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Coordinated Calculus

Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

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Appendix C Answers to Selected Exercises

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This appendix contains answers to all non- WeBWorK exercises in the text. For WeBWorK exercises, please use the HTML version of the text for access to answers and solutions.

0 PreCalculus Review
0.1 Functions
0.1.12 Exercises

0.1.12.1. Slope and Intercept.

Answer 1.
2
Answer 2.
−113

0.1.12.2. Graphs of Linear Equations.

Answer 1.
VI
Answer 2.
II
Answer 3.
I
Answer 4.
III
Answer 5.
IV
Answer 6.
V

0.1.12.3. Proportionality.

Answer 1.
h
Answer 2.
kh2

0.1.12.4. Finding Lines.

Answer.
2−43−−3(x+3)+4

0.2 Exponential and Logarithmic Functions
0.2.4 Exercises

0.2.4.1. General Exponential Functions.

Answer 1.
8
Answer 2.
e2
Answer 3.
growth

0.2.4.4. Half-Life.

Answer.
878.371 hr

0.2.4.5. Applied Half-Life.

Answer 1.
No
Answer 2.
337.461

0.3 Trigonometric Functions
0.3.6 Exercises

0.3.6.1. Period and Amplitude.

Answer 1.
2π3
Answer 2.
0.5

0.3.6.3. Finding Trigonometric Functions.

Answer.
6sin(6πx)+4

0.3.6.4. The Unit Circle.

Answer 1.
1⋅22−1⋅(2−2)2
Answer 2.
8cos(4)
Answer 3.
8sin(4)

0.3.6.5. Trigonometric Functions as Compositions.

Answer 1.
x4
Answer 2.
tan(x)
Answer 3.
tan(x)
Answer 4.
tan(x)
Answer 5.
tan(x)
Answer 6.
x4

1 Understanding the Derivative
1.1 Introduction to Continuity
1.1.5 Exercises

1.1.5.1. Types of discontinuity.

Answer.
2

1.1.5.2. Types of discontinuity.

Answer.
1

1.1.5.4. Determining continuity from a graph.

Answer 1.
is
Answer 2.
is not
Answer 3.
is
Answer 4.
is not
Answer 5.
is not
Answer 6.
is

1.1.5.5. Determining continuity from a graph.

Answer 1.
is not
Answer 2.
is
Answer 3.
is not

1.1.5.6. Interpretting continuity.

Answer 1.
7
Answer 2.
6
Answer 3.
7
Answer 4.
21
Answer 5.
7

1.1.5.7. Values that make a function continuous.

Answer.
5⋅4

1.1.5.8. Values that make a function continuous.

Answer.
−158

1.1.5.9. Values that make a function continuous.

Answer.
8⋅24

1.1.5.10. Application of the Intermediate Value Theorem.

Answer 1.
7
Answer 2.
2+3+7

1.2 Introduction to Limits
1.2.6 Exercises

1.2.6.1. Limits on a piecewise graph.

Answer 1.
5
Answer 2.
6
Answer 3.
none
Answer 4.
−5

1.2.6.2. Estimating a limit numerically.

Answer.
4

1.2.6.3. Limits for a piecewise formula.

Answer 1.
3⋅2−3
Answer 2.
3⋅2−3
Answer 3.
3⋅2−3

1.2.6.4. Calculating Limits of Rational Functions.

Answer.
0.2

1.2.6.5. One-Sided Limits.

Answer 1.
14
Answer 2.
−5
Answer 3.
DNE

1.2.6.6. Evaluating a limit algebraically.

Answer.
−14

1.3 How do we Measure Velocity?
1.3.4 Exercises

1.3.4.1. Average velocity from position.

Answer 1.
2.5 (ft(s
Answer 2.
4 (ft(s

1.3.4.2. Rate of calorie consumption.

Answer.
120

1.3.4.3. Average rate of change - quadratic function.

Answer 1.
−1
Answer 2.
−4
Answer 3.
−3
Answer 4.
D
Answer 5.
C
Answer 6.
B

1.3.4.4. Comparing average rate of change of two functions.

Answer 1.
g
Answer 2.
both have an equal rate of change
Answer 3.
f
Answer 4.
g
Answer 5.
g

1.3.4.5. Matching a distance graph to velocity.

Answer.
3

1.3.4.6. Interpretting average velocity and instantaneous velocity.

Answer.
  1. s(15)−s(0)≈−98.75.
  2. AV[0,15]=s(15)−s(0)15−0≈−6.58AV[0,2]=s(2)−s(0)2−0≈−47.63AV[1,6]=s(6)−s(1)6−1≈−13.25AV[8,10]=s(10)−s(8)10−8≈−7.35
  3. Most negative average velocity on [0,4]; most positive average velocity on [4,8].
  4. 21.31+22.252=21.78 feet per second.
  5. The average velocities are negative; the instantaneous velocity was positive. Downward motion corresponds to negative average velocity; upward motion to positive average velocity.

1.3.4.7. Graphing Velocity.

Answer.
  1. Sketch a plot where the diver’s height at time t is on the vertical axis. For instance, h(2.45)=0.
  2. AV[2.45,7]≈−3.5−07−2.45=−3.54.55=−0.7692 m/sec. The average velocity is not the same on every time interval within [2.45,7].
  3. When the diver is going upward, her velocity is positive. When she is going downward, her velocity is negative. At the peak of her dive and when her feet touch the bottom of the pool.
  4. It looks like when the position function is steep, the velocity function’s value is farther away from zero, and that whenever the height/position function is rising/increasing, the velocity function has a positive value. Similarly, whenever the position function is decreasing, the velocity is negative.

1.3.4.8. Population Growth Rate.

Answer.
  1. 15957 people.
  2. In an average year the population grew by about 798 people/year.
  3. The slope of a secant line through the points (a,f(a)) and (b,f(b)).
  4. AV[0,20]≈798 people per year.
  5. AV[5,10]≈734.50AV[5,9]≈733.06AV[5,8]≈731.62AV[5,7]≈730.19AV[5,6]≈728.7535

1.4 The Derivative of a Function at a Point
1.4.3 Exercises

1.4.3.1. Estimating derivative values graphically.

Answer 1.
0.905829
Answer 2.
−0.5
Answer 3.
3.22474
Answer 4.
0.355567
Answer 5.
0.75

1.4.3.2. Tangent line to a curve.

Answer 1.
5.4
Answer 2.
4.4
Answer 3.
5.4
Answer 4.
−1⋅(−0.02)0.07

1.4.3.3. Interpreting values and slopes from a graph.

Answer 1.
<
Answer 2.
<
Answer 3.
>
Answer 4.
>

1.4.3.4. Estimating a derivative value graphically.

Answer.
−90

1.4.3.5. Estimating a derivative from the limit definition.

Answer.
354.9

1.4.3.6. Using a graph.

Answer.
  2. AV[−3,−1]≈1.15; AV[0,2]≈−0.4.
  3. f′(−3)≈3; f′(0)≈−12.

1.4.3.7. Creating graphs with certain properties.

Answer.
  1. For instance, you could let f(−3)=3 and have f pass through the points (−3,3), (−1,−2), (0,−3), (1,−2), and (3,−1) and draw the desired tangent lines accordingly.
  2. For instance, you could draw a function g that passes through the points (−2,3), (−1,2), (1,0), (2,0), and (3,3) in such a way that the tangent line at (−1,2) is horizontal and the tangent line at (2,0) has slope 1.

1.4.3.8. Population Growth.

Answer.
  1. AV[0,7]=0.11757≈0.01679 billion people per year; P′(7)≈0.1762 billion people per year; P′(7)>AV[0,7].
  2. AV[19,29]≈0.02234 billion people/year.
  3. We will say that today’s date is July 1, 2015, which means that t=22.5;
    P′(22.5)=limh→0115(1.014)22.5+h−115(1.014)22.5h;
    P′(22.5)≈0.02186 billions of people per year.
  4. y−1.57236=0.02186(t−22.5).

1.4.3.9. Using the limit definition of the derivative.

Answer.
  1. All three approaches show that f′(2)=1.
  2. All three approaches show that f′(1)=−1.
  3. All three approaches show that f′(1)=12.
  4. All three approaches show that f′(1) does not exist.
  5. The first two approaches show that f′(π2)=0.

1.5 The Derivative Function
1.5.3 Exercises

1.5.3.2. The derivative function graphically.

Answer 1.
1
Answer 2.
−2
Answer 3.
0
Answer 4.
2

1.5.3.3. Applying the limit definition of the derivative.

Answer 1.
3(x+h)2−4−(3x2−4)
Answer 2.
3⋅2x

1.5.3.4. Sketching the derivative.

Answer.
1

1.5.3.5. Comparing function and derivative values.

Answer 1.
x6
Answer 2.
x3
Answer 3.
x5
Answer 4.
x2

1.5.3.6. Limit definition of the derivative for a rational function.

Answer 1.
−1
Answer 2.
−1
Answer 3.
−14
Answer 4.
−116

1.5.3.7. Determining functions from their derivatives.

Answer.
  1. See the figure below.
  2. See the figure below.
  3. One example of a formula for f is f(x)=12x2−1.

1.5.3.8. Algebraic and graphical connections between a function and its derivative.

Answer.
  1. g′(x)=2x−1.
  3. p′(x)=10x−4.
  4. The constants 3 and 12 don’t seem to affect the results at all. The coefficient −4 on the linear term in p(x) appears to make the ``−4’’ appear in p′(x)=10x−4. The leading coefficient 5 in (x)=5x2−4x+12 leads to the coefficient of ``10’’ in p′(x)=10x−4.

1.5.3.9. Graphing functions based on continuity and derivatives.

Answer.
  1. g is linear.
  2. On −3.5<x<−2, −2<x<0 and 2<x<3.5.
  3. At x=−2,0,2; g must have sharp corners at these points.

1.5.3.10. Graphing the Derivative Function.

Answer.

1.6 Interpreting, Estimating, and Using the Derivative
1.6.5 Exercises

1.6.5.1. A cooling cup of coffee.

Answer 1.
negative
Answer 2.
degC/min
Answer 3.
30
Answer 4.
temperature
Answer 5.
51 degC
Answer 6.
decrease
Answer 7.
1.125 degC

1.6.5.2. A cost function.

Answer 1.
gallons
Answer 2.
dollars
Answer 3.
gallons
Answer 4.
dollars/gallon

1.6.5.3. Weight as a function of calories.

Answer 1.
cal
Answer 2.
lb
Answer 3.
cal
Answer 4.
lb/cal
Answer 5.
lb
Answer 6.
cal
Answer 7.
lb/cal
Answer 8.
2600
Answer 9.
0.02

1.6.5.4. Displacement and velocity.

Answer 1.
3
Answer 2.
3.5
Answer 3.
5
Answer 4.
4.5
Answer 5.
4

1.6.5.5. Another cup of coffee.

Answer.
  1. F′(10)≈−3.33592.
  2. The coffee’s temperature is decreasing at about 3.33592 degrees per minute.
  3. F′(20).
  4. We expect F′ to get closer and closer to 0 as time goes on.

1.6.5.6. Body temperature.

Answer.
  1. If a patient takes a dose of 50 ml of a drug, the patient will experience a body temperature change of 0.75 degrees F.
  2. ``degrees Fahrenheit per milliliter.’’
  3. For a patient taking a 50 ml dose, adding one more ml to the dose leads us to expect a temperature change that is about 0.02 degrees less than the temperature change induced by a 50 ml dose.

1.6.5.7. Tossing a ball.

Answer.
  1. t=0.
  2. v′(1)=−32.
  3. ``feet per second per second’’; v′(1)=−32 tells us that the ball’s velocity is decreasing at a rate of 32 feet per second per second.
  4. The acceleration of the ball.

1.6.5.8. Value of a car.

Answer.
  1. AV[40000,55000]≈−0.153 dollars per mile.
  2. h′(55000)≈−0.147 dollars per mile. During 550001st mile, we expect the car’s value to drop by 0.147 dollars.
  3. h′(30000)<h′(80000).
  4. The graph of h might have the general shape of the graph of y=e−x for positive values of x: always positive, always decreasing, and bending upwards while tending to 0 as x increases.

1.7 The Second Derivative
1.7.6 Exercises

1.7.6.1. Comparing f,f′,f″ values.

Answer 1.
positive
Answer 2.
positive
Answer 3.
negative

1.7.6.2. Signs of f,f′,f″ values.

Answer 1.
zero
Answer 2.
negative
Answer 3.
zero
Answer 4.
negative
Answer 5.
negative
Answer 6.
negative
Answer 7.
zero
Answer 8.
positive
Answer 9.
zero
Answer 10.
positive
Answer 11.
positive
Answer 12.
positive
Answer 13.
positive
Answer 14.
positive
Answer 15.
positive

1.7.6.3. Acceleration from velocity.

Answer 1.
30 (ft(s2
Answer 2.
22 (ft(s2

1.7.6.4. Rates of change of stock values.

Answer 1.
positive
Answer 2.
negative
Answer 3.
positive
Answer 4.
positive

1.7.6.5. Interpreting a graph of f′.

Answer 1.
x6
Answer 2.
x1
Answer 3.
x4
Answer 4.
x2
Answer 5.
x3
Answer 6.
x1

1.7.6.6. Interpretting a graph of f based on the first and second derivatives.

Answer.
  1. f is increasing and concave down at x=2.
  2. Greater.
  3. Less.

1.7.6.7. Interpreting a graph of f′.

Answer.
  1. g′(2)≈1.4.
  2. At most one.
  3. 9.
  4. g″(2)≈5.5.

1.7.6.8. Using data to interpret derivatives.

Answer.
  1. h′(4.5)≈14.3; h′(5)≈21.2; h′(5.5)≈=23.9; rising most rapidly at t=5.5.
  2. h′(5)≈9.6.
  3. Acceleration of the bungee jumper in feet per second per second.
  4. 0<t<2, 6<t<10.

1.7.6.9. Sketching functions.

Answer.

1.8 Differentiability
1.8.6 Exercises

1.8.6.1. Continuity and differentiability of a graph.

Answer 1.
none
Answer 2.
2,4,6

1.8.6.2. Continuity and differentiability of a graph.

Answer.
  1. a=0.
  2. a=0,3.
  3. a=−2,0,1,2,3.

1.8.6.3. Examples of functions.

Answer.
  1. f(x)=|x−2|.
  2. Impossible.
  3. Let f be the function defined to be f(x)=1 for every value of x≠−2, and such that f(−2)=4.

1.8.6.4. Estimating the derivative at a point.

Answer.
  1. At x=0.
    g′(0)=limh→0g(0+h)−g(0)h=limh→0|h|−|0|h=limh→0|h|h
  2. h 0.1 0.01 0.001 0.0001 −0.1 −0.01 −0.001 −0.0001
    |h|/h 3.162 10 31.62 100 −3.162 −10 −31.62 −100
    g′(0) does not exist.

2 Computing Derivatives
2.1 Elementary Derivative Rules
2.1.5 Exercises

2.1.5.1. Derivative of a power function.

Answer.
1516x1516−1

2.1.5.2. Derivative of a rational function.

Answer.
−1⋅19x−1⋅(19+1)

2.1.5.3. Derivative of a root function.

Answer.
12x12−1

2.1.5.4. Derivative of a quadratic function.

Answer.
2⋅3t−6

2.1.5.5. Derivative of a sum of power functions.

Answer.
6⋅4t4−1−52t−12−14t2

2.1.5.6. Simplifying a product before differentiating.

Answer.
2⋅3+12x2⋅3−12+82x−12

2.1.5.7. Simplifying a quotient before differentiating.

Answer.
6x5x−(x6+8)x2

2.1.5.8. Finding a tangent line equation.

Answer.
32(x−2)+18

2.1.5.9. Determining where f′(x)=0.

Answer.
8,−12

2.1.5.10. Calculating derivative values from two tables.

Answer.
  1. h(2)=27; h′(2)=−19/2.
  2. L(x)=27−192(x−2).
  3. p is increasing at x=2.
  4. p(2.03)≈−11.56.

2.1.5.11. Calculating derivative values from two graphs.

Answer.
  1. p is not differentiable at x=−1 and x=1; q is not differentiable at x=−1 and x=1.
  2. r is not differentiable at x=−1 and x=1.
  3. r′(−2)=4; r′(0)=1.
  4. y=4.

2.1.5.12. Using the sum and constant multiple rules.

Answer.
  1. w′(t)=3tt(ln⁡(t)+1)+211−t2.
  2. L(t)=(32−2π3)+(32(ln⁡(12)+1)+43)(t−2).
  3. v is increasing at t=12.

2.1.5.13. Understanding the derivative of an exponential function.

Answer.
  1. f′(x)=limh→0ax+h−axh=limh→0ax⋅ah−axh=limh→0ax(ah−1)h,
  2. Since ax does not depend at all on h, we may treat ax as constant in the noted limit and thus write the value ax in front of the limit being taken.
  3. When a=2, L≈0.6931; when a=3, L≈1.0986.
  4. a≈2.71828 (for which L≈1.000)
  5. ddx[2x]=2x⋅ln⁡(2) and ddx[3x]=3x⋅ln⁡(3)
  6. ddx[ex]=ex.

2.2 The Sine and Cosine Functions
2.2.3 Exercises

2.2.3.1. Compute the Derivative.

Answer.
7cos(x)−6sin(x)

2.2.3.2. Compute the Derivative.

Answer.
2⋅9x+2sin(x)

2.2.3.3. Derivative Computation and Evaluation.

Answer 1.
2cos(x)−10sin(x)
Answer 2.
2cos(5.23599)−10sin(5.23599)

2.2.3.4. Find the Tangent Line.

Answer 1.
232
Answer 2.
1−1.73205⋅0.523599

2.2.3.5. Find the Tangent Line.

Answer.
x+−4

2.2.3.6. Finding Tangent and Normal Lines.

Answer 1.
y =
Answer 2.
4
Answer 3.
x =
Answer 4.
π2

2.2.3.7. Horizontal Tangent Lines.

Answer.
2π3,4π3

2.2.3.8. Making Two Graphs Tangent For Specific Values.

Answer 1.
sin(3π4)e3π4
Answer 2.
3π4
Answer 3.
12

2.2.3.9. An Elastic Band.

Answer 1.
−2sin(t)+3cos(t)
Answer 2.
2.55
Answer 3.
3.61

2.2.3.10. Analyzing the Value of an Investment.

Answer.
  1. V′(2)=24⋅1.072⋅ln⁡(1.07)+6cos⁡(2)≈−0.63778 thousands of dollars per year.
  2. V″(2)=24⋅1.072⋅ln⁡(1.07)2−6sin⁡(2)≈−5.33 thousands of dollars per year per year. At this moment, V′ is decreasing and we expect the derivative’s value to decrease by about 5.33 thousand dollars per year over the course of the next year.
  3. See the figure below. Adding the term 6sin⁡(t) to A to create the function V adds volatility to the value of the portfolio.

2.2.3.11. Differentiating Sine and Cosine.

Answer.
  1. f′(π4)=−5(22) .
  2. L(x)=3+2(x−π).
  3. Decreasing.
  4. The tangent line to f lies above the curve at this point.

2.2.3.12. Understanding the Derivatives of Sine and Cosine.

Answer.
  1. Hint: in the numerator of the difference quotient, combine the first and last terms and remove a factor of sin⁡(x) .
  2. Hint: divide each part of the numerator by h and consider the sum of two separate limits.
  3. limh→0(cos⁡(h)−1h)=0 and limh→0(sin⁡(h)h)=1 .
  4. f′(x)=sin⁡(x)⋅0+cos⁡(x)⋅1.
  5. Hint: cos⁡(α+β) is cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β).

2.3 The Product and Quotient Rules
2.3.6 Exercises

2.3.6.1. Derivative of a basic product.

Answer.
13x+xln(13)⋅13x

2.3.6.2. Derivative of a product.

Answer.
(9x9−1−19x−1+19)⋅6x+(x9−x19)ln(6)⋅6x

2.3.6.3. Derivative of a quotient of linear functions.

Answer.
2(6t+15)−6(2t+7)(6t+15)2

2.3.6.4. Derivative of a rational function.

Answer.
3r3−1(5r+6)−5r3(5r+6)2

2.3.6.5. Derivative of a product of trigonometric functions.

Answer.
6(−1sin(q)sin(q)+cos(q)cos(q))

2.3.6.6. Derivative of a product of power and trigonmetric functions.

Answer.
5x5−1cos(x)−x5sin(x)

2.3.6.7. Derivative of a sum that involves a product.

Answer.
sin(t)+tcos(t)+1cos2(t)

2.3.6.8. Product and quotient rules with graphs.

Answer 1.
0.666667⋅1.33333+0⋅1.33333
Answer 2.
−4⋅(−1)−−1⋅(−2)(−1)2

2.3.6.9. Product and quotient rules with given function values.

Answer 1.
3⋅4+3⋅4
Answer 2.
3⋅4−3⋅44⋅4

2.3.6.10. Tangent lines using given function values.

Answer.
  1. h(2)=−15; h′(2)=23/2.
  2. L(x)=−15+23/2(x−2).
  3. Increasing.
  4. r(2.06)≈−0.5796.

2.3.6.11. Product and quotient rule with non-basic functions.

Answer.
  1. w′(t)=tt(ln⁡t+1)⋅(arccos⁡t)+tt⋅−11−t2.
  2. L(t)≈0.740−0.589(t−0.5).
  3. Increasing.

2.3.6.12. Product and quotient rules with analysis of graphs.

Answer.
  1. r′(−2)=2 and r′(0)=0.25.
  2. At x=−1 and x=1.
  3. L(x)=2.
  4. z′(0)=116 and z′(2)=−1.
  5. At x=−1, x=1, x=−1.5, and x=1.

2.3.6.13. An application to crop yield.

Answer.
  1. C(t)=A(t)Y(t) bushels in year t.
  2. 1190000 bushels of corn.
  3. C′(t)=A(t)Y′(t)+A′(t)Y(t).
  4. C′(0)=158000 bushels per year.
  5. C(1)≈1348000bushels.

2.3.6.14. An application to fuel consumption.

Answer.
  1. g(80)=20 kilometers per liter, and g′(80)=−0.16. kilometers per liter per kilometer per hour.
  2. h(80)=4 liters per hour and h′(80)=0.082 liters per hour per kilometer per hour.
  3. Think carefully about units and how each of the three pairs of values expresses fundamentally the same facts.

2.4 Derivatives of Other Trigonometric Functions
2.4.3 Exercises

2.4.3.1. A sum and product involving tan⁡x.

Answer.
tan(t)+t1cos2(t)+−1sin(t)

2.4.3.2. A quotient involving tan⁡t.

Answer 1.
5sec2(x)x−5tan(x)x2
Answer 2.
17.1673

2.4.3.3. A quotient of trigonometric functions.

Answer 1.
sec2(x)sec(x)−(tan(x)−2)sec(x)tan(x)sec2(x)
Answer 2.
−2.16725

2.4.3.4. A quotient that involves a product.

Answer 1.
(2⋅2xtan(x)+2x2sec2(x))sec(x)−2x2tan(x)sec(x)tan(x)sec2(x)
Answer 2.
4.44649

2.4.3.5. Finding a tangent line equation.

Answer 1.
3⋅2
Answer 2.
3−6⋅0.785398

2.4.3.6. Oscillatory Motion.

Answer.
  1. h′(2)=−2sin⁡(2)−2cos⁡(2)ln⁡(1.2)1.22≈−1.1575 feet per second.
  2. h″(2)=cos⁡(2)(−2+2ln2(1.2))+4ln⁡(1.2)sin⁡(2))1.22≈1.0193 feet per second per second.
  3. The object is falling and slowing down.

2.4.3.7. A product of trigonometric functions.

Answer.
  1. f′(x)=sin⁡(x)⋅(−csc2⁡(x))+cot⁡(x)⋅cos⁡(x).
  2. False.
  3. f′(x)=−sin2⁡(x)sin⁡(x)=−sin⁡(x) for x≠π2+kπ for some integer value of k.

2.4.3.8. Combining Differentiation Rules.

Answer.
  1. p′(z)=(z2sec⁡(z)+1)(zsec2⁡(z)+tan⁡(z))−ztan⁡(z)(z2sec⁡(z)tan⁡(z)+2zsec⁡(z))(z2sec⁡(z)+1)2+3ez
  2. y−4=3(x−0).
  3. Increasing.

2.5 The Chain Rule
2.5.5 Exercises

2.5.5.1. Mixing rules: chain, product, sum.

Answer.
5e5x(x2+5x)+e5x(2x+ln(5)⋅5x)

2.5.5.2. Mixing rules: chain and product.

Answer.
6t6−1e−1ct−ct6e−1ct

2.5.5.3. Using the chain rule repeatedly.

Answer.
−1⋅5te−1⋅5t2e−1⋅5t2+7

2.5.5.4. Derivative involving arbitrary constants a and b.

Answer.
ae−1bx+12−abxe−1bx+12

2.5.5.5. Chain rule with graphs.

Answer 1.
−0.5
Answer 2.
−0.5
Answer 3.
−0.5

2.5.5.6. Chain rule with function values.

Answer 1.
1
Answer 2.
6⋅5
Answer 3.
5
Answer 4.
7⋅2
Answer 5.
2⋅3−1⋅532

2.5.5.7. A product involving a composite function.

Answer.
2sin(4x)+2⋅4xcos(4x)

2.5.5.8. Using the chain rule to compare composite functions.

Answer.
  1. h′(π4)=322.
  2. r′(0.25)=cos⁡(0.253)⋅3(0.25)2≈0.1875>h′(0.25)=3sin2⁡(0.25)⋅cos⁡(0.25)≈0.1779; r is changing more rapidly.
  3. h′(x) is periodic; r′(x) is not.

2.5.5.9. Chain rule with an arbitrary function u.

Answer.
  1. p′(x)=eu(x)⋅u′(x).
  2. q′(x)=u′(ex)⋅ex.
  3. r′(x)=−csc2⁡(u(x)⋅u′(x).
  4. s′(x)=u′(cot⁡(x))⋅(−csc2⁡(x)).
  5. a′(x)=u′(x4)⋅4x3.
  6. b′(x)=4(u(x))3⋅u′(x).

2.5.5.10. More on using the chain rule with graphs.

Answer.
  1. C′(0)=0 and C′(3)=−12.
  2. Consider C′(1). By the chain rule, we’d expect that C′(1)=p′(q(1))⋅q′(1), but we know that q′(1) does not exist since q has a corner point at x=1. This means that C′(1) does not exist either.
  3. Since Y(x)=q(q(x)), the chain rule implies that Y′(x)=q′(q(x))⋅q′(x), and thus Y′(−2)=q′(q(−2))⋅q′(−2)=q′(−1)⋅q′(−2). But q′(−1) does not exist, so Y′(−2) also fails to exist. Using Z(x)=q(p(x)) and the chain rule, we have Z′(x)=q′(p(x))⋅p′(x). Therefore Z′(0)=q′(p(0))⋅p′(0)=q′(−0.5)⋅p′(0)=0⋅0.5=0.

2.5.5.11. Applying the chain rule in a physical context.

Answer.
  1. dVdh=π(8h−h2), cubic feet per foot.
  2. dVdt=π3[12⋅2(sin⁡(πt)+1)⋅πcos⁡(πt)−3(sin⁡(πt)+1)2⋅πcos⁡(πt)] cubic feet per hour.
  3. dVdt|t=0=7π2 cubic feet per hour.
  4. In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time.

2.6 Derivatives of Inverse Functions
2.6.6 Exercises

2.6.6.1. Composite function involving logarithms and polynomials.

Answer.
3t2t3+9

2.6.6.2. Composite function involving trigonometric functions and logarithms.

Answer.
−1sin(ln(t))t

2.6.6.3. Product involving arcsin⁡(w).

Answer.
5sin−1(w)+5w1−w2

2.6.6.4. Derivative involving arctan⁡(x).

Answer.
0

2.6.6.5. Composite function from a graph.

Answer 1.
4⋅54−1(5.1−5)2.3−2
Answer 2.
2.3−25.1−5

2.6.6.6. Composite function involving an inverse trigonometric function.

Answer.
711−(x2)2⋅2x

2.6.6.7. Mixing rules: product, chain, and inverse trig.

Answer.
8⋅3x3−1tan−1(9x3)+8x3⋅9⋅3x3−11+9⋅9x2⋅3

2.6.6.8. Mixing rules: product and inverse trig.

Answer.
8cos(x)11−x2−8sin(x)sin−1(x)

2.6.6.9. Mixing rules: chain, product, logarithms, and inverse trig.

Answer.
  1. f′(x)=12arctan⁡(x)+3arcsin⁡(x)+5⋅(21+x2+31−x2).
  2. r′(z)=11+(ln⁡(arcsin⁡(z)))2⋅(1arcsin⁡(z))⋅11−z2.
  3. q′(t)=arctan2⁡(3t)⋅[4arcsin3⁡(7t)(71+(7t)2)]+arcsin4⁡(7t)⋅[2arctan⁡(3t)(31+(3t)2)].
  4. g′(v)=1arctan⁡(v)arcsin⁡(v)+v2⋅(arcsin⁡(v)+v2)⋅11+v2−arctan⁡(v)⋅(11−v2+2v)(arcsin⁡(v)+v2)2

2.6.6.10. Graphs of inverse functions.

Answer.
  1. f′(1)≈2.
  3. f′(−1)≈1/2.

2.6.6.11. Differentiating the inverse of a cubic polynomial.

Answer.
  1. f passes the horizontal line test.
  2. f−1(x)=g(x)=4x−163.
  3. f′(x)=34x2; f′(2)=3. g′(x)=13(4x−16)−2/3⋅4; g′(6)=13. These two derivative values are reciprocals.

2.6.6.12. Using a graph when a formula is unavailable.

Answer.
  1. h passes the horizontal line test.
  2. The equation y=x+sin⁡(x) can’t be solved for x in terms of y.
  3. (h−1)′(π2+1)=1.

2.7 Derivatives of Functions Given Implicitly
2.7.3 Exercises

2.7.3.1. Implicit differentiation in a polynomial equation.

Answer.
1−2xyx2−3

2.7.3.2. Implicit differentiation in an equation with logarithms.

Answer.
y(8−xln(y))x(x+3y3)

2.7.3.3. Implicit differentiation in an equation with inverse trigonometric functions.

Answer.
3x3−1y−y3−x2⋅3y3+23xy3−1+3x2⋅3+1y3+1−x3

2.7.3.4. Slope of the tangent line to an implicit curve.

Answer.
−76

2.7.3.5. Equation of the tangent line to an implicit curve.

Answer.
4x+63y=99

2.7.3.6. Finding horizontal and vertical tangency with implicit differentiation.

Answer.
Horizontal tangent lines: (0,−1), (0,−0.618), (0,1.618), (1,−1), (1,−0.618), (1,1.618), (0.5,−1.0493), (0.5,0.2104), (0.5,1.6139). Vertical tangent lines: (−0.1756,−0.379), (0.2912,−0.379), (0.7088,−0.379), (1.1756,−0.379), (−0.8437,1.235), and (1.8437,1.235).

2.7.3.7. Equation of the tangent line to an implicit trigonometric curve.

Answer.
y=π2−(x−π2).

2.7.3.8. Revisiting exponential derivatives using implicit differentiation.

Answer.
  1. x=ln⁡(y)ln⁡(a).
  2. 1=1ln⁡(a)⋅1ydydx.
  3. ddx[ax]=axln⁡(a).

2.8 Derivatives of Hyperbolic Functions
2.8.6 Exercises

2.8.6.1. Simplifying Hyperbolic Trigonometric Functions.

Answer.
t+1t2

2.8.6.2. Limits of Hyperbolic Functions.

Answer.
2

2.8.6.8. Derivatives of Hyperbolic Functions.

Answer.
−6sinh(x)(2+cosh(x))2

2.8.6.9. Derivatives of Hyperbolic Functions.

Answer.
0

2.9 The Tangent Line Approximation
2.9.4 Exercises

2.9.4.1. Approximating x.

Answer 1.
0.0714285714285714
Answer 2.
3.5
Answer 3.
7.01428571428571

2.9.4.2. Local Linearization of a Graph.

Answer 1.
4
Answer 2.
5
Answer 3.
6
Answer 4.
under

2.9.4.3. Estimating With the Local Linearization.

Answer.
68

2.9.4.4. Predicting Behavior From the Local Linearization.

Answer 1.
negative
Answer 2.
degC/min
Answer 3.
35
Answer 4.
temperature
Answer 5.
59 degC
Answer 6.
decrease
Answer 7.
0.3 degC

2.9.4.5. Using the Local Linearization to Analyze a Function.

Answer.
  1. p(3)=−1 and p′(3)=−2.
  2. p(2.79)≈−0.58.
  3. Too large.

2.9.4.6. Using the Local Linearization with Physical Context.

Answer.
  1. F′(60)≈1.56 degrees per minute.
  2. L(t)≈1.56(t−60)+324.5.
  3. F(63)≈L(63)≈=329.18 degrees F.
  4. Overestimate.

2.9.4.7. Local Linearity and the Position of a Moving Object.

Answer.
  1. s(9.34)≈L(9.34)=3.592.
  2. underestimate.
  3. The object is slowing down as it moves toward toward its starting position at t=4.

2.9.4.8. Estimating a Function Through its Derivative.

Answer.
  1. x=1.
  2. On −0.37<x<1.37; f is concave up.
  3. f(1.88)≈0.02479, and this estimate is larger than the true value of f(1.88).

2.10 The Mean Value Theorem
2.10.3 Exercises

2.10.3.1. Understanding the Statement of the Mean Value Theorem.

Answer 1.
no
Answer 2.
no
Answer 3.
there is no such point

2.10.3.2. Applying Theorems.

Answer 1.
False
Answer 2.
True
Answer 3.
False
Answer 4.
False
Answer 5.
True

2.10.3.3. Conclusion of the Mean Value Theorem.

Answer.
8

2.10.3.4. Conclusion of the Mean Value Theorem.

Answer.
2.04771

2.10.3.5. Applying the Mean Value Theorem.

Answer.
−15.5

2.10.3.6. Applying the Mean Value Theorem.

Answer.
−2.4641,4.4641

2.10.3.7. Applying the Mean Value Theorem.

Answer 1.
−31
Answer 2.
−2.38372673257701,5.38372673257701

3 Using Derivatives
3.1 Using Derivatives to Identify Extreme Values
3.1.4 Exercises

3.1.4.1. Finding critical points and inflection points.

Answer 1.
0
Answer 2.
−0.25,0.25
Answer 3.
0
Answer 4.
−0.25,0.25

3.1.4.2. Finding inflection points.

Answer.
−3,0.5

3.1.4.3. Matching graphs of f,f′,f″.

Answer 1.
A
Answer 2.
B
Answer 3.
C

3.1.4.4. Using a derivative graph to analyze a function.

Answer.
  1. f′ is positive for −1<xlt1 and for x>1; f′ is negative for all x<−1. f has a local minimum at x=−1.
  2. A possible graph of y=f″(x) is shown at right in the figure.
  3. f″(x) is negative for −0.35<x<1; f″(x) is positive everywhere else; f has points of inflection at x≈−0.35 and x=1.
  4. A possible graph of y=f(x) is shown at left in the figure.

3.1.4.5. Using derivative tests.

Answer.
  1. Neither.
  2. g″(2)=0; g″ is negative for 1<x<2 and positive for 2<x<3.
  3. g has a point of inflection at x=2.

3.1.4.6. Using a derivative graph to analyze a function.

Answer.
  1. h can have no, one, or two real zeros.
  2. One root is negative and the other positive.
  3. h will look like a line with slope 3.
  4. h is concave up everywhere; h is almost linear for large values of |x|.

3.1.4.7. Applying derivative tests.

Answer.
  1. p″(x) is negative for −1<x<2 and positive for all other values of x; p has points of inflection at x=−1 and x=2.
  2. Local maximum.
  3. Neither.

3.2 Global Optimization
3.2.4 Exercises

3.2.4.1. Finding Global Extrema.

Answer 1.
−70
Answer 2.
442

3.2.4.2. Finding Global Extrema.

Answer 1.
−254
Answer 2.
3971

3.2.4.3. Analyzing Function Behavior.

Answer 1.
−1
Answer 2.
(−∞,−1)
Answer 3.
(−1,∞)
Answer 4.
5
Answer 5.
2
Answer 6.
2
Answer 7.
Undefined

3.2.4.5. Conditions for When Global Extrema May Occur.

Answer.
  1. Not enough information is given.
  2. Global minimum at x=a.
  3. Global minimum at x=a; global maximum at x=b.
  4. Not enough information is provided.

3.2.4.6. Finding Extrema on Closed and Bounded Intervals.

Answer.
  1. Global maximum p(0)=p(a)=0; global minimum p(a3)=−2a333.
  2. Global max r(1b)≈0.368ab; global min r(2b)≈0.270ab.
  3. Global minimum g(b)=a(1−e−b2); global maximum g(3b)=a(1−e−3b2).
  4. Global max s(π2k)=1; global min s(5π6k)=12.

3.2.4.7. Conditions for Where Global Extrema May Occur.

Answer.
  1. Global maximum at x=a; global minimum at x=b.
  2. Global maximum at x=c; global minimum at either x=a or x=b.
  3. Global minimum at x=a and x=b; global maximum somewhere in (a,b).
  4. Global minimum at x=c; global maximum value at x=a.

3.2.4.8. Using the Extreme Value Theorem.

Answer.
  1. Global max s(5π12)=8; global min s(11π12)=2.
  2. Global max s(5π12)=8; global min s(0)=5−332≈2.402.
  3. Global max s(5π12)=8; global min s(11π12)=2. (There are other points at which the function achieves these values on the given interval.)
  4. Global max s(5π12)=8; global min s(π3)=6.5.

3.3 Applied Optimization
3.3.3 Exercises

3.3.3.1. Maximizing the volume of a box.

Answer 1.
10.1633×6.16333
Answer 2.
1.91833

3.3.3.2. Minimizing the cost of a container.

Answer.
$429.74

3.3.3.3. Maximizing area contained by a fence.

Answer.
9922.5 ft2

3.3.3.4. Minimizing the area of a poster.

Answer 1.
13.8489 cm
Answer 2.
55.3954 cm

3.3.3.5. Maximizing the area of a rectangle.

Answer 1.
1.15
Answer 2.
0.66667

3.3.3.6. Maximizing the volume of a closed box.

Answer.
The absolute maximum volume is V(53)=1512(53)−14(53)3≈1.07583 cubic feet.

3.3.3.7. Maximizing pasture area with limited fencing.

Answer.
Exercise Answer

3.3.3.8. Minimizing cable length.

Answer.
172.047 feet of cable.

3.3.3.9. Minimizing construction costs.

Answer.
The minimum cost is $1165.70.

3.4 Using Derivatives to Describe Families of Functions
3.4.3 Exercises

3.4.3.1. Drug dosage with a parameter.

Answer 1.
C
Answer 2.
C2

3.4.3.2. Using the graph of g′.

Answer 1.
1
Answer 2.
−2
Answer 3.
2
Answer 4.
(−10,∞)
Answer 5.
>

3.4.3.3. Using the graph of f.

Answer 1.
T
Answer 2.
F
Answer 3.
T
Answer 4.
T
Answer 5.
T
Answer 6.
1
Answer 7.
4

3.4.3.4. Sign Change.

Answer.
1

3.4.3.5. Critical and inflection points of a function with parameters.

Answer 1.
a+b2
Answer 2.
DNE

3.4.3.6. Behavior of a function with parameters.

Answer 1.
a+ln(bx)+1
Answer 2.
increasing
Answer 3.
1x
Answer 4.
up

3.4.3.7. Analyzing and curve sketching.

Answer 1.
0
Answer 2.
0
Answer 3.
0
Answer 4.
0
Answer 5.
1
Answer 6.
1
Answer 7.
0
Answer 8.
0
Answer 9.
1,2,3
Answer 10.
−∞

3.4.3.8. Analyzing families of functions.

Answer.
  2. x=0 and x=2a3.
  3. x=a2; p″(x) changes sign from negative to positive at x=a2.
  4. As we increase the value of a, both the location of the critical number and the inflection point move to the right along with a.

3.4.3.9. Analyzing families of functions.

Answer.
  1. x=c is a vertical asymptote because limx→c+e−xx−c=∞ and limx→c−e−xx−c=−∞.
  2. limx→∞e−xx−c=0; limx→−∞e−xx−c=−∞.
  3. The only critical number for q is x=c−1.
  4. When x<c−1, q′(x)>0; when x>c−1, q′(x)<0; q has a local maximum at x=c−1.

3.4.3.10. Analyzing families of functions.

Answer.
  1. x=m.
  2. E is increasing for x<m and decreasing for x>m, with a local maximum at x=m.
  3. x=m±s.
  4. limx→∞E(x)=limx→−∞E(x)=0.

3.5 Related Rates
3.5.3 Exercises

3.5.3.1. Height of a conical pile of gravel.

Answer.
2572π

3.5.3.2. Movement of a shadow.

Answer.
15

3.5.3.3. A leaking conical tank.

Answer.
295665.6697

3.5.3.4. Docking a boat.

Answer.
The boat is approaching the dock at a rate of 136≈2.167 feet per second.

3.5.3.5. Filling a swimming pool.

Answer.
The depth of the water is increasing at
dhdt|h=5=1.28
feet per minute. The depth of the water is increasing at a decreasing rate.

3.5.3.6. Baseball player and umpire.

Answer.
dθdt|x=30=−0.24 radians per second.

3.5.3.7. A conical pile of sand.

Answer.
dhdt|V=1000=10π(3000π3)2≈0.0328 feet per minute.

3.6 Using Derivatives to Evaluate Limits
3.6.5 Exercises

3.6.5.1. L’Hopital’s Rule with graphs.

Answer 1.
undefined
Answer 2.
zero

3.6.5.2. L’Hopital’s Rule to evaluate a limit.

Answer.
118

3.6.5.3. Determining if L’Hopital’s Rule applies.

Answer 1.
1.5625
Answer 2.
1.25100605884547

3.6.5.4. Using L’Hopital’s Rule multiple times.

Answer.
0

3.6.5.5. Using L’Hopital’s Rule multiple times.

Answer.
limx→3h(x)=−2.

3.6.5.6. Analyzing a family of functions.

Answer.
Horizontal asymptote: y=35; vertical asymptote: x=c; hole: (a,3(a−b)5(a−c)). R is not continuous at x=a and x=c.

3.6.5.7. An algebraic trick to use L’Hopital’s Rule.

Answer.
  1. ln⁡(x2x)=2x⋅ln⁡(x).
  2. x=11x.
  3. limx→0+h(x)=0.
  4. limx→0+g(x)=limx→0+x2x=1.

3.6.5.8. Dominance.

Answer.
  1. Show that limx→∞ln⁡(x)x=0.
  2. Show that limx→∞ln⁡(x)xn=0.
  3. Consider limx→∞p(x)ex By repeated application of LHR, the numerator will eventually be simply a constant (after n applications of LHR), and thus with ex still in the denominator, the overall limit will be 0.
  4. Show that limx→∞ln⁡(x)xn=0
  5. For example, f(x)=3x2+1 and g(x)=−0.5x2+5x−2.

3.7 Parametric Equations
3.7.4 Exercises

3.7.4.1. Evaluating Parametric Equations.

Answer 1.
(5,7)
Answer 2.
(5,455)
Answer 3.
(5,2408)

3.7.4.3. Parametric Lines.

Answer 1.
N
Answer 2.
Y
Answer 3.
Y
Answer 4.
(8,4)
Answer 5.
(−20,−12)
Answer 6.
(−13,−8)
Answer 7.
(−∞,−1.14286)

3.7.4.4. Slope in Parametric Equations.

Answer 1.
2
Answer 2.
to the right
Answer 3.
up
Answer 4.
−2
Answer 5.
to the left
Answer 6.
up

3.7.4.5. Derivatives of Parametric Equations.

Answer 1.
sin2(t)+cos2(t2)4 (ft(s
Answer 2.
3.14159 s
Answer 3.
nπ s

3.7.4.6. Tangent Lines and Parametric Equations.

Answer 1.
16+12(t−2)
Answer 2.
48+72(t−2)

3.7.4.7. Horizontal Tangent Lines and Parametric Equations.

Answer 1.
3
Answer 2.
8

3.7.4.8. Speed and Parametric Equations.

Answer.
192

3.7.4.9. Calculating Speed Using Parametric Equations.

Answer.
12

3.7.4.10. Intersecting Parametric Equations.

Answer.
B

4 The Definite Integral
4.1 Determining Distance Traveled from Velocity
4.1.6 Exercises

4.1.6.1. Estimating distance traveled from velocity data.

Answer 1.
134 ft
Answer 2.
328 ft

4.1.6.2. Distance from a linear velocity function.

Answer.
750 m

4.1.6.3. Change in position from a linear velocity function.

Answer.
8 cm

4.1.6.5. Finding average acceleration from velocity data.

Answer 1.
28.64 (ft(s2
Answer 2.
21 (ft(s2

4.1.6.6. Change in position from a quadratic velocity function.

Answer.
33.8333333333333

4.1.6.7. A piecewise velocity function.

Answer.
  1. At time t=1, 12 miles north of the lake.
  2. s(4)−s(0)=0.
  3. 40 miles.

4.1.6.8. Physical interpretations of velocity.

Answer.
  1. t=50032=1258=15.625 is when the rocket reaches its maximum height.
  2. A=3906.25, the vertical distance traveled on [0,15.625].
  3. s(t)=500t−16t2.
  4. s(15.625)−s(0)=3906.25 is the change of the rocket’s position on [0,15.625].
  5. s(5)−s(1)=1616; the rocket rose 1616 feet on [1,5].

4.1.6.9. Physical interpretations of velocity.

Answer.
  1. 12+14π≈1.285.
  2. s(5)−s(2)=−2 is the change in position of the object on [2,5].
  3. On the time interval [5,7].
  4. s is increasing on the intervals (0,2) and (5,7); the position function has a relative maximum at t=2.

4.1.6.10. Pollution Data.

Answer.
  2. Think about the product of the units involved: ``units of pollution per day’’ times ``days’’. Connect this to the area of a thin vertical rectangle whose height is given by the curve.
  3. An underestimate is 546 units of pollution.

4.2 Riemann Sums
4.2.5 Exercises

4.2.5.1. Evaluating Riemann sums for a quadratic function.

Answer 1.
18.5
Answer 2.
there is ambiguity
Answer 3.
17.7
Answer 4.
there is ambiguity

4.2.5.2. Estimating distance traveled with a Riemann sum from data.

Answer 1.
3 ft
Answer 2.
19 ft
Answer 3.
−1 ft
Answer 4.
17 ft

4.2.5.3. Writing basic Riemann sums.

Answer 1.
0.346574
Answer 2.
0.458145
Answer 3.
0.458145
Answer 4.
0.549306

4.2.5.4. Using the Middle Riemann sum.

Answer.
  1. MID(4)=43.5.
  2. A=872.
  3. The rectangles with heights that come from the midpoint have the same area as the trapezoids that are formed by the function values at the two endpoints of each subinterval.
    MID(n) will give the exact area for any value of n. Neither LEFT(n) nor RIGHT(n) will be exact for any n.
  4. For any linear function g of the form g(x)=mx+b such that g(x)≥0 on the interval of interest.

4.2.5.5. Identifying and manipulating Riemann sum components.

Answer.
  1. f(x)=x2+1 on the interval [1,3].
  2. If S is a left Riemann sum, f(x)=x2+1 on the interval [1.4,3.4]. If S is a middle Riemann sum, f(x)=x2+1 on the interval [1.2,3.2].
  3. The area under f(x)=x2+1 on [1,3].
  4. RIGHT(10)=∑i=110((1+0.2i)2+1)⋅0.2.

4.2.5.6. Evaluating Riemann sums with data.

Answer.
  2. MID(3)=99.6 feet.
  3. LEFT(6)=114,
    RIGHT(6)=84,
    and 12(LEFT(6)+RIGHT(6))=99.
  4. 114 feet.

4.2.5.7. Evaluating Riemann sums with graphs and formulas.

Answer.
  1. MID(4)≈6.4.
  2. The total tonnage of pollution escaping the scrubbing process in the time interval [0,4] weeks.
  3. LEFT(5)≈5.19620599.
  4. 6.4 tons.

4.3 The Definite Integral
4.3.5 Exercises

4.3.5.1. Evaluating definite integrals from graphical information.

Answer 1.
−55
Answer 2.
5
Answer 3.
−50
Answer 4.
60

4.3.5.2. Estimating definite integrals from a graph.

Answer 1.
−8
Answer 2.
−A

4.3.5.3. Finding the average value of a linear function.

Answer.
8

4.3.5.4. Finding the average value of a function given graphically.

Answer 1.
14
Answer 2.
38
Answer 3.
0

4.3.5.5. Estimating a definite integral and average value from a graph.

Answer.
5

4.3.5.6. Using rules to combine known integral values.

Answer 1.
2−5−2
Answer 2.
−2⋅(−5)+5⋅1−5⋅(−1)

4.3.5.7. Using definite integrals on a velocity function.

Answer.
  1. The total change in position is P=∫04v(t)dt.
  2. P=−2.625 feet.
  3. D=3.375 feet.
  4. AV=−0.65625 feet per second.
  5. s(t)=−t2+t.

4.3.5.8. Riemann sum estimates and definite integrals.

Answer.
  1. The total change in position, P, is P=∫01v(t)dt+∫13v(t)dt+∫34v(t)dt=∫04v(t)dt.
  2. P=∫04v(t)dt≈2.665.
  3. The total distance traveled, D, is D=∫01v(t)dt−∫13v(t)dt+∫34v(t)dt.
  4. D≈8.00016.
  5. vAVG[0,4]≈=0.66625
    feet per second.

4.3.5.9. Using the Sum and Constant Multiple Rules.

Answer.
  1. ∫01[f(x)+g(x)]dx=1−π4.
  2. ∫14[2f(x)−3g(x)]dx=−152−3π.
  3. hAVG[0,4]=58+3π16.
  4. c=−38+3π16.

4.3.5.10. Finding the area of a bounded region.

Answer.
  1. A1=∫−11(3−x2)dx.
  2. A2=∫−112x2dx.
  3. The exact area between the two curves is ∫−11(3−x2)dx−∫−112x2dx.
  4. Use the sum rule for definite integrals over the same interval.
  5. Think about subtracting the area under q from the area under p.

4.4 The Fundamental Theorem of Calculus
4.4.4 Exercises

4.4.4.1. Using Graphs to Evaluate.

Answer 1.
0
Answer 2.
−4
Answer 3.
−6
Answer 4.
−5.5
Answer 5.
−4
Answer 6.
−2.5
Answer 7.
−1.5
Answer 8.
−1
Answer 9.
−1
Answer 10.
(2,7)
Answer 11.
(0,2)
Answer 12.
0
Answer 13.
2

4.4.4.2. Using Graphs to Evaluate.

Answer 1.
4−(x−4)(0.5x−2)2
Answer 2.
−2t28+2t

4.4.4.3. Estimating using the FTC.

Answer 1.
B
Answer 2.
A
Answer 3.
1
Answer 4.
1.264204

4.4.4.4. Finding Values using the FTC.

Answer.
(0,−8.5),(2,−2),(6,−10),(8,−8)

4.4.4.5. Average Value.

Answer.
53.5

4.4.4.6. Average Value (Estimating from a Graph).

Answer 1.
45.6666666666667
Answer 2.
5.2

4.4.4.7. Average Value (Estimating from a Table).

Answer.
38.3333333333333

4.4.4.8. Creating and using new functions from data.

Answer.
  1. h (feet) 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000
    c (ft/min) 925 875 830 780 730 685 635 585 535 490 440
    m (min/ft) 1925 1875 1830 1780 1730 1685 1635 1585 1535 1490 1440
  2. The antiderivative function tells the total number of minutes it takes for the plane to climb to an altitude of h feet.
  3. M=∫010000m(h)dh.
  4. It takes the plane aabout M5≈15.27 minutes.

4.4.4.9. Connecting average rate of change and average value of a function.

Answer.
Yes.

5 Evaluating Integrals
5.1 Constructing Accurate Graphs of Antiderivatives
5.1.5 Exercises

5.1.5.1. Definite integral of a piecewise linear function.

Answer 1.
−2
Answer 2.
−A

5.1.5.2. A smooth function that starts out at 0.

Answer 1.
−17.8
Answer 2.
7.2

5.1.5.3. A piecewise constant function.

Answer 1.
0
Answer 2.
1
Answer 3.
1

5.1.5.4. Another piecewise linear function.

Answer 1.
1
Answer 2.
1.5
Answer 3.
1
Answer 4.
0.166666666666667
Answer 5.
−0.333333333333333
Answer 6.
−0.5

5.1.5.5. Determining graphical properties for an antiderivative.

Answer.
  1. s(1)=53, s(3)=−1, s(5)=−113, s(6)=−52.
  2. s is increasing on 0<t<1 and 5<t<6; decreasing for 1<t<5.
  3. s is concave down for t<2; concave up for t>2.
  5. s(t)=−2t+16(t−3)3+5.

5.1.5.6. Calories Burned.

Answer.
  1. C measures the total number of calories burned in the workout since t=0.
  2. C(5)=12.5, C(10)=50, C(15)=125, C(20)=187.5, C(25)=237.5, C(30)=262.5.
  4. C(t)=12.5+7.5(t−5) on this interval.

5.1.5.7. Functions defined by integrals.

Answer.
  1. B(−1)=−1, B(0)=0, B(1)=12, B(2)=0, B(3)=−1, B(4)=−32, B(5)=−1, B(6)=0. Also, A(x)=1+B(x) and C(x)=B(x)−12.
    x −1 0 1 2 3 4 5 6
    A(x) 0 1 1.5 1 0 −0.5 0 1
    B(x) −1 0 0.5 0 −1 −1.5 −1 0
    C(x) −1.5 −0.5 0 −0.5 −1.5 −2 −1.5 −0.5
  3. A, B, and C are vertical translations of each other.
  4. A′=f.

5.2 Antiderivatives from Formulas
5.2.4 Exercises

5.2.4.1. Finding Antiderivatives.

Answer.
4t14

5.2.4.2. Finding Antiderivatives (Constants).

Answer 1.
x
Answer 2.
y
Answer 3.
πz

5.2.4.3. Finding Antiderivatives (Polynomials).

Answer 1.
x22
Answer 2.
ax
Answer 3.
t2x

5.2.4.4. Finding Antiderivatives.

Answer.
85t52−2t12+C

5.2.4.5. Finding Antiderivatives (Power Functions).

Answer 1.
1.09861228866811
Answer 2.
0.833333333333333
Answer 3.
0.19917695473251
Answer 4.
9
Answer 5.
0.024390243902439

5.2.4.6. Finding Antiderivatives (nth-Roots).

Answer 1.
42.6666666666667
Answer 2.
2
Answer 3.
191.25

5.2.4.7. Definite Integrals from Antiderivatives.

Answer 1.
330
Answer 2.
114
Answer 3.
0

5.2.4.8. Definite Integrals from Antiderivatives (Trigonometric Functions).

Answer 1.
−8
Answer 2.
7
Answer 3.
4.22649730810374

5.2.4.9. Finding exact displacement.

Answer.
18

5.2.4.10. Evaluating the definite integral of a rational function.

Answer.
−111+15

5.2.4.11. Evaluating the definite integral of a linear function.

Answer.
336

5.2.4.12. Evaluating the definite integral of a quadratic function.

Answer.
1333.33333333333

5.2.4.13. Simplifying an integrand before integrating.

Answer.
107.689294642159

5.2.4.14. Evaluating the definite integral of a trigonometric function.

Answer.
4

5.2.4.15. Analyzing a velocity function.

Answer.
  1. 20 meters.
  2. vAVG[12,24]=12.5 meters per minute.
  3. The object’s maximum acceleration is 3 meters per minute per minute at the instant t=2.
  4. c=6024=2.5.

5.2.4.16. Evaluating the definite integral of a piecewise function.

Answer.
  1. −56.
  2. fAVG[0,5]=12.
  3. g(x)=f(x) for 0≤x<5 and g(x)=−54(x−5) on 5≤x≤7.

5.3 Differential Equations
5.3.4 Exercises

5.3.4.1. General Solution to a Differential Equation.

Answer.
23kt32

5.3.4.2. Setting up a Differential Equation.

Answer.
75⋅52807⋅3600

5.3.4.3. Setting up a Differential Equation.

Answer 1.
3223016
Answer 2.
15⋅121.32622

5.3.4.4. Displacement.

Answer.
58.6666666666667

5.3.4.5. Initial Value Problems.

Answer 1.
316
Answer 2.
34
Answer 3.
34
Answer 4.
3

5.3.4.6. Initial Velocity.

Answer.
−35

5.3.4.7. Finding the Position Function.

Answer.
5t−cos(t)+−6

5.3.4.8. Finding the Position Function Given Acceleration.

Answer.
t412+t36−3t22

5.3.4.9. Finding Displacement.

Answer 1.
0.416667
Answer 2.
0.472222

5.3.4.10. Finding Displacement.

Answer 1.
90
Answer 2.
ft
Answer 3.
−600
Answer 4.
ft

5.3.4.11. Finding Displacement from a Graph.

Answer 1.
−9 ft
Answer 2.
−4 ft
Answer 3.
14 ft
Answer 4.
(6,8]
Answer 5.
[0,5)
Answer 6.
[5,6]

5.3.4.12. Finding Displacement from a Graph.

Answer 1.
−7 ft
Answer 2.
3 ft
Answer 3.
17 ft
Answer 4.
(4,8]
Answer 5.
[0,3)
Answer 6.
[3,4]

5.4 The Second Fundamental Theorem of Calculus
5.4.5 Exercises

5.4.5.1. A definite integral starting at 3.

Answer 1.
0
Answer 2.
−0.44
Answer 3.
(1.25,6)
Answer 4.
3.68034

5.4.5.2. Variable in the lower limit.

Answer.
−1tan(ln(x))

5.4.5.3. Approximating a function with derivative e−x25.

Answer.
5.86716

5.4.5.4. Sketching an antiderivative function based on definite integral values.

Answer.
F is increasing on x<−1, 0.5<x<4, and 5<x<6.5; decreasing on −1<x<0.5 and 4<x<5; concave up on approximately −0.4<x<2 and 4.5<x<6; concave down on approximately 2<x<4.5 and x>6; F(2)=0; F(0.5)=−6.06; F(−1)=−1.77; F(4)=6.69; F(5)=6.33; F(6.5)=8.12.

5.4.5.5. Sand on the Beach.

Answer.
Exercise Answer

5.4.5.6. Altitude Changes.

Answer.
Exercise Answer

5.5 Integration by Substitution
5.5.7 Exercises

5.5.7.1. Practice the steps of the method of substitution.

Answer 1.
cos(7t)
Answer 2.
−7sin(7t)
Answer 3.
−u157
Answer 4.
−u16112
Answer 5.
−cos16(7t)112

5.5.7.2. Product involving a 4th power polynomial.

Answer.
(t3−6)412+C

5.5.7.3. Product involving sin⁡(x6).

Answer.
1sin(x2)+C

5.5.7.4. Fraction involving ln9⁡(x).

Answer.
0.25ln4(z)+C

5.5.7.5. Fraction involving e5x.

Answer.
0.333333ln(e3x+5)+C

5.5.7.6. Fraction involving e5y.

Answer.
8e2y2+C

5.5.7.7. Definite integral involving e−cos⁡(q).

Answer.
−1.71828182845905

5.5.7.8. Working with negative exponents.

Answer.
(x+12)−2−2

5.5.7.9. Product involving sin⁡(9sin⁡(t)).

Answer.
−18cos(8sin(t))

5.5.7.10. Fraction involving sin⁡(9/x).

Answer.
19cos(9x)

5.5.7.11. Fraction involving sums of exponential functions.

Answer.
ln(|ex+5e−x|)

5.5.7.12. Integral involving a rational function.

Answer.
0.333333(x3+(−7))−2−2+C

5.5.7.13. Integral of a partial fraction.

Answer.
0.916290731874155

5.5.7.14. Find the value of a definite integral based on another.

Answer.
−2

5.5.7.15. Using the method of substitution to derive some other trignonometric antiderivative rules.

Answer.
  1. ∫tan⁡(x)dx=ln⁡(|sec⁡(x)|)+C.
  2. ∫cot⁡(x)dx=−ln⁡(|csc⁡(x)|)+C.
  3. ∫sec2⁡(x)+sec⁡(x)tan⁡(x)sec⁡(x)+tan⁡(x)dx=ln⁡(|sec⁡(x)+tan⁡(x)|)+C.
  4. sec2⁡(x)+sec⁡(x)tan⁡(x)sec⁡(x)+tan⁡(x)=sec⁡(x).
  5. ∫sec⁡(x)dx=ln⁡(|sec⁡(x)+tan⁡(x)|)+C.
  6. ∫csc⁡(x)dx=−ln⁡(|csc⁡(x)+cot⁡(x)|)+C.

5.5.7.16. Integral of sec⁡(x).

Answer.
ln(|sec(x)+tan(x)|)

5.5.7.17. Products involving sec⁡(x) and tan⁡(x).

Answer 1.
17sec(7x)
Answer 2.
164tan8(8x)

5.5.7.18. Definite integral of tan⁡(8t).

Answer.
ln(2)6

5.5.7.19. Integrals of tan⁡(5x) and sec⁡(5x).

Answer 1.
−0.2ln(cos(5x))+C
Answer 2.
0.2ln(sec(5x)+tan(5x))+C

5.5.7.20. A clever substitution.

Answer.
  1. ∫xx−1dx=∫(u+1)udu.
  2. ∫xx−1dx=25(x−1)52+23(x−1)32+C.
  3. ∫x2x−1dx=27(x−1)72+45(x−1)52+23(x−1)32+C.
    ∫xx2−1dx=13(x2−1)32+C.

5.5.7.21. Fractions involving a square root or trigonometric functions.

Answer 1.
2(x+9)x+93−18x+9
Answer 2.
−19(9sin(t)+10)

5.5.7.22. Definite integral with a clever substitution.

Answer.
−112−415

5.5.7.23. Product involving trigonometric functions and square roots.

Answer.
−26(15−sin(2x))32

5.5.7.24. Integral involving a square root of a linear expression.

Answer.
−43(2−x)2−x+25(2−x)22−x

5.5.7.25. Re-writing a function in order to use substitution.

Answer.
  1. We don’t have a function-derivative pair.
  2. sin3⁡(x)=sin⁡(x)(1−cos2⁡(x)).
  3. u=cos⁡(x) and du=−sin⁡(x)dx.
  4. ∫sin3⁡(x)dx=13cos3⁡(x)−cos⁡(x)+C.
  5. ∫cos3⁡(x)dx=sin⁡(x)−13sin3⁡(x)+C.

5.5.7.26. Fraction involving sin⁡(4t−1).

Answer.
sec(4t−1)4

5.5.7.27. An integral that requires rewriting before integrating.

Answer.
5tan−1(x)+(−1)ln(1+x2)+C+c

5.5.7.28. Power Consumption.

Answer.
  1. The model is reasonable because it appears to be periodic and the rate of consumption seems to peak at the times of day where people are most active in their homes.
  2. The total power consumed in 24 hours, measured in megawatt-hours.
  3. ∫024r(t)dt≈95.7809 megawatts of power used in 24 hours.
  4. rAVG[0,24]≈3.99087 megawatts.

5.6 Integration by Parts
5.6.7 Exercises

5.6.7.2. Product involving cos⁡(5x).

Answer.
2(xsin(2x)+0.5cos(2x))

5.6.7.3. Product involving e8z.

Answer.
(1+z7−0.0204082)e7z+C

5.6.7.4. Definite integral of te−t.

Answer.
0.992704944275564

5.6.7.5. Evaluating using FTC.

Answer.
  1. F′(x)=xe−2x.
  2. F(x)=−12xe−2x−14e−2x+14
  3. Increasing.

5.6.7.6. Evaluating an integral using substitution and IBP.

Answer.
  1. ∫e2xcos⁡(ex)dx=∫zcos⁡(z)dz.
  2. ∫e2xcos⁡(ex)dx=exsin⁡(ex)+cos⁡(ex)+C.
  3. ∫e2xcos⁡(e2x)dx=12sin⁡(e2x+C
    • ∫e2xsin⁡(ex)dx=sin(ex)−zcos⁡(ex)+C.
    • ∫e3xsin⁡(e3x)dx=−13cos⁡(e3x)+C.
    • ∫xex2cos⁡(ex2)sin⁡(ex2)dx=14sin2⁡(ex2)+C.

5.6.7.7. Identifying when substitution and/or IBP can be used to solve various integrals.

Answer.
  1. u-substitution; ∫x2cos⁡(x3) dx=13sin⁡(x3)+C.
  2. Both are needed; ∫x5cos⁡(x3) dx=13(x3sin⁡(x3)+cos⁡(x3))+C.
  3. Integration by parts; ∫xln(x2) dx=x22ln⁡(x2)−x22+C.
  4. Neither.
  5. u-substitution; ∫x3sin⁡(x4) dx=−14cos⁡(x4)+C.
  6. Both are needed; ∫x7sin⁡(x4) dx=−14x4cos⁡(x4)+14sin⁡(x4)+C.

5.7 The Method of Partial Fractions
5.7.3 Exercises

5.7.3.1. Partial fractions: linear over difference of squares.

Answer 1.
1x−1
Answer 2.
1x+1
Answer 3.
ln(|x−1|)
Answer 4.
ln(|x+1|)
Answer 5.
1w
Answer 6.
ln(|w|)
Answer 7.
ln(|x2−1|)

5.7.3.2. Partial fractions: constant over product.

Answer.
0.5ln(|x+6|)−0.5ln(|x+8|)+C

5.7.3.3. Partial fractions: linear over quadratic.

Answer.
9ln(|x−2|)−6ln(|x−1|)+C

5.7.3.4. Partial fractions: cubic over 4th degree.

Answer 1.
4
Answer 2.
2
Answer 3.
3
Answer 4.
2
Answer 5.
−4x+2ln(x)+1.5ln(x2+9)+0.666667tan−1(x3)

5.7.3.5. Partial fractions: quadratic over factored cubic.

Answer 1.
1
Answer 2.
0
Answer 3.
2
Answer 4.
ln(|x+3|)+2tan−1(x3)3+C

5.9 Numerical Integration
5.9.6 Exercises

5.9.6.1. Various methods for ex numerically.

Answer 1.
e5−1
Answer 2.
2.5(1+e2.5)
Answer 3.
2.5(e2.5+e5)
Answer 4.
1.25(1+2e2.5+e5)
Answer 5.
2.5(e1.25+e3⋅1.25)
Answer 6.
2⋅115.029+217.2233
Answer 7.
147.413−32.9562
Answer 8.
147.413−401.489
Answer 9.
147.413−217.223
Answer 10.
147.413−115.029
Answer 11.
147.413−149.094
Answer 12.
1.25(1+e1.25+e2.5+e3⋅1.25)
Answer 13.
1.25(e1.25+e2.5+e3⋅1.25+e5)
Answer 14.
0.625(1+2e1.25+2e2.5+2e3⋅1.25+e5)
Answer 15.
1.25(e0.625+e3⋅0.625+e5⋅0.625+e7⋅0.625)
Answer 16.
2⋅138.236+166.1263
Answer 17.
147.413−73.9924
Answer 18.
147.413−258.259
Answer 19.
147.413−166.126
Answer 20.
147.413−138.236
Answer 21.
147.413−147.533
Answer 22.
114.45773.4206
Answer 23.
−254.076−110.846
Answer 24.
−69.81−18.713
Answer 25.
32.3849.177
Answer 26.
−1.681−0.12

5.9.6.2. Comparison of methods for increasing concave down function.

Answer 1.
RIGHT(n)
Answer 2.
MID(n)
Answer 3.
Exact
Answer 4.
TRAP(n)
Answer 5.
LEFT(n)

5.9.6.5. Identifying types of Riemann sums.

Answer 1.
7.6875
Answer 2.
9.6875

5.9.6.6. Identifying types of Riemann sums.

Answer 1.
midpoint
Answer 2.
10.0265

5.9.6.7. Computing left and right Riemann sums.

Answer 1.
e1⋅0.5
Answer 2.
e1.5⋅0.5
Answer 3.
e1.5⋅0.5
Answer 4.
e2⋅0.5

5.9.6.9. Comparing methods and identifying whether each is an overestimate or underestimate.

Answer.
  1. u-substitution fails since there’s not a composite function present; try showing that each of the choices of u=x and dv=tan⁡(x)dx, or u=tan⁡(x) and dv=xdx, fail to produce an integral that can be evaluated by parts.
    • LEFT(4)=0.25892
    • RIGHT(4)=0.64827
    • MID(4)=0.41550
    • TRAP(4)=LEFT(4)+RIGHT(4)2=0.45360
    • SIMP(8)=2MID(4)+TRAP(4)3=0.42820
  3. LEFT(4) and MID(4) are underestimates; RIGHT(4) and MID(4) are overestimates.

5.9.6.10. Relation of function properties to approximation methods.

Answer.
  1. Decreasing.
  2. Concave down.
  3. ∈TRAP(3)6f(x)≈7.03.

5.9.6.11. Modeling flow rate using an integral and approximating solution.

Answer.
  1. ∫060r(t)dt.
  2. ∫060r(t)dt>MID(3)=204000.
  3. ∫060r(t)dt≈SIMP(6)=6190003≈206333.33.
  4. 160SIMP(6)≈3438.89; 2000+2100+2400+3000+3900+5100+65007=250007≈3571.43. each estimates the average rate at which water flows through the dam on [0,60], and the first is more accurate.

5.10 Improper Integrals
5.10.6 Exercises

5.10.6.1. An improper integral on a finite interval.

Answer.
D

5.10.6.2. An improper integral on an infinite interval.

Answer.
2e−1⋅12

5.10.6.3. An improper integral involving a ratio of exponential functions.

Answer.
3.00000153604837

5.10.6.4. A subtle improper integral.

Answer.
diverges

5.10.6.5. An improper integral involving a ratio of trigonometric functions.

Answer.
diverges

5.11 Comparison of Improper Integrals
5.11.3 Exercises

5.11.3.1. Determining convergence or divergence of various improper integrals.

Answer.
  1. Diverges.
  2. Diverges.
  3. Converges to 1.
  4. ∫e∞1x(ln⁡(x))pdx diverges if p≤1 and converges to 1p−1 if p>1.
  5. Diverges.
  6. Converges to −1.

5.12 Using Technology and Tables to Evaluate Integrals
5.12.4 Exercises

5.12.4.1. Using a Computer Algebra System to Antidifferentiate Rational Functions.

Answer.
Exercise Answer

5.12.4.2. Using a Table of Integrals to Find Antiderivatives of Radical Functions.

Answer.
  1. ∫1x9x2+25dx=−15ln⁡|5+9x2+523x|+C.
  2. ∫x1+x4dx=12(x22x4+1+12ln⁡|x2+x4+1|)+C
  3. ∫ex4+e2xdx=ex2e2x+4+2ln⁡|ex+e2x+4|+C
  4. ∫tan⁡(x)9−cos2⁡(x)dx=13ln⁡|3+9−cos2⁡(x)cos⁡(x)|+C.

5.12.4.3. Comparing Antidifferentiation Tools.

Answer.
  1. Try u=1+x2 or u=x+1+x2.
  2. Try u=x+1+x2 and dv=1xdx.
  3. No.
  4. It appears that the function x+1+x2x does not have an elementary antiderivative.

6 Using Definite Integrals
6.1 Using Definite Integrals to Find Area and Volume
6.1.5 Exercises

6.1.5.1. Area between two power functions.

Answer.
0.0833333

6.1.5.2. Area between two trigonometric functions.

Answer.
14.560219778561

6.1.5.3. Area between two curves.

Answer.
20.8333

6.1.5.4. Area of regions between two curves.

Answer.
  1. A=∫3−323+32−2y2+6y−3 dy=3.
  2. A=∫π/43π/4sin⁡(x)−cos⁡(x) dx=2.
  3. A=∫−15/2y+12−(y2−y−2) dy=34348.
  4. A=∫m−m2+42m+m2+42mx−(x2−1) dx=−13(m+m2+42)3−13(m−m2+42)3+m2(m+m2+42)2−m2(m−m2+42)2+(m+m2+42−m−m2+42).

6.1.5.5. Setting up a definite integral for area and solving for a missing term.

Answer.
a=12.

6.1.5.6. Average value of a continuous function and relation to area.

Answer.
  1. r=43.
  3. A1=A2=4627.
  4. Yes.

6.2 Using Definite Integrals to Find Volume by Rotation and Arc Length
6.2.6 Exercises

6.2.6.1. Solid of revolution from one function about the x-axis.

Answer.
0.785134691665605

6.2.6.2. Solid of revolution from one function about the y-axis.

Answer.
2.35619449019234

6.2.6.3. Solid of revolution from two functions about the x-axis.

Answer.
2.89993168023673

6.2.6.4. Solid of revolution from two functions about a horizontal line.

Answer.
29.8278687110064

6.2.6.5. Solid of revolution from two functions about a different horizontal line.

Answer.
44.6804288510548

6.2.6.6. Solid of revolution from two functions about a vertical line.

Answer.
17.6976386152225

6.2.6.7. Arc length and area of a region, and volume of its solid of revolution.

Answer.
  1. L=∫01.842571+(−3sin⁡(x34)⋅34x2)2dx≈4.10521.
  2. A=∫01.845273cos⁡(x34)dx≈4.6623.
  3. V=∫01.84527π⋅9cos2⁡(x34)dx≈40.31965.
  4. V=∫03π(4arccos⁡(y3))2/3dy≈23.29194.

6.2.6.8. Solid of revolution from a two functions about multiple horizontal and vertical lines.

Answer.
  2. A=∫0π4(cos⁡(x)−sin⁡(x))dx.
  3. V=∫0π4π(cos2⁡(x)−sin2⁡(x))dx.
  4. V=∫022πarcsin2⁡(y)dy+∫221πarccos2⁡(y)dy
  5. V=∫0π4π[(2−sin⁡(x))2−(2−cos⁡(x))2]dx.
  6. V=∫022π[(1+arcsin⁡(y))2−12]dy+∫221π[(1+arccos⁡(y))2−12]dy

6.2.6.9. Area and perimeter of a region and volume of a solid of revolution around multiple lines.

Answer.
  1. A=∫01.51+12(x−2)2−12x2 dx=2.25.
  2. V=∫01.5π[(2+12(x−2)2)2−(1+12x2)2] dx=31532π
  3. V=∫01.125π(2y)2 dy+∫1.1253π(2−2(y−1))2 dy≈7.06858347.
  4. P=3+∫01.51+(x−2)2+1+x2 dx≈7.387234642.

6.2.6.10. Arc length of a curve.

Answer.
13.5428669014242

6.2.6.11. Length of a parametric curve.

Answer.
L≈9.429

6.3 Area and Arc Length in Polar Coordinates
6.3.6 Exercises

6.3.6.1. Converting Coordinates: Polar to Cartesian.

Answer 1.
2.5
Answer 2.
4.33012701892219
Answer 3.
−1
Answer 4.
0

6.3.6.2. Converting Coordinates: Cartesian to Polar.

Answer 1.
7
Answer 2.
1.5707963267949
Answer 3.
1
Answer 4.
0.523598775598299

6.3.6.3. Converting Coordinates: Both Directions.

Answer.

6.3.6.4. Describing a Polar Region.

Answer 1.
4.24264068711928
Answer 2.
8.48528137423857
Answer 3.
0.785398163397448
Answer 4.
1.5707963267949

6.3.6.5. Describing a Polar Region as a Function of θ.

Answer 1.
2
Answer 2.
4cos(t)
Answer 3.
0
Answer 4.
π2

6.3.6.6. Area Inside a Cardioid.

Answer.
42.4115008234622

6.3.6.7. Area in a Region Defined by Two Curves.

Answer.
14.1371669411541

6.4 Density, Mass, and Center of Mass
6.4.6 Exercises

6.4.6.1. Center of mass for a linear density function.

Answer 1.
(4+4x)Dx
Answer 2.
30 g

6.4.6.2. Center of mass for a nonlinear density function.

Answer 1.
2+8π
Answer 2.
π2

6.4.6.3. Interpreting the density of cars on a road.

Answer 1.
350(2+sin(4x+0.175))Dx
Answer 2.
18⋅350(320+40.175cos(40.175)−420+0.175cos(420+0.175)−sin(40.175)+sin(420+0.175))

6.4.6.4. Center of mass in a point-mass system.

Answer 1.
3.66667 cm
Answer 2.
to the right of the origin

6.4.6.5. Center of mass in a continuous 1-dimensional object.

Answer.
  1. a=−10ln⁡(0.7)≈3.567 cm.
  2. Left of the midpoint.
  3. x―≈50.333830≈1.687.
  4. q=−10ln⁡(0.85)≈1.625 cm.

6.4.6.6. Combining masses and centers of masses for continuous 1-dimensional object.

Answer.
  1. M1=arctan⁡(10)≈1.47113; M2=10−10e−1≈6.32121.
  2. x1―≈1.56857; x2―≈4.18023.
    1. M=∫010ρ(x)dx+∫010p(x)dx≈1.47113+6.32121=7.79234.
    2. ∫010x(ρ(x)+p(x)))dx=28.73167.
    3. False.

6.4.6.7. Mass and center of mass for a solid of revolution.

Answer.
  2. V=∫030π(2xe−1.25x+(30−x)e−0.25(30−x))2dx≈52.0666 cubic inches.
  3. W≈0.6⋅52.0666=31.23996 ounces.
  4. At a given x-location, the amount of weight concentrated there is approximately the weight density (0.6 ounces per cubic inch) times the volume of the slice, which is Vtextslice≈πf(x)2.
  5. x―≈23.21415

6.5 Physics Applications: Work, Force, and Pressure
6.5.5 Exercises

6.5.5.1. Work to empty a conical tank.

Answer.
232070.147452379 J

6.5.5.2. Work to empty a cylindrical tank.

Answer.
1.66253×106 ftlb

6.5.5.3. Work to empty a rectangular pool.

Answer.
5.7798×106 ftlb

6.5.5.4. Work to empty a cylindrical tank to differing heights.

Answer 1.
106741 lbf
Answer 2.
242594 lbf
Answer 3.
98801.8 lbf

6.5.5.5. Force due to hydrostatic pressure.

Answer 1.
9800 N
Answer 2.
3920 N
Answer 3.
3136 N

6.5.5.6. Work to fill an irregularly shaped tank.

Answer.
  1. W≈1646.79 foot-pounds.
  2. W=∫0h3744xcos⁡(x34)dx.
  3. F≈462.637 pounds.

6.5.5.7. Work to half-empty a cylindrical tank.

Answer.
  1. W=5904(19π−8)≈305179.3 foot-pounds.
  2. F≈1123.2
    pounds.

7 Sequences and Series
7.1 Sequences
7.1.3 Exercises

7.1.3.2. Formula for a Sequence, Given First Terms.

Answer.
(−1)n(2n+3)

7.1.3.3. Divergent or Convergent Sequences.

Answer 1.
4
Answer 2.
diverges
Answer 3.
0
Answer 4.
0

7.1.3.4. Terms of a Sequence from Sampling a Signal.

Answer.
0.09,0.01,0.01,0.09,0.25,0.49

7.1.3.5. Finding the Limit of a Convergent Sequence.

Answer.
  1. Unclear whether it converges or diverges.
    n 1 2 3 4 5
    ln⁡(n)n 0 0.3466 0.3663 0.3466 0.3218
    t 6 7 8 9 10
    ln⁡(n)n 0.2987 0.2781 0.2599 0.2442 0.2303
  2. If limx→∞f(x)=L, then limn→∞ln⁡(n)n=L as well.
  3. limn→∞ln⁡(n)n=limx→∞ln⁡(x)x=0.

7.1.3.6. The Formula for the Amount in a Bank Account.

Answer.
  1. P1(r12) in interest in the second month; at the end of the second month, P2=P(1+r12)2.
  2. P3=P(1+r12)3.
  3. Pn=P(1+r12)n is a pattern to these calculations.

7.1.3.7. Half-life of GLP-1.

Answer.
Exercise Answer

7.1.3.8. Sampling Continuous Data.

Answer.
  1. The data points do not appear periodic at all.
  2. At least 13 samples, so at least every 10/13≈0.76923 seconds.
  3. 44100 Hz is slightly more than double 20 KHz.

7.2 Geometric Series
7.2.3 Exercises

7.2.3.1. Seventh term of a geometric sequence.

Answer.
−166083.84375

7.2.3.2. A geometric series.

Answer.
−5.33333333333333

7.2.3.3. Two sums of geometric sequences.

Answer 1.
−4.00003051757812
Answer 2.
0.00130208333333333

7.2.3.4. A series that is not geometric.

Answer.
2.33333333333333

7.2.3.6. Do geometric series grow quickly?

Answer.
  1. 30⋅500=1500 dollars.
  2. Day Pay on this day Total amount paid to date
    1 $0.01 $0.01
    2 $0.02 $0.03
    3 $0.04 $0.07
    4 $0.08 $0.15
    5 $0.16 $0.31
    6 $0.32 $0.63
    7 $0.64 $1.27
    8 $1.28 $2.55
    9 $2.56 $5.11
    10 $5.12 $10.23
  3. $0.01(230−1)=$10,737,418.23.

7.2.3.7. Application to model behavior of a ball drop.

Answer.
  1. h1=(34)h.
  2. h2=(34)h1=(34)2h.
  3. h3=(34)h2=(34)3h.
  4. hn=(34)hn−1=(34)nh.
  5. The distance traveled by the ball is 7h, which is finite.

7.2.3.8. Computing probabilities using geometric series.

Answer.
  1. There are 6 equally possible outcomes when we roll one die.
  2. The three rolls are independent so the probability of the overall outcome is the product of the three probabilities.
  3. See (b).
  4. P=611.

7.2.3.9. Application to economics.

Answer.
  1. 0.75P dollars spent.
  2. 0.75P+0.75(0.75P)=0.75P(1+0.75) dollars.
  3. 0.75P+0.752P+0.752P+⋯=0.75P(1+0.75+0.752+⋯) dollars.
  4. A stimulus of 200 billion dollars adds 600 billion dollars to the economy.

7.2.3.10. Computing loan payments using geometric series.

Answer.
  1. (r12)P1 dollars.
  2. P2=(1+r12)P1−M.
  3. P2=(1+r12)2P−[1+(1+r12)]M.
  4. P3=(1+r12)P2−M.
    P3=(1+r12)3P−[1+(1+r12)+(1+r12)2]M.
  5. Pn=P(1+r12)n−(12Mr)((1+r12)n−1).
  6. P(t)=P(1+r12)12t−(12Mr)((1+r12)12t−1)
  7. A(t)=(1000−12(25)0.2)(1+0.212)12t+12(25)0.2.
    t≈5.5.
    We pay $659 dollars in interest on our $1000 loan.
  8. $291.74 each month to complete the loan in 5 years; we pay $2,504.40 in interest.

7.3 Convergence of Series
7.3.5 Exercises

7.3.5.1. Convergence of a sequence and its series.

Answer 1.
11.6666666666667
Answer 2.
0

7.3.5.2. Two partial sums.

Answer 1.
10.2666666666667
Answer 2.
14.631746031746

7.3.5.3. Convergence of a series and its sequence.

Answer 1.
∞
Answer 2.
98

7.3.5.4. Convergence of an integral and a related series.

Answer 1.
9π4
Answer 2.
converges

7.3.5.5. Adding two series together.

Answer.
    1. an=1+12n→1≠0 and bn=−1→−1≠0.
    2. The series is geometric with r=12.
    3. Since the two individual series diverge, neither sum is a finite number, so it doesn’t make any sense to add them.
    1. Note that An+Bn=(a1+b1)+(a2+b2)+⋯+(an+bn).
    2. Note that limn→∞∑k=1n(ak+bk)=limn→∞(∑k=1nak+∑k=1nbk) .
  3. ∑k=0∞2k+3k5k=256.

7.3.5.6. Using the integral test on a series involving a logarithm.

Answer 1.
0.360673760222241
Answer 2.
C

7.3.5.7. Using the integral test on a series involving an exponential.

Answer 1.
33e
Answer 2.
converges

7.4 Comparison Tests
7.4.4 Exercises

7.4.4.1. The direct comparison test.

Answer.
    1. S1=1 and T1=12.
    2. S2>T2.
    3. S3>T3.
    4. Sn>Tn.
    5. ∑1k2>∑1k2+k; ∑1k2+k converges.
  2. If ∑bk diverges, then ∑bk is infinite, and anything larger must also be infinite; if ∑ak is convergent then anything smaller and positive must also be finite.
    1. Note that 0<1k<1k−1.
    2. Note that 1k3>1k3+1.

7.4.4.3. Determining series convergence and which test(s) to use.

Answer 1.
Converges
Answer 2.
Converges

7.5 Ratio Test and Alternating Series
7.5.4 Exercises

7.5.4.1. Quick check of understanding for the ratio test.

Answer 1.
Convergent
Answer 2.
Inconclusive
Answer 3.
Divergent

7.5.4.2. Convergence of a series using the ratio test.

Answer.
  1. ∑10kk! converges.
  2. ∑10kk! converges.
  3. The sequence {bnn!} has to converge to 0.

7.5.4.3. The root test for convergence.

Answer.
  1. ann≈r for large n.
  2. an+1an≈r.
  3. 0<r<1.

7.5.4.5. Determining series convergence and which test(s) to use.

Answer 1.
Converges
Answer 2.
Converges

7.5.4.7. A closer look at a condition of the Alternating Series Test.

Answer.
  1. {1n} and {−1n2} converge to 0.
  2. Notice that 1k−1k2=k−1k2 and compare to the Harmonic series.
  3. It is possible for a series to alternate, have the terms go to zero, have the terms not decrease to zero, and the series diverge.

7.5.4.9. Practice determining series convergence.

Answer 1.
Converges
Answer 2.
Converges
Answer 3.
Converges

7.6 Absolute Convergence and Error Bounds
7.6.5 Exercises

7.6.5.1. Estimating the sum of an alternating series.

Answer.
4

7.6.5.2. Estimating the sum of a different alternating series.

Answer.
4

7.6.5.3. Estimating the sum of one more alternating series.

Answer.
6

7.6.5.4. A series that converges conditionally and slowly.

Answer.
  1. ∑k=0∞12k+1 diverges by comparison to the Harmonic series.
  2. |S100−∑k=0∞(−1)k12k+1|<≈0.0049.
  3. n>4,999,999,999.5

7.6.5.5. A alternative approximation method for convergent alternating series.

Answer.
  1. Sn+Sn+12=Sn+Sn+(−1)n+2an+12.
  2. S20=0.668771403…; S20+S212=161227687232792560=0.692580926…, accurate to within about 0.0006.

7.6.5.7. Determine whether a series is absolutely convergent, conditionally convergent, or divergent.

Answer 1.
Conditionally Convergent
Answer 2.
Divergent
Answer 3.
Absolutely Convergent

7.6.5.9. Practice determining series convergence.

Answer 1.
Diverges
Answer 2.
Diverges
Answer 3.
Converges

7.7 Power Series
7.7.4 Exercises

7.7.4.1. Radius of convergence of a a power series.

Answer.
∞

7.7.4.3. Interval of convergence of a power series.

Answer.
(0,4)

7.7.4.4. Interval of convergence of a power series.

Answer.
[−0.2,0.2]

7.7.4.5. Interval of convergence of a power series.

Answer.
(1−2,12)

7.7.4.6. Radius and interval of convergence of a power series.

Answer 1.
5
Answer 2.
[−13,−3)

7.7.4.7. Radius and interval of convergence of a power series.

Answer 1.
∞
Answer 2.
(−∞,∞)

7.8 Taylor Polynomials
7.8.3 Exercises

7.8.3.1. Determining Taylor polynomials from a function formula.

Answer 1.
1−422x2
Answer 2.
1−422x2+444!x4
Answer 3.
1−422!x2+444!x4−466!x6

7.8.3.2. Determining Taylor polynomials from given derivative values.

Answer 1.
−2+1(x−7)+12!⋅4(x−7)2
Answer 2.
−2+1(x−7)+12!⋅4(x−7)2+13!⋅1(x−7)3
Answer 3.
−2.08
Answer 4.
−2.08017

7.9 Taylor Series
7.9.6 Exercises

7.9.6.1. Finding the Taylor series for a given rational function.

Answer 1.
54
Answer 2.
−5(x−4)42
Answer 3.
5(x−4)243
Answer 4.
−5(x−4)344

7.9.6.2. Finding the Taylor series for a given trigonometric function.

Answer 1.
12
Answer 2.
−1(x−π4)2
Answer 3.
−1(x−π4)222
Answer 4.
1(x−π4)362

7.9.6.3. Finding the Taylor series for a given logarithmic function.

Answer 1.
ln(3)
Answer 2.
13(x−3)
Answer 3.
−(13)22(x−3)2
Answer 4.
(13)33(x−3)3
Answer 5.
−(13)44(x−3)4

7.9.6.4. Finding the Taylor series for a polynomial about a=1.

Answer 1.
−8
Answer 2.
12
Answer 3.
−6
Answer 4.
1
Answer 5.
0

7.9.6.5. Finding the Taylor series for a given exponential function.

Answer.
e(−1)k(x−−1)kk!

7.9.6.6. Using a Taylor series to find high-order derivatives.

Answer.
−4480

7.9.6.7. Taylor series of polynomials.

Answer.
  1. P3(x)=−1+3x−42!x2+63!x3, which is the same polynomial as f(x).
  2. For n≥3, Pn(x)=f(x).
  3. For n≥m, Pn(x)=f(x).

7.9.6.9. Finding coefficients in a power series expansion of a rational function.

Answer 1.
7
Answer 2.
70
Answer 3.
700
Answer 4.
7000
Answer 5.
70000
Answer 6.
0.1

7.9.6.10. Finding coefficients in a power series expansion of a function with arctan⁡(x).

Answer 1.
0
Answer 2.
0
Answer 3.
40
Answer 4.
0
Answer 5.
−333.333333333333
Answer 6.
0.2

7.10 Applications of Taylor Series
7.10.5 Exercises

7.10.5.4. Finding a limit using Taylor series.

Answer.
−0.0238095238095238

7.10.5.6. Estimating a function value with Taylor polynomials.

Answer 1.
3.33333
Answer 2.
3.31481
Answer 3.
3.31687

7.10.5.7. Using a Taylor series to estimate an integral.

Answer 1.
5x33−53x742
Answer 2.
0.284767654

7.10.5.8. Using a Taylor series to estimate a definite integral within a specific accuracy.

Answer.
0.291243

8 Differential Equations
8.1 An Introduction to Differential Equations
8.1.3 Exercises

8.1.3.2. Finding constant to complete solution.

Answer.
3

8.1.3.4. Analyzing Newton’s Law of Cooling.

Answer.
  1. dTdt|T=105=−2; when T=105, the coffee’s temperature is decreasing at an instantaneous rate of −2 degrees F per minute.
  2. T decreasing at t=0.
  3. T(1)≈103 degrees F.
  5. For T<75, T increases. For T>75, T decreases.
  6. Room temperature is 75 degrees F.
  7. Substitute T(t)=75+30e−t/15 in for T in the differential equation dTdt=−115T+5 and verify the equality holds; T(0)=75+30e0=75+30=105; T(t)=75+30e−t/15→75 as t→∞.

8.1.3.5. Population growth.

Answer.
  1. 1<P<3.
  2. P<1 and 3<P<4.
  3. P will not change at all.
  4. The population will decrease toward P=0 with P always being positive.
  5. The population will increase toward P=3 with P always being between 1 and 3.
  6. The population will decrease toward P=3 with P always being above 3.
  7. There’s a maximum threshold of P=3.

8.1.3.6. A look at solutions to differential equations.

Answer.
    1. y(t)=t+1+2et is a solution to the DE.
    2. y(t)=t+1 is a solution to the DE.
    3. y(t)=t+2 is a not solution to the DE.
  2. k=9.

8.2 Qualitative Behavior of Solutions to DEs
8.2.4 Exercises

8.2.4.1. Graphing equilibrium solutions.

Answer.
−3,2

8.2.4.2. Sketching solution curves.

Answer 1.
positive
Answer 2.
negative
Answer 3.
negative
Answer 4.
negative

8.2.4.4. Describing equilibrium solutions.

Answer 1.
−2
Answer 2.
1
Answer 3.
3
Answer 4.
5

8.2.4.5. A look at dydt=t−y.

Answer.
  2. Sketch curves through appropriate points in the slope field above.
  3. y(t)=t−1.
  4. t and y are equal.

8.2.4.6. Slope field and equilibrium solutions of a population growth problem.

Answer.
  2. Any solution curve that starts with P(0)>3 will decrease to P(t)=3 as t→∞; any curve that starts with 1<P(0)<3 will increase to P(t)=3; any curve that starts with 0<P(0)<1 will decrease to P(t)=0.
  3. P=0, P=1, and P=3. P=1 is unstable; P=0 and P=3 are stable.
  4. The population will stabilize either at the value P=3 or at P=0.
  5. P(t)=1 is the threshold.

8.2.4.7. Stable and unstable solutions to a fish population problem.

Answer.
  1. A graph of f against P is given in blue in the figure below. The equilibrium solutions are P=0 (unstable) and P=6 (stable).
  2. dPdt=g(P)=P(6−P)−1 ; the equilibrium at P≈0.172 is unstable; the equilibrium at P≈5.83 is stable.
  3. If P<6−322, then the fish population will die out. If 6−322<, then the fish population will approach 6+322 thousand fish.
  4. dPdt=g(P)=P(6−P)−h; equilibrium solutions P=6+36−4h2, 6−36−4h2.
  5. 9000 fish; harvesting at that rate will maintain the number of fish we start with, provided it’s at least 3000.

8.2.4.8. Setting up and analyzing a differential equation modeling a mice population.

Answer.
  1. dydt=20y.
  2. dydt=20y−Cy2+y
  3. For positive y near 0, M(y)=y2+y≈0; for large values of y, M(y)=y2+y≈1.
  4. The only equilibrium solution is y=0, which is unstable.
  5. The equilibrium solutions are y=0 (stable) and y=1 (unstable).
  6. At least 41 cats.

8.3 Euler’s Method
8.3.4 Exercises

8.3.4.1. A few steps of Euler’s method.

Answer 1.
0.2561
Answer 2.
−1.4

8.3.4.2. Using Euler’s method for a solution of y′=4y.

Answer 1.
0.2
Answer 2.
0.08
Answer 3.
0.032
Answer 4.
0.0128
Answer 5.
0.00512

8.3.4.3. Using Euler’s method with different time steps.

Answer 1.
1352
Answer 2.
1352.52
Answer 3.
1352.785213

8.3.4.4. Using Euler’s method to approximate temperature change.

Answer.
  1. Alice’s coffee: dTAdt|T=100=−0.5(30)=−15 degrees per minute; Bob’s coffee: dTBdt|T=100=−0.1(30)=−3 degrees per minute.
  2. Consider the insulation of the containers.
  3. Alice’s coffee:
    dTAdt=−0.5(TA−(70+10sin⁡t)),
    with the inital condition TA(0)=100.
    t TA(t)
    0.0 100
    0.1 98.5
    0.2 97.12492
    0.3 95.86801
    0.4 94.72237
    ⋮ ⋮
    49.6 65.56715
    49.7 65.48008
    49.8 65.43816
    49.9 65.44183
    50 65.49103
  4. t TA(t)
    0.0 100
    0.1 99.7
    0.2 99.41298
    0.3 99.13872
    0.4 98.87689
    ⋮ ⋮
    49.6 69.39515
    49.7 69.33946
    49.8 69.29248
    49.9 69.25467
    50 69.22638
  5. Compare the rate of initial decrease and amplitude of oscillation.

8.3.4.5. Accelerated convergence.

Answer.
  1. K=1.054; y(1)=2.6991.
  2. K=1.272; y(1)=2.7169.
  3. K=0.122 and y(0.3)=0.0412.

8.3.4.6. The Improved Euler’s Method.

Answer.
  1. y(1)≈y5=2.7027.
  2. y(1)≈y10=2.7141.
  3. The square of Δt.

8.4 Separable differential equations
8.4.3 Exercises

8.4.3.1. Initial value problem for dy/dx=x8y.

Answer.
y=3e15ex55

8.4.3.2. Initial value problem for dy/dt=0.9(y−300).

Answer.
−245e0.5t+300

8.4.3.3. Initial value problem for dy/dt=y2(8+t).

Answer.
122+(1−t)(13+t)⋅6

8.4.3.4. Initial value problem for du/dt=e6u+10t.

Answer.
−ln(e−8⋅3+38−38e8t)3

8.4.3.5. Initial value problem for dy/dx=170yx16.

Answer.
6exp(9x16)

8.4.3.6. Radioactive decay.

Answer.
  1. dMdt=kM.
  2. M(t)=M0ekt.
  3. M(t)=M0e−ln⁡(2)5730t≈M0e−0.000121t
  4. t=5730ln⁡(4)ln⁡(2)≈11460 years.
  5. t=−5730ln⁡(0.3)ln⁡(2)≈9952.8 years.

8.4.3.7. Initial value problem for dy/dt=−ty.

Answer.
  1. y=64−t2.
  2. −8≤t≤8.
  3. y(8)=0.
  4. dydt=−ty is not defined when y=0.

8.4.3.8. Torricelli’s Law.

Answer.
  1. dhdt=kh.
  2. The tank with k=−10 has water leaving the tank much more rapidly.
  3. k=−5.
  4. h(t)=(5−2.5t)2.
  5. 2.5 minutes.
  6. No.

8.4.3.9. The Gompertz equation.

Answer.
  2. P=3 is stable.
  3. P(t)=3eln⁡(13)e−t.
  4. P(t)=3eln⁡(2)e−t.
  5. Yes.

8.5 Modeling with Differential Equations
8.5.4 Exercises

8.5.4.1. Mixing problem.

Answer 1.
0 kg
Answer 2.
0.24−4S2960
Answer 3.
177.6(1−e−0.0013514t)
Answer 4.
9.1176 kg

8.5.4.2. Mixing problem.

Answer 1.
0.05
Answer 2.
27.2009208145593
Answer 3.
0.025

8.5.4.3. Population growth problem.

Answer 1.
260⋅2.718280.298627t
Answer 2.
472.451354136356
Answer 3.
7.13591939674729

8.5.4.4. Radioactive decay problem.

Answer 1.
393.040675410335
Answer 2.
194.204962705311

8.5.4.5. Investment problem.

Answer 1.
k0.06(exp(0.06t)−1)
Answer 2.
28236.5578786596

8.5.4.6. Comparing lottery investment options.

Answer.
  1. dAdt=1+0.05A
  2. A(25)=49.80686 million dollars.
  3. A(25)=34.90343 million dollars.
  4. The first.
  5. t=20ln⁡(2)≈13.86 years.

8.5.4.7. Velocity of a skydiver.

Answer.
  1. dvdt=9.8−kv
  2. v=9.8k is a stable equilibrium.
  3. v(t)=9.8−9.8e−ktk
  4. k=9.8/54≈0.181481.
  5. t=ln⁡(0.5)−0.181481≈3.1894 seconds.

8.5.4.8. Weight gain.

Answer.
  1. dwdt=kw.
  2. w(t)=7t+64; w(12)=148≈12.17 pounds.
  3. The model is unrealistic.

8.5.4.9. Mixing problem and equilibrium solutions.

Answer.
  1. The inflow and outflow are at the same rate.
  2. 60 grams per minute.
  3. S(t)100gramsgallon
  4. 3S(t)100gramsminute.
  5. dSdt=60−3100S.
  6. S=2000 is a stable equilibrium solution.
  7. S(t)=2000−2000e−3100t.
  8. S(t)→2000.

8.5.4.10. Finding the coefficients of the solution to Airy’s Equation.

Answer.
The results from the various part of this exercise show that
y=a0(1+∑k=1∞x3k(2)(3)(5)(6)⋯(3k−1)(3k))=+a1(x+∑k=1∞x3k+1(3)(4)(6)(7)⋯(3k)(3k+1)).

8.6 Population Growth and the Logistic Equation
8.6.4 Exercises

8.6.4.1. Analyzing a logistic equation.

Answer 1.
0.75
Answer 2.
1.5
Answer 3.
2.5
Answer 4.
1.5
Answer 5.
(0,1.5)
Answer 6.
(1.5,∞)
Answer 7.
1.5
Answer 8.
0.75
Answer 9.
(0,1.5)
Answer 10.
(1.5,∞)
Answer 11.
0,1.5
Answer 12.
1.52

8.6.4.2. Analyzing a logistic model.

Answer 1.
0.65
Answer 2.
12.7
Answer 3.
3.65
Answer 4.
1985
Answer 5.
20.8
Answer 6.
72
Answer 7.
75

8.6.4.3. Finding a logistic function for an infection model.

Answer 1.
47501+(4750−20)e−1.6t20
Answer 2.
47502

8.6.4.4. Analyzing a population growth model.

Answer 1.
(0,1.74796)
Answer 2.
(1.74796,∞)
Answer 3.
∞
Answer 4.
ba
Answer 5.
0

8.6.4.5. A logistic equation modeling the spread of a rumor.

Answer.
  1. p(t)→1 as t→∞ provided p(0)>0.
  2. p(t)=19e−0.2t+1.
  3. t=−5ln⁡(1/9)≈10.986 days.
  4. t=−5ln⁡(0.2/9)≈19.033 days.

8.6.4.6. Per capita growth rate of bacteria.

Answer.
  1. dbdt=13000b(15000−b)
  2. b=15000.
  3. When b=7500.
  4. t=−15ln⁡(1/70)≈0.8497 days.

8.6.4.7. A logistic equation modeling a fish population.

Answer.
  1. 10000 fish.
  2. dPdt=0.1P(10−P)−0.2P.
  3. 8000 fish.
  4. P(1)≈8.7899 thousand fish.
  5. t=−1.2ln⁡(5/11)≈0.9461 years.
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