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Applied Calculus

Kevin Gonzales, Eric Hopkins, Catherine Zimmitti, Cheryl Kane, Modified to fit Applied Calculus from Coordinated Calculus by Nathan Wakefield et. al., Based upon Active Calculus by Matthew Boelkins

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Appendix B 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
0.1.11 Exercises

0.1.11.1. Slope and Intercept.

Answer 1.
2
Answer 2.
−113

0.1.11.2. Graphs of Linear Equations.

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

0.1.11.3. Proportionality.

Answer 1.
h
Answer 2.
kh2

0.1.11.4. Finding Lines.

Answer.
2−43−−3(x+3)+4

1 Limits
1.1 Introduction to Limits
1.1.3 Exercises

1.1.3.1. Limits on a piecewise graph.

Answer 1.
5
Answer 2.
9
Answer 3.
none
Answer 4.
−4

1.1.3.2. Estimating a limit numerically.

Answer.
9

1.1.3.3. Limits for a piecewise formula.

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

1.1.3.4. Calculating Limits of Rational Functions.

Answer.
0.5

1.1.3.5. One-Sided Limits.

Answer 1.
6
Answer 2.
−3
Answer 3.
DNE

1.2 Algebraic Limits
1.2.4 Exercises

1.2.4.1. Limits on a piecewise graph.

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

1.2.4.2. Limits for a piecewise formula.

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

1.2.4.3. Calculating Limits of Rational Functions.

Answer.
0.2

1.2.4.4. One-Sided Limits.

Answer 1.
16
Answer 2.
0
Answer 3.
DNE

1.2.4.5. Evaluating a limit algebraically.

Answer.
−14

1.3 Limits to Infinity
1.3.3 Exercises

1.3.3.1. Calculating Limits of Rational Functions.

Answer.
0.125

1.3.3.2. Evaluating a limit algebraically.

Answer.
−16

1.4 Continuous Functions
1.4.6 Exercises

1.4.6.1. Types of discontinuity.

Answer.
2

1.4.6.2. Types of discontinuity.

Answer.
1

1.4.6.4. 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

1.4.6.5. Determining continuity from a graph.

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

1.4.6.6. Interpretting continuity.

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

1.4.6.7. Values that make a function continuous.

Answer.
5⋅4

1.4.6.8. Values that make a function continuous.

Answer.
−158

1.4.6.9. Values that make a function continuous.

Answer.
8⋅24

2 Derivatives
2.1 The Average Rate of Change
2.1.3 Exercises

2.1.3.1. Estimating derivative values graphically.

Answer 1.
0.905829
Answer 2.
−0.5
Answer 3.
3.22474
Answer 4.
0.355567
Answer 5.
2.25

2.1.3.2. Tangent line to a curve.

Answer 1.
1.2
Answer 2.
8
Answer 3.
1.2
Answer 4.
−1⋅(−0.05)0.03

2.1.3.3. Interpreting values and slopes from a graph.

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

2.1.3.4. Estimating a derivative value graphically.

Answer.
36

2.1.3.5. Estimating a derivative from the limit definition.

Answer.
354.9

2.1.3.6. Using a graph.

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

2.1.3.7. 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.

2.1.3.8. 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).

2.1.3.9. 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.

2.2 The Derivative Function
2.2.5 Exercises

2.2.5.1. Applying the limit definition of the derivative.

Answer 1.
4(x+h)2−6−(4x2−6)
Answer 2.
4⋅2x

2.2.5.2. Sketching the derivative.

Answer.
7

2.2.5.3. Comparing function and derivative values.

Answer 1.
x6
Answer 2.
x4
Answer 3.
x5
Answer 4.
x3

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

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

2.2.5.5. Derivative from Tangent Line.

Answer 1.
−5
Answer 2.
9−−5−7−−8

2.2.5.6. 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.

2.2.5.7. 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.

2.2.5.8. 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.

2.2.5.9. Graphing the Derivative Function.

Answer.

2.3 Differentiability
2.3.5 Exercises

2.3.5.1. Continuity and differentiability of a graph.

Answer 1.
3
Answer 2.
3

2.3.5.4. Continuity and differentiability of a graph.

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

2.3.5.5. 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.

2.3.5.6. 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.

2.4 Derivative Rules
2.4.5 Exercises

2.4.5.1. Derivative of a power function.

Answer.
1112x1112−1

2.4.5.2. Derivative of a rational function.

Answer.
−1⋅13x−1⋅(13+1)

2.4.5.3. Derivative of a root function.

Answer.
18x18−1

2.4.5.4. Derivative of a quadratic function.

Answer.
2⋅8t−8

2.4.5.5. Derivative of a sum of power functions.

Answer.
15⋅8t8−1−92t−12−11t2

2.4.5.6. Simplifying a product before differentiating.

Answer.
2⋅1+12x2⋅1−12+22x−12

2.4.5.7. Simplifying a quotient before differentiating.

Answer.
3x2x−(x3+12)x2

2.4.5.8. Determining where f′(x)=0.

Answer.
4,−6

2.5 The Tangent Line
2.5.4 Exercises

2.5.4.1. Sign of tangent line slope.

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

2.5.4.2. Sign of tangent line slope.

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

2.5.4.3. Equation of tangent line for polynomial.

Answer 1.
6⋅3x2−3⋅2x−7
Answer 2.
77
Answer 3.
77(x−(−2))+(−55)

2.5.4.4. Equation of tangent line for polynomial.

Answer 1.
7⋅4x3−2x
Answer 2.
9590
Answer 3.
9590(x−7)+16757

2.5.4.5. Equation of tangent line for non-polynomial.

Answer 1.
8⋅1.5x0.5−512x
Answer 2.
19.3412
Answer 3.
19.3412(x−3)+37.909

2.5.4.6. Equation of tangent line for non-polynomial.

Answer.
−0.888889(x−(−3))+(−3.16667)

2.5.4.7. Equation of tangent line.

Answer.
−11(x−1)+11.3333

2.6 The Product and Quotient Rules
2.6.5 Exercises

2.6.5.1. Derivative of a quotient of linear functions.

Answer.
3(7t+2)−7(3t+10)(7t+2)2

2.6.5.2. Derivative of a rational function.

Answer.
4r4−1(6r+8)−6r4(6r+8)2

2.6.5.3. Product and quotient rules with graphs.

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

2.6.5.4. Product and quotient rules with given function values.

Answer 1.
3⋅1+2⋅5
Answer 2.
2⋅5−3⋅15⋅5

2.6.5.5. Product rule.

Answer.
(2t+5)(6t−2+4t−3)+(t2+5t+3)(−2⋅6t−3−3⋅4t−4)

2.6.5.6. Product rule and tangent lines.

Answer.
288(x−2)+−93

2.6.5.7. Quotient rule and tangent lines.

Answer.
−0.177515x+0.961538−−0.177515⋅5

2.6.5.8. 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.

2.6.5.9. 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.

2.6.5.10. 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.

2.6.5.11. 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.

2.6.5.12. 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.

2.7 The Chain Rule
2.7.6 Exercises

2.7.6.1. Chain rule with graphs.

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

2.7.6.2. Chain rule with function values.

Answer 1.
4
Answer 2.
4⋅1
Answer 3.
4
Answer 4.
3⋅1
Answer 5.
1⋅4−3⋅142

2.7.6.4. Evaluating the derivative using chain rule.

Answer 1.
−3(3x+3)−4⋅3
Answer 2.
−3⋅(3⋅5+3)−3−1⋅3

2.7.6.8. 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.

2.7.6.9. 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.

2.7.6.10. 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.

2.8 Higher-Order Derivatives
2.8.4 Exercises

2.8.4.1. Comparing f,f′,f″ values.

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

2.8.4.2. Signs of f,f′,f″ values.

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

2.8.4.3. Acceleration from velocity.

Answer 1.
34.09 (ft(s2
Answer 2.
25 (ft(s2

2.8.4.4. Rates of change of stock values.

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

2.8.4.5. Interpreting a graph of f′.

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

2.8.4.6. Higher order derivatives.

Answer 1.
24x7−6x5+4x3
Answer 2.
168x6−30x4+12x2
Answer 3.
1008x5−120x3+24x
Answer 4.
5040x4−360x2+24

2.8.4.7. Chain rule with higher order derivatives.

Answer 1.
9t9t2+11
Answer 2.
99(9t2+11)1.5

2.8.4.8. Quotient rule with higher order derivatives.

Answer 1.
36(4−5x)2
Answer 2.
360(4−5x)3

2.8.4.9. Computing velocity and acceleration.

Answer 1.
4t3−196t
Answer 2.
12t2−196
Answer 3.
1600
Answer 4.
−480
Answer 5.
−88

2.8.4.10. 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.

2.8.4.11. Interpreting a graph of f′.

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

2.8.4.12. 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.

2.8.4.13. Sketching functions.

Answer.

3 Exponentials and Logarithms
3.1 Exponential Functions
3.1.4 Exercises

3.1.4.1. General Exponential Functions.

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

3.1.4.4. Half-Life.

Answer.
878.371 hr

3.1.4.5. Applied Half-Life.

Answer 1.
No
Answer 2.
346.076

3.1.4.7. Mixing rules: chain, product, sum.

Answer.
5e5x(x2+6x)+e5x(2x+ln(6)⋅6x)

3.1.4.8. Mixing rules: chain and product.

Answer.
6t6−1e−1ct−ct6e−1ct

3.1.4.9. Using the chain rule repeatedly.

Answer.
−1⋅5te−1⋅5t2e−1⋅5t2+8

3.3 Derivatives of Log Functions
3.3.3 Exercises

3.3.3.1. Composite function involving logarithms and polynomials.

Answer.
9t8t9+7

3.3.3.2. Composite function from a graph.

Answer 1.
4⋅44−1(4.4−4)1.3−1
Answer 2.
1.3−14.4−4

3.3.3.3. Derivatives and properties of logs.

Answer 1.
3x
Answer 2.
3x

3.3.3.4. Derivatives using properties of logs.

Answer.
44x+(−7)−99x+62

3.3.3.5. Tangent lines using logs.

Answer.
0.75(x−4)+3ln(4)

4 Applications of the Derivative
4.1 First Derivative Test
4.1.5 Exercises

4.1.5.1. Increasing and decreasing.

Answer 1.
(−14,−8)∪(−2,4)
Answer 2.
(−8,−2)
Answer 3.
−14,−8,−2,4
Answer 4.
(−14,−11)∪(−5,1)
Answer 5.
(−11,−5)∪(1,4)

4.1.5.2. Finding critical values.

Answer.
45

4.1.5.5. Identifying relative extrema.

Answer 1.
1,−2
Answer 2.
0

4.1.5.6. Determining intervals of increasing and decreasing and finding extrema.

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

4.2 Second Derivative Test
4.2.4 Exercises

4.2.4.2. Finding critical points and inflection points.

Answer 1.
0
Answer 2.
−0.447213595499958,0.447213595499958
Answer 3.
0
Answer 4.
−0.447213595499958,0.447213595499958

4.2.4.3. Finding inflection points.

Answer.
−7,0.125

4.2.4.4. Matching graphs of f,f′,f″.

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

4.2.4.7. 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.

4.2.4.8. 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.

4.2.4.9. 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|.

4.2.4.10. 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.

4.3 Absolute Extrema
4.3.3 Exercises

4.3.3.1. Finding absolute Extrema.

Answer 1.
−122
Answer 2.
1606

4.3.3.2. Finding absolute Extrema.

Answer 1.
−80
Answer 2.
1520

4.3.3.3. Analyzing Function Behavior.

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

4.3.3.5. Conditions for When absolute Extrema May Occur.

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

4.3.3.6. Finding Extrema on Closed and Bounded Intervals.

Answer.
  1. absolute maximum p(0)=p(a)=0; absolute minimum p(a3)=−2a333.
  2. absolute max r(1b)≈0.368ab; absolute min r(2b)≈0.270ab.
  3. absolute minimum g(b)=a(1−e−b2); absolute maximum g(3b)=a(1−e−3b2).

4.3.3.7. Conditions for Where absolute Extrema May Occur.

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

4.4 Applied Optimization
4.4.5 Exercises

4.4.5.1. Maximizing the volume of a box.

Answer 1.
10.7218×4.72184
Answer 2.
1.63908

4.4.5.2. Minimizing the cost of a container.

Answer.
$321.85

4.4.5.3. Maximizing area contained by a fence.

Answer.
4410 ft2

4.4.5.4. Minimizing the area of a poster.

Answer 1.
15.3137 cm
Answer 2.
45.9411 cm

4.4.5.5. Maximizing the area of a rectangle.

Answer 1.
3.83
Answer 2.
7.33333

4.4.5.6. Maximizing the volume of a closed box.

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

4.4.5.7. Maximizing pasture area with limited fencing.

Answer.
Exercise Answer

4.4.5.8. Minimizing cable length.

Answer.
172.047 feet of cable.

4.4.5.9. Minimizing construction costs.

Answer.
The minimum cost is $1165.70.

4.5 Tangent Line Approximations
4.5.6 Exercises

4.5.6.1. Approximating value of unknown function.

Answer 1.
3
Answer 2.
3
Answer 3.
8
Answer 4.
2.2

4.5.6.2. Approximating value of square root.

Answer 1.
0.142857x+7
Answer 2.
7.02857
Answer 3.
x2+2x+49−0.142857x−7
Answer 4.
0.00278693

4.5.6.3. Approximating value of natural log.

Answer 1.
−5x+0
Answer 2.
0.4
Answer 3.
log(−5x+1)−(−5)x−0
Answer 4.
−0.0635278

4.5.6.6. Using marginal profit.

Answer 1.
1155+12
Answer 2.
1155+4⋅12
Answer 3.
1155−12

4.6 Elasticity of Demand
4.6.5 Exercises

4.6.5.2. Maximizing revenue.

Answer 1.
450p2225(150−p2)
Answer 2.
1503

4.6.5.3. Interpreting elasticity and revenue.

Answer 1.
0.01p
Answer 2.
0.07
Answer 3.
10.01

4.7 Derivatives of Functions Given Implicitly
4.7.3 Exercises

4.7.3.1. Implicit differentiation in a polynomial equation.

Answer.
1−5x4yx5−3

4.7.3.2. Implicit differentiation in an equation with logarithms.

Answer.
y(8−xln(y))x(x+2y2)

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

Answer.
−711

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

Answer.
x+7y=10

4.8 Related Rates
4.8.3 Exercises

4.8.3.1. Related rates with polynomial equations.

Answer.
−2⋅4⋅53+2⋅(−2)(−3)42⋅5⋅4⋅(−3)3

4.8.3.2. Height of a conical pile of gravel.

Answer.
120289π

4.8.3.3. Movement of a shadow.

Answer.
6.4

4.8.3.4. Docking a boat.

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

4.8.3.5. A leaking conical tank.

Answer.
704196.4446

4.8.3.6. 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.

5 Integration
5.1 Antiderivatives from Formulas
5.1.4 Exercises

5.1.4.1. Finding Antiderivatives.

Answer.
5t15

5.1.4.2. Finding Antiderivatives (Constants).

Answer 1.
x
Answer 2.
y
Answer 3.
πz

5.1.4.3. Finding Antiderivatives (Polynomials).

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

5.1.4.4. Finding Antiderivatives (Exponential).

Answer.
13e3x

5.1.4.5. Finding Antiderivatives.

Answer.
25t52−21t12+C

5.1.4.6. Finding Antiderivatives (Simplifying).

Answer.
−0.666667x3+1.5x2+(−1)x+(−6)ln(x)+C

5.1.4.8. Applying Antiderivatives.

Answer 1.
8t−2t2+72
Answer 2.
8⋅5−2⋅52+72

5.2 Determining Area Under a Curve
5.2.9 Exercises

5.2.9.1. Evaluating definite integrals from graphical information.

Answer 1.
−21
Answer 2.
7
Answer 3.
−14
Answer 4.
28

5.2.9.2. Estimating definite integrals from a graph.

Answer 1.
−4
Answer 2.
−A

5.2.9.3. 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.

5.2.9.4. 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.

5.3 The Definite Integral
5.3.4 Exercises

5.3.4.1. Evaluating definite integrals.

Answer 1.
42.6666666666667
Answer 2.
6
Answer 3.
48.75

5.3.4.3. Area under a curve with no bounds.

Answer.
4.89897948556636

5.3.4.4. Application of definite integrals.

Answer.
3⋅(9500−2000)−0.005⋅(95002−20002)+2×10−6(95003−20003)

5.3.4.5. Another application of definite integrals.

Answer.
11520

5.4 Properties of Integration
5.4.5 Exercises

5.4.5.1. Finding the average value of a linear function.

Answer.
29.5

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

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

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

Answer.
5

5.4.5.4. Using rules to combine known integral values.

Answer 1.
2−9−6
Answer 2.
−2⋅(−13)+9⋅13−9⋅10

5.4.5.6. Area between two curves.

Answer.
23+2⋅2−2⋅33

5.4.5.7. 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.

5.4.5.8. 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.

5.6 Integration by Substitution
5.6.5 Exercises

5.6.5.1. Product involving a 4th power polynomial.

Answer.
(t4−6)416+C

5.6.5.2. Fraction involving ln9⁡(x).

Answer.
0.166667ln6(z)+C

5.6.5.3. Fraction involving e5x.

Answer.
0.25ln(e4x+7)+C

5.6.5.4. Fraction involving e5y.

Answer.
12e3y3+C

5.6.5.5. Working with negative exponents.

Answer.
(x+13)−1−1

5.6.5.6. Fraction involving sums of exponential functions.

Answer.
ln(|10ex−7e−x|)

5.6.5.7. Integral involving a rational function.

Answer.
0.2(x5+(−9))−4−4+C

5.6.5.8. Integral of a partial fraction.

Answer.
1.02961941718116

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

Answer.
−2.5

5.6.5.10. 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.

5.6.5.11. Definite integral with a clever substitution.

Answer.
−378−12815

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

Answer.
−83(4−x)4−x+25(4−x)24−x
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