ax2 + bx + c, a ≠ 0 y = |x|

70 60 50 40 30 20

b. Increasing fast b. Decreasing slowly

10

x (s) 1

2

3

4

5

6

7

8

Problem Set 1-2

1

x = 2: rate ≈ 18 (ft/s)/s x = 5: rate = 0 (ft/s)/s x = 6: rate ≈ −11 (ft/s)/s b. Units are (ft/s)/s, sometimes written as ft/s2. The physical quantity is acceleration. 13. a. 18

x

2 3 4

7

• Increasing at x = 3 • Decreasing at x = 7 b. h (3) = 17, h(3.1) = 17.19 0.19 = 1.9 ft/s Average rate = 0.1 c. From 3 to 3.01: 0.0199 average rate = = 1.99 ft/s 0.01 From 3 to 3.001: 0.001999 average rate = = 1.99 ft/s 0.001 The limit appears to be 2 ft/s. d. h (7) = 9, h(7.001) = 8.993999 −0.006001 = −6.001 ft/s Average rate = 0.001 The derivative at x = 7 appears to be −6 ft/s. The derivative is negative because h(x) is decreasing at x = 7. 14. a. f (t )

Not much

500

300 Decrease

Increase

100

t 10

b. Enter y2 =

y1 ( x ) − y1 (1) x −1 r(t) = y2 (foxes/year)

0.98 0.99 1 1.01 1.02 1.03

2

f ( 4.01) − f ( 4) = −129.9697K 0.01 f ( 4) − f (3.99) = −131.4833K −0.01 Instantaneous rate = (−129.9697… − 131.4833…)/2 = −130.7265… foxes/year (actual: −130.7287…) The answer is negative because the number of foxes is decreasing. a(2.1) − a(2) 15. a. Average rate = = 0.1 52.9902… bacteria/h d.

h (x )

t 0.97

c. Substituting 1 for t causes division by zero, so r(1) is undefined. Estimate: r approaches the average of r(0.99) and r(1.01), 108.0586… foxes/year. (Actual is 108.0604… .) The instantaneous rate is called the derivative.

Problem Set 1-2

110.5684… 109.7361… 108.9001… undefined 107.2171… 106.3703… 105.5200…

b. r (t ) =

200(1.2 t ) − 200(1.2 2 ) t−2

r (t ) (mm2/hr) 60 40 20

t (mm) 2

r(2) is undefined. c. r(2.01) = 52.556504… 52.556504… − 52.508608… = 0.04789… Use the solver to find t when r(t) = 52.508608… + 0.01 = 52.518608… . t = 2.002088… , so keep t within 0.002 unit of 2. 4 16. a. v( x ) = πx 3 → v(6) = 288π 3 4 π (6.13 − 6 3 ) b. 6 to 6.1: average rate = 3 = 0.1 146.4133…π 4 π (6 3 − 5.9 3 ) 5.9 to 6: average rate = 3 = 0.1 141.6133…π Estimate of instantaneous rate is (146.4133…π + 141.6133…π)/2 = 144.0133…π = 452.4312… cm3/cm. c. r ( x ) =

4 3

π x 3 − 43 π 6 3 x−6

r (x ) (cm3/cm)

144π 48π

x 6

r(6) is undefined. Calculus Solutions Manual © 2005 Key Curriculum Press

17. 18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

d. r(6.1) = 146.4133…π = 459.9710… r(6.1) is 7.5817… units from the derivative. Use the solver feature to find x if r(x) = 144π + 0.1. x = 6.001326… , so keep x within 0.00132… unit of 6. a. i. −1.0 in./s ii. 0.0 in./s iii. 1.15 in./s b. 1.7 s, because y = 0 at that time a. i. −0.395 in./min ii. −0.14 in./min iii. −0.105 in./min b. The rate is negative, because y is decreasing as the tire goes down. a. Quadratic (or polynomial) b. f (3) = 30 c. Increasing at about 11.0 (2.99 to 3.01) a. Quadratic (or polynomial) b. f (1) = 12 c. Increasing at about 6.0 (0.99 to 1.01) a. Exponential b. Increasing, because the rate of change from 1.99 to 2.01 is positive. a. Exponential b. Increasing, because the rate of change from −3.01 to −2.99 is positive. a. Rational algebraic b. Decreasing, because the rate of change from 3.99 to 4.01 is negative. a. Rational algebraic b. Increasing, because the rate of change from −2.01 to −1.99 is positive. a. Linear (or polynomial) b. Decreasing, because the rate of change from 4.99 to 5.01 is negative. a. Linear (or polynomial) b. Increasing, because the rate of change from 7.99 to 8.01 is positive. a. Circular (or trigonometric) b. Decreasing, because the rate of change from 1.99 to 2.01 is negative. a. Circular (or trigonometric) b. Decreasing, because the rate of change from 0.99 to 1.01 is negative. • Physical meaning of a derivative: instantaneous rate of change • To estimate a derivative graphically: Draw a tangent line at the point on the graph and measure its slope. • To estimate a derivative numerically: Take a small change in x, find the corresponding

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change in f (x), then divide. Repeat, using a smaller change in x. See what number these average rates approach as the change in x approaches zero. • The numerical method illustrates the fact that the derivative is a limit. 30. See the text for the definition of limit.

Problem Set 1-3 Q2. y = cos x Q4. y = 1/x Q6. f (5) = 4 Q8.

Q1. 72 ft2 Q3. y = 2x Q5. y = x 2 Q7. y

y x

x

Q10. x = 3

Q9. y

x

1. f (x) = −0.1x 2 + 7 2. f (x) = −0.2x 2 + 8 a. Approximately 30.8 a. Approximately 22.2 b. Approximately 41.8 b. Approximately 47.1 7

8 f( x)

f (x )

x x

5 6

–1

–2

3

5

3. h(x) = sin x 4. g( x ) = 2 x + 5 a. Approximately 2.0 a. Approximately 7.9 b. Approximately 1.0 b. Approximately 12.2 g ( x)

h (x )

1 6

x 3

x –1

1

Problem Set 1-3

2

3

5. There are approximately 6.8 squares between the curve and the x-axis. Each square represents (5)(20) = 100 feet. So the distance is about (6.8)(100) = 680 feet. 6. There are approximately 53.3 squares between the curve and the x-axis. Each square represents (0.5)(10) = 5 miles. So the distance is about (53.3)(5) = 266.5 miles. tan 1.01 – tan 0.99 = 3.42K 7. Derivative ≈ 1.01 – 0.99 8. Derivative = −7 (exactly, because that is the slope of the linear function) 9. a. v (t )

v(3.01) – v(2.99) = 1.8648K 3.01 − 2.99 About 1.86 (ft/s)/s The derivative represents the acceleration. From t = 0 to t = 5, the object travels about 11.4 cm. From t = 5 to t = 9, the object travels back about 4.3 cm. So the object is located about 11.4 − 4.3 = 7.1 cm from its starting point. See the text for the meaning of derivative. See the text for the meaning of definite integral. See the text for the meaning of limit. d. Rate ≈

11.

12. 13. 14.

Problem Set 1-4 Q1. y changes at 30 Q3.

100

Q2. Derivative ≈ −500 Q4. f (3) = 9

y 60

x

t 5

8.7 10

The range is 0 ≤ y ≤ 32.5660… . b. Using the solver, x = 8.6967… ≈ 8.7 s. c. By counting squares, distance ≈ 150 ft. The concept used is the definite integral. v(5.01) − v( 4.99) d. Rate ≈ = 3.1107K 5.01 − 4.99 About 3.1 (ft/s)/s The concept is the derivative. The rate of change of velocity is called acceleration. 10. a. 10

v (t)

5

t 1

2

3

4

5

b. v(4) = 9.3203… ≈ 9.3 ft/s Domain: 0 ≤ t ≤ 4 Range: 0 ≤ v(t) ≤ 9.3203… c. By counting squares, the integral from t = 0 to t = 4 is about 21.3 ft. The units of the integral are (ft/s) · s = ft. The integral tells the length of the slide.

4

Problem Set 1-4

Q5. Q7. Q9. 1.

100 366 days Definite integral a.

Q6. sin (π/2) = 1 Q8. Derivative Q10. f (x) = 0 at x = 4

v (t ) 20,000

t 30

b. Integral ≈ 5(0.5v(0) + v(5) + v(10) + v(15) + v(20) + v(25) + 0.5v(30)) = 5(56269.45…) = 281347.26… ≈ 281,000 ft The sum overestimates the integral because the trapezoids are circumscribed about the region and thus include more area. c. The units are (ft/s)(s), which equals feet, so the integral represents the distance the spaceship has traveled. d. Yes, it will be going fast enough, because v(30) = 27,919.04… , which is greater than 27,000. 2. a. v(t) = 4 + sin 1.4t v (t ) 5

t 3

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3. 4.

5.

6.

7.

b. A definite integral has the units of the x-variable times the y-variable. Distance = rate × time. Because v(t) is distance/time and t is time, their product is expressed in units of distance. c. See graph in part a. Distance ≈ 0.5(0.5v(0) + v(0.5) + v(1) + v(1.5) + v(2) + v(2.5) + 0.5v(3)) = 0.5(26.041…) = 13.02064… ≈ 13.0 ft d. v(3) = 3.128… ≈ 3.1 mi/h Maximum speed was 5 mi/h at about 1.12 h. Distance ≈ 0.6(150 + 230 + 150 + 90 + 40 + 0) = 396 ft Volume ≈ 3(2500 + 8000 + 12000 + 13000 + 11000 + 7000 + 4000 + 6000 + 4500) = 204,000 ft3 Programs will vary depending on calculator. See the program TRAPRULE in the Instructor’s Resource Book for an example. The program gives T20 = 23.819625. See the program TRAPDATA in the Instructor’s Resource Book for an example. The program gives T7 = 33, as in Example 2. a.

9.

f (x ) 7

x 1

4

b. T10 = 18.8955 T20 = 18.898875 T50 = 18.89982 These values underestimate the integral, because the trapezoids are inscribed in the region. c. T10: 0.0045 unit from the exact answer T20: 0.001125 unit from the exact answer T50: 0.00018 unit from the exact answer Tn is first within 0.01 unit of 18.9 when n = 7. T7 = 18.8908… , which is 0.0091… unit from 18.9. Because Tn is getting closer to 18.9 as n increases, Tn is within 0.01 unit of 18.9 for all n ≥ 7. 8. a.

10.

11.

12.

13.

b. T10 = 8.6700… T20 = 8.6596… Τ50 = 8.65672475… These values overestimate the integral, because the trapezoids are circumscribed about the region. c. T10: 0.01385… unit from answer T20: 0.003465… unit from answer T50: 0.0005545… unit from answer Tn is first within 0.01 of 8.65617024… when n = 12. T12 = 8.665795… , which is 0.009624… unit from 8.65617024… . Because Tn is getting closer to the exact answer as n increases, Tn is within 0.01 unit of the answer for all n ≥ 12. From the given equation, y = ±( 40/110) 110 2 – x 2 . Using the trapezoidal rule program on the positive branch with n = 100 increments gives 6904.190… for the top half of the ellipse. Doubling this gives an area of 13,808.38… cm 2 . The estimate is too low because the trapezoids are inscribed within the ellipse. The area of an ellipse is πab, where a and b are the x- and y-radii, respectively. So the exact area is π (110)(40) = 4400π = 13,823.007… cm 2 , which agrees both with the answer and with the conclusion that the trapezoidal rule underestimates the area. Integral = 1(0.0 + 2.1 + 7.9 + 15.9 + 23.8 + 29.7 + 31.8 + 29.7 + 23.8 + 15.9 + 7.9 + 2.1 + 0) = 190.6 The integral will have the units (in.2)(in.) = in.3, representing the volume of the football. n = 10: integral ≈ 21.045 n = 100: integral ≈ 21.00045 n = 1000: integral ≈ 21.0000045 Conjecture: integral = 21 The word is limit. The trapezoidal rule with n = 100 gives integral ≈ 156.0096. Conjecture: integral = 156 If the trapezoids are inscribed (graph concave down), the rule underestimates the integral. If the trapezoids are circumscribed (graph concave up), the rule overestimates the integral.

g (x )

x

1 1

Calculus Solutions Manual © 2005 Key Curriculum Press

3

Concave down Inscribed trapezoids Underestimates integral

Concave up Circumscribed trapezoids Overestimates integral

Problem Set 1-4

5

Problem Set 1-5 1. Answers will vary.

Problem Set 1-6

R3. By counting squares, the integral is approximately 23.2. Distance ≈ 23.2 ft (exact answer: 23.2422…) Concept: definite integral R4. a. f (x )

Review Problems

5

R1. a. When t = 4, d = 90 − 80 sin [1, 2(4 − 3)] ≈ 15.4 ft. b. From 3.9 to 4: average rate ≈ –40.1 ft/s From 4 to 4.1: average rate ≈ −29.3 ft/s Instantaneous rate ≈ −34.7 ft/s The distance from water is decreasing, so he is going down. d (5.01) − d ( 4.99) ≈ 70.8 0.02 d. Going up at about 70.8 ft/s e. Derivative

c. Instantaneous rate ≈

R2. a. Physical meaning: instantaneous rate of change of a function Graphical meaning: slope of a tangent line to a function at a given point b. x = − 4: decreasing fast x = 1: increasing slowly x = 3: increasing fast x = 5: neither increasing nor decreasing c. From 2 to 2.1: 52.1 − 52 average rate = = 43.6547K 0.1 From 2 to 2.01: 52.01 − 52 average rate = = 40.5614K 0.01 From 2 to 2.001: 52.001 − 52 average rate = = 40.2683K 0.001 Differences between average rates and instantaneous rates, respectively: 43.6547… − 40.235947… = 3.4187… 40.5617… − 40.235947… = 0.3255… 40.2683… − 40.235947… = 0.03239… The average rates are approaching the instantaneous rate as x approaches 2. The concept is the derivative. The concept used is the limit. d. t = 2: 3.25 m/s t = 18: 8.75 m/s t = 24: 11.5 m/s Her velocity stays constant, 7 m/s, from 6 s to 16 s. At t = 24, Mary is in her final sprint toward the finish line.

6

Problem Set 1-6

x 1

4

The graph agrees with Figure 1-6c. b. By counting squares, integral ≈ 15.0. (Exact answer is 15.) c. T 6 = 0.5(2.65 + 5.575 + 5.6 + 5.375 + 4.9 + 4.175 + 1.6) = 14.9375 The trapezoidal sum underestimates the integral because the trapezoids are inscribed in the region. d. T50 = 14.9991; Difference = 0.0009 T100 = 14.999775; Difference = 0.000225 The trapezoidal sums are getting closer to 15. Concept: limit R5. Answers will vary. Concept Problems C1. a. f (3) = 32 − 7.3 + 11 − 1 b. f (x) − f (3) = x2 − 7x + 11 + 1 = x 2 − 7x + 12 2 c. f ( x ) – f (3) = x – 7 x + 12 = ( x – 4)( x – 3) = x–3 x–3 x–3

x − 4, if x ≠ 3 d. The limit is found by substituting 3 for x in (x − 4). Limit = exact rate = 3 − 4 = −1 C2. The line through (3, f (3)) with slope −1 is y = −x + 2. f (x )

2

x 3

The line is tangent to the graph. Zooming in by a factor of 10 on the point (3, 2) shows that the graph becomes straighter and looks almost like the tangent line. (Soon students will learn that this property is called local linearity.)

Calculus Solutions Manual © 2005 Key Curriculum Press

T5. Concept: definite integral By counting squares, distance ≈ 466. (Exact answer is 466.3496… .) 2

T6. 3 Speed (ft/s)

4 x 2 − 19 x + 21 ( 4 x − 7)( x − 3) = = x −3 x −3 4x − 7, x ≠ 3 When x = 3, 4x − 7 = 4 ⋅ 3 − 7 = 5.

C3. a. f ( x ) =

25 20 15 10 5

b.

Time (s) 5

f (x ) (ft)

10

15

20

25

30

35

40

6 5.8 5

T 7 = 5(2.5 + 5 + 5 + 10 + 20 + 25 + 20 + 5) = 462.5 Trapezoidal rule probably underestimates the integral, but some trapezoids are inscribed and some circumscribed.

4.2 4 3 2

T7. Concept: derivative

1 2.8 1

2

x (s)

3.2 3

c. 5.8 = 4(3 + δ ) − 7 5.8 = 12 + 4δ − 7 4δ = 0.8 δ = 0.2 d. 4(3 + δ ) − 7 = 5 + ε 12 + 4δ − 7 = 5 + ε 4δ = ε δ = 14 ε

4

5

4.2 4.2 −4δ δ

6

= 4(3 − δ ) − 7 = 12 − 4δ − 7 = −0.8 = 0.2

There is a positive value of δ, namely 14 ε , for each positive value of ε, no matter how small ε is. e. L = 5, c = 3. “. . . but not equal to 3” is needed so that you can cancel the (x − 3) factors without dividing by zero. Chapter Test T1. Limit, derivative, definite integral, indefinite integral T2. See the text for the definition of limit. T3. Physical meaning: instantaneous rate T4. y 6

3

Speed (ft/s) 25 20 15 10 5 Time (s) 5

10

15

20

25

30

35

40

Slope ≈ −1.8 (ft/s)/s (Exact answer is −1.8137… .) Name: acceleration T8. The roller coaster is at the bottom of the hill at 25 s because that’s where it is going the fastest. The graph is horizontal between 0 and 10 seconds because the velocity stays constant, 5 ft/s, as the roller coaster climbs the ramp. T9. Distance = (rate)(time) = 5(10) = 50 ft T10. T5 = 412.5; T50 = 416.3118… ; T100 = 416.340219… T11. The differences between the trapezoidal sum and the exact sum are: For T5: difference = 3.8496… For T50: difference = 0.03779… For T100: difference = 0.009447… The differences are getting smaller, so Tn is getting closer to 416.349667… .

x 2

5

Calculus Solutions Manual © 2005 Key Curriculum Press

Problem Set 1-6

7

T12. From 30 to 31: y(31) − y(30) average rate = = −1.9098K 1 From 30 to 30.1: y(30.1) − y(30) = −1.8246K 0.1 From 30 to 30.01: y(30.01) − y(30) average rate = = −1.8148K 0.01 T13. The rates are negative because the roller coaster is slowing down. T14. The differences between the average rates and instantaneous rate are: For 30 to 31: difference = 0.096030… For 30 to 31.1: difference = 0.010833… For 30 to 30.01: difference = 0.001095… average rate =

8

Problem Set 1-6

The differences are getting smaller, so the average rates are getting closer to the instantaneous rate. y( x ) − y(30) = −1.81379936 + 1, getting x − 30 x = 30.092220… . So keep x within 0.092… unit of 30, on the positive side. T16. Concept: derivative f ( 4.3) − f (3.7) 35 − 29 T17. f ′( 4) ≈ = = 10 4.3 − 3.7 0.6 T18. Answers will vary. T15. Solve

Calculus Solutions Manual © 2005 Key Curriculum Press