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Coordinated Calculus

Appendix C Answers to Selected Exercises

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.
\(\frac{-11}{3}\)

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.
\(kh^{2}\)

0.1.12.4. Finding Lines.

Answer.
\(\frac{2-4}{3--3}\!\left(x+3\right)+4\)

0.2 Exponential and Logarithmic Functions
0.2.4 Exercises

0.2.4.1. General Exponential Functions.

Answer 1.
\(8\)
Answer 2.
\(e^{2}\)
Answer 3.
\(\text{growth}\)

0.2.4.4. Half-Life.

Answer.
\(878.371\ {\rm hr}\)

0.2.4.5. Applied Half-Life.

Answer 1.
\(\text{No}\)
Answer 2.
\(337.461\)

0.3 Trigonometric Functions
0.3.6 Exercises

0.3.6.1. Period and Amplitude.

Answer 1.
\(\frac{2\pi }{3}\)
Answer 2.
\(0.5\)

0.3.6.3. Finding Trigonometric Functions.

Answer.
\(6\sin\!\left(6\pi x\right)+4\)

0.3.6.4. The Unit Circle.

Answer 1.
\(1\cdot 2^{2}-1\cdot \left(2-2\right)^{2}\)
Answer 2.
\(8\cos\!\left(4\right)\)
Answer 3.
\(8\sin\!\left(4\right)\)

0.3.6.5. Trigonometric Functions as Compositions.

Answer 1.
\(x^{4}\)
Answer 2.
\(\tan\!\left(x\right)\)
Answer 3.
\(\tan\!\left(x\right)\)
Answer 4.
\(\tan\!\left(x\right)\)
Answer 5.
\(\tan\!\left(x\right)\)
Answer 6.
\(x^{4}\)

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.
\(5\cdot 4\)

1.1.5.8. Values that make a function continuous.

Answer.
\(\frac{-15}{8}\)

1.1.5.9. Values that make a function continuous.

Answer.
\(8\cdot 2^{4}\)

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.
\(\text{none}\)
Answer 4.
\(-5\)

1.2.6.2. Estimating a limit numerically.

Answer.
\(4\)

1.2.6.3. Limits for a piecewise formula.

Answer 1.
\(3\cdot 2-3\)
Answer 2.
\(3\cdot 2-3\)
Answer 3.
\(3\cdot 2-3\)

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.
\(\text{DNE}\)

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.
\(2.5\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}\)
Answer 2.
\(4\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}\)

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.
\(\text{g}\)
Answer 2.
\(\text{both have an equal rate of change}\)
Answer 3.
\(\text{f}\)
Answer 4.
\(\text{g}\)
Answer 5.
\(\text{g}\)

1.3.4.5. Matching a distance graph to velocity.

Answer.
\(3\)

1.3.4.6. Interpretting average velocity and instantaneous velocity.

Answer.
  1. \(s(15)-s(0) \approx -98.75\text{.}\)
  2. \begin{align*} AV_{[0,15]} &= \frac{s(15)-s(0)}{15-0} \approx -6.58\\ AV_{[0,2]} &= \frac{s(2)-s(0)}{2-0} \approx -47.63\\ AV_{[1,6]} &= \frac{s(6)-s(1)}{6-1} \approx -13.25\\ AV_{[8,10]} &= \frac{s(10)-s(8)}{10-8} \approx -7.35 \end{align*}
  3. Most negative average velocity on \([0,4]\text{;}\) most positive average velocity on \([4,8]\text{.}\)
  4. \(\frac{21.31+22.25}{2} = 21.78\) 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, \(h(2.45) = 0\text{.}\)
  2. \(AV_{[2.45,7]} \approx \frac{-3.5-0}{7-2.45}=\frac{-3.5}{4.55}=-0.7692\) m/sec. The average velocity is not the same on every time interval within \([2.45,7]\text{.}\)
  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. \(15 957\) 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 \((b,f(b))\text{.}\)
  4. \(AV_{[0,20]} \approx 798\) people per year.
  5. \begin{align*} AV_{[5,10]} & \approx 734.50\\ AV_{[5,9]} & \approx 733.06\\ AV_{[5,8]} & \approx 731.62\\ AV_{[5,7]} & \approx 730.19\\ AV_{[5,6]} & \approx 728.7535 \end{align*}

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.
\(\frac{-1\cdot \left(-0.02\right)}{0.07}\)

1.4.3.3. Interpreting values and slopes from a graph.

Answer 1.
\({\verb!<!}\)
Answer 2.
\({\verb!<!}\)
Answer 3.
\({\verb!>!}\)
Answer 4.
\({\verb!>!}\)

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.
  1. \(AV_{[-3,-1]} \approx 1.15\text{;}\) \(AV_{[0,2]} \approx -0.4\text{.}\)
  2. \(f'(-3) \approx 3\text{;}\) \(f'(0) \approx -\frac{1}{2}\text{.}\)

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 \((-3,3)\text{,}\) \((-1,-2)\text{,}\) \((0,-3)\text{,}\) \((1,-2)\text{,}\) and \((3,-1)\) and draw the desired tangent lines accordingly.
  2. For instance, you could draw a function \(g\) that passes through the points \((-2,3)\text{,}\) \((-1,2)\text{,}\) \((1,0)\text{,}\) \((2,0)\text{,}\) 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 \(1\text{.}\)

1.4.3.8. Population Growth.

Answer.
  1. \(AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679\) billion people per year; \(P'(7) \approx 0.1762\) billion people per year; \(P'(7) \gt AV_{[0,7]}\text{.}\)
  2. \(AV_{[19,29]} \approx 0.02234\) billion people/year.
  3. We will say that today’s date is July 1, 2015, which means that \(t = 22.5\text{;}\)
    \begin{equation*} P'(22.5) = \lim_{h \to 0} \frac{115(1.014)^{22.5+h}-115(1.014)^{22.5}}{h}; \end{equation*}
    \(P'(22.5) \approx 0.02186\) billions of people per year.
  4. \(y - 1.57236 = 0.02186(t-22.5)\text{.}\)

1.4.3.9. Using the limit definition of the derivative.

Answer.
  1. All three approaches show that \(f'(2) = 1\text{.}\)
  2. All three approaches show that \(f'(1) = -1\text{.}\)
  3. All three approaches show that \(f'(1) = \frac{1}{2}\text{.}\)
  4. All three approaches show that \(f'(1)\) does not exist.
  5. The first two approaches show that \(f'(\frac{\pi}{2}) = 0\text{.}\)

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.
\(3\!\left(x+h\right)^{2}-4-\left(3x^{2}-4\right)\)
Answer 2.
\(3\cdot 2x\)

1.5.3.4. Sketching the derivative.

Answer.
\(\text{1}\)

1.5.3.5. Comparing function and derivative values.

Answer 1.
\(\text{x6}\)
Answer 2.
\(\text{x3}\)
Answer 3.
\(\text{x5}\)
Answer 4.
\(\text{x2}\)

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

Answer 1.
\(-1\)
Answer 2.
\(-1\)
Answer 3.
\(\frac{-1}{4}\)
Answer 4.
\(\frac{-1}{16}\)

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 \(f(x) = \frac{1}{2}x^2 - 1\text{.}\)

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

Answer.
  1. \(g'(x) = 2x - 1\text{.}\)
  2. \(p'(x) = 10x - 4\text{.}\)
  3. 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 \(p'(x)= 10x - 4\text{.}\) The leading coefficient \(5\) in \((x) = 5x^2 - 4x + 12\) leads to the coefficient of ``\(10\)’’ in \(p'(x) = 10x -4\text{.}\)

1.5.3.9. Graphing functions based on continuity and derivatives.

Answer.
  1. \(g\) is linear.
  2. On \(-3.5 \lt x \lt -2\text{,}\) \(-2 \lt x \lt 0\) and \(2 \lt x \lt 3.5\text{.}\)
  3. At \(x = -2, 0, 2\text{;}\) \(g\) 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.
\(\text{negative}\)
Answer 2.
\({\text{degC/min}}\)
Answer 3.
\(30\)
Answer 4.
\(\text{temperature}\)
Answer 5.
\(51\ {\rm degC}\)
Answer 6.
\(\text{decrease}\)
Answer 7.
\(1.125\ {\rm degC}\)

1.6.5.2. A cost function.

Answer 1.
\(\text{gallons}\)
Answer 2.
\(\text{dollars}\)
Answer 3.
\(\text{gallons}\)
Answer 4.
\({\text{dollars/gallon}}\)

1.6.5.3. Weight as a function of calories.

Answer 1.
\(\text{cal}\)
Answer 2.
\(\text{lb}\)
Answer 3.
\(\text{cal}\)
Answer 4.
\({\text{lb/cal}}\)
Answer 5.
\(\text{lb}\)
Answer 6.
\(\text{cal}\)
Answer 7.
\({\text{lb/cal}}\)
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. \(F'(10) \approx -3.33592\text{.}\)
  2. The coffee’s temperature is decreasing at about \(3.33592\) degrees per minute.
  3. \(F'(20)\text{.}\)
  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. \(t=0\text{.}\)
  2. \(v'(1) = -32\text{.}\)
  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. \(AV_{[40000,55000]} \approx -0.153\) dollars per mile.
  2. \(h'(55000) \approx -0.147\) dollars per mile. During \(55 0001\)st mile, we expect the car’s value to drop by \(0.147\) dollars.
  3. \(h'(30000) \lt h'(80000)\text{.}\)
  4. The graph of \(h\) might have the general shape of the graph of \(y = e^{-x}\) for positive values of \(x\text{:}\) 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.
\(\text{positive}\)
Answer 2.
\(\text{positive}\)
Answer 3.
\(\text{negative}\)

1.7.6.2. Signs of \(f, f', f''\) values.

Answer 1.
\(\text{zero}\)
Answer 2.
\(\text{negative}\)
Answer 3.
\(\text{zero}\)
Answer 4.
\(\text{negative}\)
Answer 5.
\(\text{negative}\)
Answer 6.
\(\text{negative}\)
Answer 7.
\(\text{zero}\)
Answer 8.
\(\text{positive}\)
Answer 9.
\(\text{zero}\)
Answer 10.
\(\text{positive}\)
Answer 11.
\(\text{positive}\)
Answer 12.
\(\text{positive}\)
Answer 13.
\(\text{positive}\)
Answer 14.
\(\text{positive}\)
Answer 15.
\(\text{positive}\)

1.7.6.3. Acceleration from velocity.

Answer 1.
\(30\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s^{2}}}\)
Answer 2.
\(22\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s^{2}}}\)

1.7.6.4. Rates of change of stock values.

Answer 1.
\(\text{positive}\)
Answer 2.
\(\text{negative}\)
Answer 3.
\(\text{positive}\)
Answer 4.
\(\text{positive}\)

1.7.6.5. Interpreting a graph of \(f'\).

Answer 1.
\(\text{x6}\)
Answer 2.
\(\text{x1}\)
Answer 3.
\(\text{x4}\)
Answer 4.
\(\text{x2}\)
Answer 5.
\(\text{x3}\)
Answer 6.
\(\text{x1}\)

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 \(x=2\text{.}\)
  2. Greater.
  3. Less.

1.7.6.7. Interpreting a graph of \(f'\).

Answer.
  1. \(g'(2) \approx 1.4\text{.}\)
  2. At most one.
  3. \(9\text{.}\)
  4. \(g''(2) \approx 5.5 \text{.}\)

1.7.6.8. Using data to interpret derivatives.

Answer.
  1. \(h'(4.5) \approx 14.3\text{;}\) \(h'(5) \approx 21.2\text{;}\) \(h'(5.5) \approx = 23.9\text{;}\) rising most rapidly at \(t = 5.5\text{.}\)
  2. \(h'(5) \approx 9.6 \text{.}\)
  3. Acceleration of the bungee jumper in feet per second per second.
  4. \(0 \lt t \lt 2\text{,}\) \(6 \lt t \lt 10\text{.}\)

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.
\(\text{none}\)
Answer 2.
\(2, 4, 6\)

1.8.6.2. Continuity and differentiability of a graph.

Answer.
  1. \(a = 0\text{.}\)
  2. \(a = 0, 3\text{.}\)
  3. \(a = -2, 0, 1, 2, 3\text{.}\)

1.8.6.3. Examples of functions.

Answer.
  1. \(f(x) = |x-2|\text{.}\)
  2. Impossible.
  3. Let \(f\) be the function defined to be \(f(x) = 1\) for every value of \(x \ne -2\text{,}\) and such that \(f(-2) = 4\text{.}\)

1.8.6.4. Estimating the derivative at a point.

Answer.
  1. At \(x = 0\text{.}\)
    \begin{align*} g'(0) & = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}\\ & = \lim_{h \to 0} \frac{\sqrt{|h|} - \sqrt{|0|}}{h}\\ & = \lim_{h \to 0} \frac{\sqrt{|h|}}{h} \end{align*}
  2. \(h\) \(0.1\) \(0.01\) \(0.001\) \(0.0001\) \(-0.1\) \(-0.01\) \(-0.001\) \(-0.0001\)
    \(\sqrt{|h|}/h\) \(3.162\) \(10\) \(31.62\) \(100\) \(-3.162\) \(-10\) \(-31.62\) \(-100\)
    \(g'(0)\) 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.
\(\frac{15}{16}x^{\frac{15}{16}-1}\)

2.1.5.2. Derivative of a rational function.

Answer.
\(-1\cdot 19x^{-1\cdot \left(19+1\right)}\)

2.1.5.3. Derivative of a root function.

Answer.
\(\frac{1}{2}x^{\frac{1}{2}-1}\)

2.1.5.4. Derivative of a quadratic function.

Answer.
\(2\cdot 3t-6\)

2.1.5.5. Derivative of a sum of power functions.

Answer.
\(6\cdot 4t^{4-1}-\frac{5}{2}t^{\frac{-1}{2}}-\frac{14}{t^{2}}\)

2.1.5.6. Simplifying a product before differentiating.

Answer.
\(\frac{2\cdot 3+1}{2}x^{\frac{2\cdot 3-1}{2}}+\frac{8}{2}x^{\frac{-1}{2}}\)

2.1.5.7. Simplifying a quotient before differentiating.

Answer.
\(\frac{6x^{5}x-\left(x^{6}+8\right)}{x^{2}}\)

2.1.5.8. Finding a tangent line equation.

Answer.
\(32\!\left(x-2\right)+18\)

2.1.5.9. Determining where \(f'(x) = 0\).

Answer.
\(8, -12\)

2.1.5.10. Calculating derivative values from two tables.

Answer.
  1. \(h(2) = 27\text{;}\) \(h'(2) = -19/2\text{.}\)
  2. \(L(x) = 27 - \frac{19}{2}(x-2)\text{.}\)
  3. \(p\) is increasing at \(x=2\text{.}\)
  4. \(p(2.03) \approx -11.56\text{.}\)

2.1.5.11. Calculating derivative values from two graphs.

Answer.
  1. \(p\) is not differentiable at \(x=-1\) and \(x=1\text{;}\) \(q\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)
  2. \(r\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)
  3. \(r'(-2) = 4\text{;}\) \(r'(0) = 1\text{.}\)
  4. \(y = 4\text{.}\)

2.1.5.12. Using the sum and constant multiple rules.

Answer.
  1. \(w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\text{.}\)
  2. \(L(t) = (\frac{3}{\sqrt{2}} - \frac{2\pi}{3}) + (\frac{3}{\sqrt{2}}(\ln(\frac{1}{2}) + 1) + \frac{4}{\sqrt{3}})(t-2)\text{.}\)
  3. \(v\) is increasing at \(t = \frac{1}{2}\text{.}\)

2.1.5.13. Understanding the derivative of an exponential function.

Answer.
  1. \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a^{x+h}-a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x} \cdot a^{h} - a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h}\text{,} \end{align*}
  2. Since \(a^x\) does not depend at all on \(h\text{,}\) we may treat \(a^x\) as constant in the noted limit and thus write the value \(a^x\) in front of the limit being taken.
  3. When \(a = 2\text{,}\) \(L \approx 0.6931\text{;}\) when \(a = 3\text{,}\) \(L \approx 1.0986\text{.}\)
  4. \(a \approx 2.71828\) (for which \(L \approx 1.000\))
  5. \(\frac{d}{dx}[2^x] = 2^x \cdot \ln(2)\) and \(\frac{d}{dx}[3^x] = 3^x \cdot \ln(3)\)
  6. \begin{equation*} \frac{d}{dx}[e^x] = e^x\text{.} \end{equation*}

2.2 The Sine and Cosine Functions
2.2.3 Exercises

2.2.3.1. Compute the Derivative.

Answer.
\(7\cos\!\left(x\right)-6\sin\!\left(x\right)\)

2.2.3.2. Compute the Derivative.

Answer.
\(2\cdot 9x+2\sin\!\left(x\right)\)

2.2.3.3. Derivative Computation and Evaluation.

Answer 1.
\(2\cos\!\left(x\right)-10\sin\!\left(x\right)\)
Answer 2.
\(2\cos\!\left(5.23599\right)-10\sin\!\left(5.23599\right)\)

2.2.3.4. Find the Tangent Line.

Answer 1.
\(\frac{2\sqrt{3}}{2}\)
Answer 2.
\(1-1.73205\cdot 0.523599\)

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.
\(\frac{\pi }{2}\)

2.2.3.7. Horizontal Tangent Lines.

Answer.
\(\frac{2\pi }{3}, \frac{4\pi }{3}\)

2.2.3.8. Making Two Graphs Tangent For Specific Values.

Answer 1.
\(\sin\!\left(\frac{3\pi }{4}\right)e^{\frac{3\pi }{4}}\)
Answer 2.
\(\frac{3\pi }{4}\)
Answer 3.
\(\frac{1}{\sqrt{2}}\)

2.2.3.9. An Elastic Band.

Answer 1.
\(-2\sin\!\left(t\right)+3\cos\!\left(t\right)\)
Answer 2.
\(2.55\)
Answer 3.
\(3.61\)

2.2.3.10. Analyzing the Value of an Investment.

Answer.
  1. \(V'(2) = 24 \cdot 1.07^2 \cdot \ln(1.07) + 6 \cos(2) \approx -0.63778\) thousands of dollars per year.
  2. \(V''(2) = 24 \cdot 1.07^2 \cdot \ln(1.07)^2 - 6 \sin(2) \approx -5.33\) 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 \(6\sin(t)\) 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. \(f'\left(\frac{\pi}{4}\right) = -5\left(\frac{\sqrt{2}}{2}\right)\) .
  2. \(L(x) = 3+2(x-\pi)\text{.}\)
  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. \(\lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) = 0\) and \(\lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = 1\) .
  4. \(f'(x) = \sin(x) \cdot 0 + \cos(x) \cdot 1\text{.}\)
  5. Hint: \(\cos(\alpha + \beta)\) is \(\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)\text{.}\)

2.3 The Product and Quotient Rules
2.3.6 Exercises

2.3.6.1. Derivative of a basic product.

Answer.
\(13^{x}+x\ln\!\left(13\right)\cdot 13^{x}\)

2.3.6.2. Derivative of a product.

Answer.
\(\left(9x^{9-1}-\frac{1}{9}x^{-1+\frac{1}{9}}\right)\cdot 6^{x}+\left(x^{9}-x^{\frac{1}{9}}\right)\ln\!\left(6\right)\cdot 6^{x}\)

2.3.6.3. Derivative of a quotient of linear functions.

Answer.
\(\frac{2\!\left(6t+15\right)-6\!\left(2t+7\right)}{\left(6t+15\right)^{2}}\)

2.3.6.4. Derivative of a rational function.

Answer.
\(\frac{3r^{3-1}\!\left(5r+6\right)-5r^{3}}{\left(5r+6\right)^{2}}\)

2.3.6.5. Derivative of a product of trigonometric functions.

Answer.
\(6\!\left(-1\sin\!\left(q\right)\sin\!\left(q\right)+\cos\!\left(q\right)\cos\!\left(q\right)\right)\)

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

Answer.
\(5x^{5-1}\cos\!\left(x\right)-x^{5}\sin\!\left(x\right)\)

2.3.6.7. Derivative of a sum that involves a product.

Answer.
\(\sin\!\left(t\right)+t\cos\!\left(t\right)+\frac{1}{\cos^{2}\!\left(t\right)}\)

2.3.6.8. Product and quotient rules with graphs.

Answer 1.
\(0.666667\cdot 1.33333+0\cdot 1.33333\)
Answer 2.
\(\frac{-4\cdot \left(-1\right)--1\cdot \left(-2\right)}{\left(-1\right)^{2}}\)

2.3.6.9. Product and quotient rules with given function values.

Answer 1.
\(3\cdot 4+3\cdot 4\)
Answer 2.
\(\frac{3\cdot 4-3\cdot 4}{4\cdot 4}\)

2.3.6.10. Tangent lines using given function values.

Answer.
  1. \(h(2) = -15\text{;}\) \(h'(2) = 23/2\text{.}\)
  2. \(L(x) = -15 + 23/2(x-2)\text{.}\)
  3. Increasing.
  4. \(r(2.06) \approx -0.5796\text{.}\)

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

Answer.
  1. \(w'(t) = t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}} \text{.}\)
  2. \(L(t) \approx 0.740-0.589(t-0.5)\text{.}\)
  3. Increasing.

2.3.6.12. Product and quotient rules with analysis of graphs.

Answer.
  1. \(r'(-2) = 2\) and \(r'(0) = 0.25\text{.}\)
  2. At \(x = -1\) and \(x = 1\text{.}\)
  3. \(L(x) = 2\text{.}\)
  4. \(z'(0) = \frac{1}{16}\) and \(z'(2) = -1\text{.}\)
  5. At \(x = -1\text{,}\) \(x = 1\text{,}\) \(x = -1.5\text{,}\) and \(x = 1\text{.}\)

2.3.6.13. An application to crop yield.

Answer.
  1. \(C(t) = A(t)Y(t)\) bushels in year \(t\text{.}\)
  2. \(1 190 000\) bushels of corn.
  3. \(C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}\)
  4. \(C'(0) = 158 000\) bushels per year.
  5. \(C(1) \approx 1 348 000\)bushels.

2.3.6.14. An application to fuel consumption.

Answer.
  1. \(g(80) = 20\) kilometers per liter, and \(g'(80) = -0.16\text{.}\) 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.
\(\tan\!\left(t\right)+t\frac{1}{\cos^{2}\!\left(t\right)}+-1\sin\!\left(t\right)\)

2.4.3.2. A quotient involving \(\tan t\).

Answer 1.
\(\frac{5\sec^{2}\!\left(x\right)x-5\tan\!\left(x\right)}{x^{2}}\)
Answer 2.
\(17.1673\)

2.4.3.3. A quotient of trigonometric functions.

Answer 1.
\(\frac{\sec^{2}\!\left(x\right)\sec\!\left(x\right)-\left(\tan\!\left(x\right)-2\right)\sec\!\left(x\right)\tan\!\left(x\right)}{\sec^{2}\!\left(x\right)}\)
Answer 2.
\(-2.16725\)

2.4.3.4. A quotient that involves a product.

Answer 1.
\(\frac{\left(2\cdot 2x\tan\!\left(x\right)+2x^{2}\sec^{2}\!\left(x\right)\right)\sec\!\left(x\right)-2x^{2}\tan\!\left(x\right)\sec\!\left(x\right)\tan\!\left(x\right)}{\sec^{2}\!\left(x\right)}\)
Answer 2.
\(4.44649\)

2.4.3.5. Finding a tangent line equation.

Answer 1.
\(3\cdot 2\)
Answer 2.
\(3-6\cdot 0.785398\)

2.4.3.6. Oscillatory Motion.

Answer.
  1. \(h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575\) feet per second.
  2. \(h''(2) = \frac{\cos(2)(-2 + 2ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193\) feet per second per second.
  3. The object is falling and slowing down.

2.4.3.7. A product of trigonometric functions.

Answer.
  1. \(f'(x) = \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x)\text{.}\)
  2. False.
  3. \(f'(x) = \frac{-\sin^2(x)}{\sin(x)} = -\sin(x)\) for \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\text{.}\)

2.4.3.8. Combining Differentiation Rules.

Answer.
  1. \(\displaystyle p'(z) = \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z\)
  2. \(y - 4 = 3(x-0)\text{.}\)
  3. Increasing.

2.5 The Chain Rule
2.5.5 Exercises

2.5.5.1. Mixing rules: chain, product, sum.

Answer.
\(5e^{5x}\!\left(x^{2}+5^{x}\right)+e^{5x}\!\left(2x+\ln\!\left(5\right)\cdot 5^{x}\right)\)

2.5.5.2. Mixing rules: chain and product.

Answer.
\(6t^{6-1}e^{-1ct}-ct^{6}e^{-1ct}\)

2.5.5.3. Using the chain rule repeatedly.

Answer.
\(\frac{-1\cdot 5te^{-1\cdot 5t^{2}}}{\sqrt{e^{-1\cdot 5t^{2}}+7}}\)

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.
\(6\cdot 5\)
Answer 3.
\(5\)
Answer 4.
\(7\cdot 2\)
Answer 5.
\(\frac{2\cdot 3-1\cdot 5}{3^{2}}\)

2.5.5.7. A product involving a composite function.

Answer.
\(2\sin\!\left(4x\right)+2\cdot 4x\cos\!\left(4x\right)\)

2.5.5.8. Using the chain rule to compare composite functions.

Answer.
  1. \(h'\left( \frac{\pi}{4} \right) = \frac{3}{2\sqrt{2}}\text{.}\)
  2. \(r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875 \gt h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779\text{;}\) \(r\) 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. \(p'(x) = e^{u(x)} \cdot u'(x)\text{.}\)
  2. \(q'(x) = u'(e^x) \cdot e^x\text{.}\)
  3. \(r'(x) = -\csc^2(u(x) \cdot u'(x)\text{.}\)
  4. \(s'(x) = u'(\cot(x)) \cdot (-\csc^2(x))\text{.}\)
  5. \(a'(x) = u'(x^4) \cdot 4x^3\text{.}\)
  6. \(b'(x) = 4(u(x))^3 \cdot u'(x)\text{.}\)

2.5.5.10. More on using the chain rule with graphs.

Answer.
  1. \(C'(0) = 0\) and \(C'(3) = -\frac{1}{2}\text{.}\)
  2. Consider \(C'(1)\text{.}\) By the chain rule, we’d expect that \(C'(1) = p'(q(1)) \cdot q'(1)\text{,}\) but we know that \(q'(1)\) does not exist since \(q\) has a corner point at \(x = 1\text{.}\) This means that \(C'(1)\) does not exist either.
  3. Since \(Y(x) = q(q(x))\text{,}\) the chain rule implies that \(Y'(x) = q'(q(x)) \cdot q'(x)\text{,}\) and thus \(Y'(-2) = q'(q(-2)) \cdot q'(-2) = q'(-1) \cdot q'(-2)\text{.}\) 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 \(Z'(x) = q'(p(x)) \cdot p'(x)\text{.}\) Therefore \(Z'(0) = q'(p(0)) \cdot p'(0) = q'(-0.5) \cdot p'(0) = 0 \cdot 0.5 = 0\text{.}\)

2.5.5.11. Applying the chain rule in a physical context.

Answer.
  1. \(\frac{dV}{dh} = \pi \left(8h-h^2 \right)\text{,}\) cubic feet per foot.
  2. \(\frac{dV}{dt} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(\pi t) + 1) \cdot \pi \cos(\pi t) - 3 (\sin(\pi t) + 1)^2 \cdot \pi \cos(\pi t) \right]\) cubic feet per hour.
  3. \(\left. \frac{dV}{dt} \right|_{t=0} = 7 \pi^2\) 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.
\(\frac{3t^{2}}{t^{3}+9}\)

2.6.6.2. Composite function involving trigonometric functions and logarithms.

Answer.
\(\frac{-1\sin\!\left(\ln\!\left(t\right)\right)}{t}\)

2.6.6.3. Product involving \(\arcsin(w)\).

Answer.
\(5\sin^{-1}\!\left(w\right)+\frac{5w}{\sqrt{1-w^{2}}}\)

2.6.6.4. Derivative involving \(\arctan(x)\).

Answer.
\(0\)

2.6.6.5. Composite function from a graph.

Answer 1.
\(\frac{4\cdot 5^{4-1}\!\left(5.1-5\right)}{2.3-2}\)
Answer 2.
\(\frac{2.3-2}{5.1-5}\)

2.6.6.6. Composite function involving an inverse trigonometric function.

Answer.
\(7\frac{1}{\sqrt{1-\left(x^{2}\right)^{2}}}\cdot 2x\)

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

Answer.
\(8\cdot 3x^{3-1}\tan^{-1}\!\left(9x^{3}\right)+\frac{8x^{3}\cdot 9\cdot 3x^{3-1}}{1+9\cdot 9x^{2\cdot 3}}\)

2.6.6.8. Mixing rules: product and inverse trig.

Answer.
\(8\cos\!\left(x\right)\frac{1}{\sqrt{1-x^{2}}}-8\sin\!\left(x\right)\sin^{-1}\!\left(x\right)\)

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

Answer.
  1. \(f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\text{.}\)
  2. \(r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\text{.}\)
  3. \(q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1+(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\text{.}\)
  4. \(\displaystyle g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2}\)

2.6.6.10. Graphs of inverse functions.

Answer.
  1. \(f'(1) \approx 2\text{.}\)
  2. \(f'(-1) \approx 1/2\text{.}\)

2.6.6.11. Differentiating the inverse of a cubic polynomial.

Answer.
  1. \(f\) passes the horizontal line test.
  2. \(f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\text{.}\)
  3. \(f'(x) = \frac{3}{4}x^2\text{;}\) \(f'(2) = 3\text{.}\) \(g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\text{;}\) \(g'(6) = \frac{1}{3}\text{.}\) 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 \(y\text{.}\)
  3. \((h^{-1})'(\frac{\pi}{2} + 1) = 1\text{.}\)

2.7 Derivatives of Functions Given Implicitly
2.7.3 Exercises

2.7.3.1. Implicit differentiation in a polynomial equation.

Answer.
\(\frac{1-2xy}{x^{2}-3}\)

2.7.3.2. Implicit differentiation in an equation with logarithms.

Answer.
\(\frac{y\!\left(8-x\ln\!\left(y\right)\right)}{x\!\left(x+3y^{3}\right)}\)

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

Answer.
\(\frac{3x^{3-1}y-y^{3}-x^{2\cdot 3}y^{3+2}}{3xy^{3-1}+3x^{2\cdot 3+1}y^{3+1}-x^{3}}\)

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

Answer.
\(-{\frac{7}{6}}\)

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: \((0,-1)\text{,}\) \((0,-0.618)\text{,}\) \((0,1.618)\text{,}\) \((1,-1)\text{,}\) \((1,-0.618)\text{,}\) \((1,1.618)\text{,}\) \((0.5,-1.0493)\text{,}\) \((0.5,0.2104)\text{,}\) \((0.5, 1.6139)\text{.}\) Vertical tangent lines: \((-0.1756,-0.379)\text{,}\) \((0.2912,-0.379)\text{,}\) \((0.7088,-0.379)\text{,}\) \((1.1756,-0.379)\text{,}\) \((-0.8437, 1.235)\text{,}\) and \((1.8437, 1.235)\text{.}\)

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

Answer.
\(y = \frac{\pi}{2} - \left(x-\frac{\pi}{2}\right)\text{.}\)

2.7.3.8. Revisiting exponential derivatives using implicit differentiation.

Answer.
  1. \(x = \frac{\ln(y)}{\ln(a)}\text{.}\)
  2. \(1 = \frac{1}{\ln(a)} \cdot \frac{1}{y}\frac{dy}{dx}\text{.}\)
  3. \(\frac{d}{dx}[a^x] = a^x \ln(a)\text{.}\)

2.8 Derivatives of Hyperbolic Functions
2.8.6 Exercises

2.8.6.1. Simplifying Hyperbolic Trigonometric Functions.

Answer.
\(\frac{t+\frac{1}{t}}{2}\)

2.8.6.2. Limits of Hyperbolic Functions.

Answer.
\(2\)

2.8.6.8. Derivatives of Hyperbolic Functions.

Answer.
\(\frac{-6\sinh\!\left(x\right)}{\left(2+\cosh\!\left(x\right)\right)^{2}}\)

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 \(\sqrt{x}\).

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.
\(\text{under}\)

2.9.4.3. Estimating With the Local Linearization.

Answer.
\(68\)

2.9.4.4. Predicting Behavior From the Local Linearization.

Answer 1.
\(\text{negative}\)
Answer 2.
\({\text{degC/min}}\)
Answer 3.
\(35\)
Answer 4.
\(\text{temperature}\)
Answer 5.
\(59\ {\rm degC}\)
Answer 6.
\(\text{decrease}\)
Answer 7.
\(0.3\ {\rm degC}\)

2.9.4.5. Using the Local Linearization to Analyze a Function.

Answer.
  1. \(p(3) = -1\) and \(p'(3) = -2\text{.}\)
  2. \(p(2.79) \approx -0.58\text{.}\)
  3. Too large.

2.9.4.6. Using the Local Linearization with Physical Context.

Answer.
  1. \(F'(60)\approx 1.56\) degrees per minute.
  2. \(L(t) \approx 1.56(t-60)+324.5\text{.}\)
  3. \(F(63)\approx L(63)\approx = 329.18\) degrees F.
  4. Overestimate.

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

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

2.9.4.8. Estimating a Function Through its Derivative.

Answer.
  1. \(x=1\text{.}\)
  2. On \(-0.37 \lt x \lt 1.37\text{;}\) \(f\) is concave up.
  3. \(f(1.88) \approx 0.02479\text{,}\) and this estimate is larger than the true value of \(f(1.88)\text{.}\)

2.10 The Mean Value Theorem
2.10.3 Exercises

2.10.3.1. Understanding the Statement of the Mean Value Theorem.

Answer 1.
\(\text{no}\)
Answer 2.
\(\text{no}\)
Answer 3.
\(\text{there is no such point}\)

2.10.3.2. Applying Theorems.

Answer 1.
\(\text{False}\)
Answer 2.
\(\text{True}\)
Answer 3.
\(\text{False}\)
Answer 4.
\(\text{False}\)
Answer 5.
\(\text{True}\)

2.10.3.3. Conclusion of the Mean Value Theorem.

Answer.
\(\sqrt{8}\)

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 \(-1 \lt x lt 1\) and for \(x \gt 1\text{;}\) \(f'\) is negative for all \(x \lt -1\text{.}\) \(f\) has a local minimum at \(x = -1\text{.}\)
  2. A possible graph of \(y = f''(x)\) is shown at right in the figure.
  3. \(f''(x)\) is negative for \(-0.35 \lt x \lt 1\text{;}\) \(f''(x)\) is positive everywhere else; \(f\) has points of inflection at \(x \approx -0.35\) and \(x = 1\text{.}\)
  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. \(g''(2) = 0\text{;}\) \(g''\) is negative for \(1 \lt x \lt 2\) and positive for \(2 \lt x \lt 3\text{.}\)
  3. \(g\) has a point of inflection at \(x = 2\text{.}\)

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 \(3\text{.}\)
  4. \(h\) is concave up everywhere; \(h\) is almost linear for large values of \(|x|\text{.}\)

3.1.4.7. Applying derivative tests.

Answer.
  1. \(p''(x)\) is negative for \(-1 \lt x \lt 2\) and positive for all other values of \(x\text{;}\) \(p\) has points of inflection at \(x = -1\) and \(x = 2\text{.}\)
  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.
\(\left(-\infty ,-1\right)\)
Answer 3.
\(\left(-1,\infty \right)\)
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 \(x = a\text{.}\)
  3. Global minimum at \(x = a\text{;}\) global maximum at \(x = b\text{.}\)
  4. Not enough information is provided.

3.2.4.6. Finding Extrema on Closed and Bounded Intervals.

Answer.
  1. Global maximum \(p(0) = p(a) = 0\text{;}\) global minimum \(p\left( \frac{a}{\sqrt{3}} \right) = -\frac{2a^3}{3\sqrt{3}}\text{.}\)
  2. Global max \(r\left( \frac{1}{b} \right) \approx 0.368 \frac{a}{b}\text{;}\) global min \(r\left( \frac{2}{b} \right) \approx 0.270 \frac{a}{b}\text{.}\)
  3. Global minimum \(g(b) = a(1-e^{-b^2})\text{;}\) global maximum \(g(3b) = a(1-e^{-3b^2})\text{.}\)
  4. Global max \(s\left( \frac{\pi}{2k} \right) = 1\text{;}\) global min \(s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}\text{.}\)

3.2.4.7. Conditions for Where Global Extrema May Occur.

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

3.2.4.8. Using the Extreme Value Theorem.

Answer.
  1. Global max \(s(\frac{5\pi}{12}) = 8\text{;}\) global min \(s(\frac{11\pi}{12}) = 2\text{.}\)
  2. Global max \(s(\frac{5\pi}{12}) = 8\text{;}\) global min \(s(0) = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\text{.}\)
  3. Global max \(s(\frac{5\pi}{12}) = 8\text{;}\) global min \(s(\frac{11\pi}{12}) = 2\text{.}\) (There are other points at which the function achieves these values on the given interval.)
  4. Global max \(s(\frac{5\pi}{12}) = 8\text{;}\) global min \(s(\frac{\pi}{3}) = 6.5\text{.}\)

3.3 Applied Optimization
3.3.3 Exercises

3.3.3.1. Maximizing the volume of a box.

Answer 1.
\(10.1633\times6.16333\)
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.
\(9922.5\ {\rm ft^{2}}\)

3.3.3.4. Minimizing the area of a poster.

Answer 1.
\(13.8489\ {\rm cm}\)
Answer 2.
\(55.3954\ {\rm cm}\)

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 \(V\left( \sqrt{\frac{5}{3}} \right) = \frac{15}{12}\left( \sqrt{\frac{5}{3}} \right) - \frac{1}{4}\left( \sqrt{\frac{5}{3}} \right)^3 \approx 1.07583\) 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.
\(\frac{C}{2}\)

3.4.3.2. Using the graph of \(g'\).

Answer 1.
\(1\)
Answer 2.
\(-2\)
Answer 3.
\(2\)
Answer 4.
\(\left(-10,\infty \right)\)
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.
\(\frac{a+b}{2}\)
Answer 2.
\(\text{DNE}\)

3.4.3.6. Behavior of a function with parameters.

Answer 1.
\(a+\ln\!\left(bx\right)+1\)
Answer 2.
increasing
Answer 3.
\(\frac{1}{x}\)
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.
\(-\infty \)

3.4.3.8. Analyzing families of functions.

Answer.
  1. \(x = 0\) and \(x = \frac{2a}{3}\text{.}\)
  2. \(x = \frac{a}{2}\text{;}\) \(p''(x)\) changes sign from negative to positive at \(x = \frac{a}{2}\text{.}\)
  3. As we increase the value of \(a\text{,}\) both the location of the critical number and the inflection point move to the right along with \(a\text{.}\)

3.4.3.9. Analyzing families of functions.

Answer.
  1. \(x=c\) is a vertical asymptote because \(\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty\) and \(\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty\text{.}\)
  2. \(\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0\text{;}\) \(\lim_{x \to -\infty} \frac{e^{-x}}{x-c} = -\infty\text{.}\)
  3. The only critical number for \(q\) is \(x=c-1\text{.}\)
  4. When \(x \lt c-1\text{,}\) \(q'(x) \gt 0\text{;}\) when \(x \gt c-1\text{,}\) \(q'(x) \lt 0\text{;}\) \(q\) has a local maximum at \(x = c-1\text{.}\)

3.4.3.10. Analyzing families of functions.

Answer.
  1. \(x = m\text{.}\)
  2. \(E\) is increasing for \(x \lt m\) and decreasing for \(x \gt m\text{,}\) with a local maximum at \(x = m\text{.}\)
  3. \(x = m \pm s\text{.}\)
  4. \(\lim_{x \to \infty} E(x) = \lim_{x \to -\infty} E(x) = 0\text{.}\)

3.5 Related Rates
3.5.3 Exercises

3.5.3.1. Height of a conical pile of gravel.

Answer.
\(\frac{25}{72\pi }\)

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 \(\frac{13}{6} \approx 2.167\) feet per second.

3.5.3.5. Filling a swimming pool.

Answer.
The depth of the water is increasing at
\begin{equation*} \left. \frac{dh}{dt}\right|_{h = 5} = 1.28 \end{equation*}
feet per minute. The depth of the water is increasing at a decreasing rate.

3.5.3.6. Baseball player and umpire.

Answer.
\(\left. \frac{d\theta}{dt} \right|_{x = 30} = -0.24\) radians per second.

3.5.3.7. A conical pile of sand.

Answer.
\(\left. \frac{dh}{dt}\right|_{V=1000} = \frac{10}{\pi \left( \sqrt[3]{\frac{3000}{\pi}} \right)^2} \approx 0.0328\) 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.
\(\text{undefined}\)
Answer 2.
\(\text{zero}\)

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

Answer.
\(\frac{1}{18}\)

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.
\(\lim\limits_{x \to 3} h(x) = -2\text{.}\)

3.6.5.6. Analyzing a family of functions.

Answer.
Horizontal asymptote: \(y = \frac{3}{5}\text{;}\) vertical asymptote: \(x = c\text{;}\) hole: \((a, \frac{3(a-b)}{5(a-c)})\text{.}\) \(R\) is not continuous at \(x = a\) and \(x = c\text{.}\)

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

Answer.
  1. \(\ln(x^{2x}) = 2x \cdot \ln(x)\text{.}\)
  2. \(x = \frac{1}{\frac{1}{x}}\text{.}\)
  3. \(\lim\limits_{x \to 0^+} h(x) = 0\text{.}\)
  4. \(\lim\limits_{x \to 0^+} g(x) = \lim\limits_{x \to 0^+} x^{2x} = 1\text{.}\)

3.6.5.8. Dominance.

Answer.
  1. Show that \(\lim\limits_{x \to \infty}\frac{\ln(x)}{\sqrt{x}} = 0\text{.}\)
  2. Show that \(\lim\limits_{x \to \infty}\frac{\ln(x)}{\sqrt[n]{x}} = 0\text{.}\)
  3. Consider \(\lim\limits_{x \to \infty} \frac{p(x)}{e^x}\) By repeated application of LHR, the numerator will eventually be simply a constant (after \(n\) applications of LHR), and thus with \(e^x\) still in the denominator, the overall limit will be \(0\text{.}\)
  4. Show that \(\lim\limits_{x \to \infty} \frac{\ln(x)}{x^n} = 0\)
  5. For example, \(f(x) = 3x^2 + 1\) and \(g(x) = -0.5x^2 + 5x - 2\text{.}\)

3.7 Parametric Equations
3.7.4 Exercises

3.7.4.1. Evaluating Parametric Equations.

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

3.7.4.3. Parametric Lines.

Answer 1.
\(\text{N}\)
Answer 2.
\(\text{Y}\)
Answer 3.
\(\text{Y}\)
Answer 4.
\(\left(8,4\right)\)
Answer 5.
\(\left(-20,-12\right)\)
Answer 6.
\(\left(-13,-8\right)\)
Answer 7.
\(\left(-\infty ,-1.14286\right)\)

3.7.4.4. Slope in Parametric Equations.

Answer 1.
\(2\)
Answer 2.
\(\text{to the right}\)
Answer 3.
\(\text{up}\)
Answer 4.
\(-2\)
Answer 5.
\(\text{to the left}\)
Answer 6.
\(\text{up}\)

3.7.4.5. Derivatives of Parametric Equations.

Answer 1.
\(\sqrt{\sin^{2}\!\left(t\right)+\frac{\cos^{2}\!\left(\frac{t}{2}\right)}{4}}\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}\)
Answer 2.
\(3.14159\ {\rm s}\)
Answer 3.
\(n\pi \ {\rm s}\)

3.7.4.6. Tangent Lines and Parametric Equations.

Answer 1.
\(16+12\!\left(t-2\right)\)
Answer 2.
\(48+72\!\left(t-2\right)\)

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.
\(134\ {\rm ft}\)
Answer 2.
\(328\ {\rm ft}\)

4.1.6.2. Distance from a linear velocity function.

Answer.
\(750\ {\rm m}\)

4.1.6.3. Change in position from a linear velocity function.

Answer.
\(8\ {\rm cm}\)

4.1.6.5. Finding average acceleration from velocity data.

Answer 1.
\(28.64\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s^{2}}}\)
Answer 2.
\(21\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s^{2}}}\)

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 \(t = 1\text{,}\) \(12\) miles north of the lake.
  2. \(s(4) - s(0) = 0\text{.}\)
  3. \(40\) miles.

4.1.6.8. Physical interpretations of velocity.

Answer.
  1. \(t = \frac{500}{32} = \frac{125}{8} = 15.625\) is when the rocket reaches its maximum height.
  2. \(A = 3906.25\text{,}\) the vertical distance traveled on \([0, 15.625]\text{.}\)
  3. \(s(t) = 500t - 16t^2\text{.}\)
  4. \(s(15.625) - s(0) = 3906.25\) is the change of the rocket’s position on \([0,15.625]\text{.}\)
  5. \(s(5) - s(1) = 1616\text{;}\) the rocket rose \(1616\) feet on \([1,5]\text{.}\)

4.1.6.9. Physical interpretations of velocity.

Answer.
  1. \(\frac{1}{2} + \frac{1}{4} \pi \approx 1.285\text{.}\)
  2. \(s(5)-s(2) = -2\) is the change in position of the object on \([2,5]\text{.}\)
  3. On the time interval \([5,7]\text{.}\)
  4. \(s\) is increasing on the intervals \((0,2)\) and \((5,7)\text{;}\) the position function has a relative maximum at \(t=2\text{.}\)

4.1.6.10. Pollution Data.

Answer.
  1. 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.
  2. 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.
\(\text{there is ambiguity}\)
Answer 3.
\(17.7\)
Answer 4.
\(\text{there is ambiguity}\)

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

Answer 1.
\(3\ {\rm ft}\)
Answer 2.
\(19\ {\rm ft}\)
Answer 3.
\(-1\ {\rm ft}\)
Answer 4.
\(17\ {\rm ft}\)

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. \(\text{MID}(4) = 43.5\text{.}\)
  2. \(A = \frac{87}{2}\text{.}\)
  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.
    \(\text{MID}(n)\) will give the exact area for any value of \(n\text{.}\) Neither \(\text{LEFT}(n)\) nor \(\text{RIGHT}(n)\) will be exact for any \(n\text{.}\)
  4. For any linear function \(g\) of the form \(g(x) = mx + b\) such that \(g(x) \ge 0\) on the interval of interest.

4.2.5.5. Identifying and manipulating Riemann sum components.

Answer.
  1. \(f(x) = x^2 + 1\) on the interval \([1,3]\text{.}\)
  2. If \(S\) is a left Riemann sum, \(f(x) = x^2 + 1\) on the interval \([1.4, 3.4]\text{.}\) If \(S\) is a middle Riemann sum, \(f(x) = x^2 + 1\) on the interval \([1.2, 3.2]\text{.}\)
  3. The area under \(f(x) = x^2 + 1\) on \([1,3]\text{.}\)
  4. \(\text{RIGHT}(10) = \sum_{i=1}^{10} \left( (1+0.2i)^2 + 1 \right) \cdot 0.2\text{.}\)

4.2.5.6. Evaluating Riemann sums with data.

Answer.
  1. \(\text{MID}(3) = 99.6\) feet.
  2. \(\text{LEFT}(6) = 114 \text{,}\)
    \begin{equation*} \text{RIGHT}(6) = 84\text{,} \end{equation*}
    and \(\frac{1}{2}(\text{LEFT}(6) + \text{RIGHT}(6)) = 99\text{.}\)
  3. \(114\) feet.

4.2.5.7. Evaluating Riemann sums with graphs and formulas.

Answer.
  1. \(\text{MID}(4) \approx 6.4\text{.}\)
  2. The total tonnage of pollution escaping the scrubbing process in the time interval \([0,4]\) weeks.
  3. \(\text{LEFT}(5) \approx 5.19620599\text{.}\)
  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.
\(\frac{1}{4}\)
Answer 2.
\(\frac{3}{8}\)
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.
\(-2\cdot \left(-5\right)+5\cdot 1-5\cdot \left(-1\right)\)

4.3.5.7. Using definite integrals on a velocity function.

Answer.
  1. The total change in position is \(P = \int_0^{4} v(t) dt\text{.}\)
  2. \(P = -2.625 \) feet.
  3. \(D = 3.375 \) feet.
  4. \(AV = -0.65625 \) feet per second.
  5. \(s(t) = -t^2+t\text{.}\)

4.3.5.8. Riemann sum estimates and definite integrals.

Answer.
  1. The total change in position, \(P\text{,}\) is \(P = \int_0^1 v(t) \, dt + \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt = \int_0^4 v(t) \, dt\text{.}\)
  2. \(P = \int_0^4 v(t) \, dt \approx 2.665\text{.}\)
  3. The total distance traveled, \(D\text{,}\) is \(D = \int_0^1 v(t) \, dt - \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt\text{.}\)
  4. \(D \approx 8.00016\text{.}\)
  5. \begin{equation*} v_{\operatorname{AVG} [0,4]} \approx = 0.66625 \end{equation*}
    feet per second.

4.3.5.9. Using the Sum and Constant Multiple Rules.

Answer.
  1. \(\int_0^1 [f(x) + g(x)] \,dx = 1-\frac{\pi}{4}\text{.}\)
  2. \(\int_1^4 [2f(x) - 3g(x)] \, dx = -\frac{15}{2} - 3\pi\text{.}\)
  3. \(h_{\operatorname{AVG} [0,4]} = \frac{5}{8} + \frac{3\pi}{16}\text{.}\)
  4. \(c = -\frac{3}{8} + \frac{3\pi}{16}\text{.}\)

4.3.5.10. Finding the area of a bounded region.

Answer.
  1. \(A_1 = \int_{-1}^{1} (3-x^2) \, dx\text{.}\)
  2. \(A_2 = \int_{-1}^{1} 2x^2 \, dx\text{.}\)
  3. The exact area between the two curves is \(\int_{-1}^{1} (3-x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx\text{.}\)
  4. Use the sum rule for definite integrals over the same interval.
  5. Think about subtracting the area under \(q\) from the area under \(p\text{.}\)

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.
\(\left(2,7\right)\)
Answer 11.
\(\left(0,2\right)\)
Answer 12.
\(0\)
Answer 13.
\(2\)

4.4.4.2. Using Graphs to Evaluate.

Answer 1.
\(4-\frac{\left(x-4\right)\!\left(0.5x-2\right)}{2}\)
Answer 2.
\(\frac{-2t^{2}}{8}+2t\)

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.
\(\left(0,-8.5\right), \left(2,-2\right), \left(6,-10\right), \left(8,-8\right)\)

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) \(\frac{1}{925}\) \(\frac{1}{875}\) \(\frac{1}{830}\) \(\frac{1}{780}\) \(\frac{1}{730}\) \(\frac{1}{685}\) \(\frac{1}{635}\) \(\frac{1}{585}\) \(\frac{1}{535}\) \(\frac{1}{490}\) \(\frac{1}{440}\)
  2. The antiderivative function tells the total number of minutes it takes for the plane to climb to an altitude of \(h\) feet.
  3. \(M = \int_{0}^{10000} m(h) \, dh \text{.}\)
  4. It takes the plane aabout \(M_5 \approx 15.27\) 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. \(s(1) = \frac{5}{3}\text{,}\) \(s(3) = -1\text{,}\) \(s(5) = -\frac{11}{3}\text{,}\) \(s(6) = -\frac{5}{2}\text{.}\)
  2. \(s\) is increasing on \(0 \lt t \lt 1\) and \(5 \lt t \lt 6\text{;}\) decreasing for \(1 \lt t \lt 5\text{.}\)
  3. \(s\) is concave down for \(t \lt 2\text{;}\) concave up for \(t \gt 2\text{.}\)
  4. \(s(t) = -2t + \frac{1}{6}(t-3)^3 + 5\text{.}\)

5.1.5.6. Calories Burned.

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

5.1.5.7. Functions defined by integrals.

Answer.
  1. \(B(-1) = -1\text{,}\) \(B(0) = 0\text{,}\) \(B(1) = \frac{1}{2}\text{,}\) \(B(2) = 0\text{,}\) \(B(3) = -1\text{,}\) \(B(4) = -\frac{3}{2}\text{,}\) \(B(5) = -1\text{,}\) \(B(6) = 0\text{.}\) Also, \(A(x) = 1+B(x)\) and \(C(x) = B(x) - \frac{1}{2}\text{.}\)
    \(x\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)
    \(A(x)\) \(0\) \(1\) \(1.5\) \(1\) \(0\) \(-0.5\) \(0\) \(1\)
    \(B(x)\) \(-1\) \(0\) \(0.5\) \(0\) \(-1\) \(-1.5\) \(-1\) \(0\)
    \(C(x)\) \(-1.5\) \(-0.5\) \(0\) \(-0.5\) \(-1.5\) \(-2\) \(-1.5\) \(-0.5\)
  2. \(A\text{,}\) \(B\text{,}\) and \(C\) are vertical translations of each other.
  3. \(A' = f\text{.}\)

5.2 Antiderivatives from Formulas
5.2.4 Exercises

5.2.4.1. Finding Antiderivatives.

Answer.
\(4t^{\frac{1}{4}}\)

5.2.4.2. Finding Antiderivatives (Constants).

Answer 1.
\(x\)
Answer 2.
\(y\)
Answer 3.
\(\pi z\)

5.2.4.3. Finding Antiderivatives (Polynomials).

Answer 1.
\(\frac{x^{2}}{2}\)
Answer 2.
\(ax\)
Answer 3.
\(t^{2}x\)

5.2.4.4. Finding Antiderivatives.

Answer.
\(\frac{8}{5}t^{\frac{5}{2}}-\frac{2}{t^{\frac{1}{2}}}+C\)

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 (\(n^{th}\)-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.
\(\frac{-1}{11}+\frac{1}{5}\)

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. \(\displaystyle v_{\operatorname{AVG} [12,24]} = 12.5 \) meters per minute.
  3. The object’s maximum acceleration is \(3\) meters per minute per minute at the instant \(t = 2\text{.}\)
  4. \(c = \frac{60}{24} = 2.5\text{.}\)

5.2.4.16. Evaluating the definite integral of a piecewise function.

Answer.
  1. \(-\frac{5}{6}\text{.}\)
  2. \(\displaystyle f_{\operatorname{AVG} [0,5]} = \frac{1}{2}\text{.}\)
  3. \(g(x) = f(x)\) for \(0 \le x \lt 5\) and \(g(x) = -\frac{5}{4}(x - 5)\) on \(5 \le x \le 7\text{.}\)

5.3 Differential Equations
5.3.4 Exercises

5.3.4.1. General Solution to a Differential Equation.

Answer.
\(\frac{2}{3}kt^{\frac{3}{2}}\)

5.3.4.2. Setting up a Differential Equation.

Answer.
\(\frac{75\cdot 5280}{7\cdot 3600}\)

5.3.4.3. Setting up a Differential Equation.

Answer 1.
\(32\sqrt{\frac{230}{16}}\)
Answer 2.
\(\frac{15\cdot 121.326}{22}\)

5.3.4.4. Displacement.

Answer.
\(58.6666666666667\)

5.3.4.5. Initial Value Problems.

Answer 1.
\(\frac{3}{16}\)
Answer 2.
\(\frac{3}{4}\)
Answer 3.
\(\frac{3}{4}\)
Answer 4.
\(3\)

5.3.4.6. Initial Velocity.

Answer.
\(-35\)

5.3.4.7. Finding the Position Function.

Answer.
\(5t-\cos\!\left(t\right)+-6\)

5.3.4.8. Finding the Position Function Given Acceleration.

Answer.
\(\frac{t^{4}}{12}+\frac{t^{3}}{6}-\frac{3t^{2}}{2}\)

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.
\(-9\ {\rm ft}\)
Answer 2.
\(-4\ {\rm ft}\)
Answer 3.
\(14\ {\rm ft}\)
Answer 4.
\(\left(6,8\right]\)
Answer 5.
\(\left[0,5\right)\)
Answer 6.
\(\left[5,6\right]\)

5.3.4.12. Finding Displacement from a Graph.

Answer 1.
\(-7\ {\rm ft}\)
Answer 2.
\(3\ {\rm ft}\)
Answer 3.
\(17\ {\rm ft}\)
Answer 4.
\(\left(4,8\right]\)
Answer 5.
\(\left[0,3\right)\)
Answer 6.
\(\left[3,4\right]\)

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.
\(\left(1.25,6\right)\)
Answer 4.
\(3.68034\)

5.4.5.2. Variable in the lower limit.

Answer.
\(-1\tan\!\left(\ln\!\left(x\right)\right)\)

5.4.5.3. Approximating a function with derivative \(e^{-\frac{x^2}{5}}\).

Answer.
\(5.86716\)

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

Answer.
\(F\) is increasing on \(x \lt -1\text{,}\) \(0.5 \lt x \lt 4\text{,}\) and \(5 \lt x \lt 6.5\text{;}\) decreasing on \(-1 \lt x \lt 0.5\) and \(4 \lt x \lt 5\text{;}\) concave up on approximately \(-0.4 \lt x \lt 2\) and \(4.5 \lt x \lt 6\text{;}\) concave down on approximately \(2 \lt x \lt 4.5\) and \(x \gt 6\text{;}\) \(F(2) = 0\text{;}\) \(F(0.5) = -6.06\text{;}\) \(F(-1) = -1.77\text{;}\) \(F(4) = 6.69\text{;}\) \(F(5) = 6.33\text{;}\) \(F(6.5) = 8.12\text{.}\)

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.
\(\cos\!\left(7t\right)\)
Answer 2.
\(-7\sin\!\left(7t\right)\)
Answer 3.
\(\frac{-u^{15}}{7}\)
Answer 4.
\(\frac{-u^{16}}{112}\)
Answer 5.
\(\frac{-\cos^{16}\!\left(7t\right)}{112}\)

5.5.7.2. Product involving a 4th power polynomial.

Answer.
\(\frac{\left(t^{3}-6\right)^{4}}{12}+C\)

5.5.7.3. Product involving \(\sin(x^6)\).

Answer.
\(1\sin\!\left(x^{2}\right)+C\)

5.5.7.4. Fraction involving \(\ln^9(x)\).

Answer.
\(0.25\ln^{4}\!\left(z\right)+C\)

5.5.7.5. Fraction involving \(e^{5 x}\).

Answer.
\(0.333333\ln\!\left(e^{3x}+5\right)+C\)

5.5.7.6. Fraction involving \(e^{5 \sqrt{y}}\).

Answer.
\(\frac{8e^{2\sqrt{y}}}{2}+C\)

5.5.7.7. Definite integral involving \(e^{-\cos(q)}\).

Answer.
\(-1.71828182845905\)

5.5.7.8. Working with negative exponents.

Answer.
\(\frac{\left(x+12\right)^{-2}}{-2}\)

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

Answer.
\(\frac{-1}{8}\cos\!\left(8\sin\!\left(t\right)\right)\)

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

Answer.
\(\frac{1}{9}\cos\!\left(\frac{9}{x}\right)\)

5.5.7.11. Fraction involving sums of exponential functions.

Answer.
\(\ln\!\left(\left|e^{x}+5e^{-x}\right|\right)\)

5.5.7.12. Integral involving a rational function.

Answer.
\(\frac{0.333333\!\left(x^{3}+\left(-7\right)\right)^{-2}}{-2}+C\)

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. \(\int \tan(x) \, dx = \ln\left(|\sec(x)|\right) + C\text{.}\)
  2. \(\int \cot(x) \, dx = -\ln\left(|\csc(x)|\right) + C\text{.}\)
  3. \(\int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx = \ln\left(|\sec(x) + \tan(x)|\right) + C\text{.}\)
  4. \(\frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} = \sec(x)\text{.}\)
  5. \(\int \sec(x) \, dx = \ln\left(|\sec(x) + \tan(x)|\right) + C\text{.}\)
  6. \(\int \csc(x) \, dx = -\ln\left(|\csc(x) + \cot(x)|\right) + C\text{.}\)

5.5.7.16. Integral of \(\sec(x) \).

Answer.
\(\ln\!\left(\left|\sec\!\left(x\right)+\tan\!\left(x\right)\right|\right)\)

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

Answer 1.
\(\frac{1}{7}\sec\!\left(7x\right)\)
Answer 2.
\(\frac{1}{64}\tan^{8}\!\left(8x\right)\)

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

Answer.
\(\frac{\ln\!\left(2\right)}{6}\)

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

Answer 1.
\(-0.2\ln\!\left(\cos\!\left(5x\right)\right)+C\)
Answer 2.
\(0.2\ln\!\left(\sec\!\left(5x\right)+\tan\!\left(5x\right)\right)+C\)

5.5.7.20. A clever substitution.

Answer.
  1. \(\int x \sqrt{x-1} \, dx = \int (u+1) \sqrt{u} \, du\text{.}\)
  2. \(\int x \sqrt{x-1} \, dx = \frac{2}{5} (x-1)^{\frac{5}{2}} + \frac{2}{3} (x-1)^{\frac{3}{2}} + C\text{.}\)
  3. \(\int x^2 \sqrt{x-1} \, dx = \frac{2}{7} (x-1)^{\frac{7}{2}} + \frac{4}{5} (x-1)^{\frac{5}{2}} + \frac{2}{3} (x-1)^{\frac{3}{2}} + C\text{.}\)
    \(\int x \sqrt{x^2 - 1} \, dx = \frac{1}{3} (x^2-1)^{\frac{3}{2}} + C\text{.}\)

5.5.7.21. Fractions involving a square root or trigonometric functions.

Answer 1.
\(\frac{2\!\left(x+9\right)\sqrt{x+9}}{3}-18\sqrt{x+9}\)
Answer 2.
\(\frac{-1}{9\!\left(9\sin\!\left(t\right)+10\right)}\)

5.5.7.22. Definite integral with a clever substitution.

Answer.
\(\frac{-112-4}{15}\)

5.5.7.23. Product involving trigonometric functions and square roots.

Answer.
\(\frac{-2}{6}\!\left(15-\sin\!\left(2x\right)\right)^{\frac{3}{2}}\)

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

Answer.
\(\frac{-4}{3}\!\left(2-x\right)\sqrt{2-x}+\frac{2}{5}\!\left(2-x\right)^{2}\sqrt{2-x}\)

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

Answer.
  1. We don’t have a function-derivative pair.
  2. \(\sin^3(x) = \sin(x) (1-\cos^2(x))\text{.}\)
  3. \(u = \cos(x)\) and \(du = -\sin(x) \, dx\text{.}\)
  4. \(\int \sin^3(x) \, dx = \frac{1}{3}\cos^3(x) - \cos(x) + C\text{.}\)
  5. \(\int \cos^3(x) \, dx = \sin(x) - \frac{1}{3}\sin^3(x) + C\text{.}\)

5.5.7.26. Fraction involving \(\sin(4t-1) \).

Answer.
\(\frac{\sec\!\left(4t-1\right)}{4}\)

5.5.7.27. An integral that requires rewriting before integrating.

Answer.
\(5\tan^{-1}\!\left(x\right)+\left(-1\right)\ln\!\left(1+x^{2}\right)+C+c\)

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. \(\int_0^{24} r(t) \, dt \approx 95.7809 \) megawatts of power used in \(24\) hours.
  4. \(\displaystyle r_{\operatorname{AVG} [0,24]} \approx 3.99087 \) megawatts.

5.6 Integration by Parts
5.6.7 Exercises

5.6.7.2. Product involving \(\cos(5 x)\).

Answer.
\(2\!\left(x\sin\!\left(2x\right)+0.5\cos\!\left(2x\right)\right)\)

5.6.7.3. Product involving \(e^{8 z}\).

Answer.
\(\left(\frac{1+z}{7}-0.0204082\right)e^{7z}+C\)

5.6.7.4. Definite integral of \(t e^{-t}\).

Answer.
\(0.992704944275564\)

5.6.7.5. Evaluating using FTC.

Answer.
  1. \(F'(x) = xe^{-2x}\text{.}\)
  2. \(\displaystyle F(x) = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + \frac{1}{4}\)
  3. Increasing.

5.6.7.6. Evaluating an integral using substitution and IBP.

Answer.
  1. \(\int e^{2x} \cos(e^x) \, dx = \int z \cos(z) \, dz\text{.}\)
  2. \(\int e^{2x} \cos(e^x) \, dx = e^x \sin(e^x) + \cos(e^x) + C\text{.}\)
  3. \(\displaystyle \int e^{2x} \cos(e^{2x}) \, dx = \frac{1}{2} \sin(e^{2x} + C\)
    • \(\int e^{2x} \sin(e^x) \, dx = sin(e^x) - z\cos(e^x) + C\text{.}\)
    • \(\int e^{3x} \sin(e^{3x}) \, dx = -\frac{1}{3} \cos(e^{3x}) + C\text{.}\)
    • \(\int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx = \frac{1}{4} \sin^2(e^{x^2}) + C\text{.}\)

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

Answer.
  1. \(u\)-substitution; \(\int x^2 \cos(x^3) \ dx = \frac{1}{3} \sin(x^3) + C\text{.}\)
  2. Both are needed; \(\int x^5 \cos(x^3) \ dx = \frac{1}{3} \left(x^3\sin(x^3) + \cos(x^3) \right) + C\text{.}\)
  3. Integration by parts; \(\int x ln(x^2) \ dx = \frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C\text{.}\)
  4. Neither.
  5. \(u\)-substitution; \(\int x^3 \sin(x^4) \ dx = -\frac{1}{4} \cos(x^4) + C\text{.}\)
  6. Both are needed; \(\int x^7 \sin(x^4) \ dx = -\frac{1}{4} x^4 \cos(x^4) + \frac{1}{4} \sin(x^4) + C\text{.}\)

5.7 The Method of Partial Fractions
5.7.3 Exercises

5.7.3.1. Partial fractions: linear over difference of squares.

Answer 1.
\(\frac{1}{x-1}\)
Answer 2.
\(\frac{1}{x+1}\)
Answer 3.
\(\ln\!\left(\left|x-1\right|\right)\)
Answer 4.
\(\ln\!\left(\left|x+1\right|\right)\)
Answer 5.
\(\frac{1}{w}\)
Answer 6.
\(\ln\!\left(\left|w\right|\right)\)
Answer 7.
\(\ln\!\left(\left|x^{2}-1\right|\right)\)

5.7.3.2. Partial fractions: constant over product.

Answer.
\(0.5\ln\!\left(\left|x+6\right|\right)-0.5\ln\!\left(\left|x+8\right|\right)+C\)

5.7.3.3. Partial fractions: linear over quadratic.

Answer.
\(9\ln\!\left(\left|x-2\right|\right)-6\ln\!\left(\left|x-1\right|\right)+C\)

5.7.3.4. Partial fractions: cubic over 4th degree.

Answer 1.
\(4\)
Answer 2.
\(2\)
Answer 3.
\(3\)
Answer 4.
\(2\)
Answer 5.
\(\frac{-4}{x}+2\ln\!\left(x\right)+1.5\ln\!\left(x^{2}+9\right)+0.666667\tan^{-1}\!\left(\frac{x}{3}\right)\)

5.7.3.5. Partial fractions: quadratic over factored cubic.

Answer 1.
\(1\)
Answer 2.
\(0\)
Answer 3.
\(2\)
Answer 4.
\(\ln\!\left(\left|x+3\right|\right)+\frac{2\tan^{-1}\!\left(\frac{x}{3}\right)}{3}+C\)

5.9 Numerical Integration
5.9.6 Exercises

5.9.6.1. Various methods for \(e^x\) numerically.

Answer 1.
\(e^{5}-1\)
Answer 2.
\(2.5\!\left(1+e^{2.5}\right)\)
Answer 3.
\(2.5\!\left(e^{2.5}+e^{5}\right)\)
Answer 4.
\(1.25\!\left(1+2e^{2.5}+e^{5}\right)\)
Answer 5.
\(2.5\!\left(e^{1.25}+e^{3\cdot 1.25}\right)\)
Answer 6.
\(\frac{2\cdot 115.029+217.223}{3}\)
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.
\(1.25\!\left(1+e^{1.25}+e^{2.5}+e^{3\cdot 1.25}\right)\)
Answer 13.
\(1.25\!\left(e^{1.25}+e^{2.5}+e^{3\cdot 1.25}+e^{5}\right)\)
Answer 14.
\(0.625\!\left(1+2e^{1.25}+2e^{2.5}+2e^{3\cdot 1.25}+e^{5}\right)\)
Answer 15.
\(1.25\!\left(e^{0.625}+e^{3\cdot 0.625}+e^{5\cdot 0.625}+e^{7\cdot 0.625}\right)\)
Answer 16.
\(\frac{2\cdot 138.236+166.126}{3}\)
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.
\(\frac{114.457}{73.4206}\)
Answer 23.
\(\frac{-254.076}{-110.846}\)
Answer 24.
\(\frac{-69.81}{-18.713}\)
Answer 25.
\(\frac{32.384}{9.177}\)
Answer 26.
\(\frac{-1.681}{-0.12}\)

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.
\(\text{midpoint}\)
Answer 2.
\(10.0265\)

5.9.6.7. Computing left and right Riemann sums.

Answer 1.
\(e^{1}\cdot 0.5\)
Answer 2.
\(e^{1.5}\cdot 0.5\)
Answer 3.
\(e^{1.5}\cdot 0.5\)
Answer 4.
\(e^{2}\cdot 0.5\)

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 \(dv = \tan(x) \, dx\text{,}\) or \(u = \tan(x)\) and \(dv = x \, dx\text{,}\) fail to produce an integral that can be evaluated by parts.
    • \(\displaystyle \text{LEFT}(4) = 0.25892\)
    • \(\displaystyle \text{RIGHT}(4) = 0.64827\)
    • \(\displaystyle \text{MID}(4) = 0.41550\)
    • \(\displaystyle \text{TRAP}(4) = \frac{\text{LEFT}(4) + \text{RIGHT}(4)}{2} = 0.45360\)
    • \(\displaystyle \text{SIMP}(8) = \frac{2\text{MID}(4) + \text{TRAP}(4)}{3} = 0.42820\)
  2. \(\text{LEFT}(4)\) and \(\text{MID}(4)\) are underestimates; \(\text{RIGHT}(4)\) and \(\text{MID}(4)\) are overestimates.

5.9.6.10. Relation of function properties to approximation methods.

Answer.
  1. Decreasing.
  2. Concave down.
  3. \(\in\text{TRAP}(3)^6 f(x) \approx 7.03\text{.}\)

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

Answer.
  1. \(\int_0^{60} r(t) \, dt \text{.}\)
  2. \(\int_0^{60} r(t) \, dt \gt \text{MID}(3) = 204000\text{.}\)
  3. \(\int_0^{60} r(t) \, dt \approx \text{SIMP}(6) = \frac{619000}{3} \approx 206333.33\text{.}\)
  4. \(\frac{1}{60} \text{SIMP}(6) \approx 3438.89 \text{;}\) \(\frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43 \text{.}\) each estimates the average rate at which water flows through the dam on \([0,60]\text{,}\) 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.
\(\text{D}\)

5.10.6.2. An improper integral on an infinite interval.

Answer.
\(\frac{2e^{-1\cdot 1}}{2}\)

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.
\(\text{diverges}\)

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

Answer.
\(\text{diverges}\)

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 \(1\text{.}\)
  4. \(\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx\) diverges if \(p \leq 1\) and converges to \(\frac{1}{p-1}\) if \(p \gt 1\text{.}\)
  5. Diverges.
  6. Converges to \(-1\text{.}\)

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. \(\int \frac{1}{x \sqrt{9x^2 + 25}} \, dx = -\frac{1}{5} \ln \left| \frac{5+\sqrt{9x^2 + 5^2}}{3x} \right| + C\text{.}\)
  2. \(\displaystyle \int x \sqrt{1 + x^4} \, dx = \frac{1}{2} \left( \frac{x^2}{2}\sqrt{x^4 + 1} + \frac{1}{2}\ln|x^2 + \sqrt{x^4 + 1}| \right) + C \)
  3. \(\displaystyle \int e^x \sqrt{4 + e^{2x}} \, dx = \frac{e^x}{2}\sqrt{e^{2x} + 4} + 2\ln|e^x + \sqrt{e^{2x} + 4}| + C \)
  4. \(\int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx = \frac{1}{3} \ln \left| \frac{3 + \sqrt{9 - \cos^2(x)}}{\cos(x)} \right| + C\text{.}\)

5.12.4.3. Comparing Antidifferentiation Tools.

Answer.
  1. Try \(u = 1+x^2\) or \(u = x + \sqrt{1+x^2}\text{.}\)
  2. Try \(u = \sqrt{x+\sqrt{1+x^2}}\) and \(dv = \frac{1}{x} \, dx\text{.}\)
  3. No.
  4. It appears that the function \(\frac{\sqrt{x+\sqrt{1+x^2}}}{x}\) 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. \(A = \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy = \sqrt{3} \text{.}\)
  2. \(A = \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx = \sqrt{2} \text{.}\)
  3. \(A = \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy = \frac{343}{48} \text{.}\)
  4. \(A = \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx = - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right) \text{.}\)

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

Answer.
\(a = \frac{1}{2}\text{.}\)

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

Answer.
  1. \(r = \frac{4}{3} \text{.}\)
  2. \(A_1 = A_2 = \frac{4 \sqrt{6}}{27}\text{.}\)
  3. 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. \(L = \int_0^{1.84257} \sqrt{1+\left( -3 \sin \left(\frac{x^3}{4}\right) \cdot \frac{3}{4}x^2 \right)^2} \, dx \approx 4.10521\text{.}\)
  2. \(A = \int_0^{1.84527} 3 \cos\left( \frac{x^3}{4} \right) \, dx \approx 4.6623 \text{.}\)
  3. \(V = \int_0^{1.84527} \pi \cdot 9 \cos^2 \left( \frac{x^3}{4} \right) \, dx \approx 40.31965 \text{.}\)
  4. \(V = \int_0^3 \pi \left( 4\arccos \left(\frac{y}{3} \right) \right)^{2/3} \, dy \approx 23.29194 \text{.}\)

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

Answer.
  1. \(A = \int_0^{\frac{\pi}{4}} ( \cos(x) - \sin(x) ) \, dx \text{.}\)
  2. \(V = \int_0^{\frac{\pi}{4}} \pi (\cos^2(x) - \sin^2(x)) \, dx \text{.}\)
  3. \(\displaystyle V = \int_0^{\frac{\sqrt{2}}{2}} \pi \arcsin^2(y) \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi \arccos^2(y) \, dy \)
  4. \(V = \int_0^{\frac{\pi}{4}} \pi [(2 - \sin(x))^2 - (2 - \cos(x))^2] \, dx \text{.}\)
  5. \(\displaystyle V = \int_0^{\frac{\sqrt{2}}{2}} \pi [ (1+\arcsin(y))^2 - 1^2 ] \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi [ (1+\arccos(y))^2 - 1^2 ] \, dy \)

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

Answer.
  1. \(A = \int_0^{1.5} 1+\frac{1}{2}(x-2)^2 - \frac{1}{2}x^2 \ dx = 2.25\text{.}\)
  2. \(\displaystyle V = \int_0^{1.5} \pi\left[\left(2+\frac{1}{2}(x-2)^2\right)^2 - \left(1+\frac{1}{2}x^2\right)^2 \right] \ dx = \frac{315}{32} \pi \)
  3. \(V = \int_{0}^{1.125} \pi \left(\sqrt{2y}\right)^2 \ dy + \int_{1.125}^3 \pi \left(2 - \sqrt{2(y-1)}\right)^2 \ dy \approx 7.06858347 \text{.}\)
  4. \(P = 3 + \int_0^{1.5} \sqrt{1+(x-2)^2} + \sqrt{1+x^2} \ dx \approx 7.387234642 \text{.}\)

6.2.6.10. Arc length of a curve.

Answer.
\(13.5428669014242\)

6.2.6.11. Length of a parametric curve.

Answer.
\(L \approx 9.429\)

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 \(\theta\).

Answer 1.
\(2\)
Answer 2.
\(\frac{4}{\cos\!\left(t\right)}\)
Answer 3.
\(0\)
Answer 4.
\(\frac{\pi }{2}\)

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.
\(\left(4+4x\right)Dx\)
Answer 2.
\(30\ {\rm g}\)

6.4.6.2. Center of mass for a nonlinear density function.

Answer 1.
\(2+8\pi \)
Answer 2.
\(\frac{\pi }{2}\)

6.4.6.3. Interpreting the density of cars on a road.

Answer 1.
\(350\!\left(2+\sin\!\left(4\sqrt{x+0.175}\right)\right)Dx\)
Answer 2.
\(\frac{1}{8}\cdot 350\!\left(320+4\sqrt{0.175}\cos\!\left(4\sqrt{0.175}\right)-4\sqrt{20+0.175}\cos\!\left(4\sqrt{20+0.175}\right)-\sin\!\left(4\sqrt{0.175}\right)+\sin\!\left(4\sqrt{20+0.175}\right)\right)\)

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

Answer 1.
\(3.66667\ {\rm cm}\)
Answer 2.
\(\text{to the right of the origin}\)

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

Answer.
  1. \(a = -10 \ln(0.7) \approx 3.567 \text{ cm}\text{.}\)
  2. Left of the midpoint.
  3. \(\overline{x} \approx \frac{50.3338}{30} \approx 1.687\text{.}\)
  4. \(q = -10\ln(0.85) \approx 1.625\) cm.

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

Answer.
  1. \(M_1 = \arctan(10) \approx 1.47113\text{;}\) \(M_2 = 10 - 10e^{-1} \approx 6.32121\text{.}\)
  2. \(\overline{x_1} \approx 1.56857 \text{;}\) \(\overline{x_2} \approx 4.18023 \text{.}\)
    1. \begin{equation*} M = \int_0^{10} \rho(x) \, dx + \int_0^{10} p(x) \, dx \approx 1.47113 + 6.32121 = 7.79234\text{.} \end{equation*}
    2. \begin{equation*} \int_0^{10} x(\rho(x) + p(x))) \, dx = 28.73167\text{.} \end{equation*}
    3. False.

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

Answer.
  1. \(V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx \approx 52.0666 \) cubic inches.
  2. \(W \approx 0.6 \cdot 52.0666 = 31.23996 \) ounces.
  3. 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 \(V_{text{slice}} \approx \pi f(x)^2\text{.}\)
  4. \(\displaystyle \overline{x} \approx 23.21415 \)

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

6.5.5.1. Work to empty a conical tank.

Answer.
\(232070.147452379\ {\rm J}\)

6.5.5.2. Work to empty a cylindrical tank.

Answer.
\(1.66253\times 10^{6}\ {\rm ft\,lb}\)

6.5.5.3. Work to empty a rectangular pool.

Answer.
\(5.7798\times 10^{6}\ {\rm ft\,lb}\)

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

Answer 1.
\(106741\ {\rm lbf}\)
Answer 2.
\(242594\ {\rm lbf}\)
Answer 3.
\(98801.8\ {\rm lbf}\)

6.5.5.5. Force due to hydrostatic pressure.

Answer 1.
\(9800\ {\rm N}\)
Answer 2.
\(3920\ {\rm N}\)
Answer 3.
\(3136\ {\rm N}\)

6.5.5.6. Work to fill an irregularly shaped tank.

Answer.
  1. \(W \approx 1646.79\) foot-pounds.
  2. \(W = \int_0^h 3744 x \cos \left( \frac{x^3}{4} \right) \, dx \text{.}\)
  3. \(F \approx 462.637 \) pounds.

6.5.5.7. Work to half-empty a cylindrical tank.

Answer.
  1. \(W = 5904(19\pi - 8) \approx 305179.3 \) foot-pounds.
  2. \begin{equation*} F \approx 1123.2 \end{equation*}
    pounds.

7 Sequences and Series
7.1 Sequences
7.1.3 Exercises

7.1.3.2. Formula for a Sequence, Given First Terms.

Answer.
\(\left(-1\right)^{n}\!\left(2n+3\right)\)

7.1.3.3. Divergent or Convergent Sequences.

Answer 1.
\(4\)
Answer 2.
\(\text{diverges}\)
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.
    \(n\) \(1\) \(2\) \(3\) \(4\) \(5\)
    \(\frac{\ln(n)}{n}\) \(0\) \(0.3466\) \(0.3663\) \(0.3466\) \(0.3218\)
    \(t\) \(6\) \(7\) \(8\) \(9\) \(10\)
    \(\frac{\ln(n)}{n}\) \(0.2987\) \(0.2781\) \(0.2599\) \(0.2442\) \(0.2303\)
  2. If \(\lim_{x \to \infty} f(x) = L\text{,}\) then \(\lim_{n \to \infty} \frac{\ln(n)}{n} = L\) as well.
  3. \begin{equation*} \lim_{n \to \infty} \frac{\ln(n)}{n} = \lim_{x \to \infty} \frac{\ln(x)}{x} = 0\text{.} \end{equation*}

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

Answer.
  1. \(P_1\left(\frac{r}{12}\right)\) in interest in the second month; at the end of the second month, \(P_2 = P\left( 1 + \frac{r}{12}\right)^2 \text{.}\)
  2. \(P_3 = P\left( 1 + \frac{r}{12}\right)^3 \text{.}\)
  3. \(P_n = P\left( 1 + \frac{r}{12}\right)^n \) 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 \(10/13 \approx 0.76923\) 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. \(30 \cdot 500 = 1500\) 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. \(\$0.01\left(2^{30}-1\right) = \$10,737,418.23 \text{.}\)

7.2.3.7. Application to model behavior of a ball drop.

Answer.
  1. \(h_1 = \left(\frac{3}{4}\right)h \text{.}\)
  2. \(h_2 = \left(\frac{3}{4}\right)h_1 = \left(\frac{3}{4}\right)^2h \text{.}\)
  3. \(h_3 = \left(\frac{3}{4}\right)h_2 = \left(\frac{3}{4}\right)^3h \text{.}\)
  4. \(h_n = \left(\frac{3}{4}\right)h_{n-1} = \left(\frac{3}{4}\right)^nh \text{.}\)
  5. The distance traveled by the ball is \(7h\text{,}\) 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. \(P = \frac{6}{11} \text{.}\)

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. \(0.75P + 0.75^2P + 0.75^2P + \cdots = 0.75P(1+0.75+0.75^2+ \cdots ) \) 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. \(\left(\frac{r}{12}\right)P_1\) dollars.
  2. \(P_2 = \left(1 + \frac{r}{12}\right)P_1 - M \text{.}\)
  3. \(P_2 = \left(1 + \frac{r}{12}\right)^2P - \left[1 + \left(1+\frac{r}{12}\right)\right] M \text{.}\)
  4. \(P_3 = \left(1 + \frac{r}{12}\right)P_2 - M \text{.}\)
    \(P_3 = \left(1 + \frac{r}{12}\right)^3P - \left[1 + \left(1+\frac{r}{12}\right) + \left(1+\frac{r}{12}\right)^2 \right] M \text{.}\)
  5. \(P_n = P \left(1+\frac{r}{12}\right)^n - \left(\frac{12M}{r}\right) \left( \left(1+\frac{r}{12}\right)^n - 1\right) \text{.}\)
  6. \(\displaystyle P(t) = P \left(1+\frac{r}{12}\right)^{12t} - \left(\frac{12M}{r}\right) \left( \left(1+\frac{r}{12}\right)^{12t} - 1\right) \)
  7. \(A(t) = \left(1000 - \frac{12(25)}{0.2}\right)\left(1+\frac{0.2}{12}\right)^{12t} + \frac{12(25)}{0.2} \text{.}\)
    \(t \approx 5.5 \text{.}\)
    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.
\(\infty \)
Answer 2.
\({\frac{9}{8}}\)

7.3.5.4. Convergence of an integral and a related series.

Answer 1.
\(\frac{9\pi }{4}\)
Answer 2.
\(\text{converges}\)

7.3.5.5. Adding two series together.

Answer.
    1. \(a_n = 1 + \frac{1}{2^n} \to 1 \ne 0\) and \(b_n = -1 \to -1 \ne 0\text{.}\)
    2. The series is geometric with \(r = \frac{1}{2}\text{.}\)
    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 \(A_n + B_n = (a_1 + b_1) + (a_2 + b_2) + \cdots + (a_n + b_n) \text{.}\)
    2. Note that \(\lim_{n \to \infty} \sum_{k=1}^n (a_k+b_k) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \right) \) .
  1. \(\sum_{k=0}^{\infty} \frac{2^k+3^k}{5^k} = \frac{25}{6} \text{.}\)

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.
\(\frac{3}{3e}\)
Answer 2.
\(\text{converges}\)

7.4 Comparison Tests
7.4.4 Exercises

7.4.4.1. The direct comparison test.

Answer.
    1. \(S_1 = 1\) and \(T_1 = \frac{1}{2}\text{.}\)
    2. \(S_2 \gt T_2\text{.}\)
    3. \(S_3 \gt T_3\text{.}\)
    4. \(S_n \gt T_n\text{.}\)
    5. \(\sum \frac{1}{k^2} \gt \sum \frac{1}{k^2+k} \text{;}\) \(\sum \frac{1}{k^2+k}\) converges.
  1. If \(\sum b_k\) diverges, then \(\sum b_k\) is infinite, and anything larger must also be infinite; if \(\sum a_k\) is convergent then anything smaller and positive must also be finite.
    1. Note that \(0 \lt \frac{1}{k} \lt \frac{1}{k-1}\text{.}\)
    2. Note that \(\frac{1}{k^3} \gt \frac{1}{k^3+1}\text{.}\)

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

Answer 1.
\(\text{Converges}\)
Answer 2.
\(\text{Converges}\)

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.
\(\text{Convergent}\)
Answer 2.
\(\text{Inconclusive}\)
Answer 3.
\(\text{Divergent}\)

7.5.4.2. Convergence of a series using the ratio test.

Answer.
  1. \(\sum \frac{10^k}{k!}\) converges.
  2. \(\sum \frac{10^k}{k!}\) converges.
  3. The sequence \(\left\{\frac{b^n}{n!}\right\}\) has to converge to 0.

7.5.4.3. The root test for convergence.

Answer.
  1. \(\sqrt[n]{a_n} \approx r\) for large \(n\text{.}\)
  2. \(\frac{a_{n+1}}{a_n} \approx r\text{.}\)
  3. \(0 \lt r \lt 1\text{.}\)

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

Answer 1.
\(\text{Converges}\)
Answer 2.
\(\text{Converges}\)

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

Answer.
  1. \(\left\{\frac{1}{n}\right\}\) and \(\left\{-\frac{1}{n^2}\right\}\) converge to 0.
  2. Notice that \(\frac{1}{k} - \frac{1}{k^2} = \frac{k-1}{k^2}\) 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.
\(\text{Converges}\)
Answer 2.
\(\text{Converges}\)
Answer 3.
\(\text{Converges}\)

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. \(\sum_{k=0}^{\infty} \frac{1}{2k+1}\) diverges by comparison to the Harmonic series.
  2. \(|S_{100} - \sum_{k=0}^{\infty} (-1)^{k} \frac{1}{2k+1}| \lt \approx 0.0049 \text{.}\)
  3. \(\displaystyle n \gt 4{,}999{,}999{,}999.5\)

7.6.5.5. A alternative approximation method for convergent alternating series.

Answer.
  1. \(\frac{S_n+S_{n+1}}{2} = \frac{S_n+S_{n} + (-1)^{n+2}a_{n+1}}{2} \text{.}\)
  2. \(S_{20} = 0.668771403 \ldots \text{;}\) \(\frac{S_{20} + S_{21}}{2} = \frac{161227687}{232792560} = 0.692580926 \ldots \text{,}\) accurate to within about \(0.0006\text{.}\)

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

Answer 1.
\(\text{Conditionally Convergent}\)
Answer 2.
\(\text{Divergent}\)
Answer 3.
\(\text{Absolutely Convergent}\)

7.6.5.9. Practice determining series convergence.

Answer 1.
\(\text{Diverges}\)
Answer 2.
\(\text{Diverges}\)
Answer 3.
\(\text{Converges}\)

7.7 Power Series
7.7.4 Exercises

7.7.4.1. Radius of convergence of a a power series.

Answer.
\(\infty \)

7.7.4.3. Interval of convergence of a power series.

Answer.
\(\left(0,4\right)\)

7.7.4.4. Interval of convergence of a power series.

Answer.
\(\left[-0.2,0.2\right]\)

7.7.4.5. Interval of convergence of a power series.

Answer.
\(\left(\frac{1}{-2},\frac{1}{2}\right)\)

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

Answer 1.
\(5\)
Answer 2.
\(\left[-13,-3\right)\)

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

Answer 1.
\(\infty \)
Answer 2.
\(\left(-\infty ,\infty \right)\)

7.8 Taylor Polynomials
7.8.3 Exercises

7.8.3.1. Determining Taylor polynomials from a function formula.

Answer 1.
\(1-\frac{4^{2}}{2}x^{2}\)
Answer 2.
\(1-\frac{4^{2}}{2}x^{2}+\frac{4^{4}}{4!}x^{4}\)
Answer 3.
\(1-\frac{4^{2}}{2!}x^{2}+\frac{4^{4}}{4!}x^{4}-\frac{4^{6}}{6!}x^{6}\)

7.8.3.2. Determining Taylor polynomials from given derivative values.

Answer 1.
\(-2+1\!\left(x-7\right)+\frac{1}{2!}\cdot 4\!\left(x-7\right)^{2}\)
Answer 2.
\(-2+1\!\left(x-7\right)+\frac{1}{2!}\cdot 4\!\left(x-7\right)^{2}+\frac{1}{3!}\cdot 1\!\left(x-7\right)^{3}\)
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.
\(\frac{5}{4}\)
Answer 2.
\(\frac{-5\!\left(x-4\right)}{4^{2}}\)
Answer 3.
\(\frac{5\!\left(x-4\right)^{2}}{4^{3}}\)
Answer 4.
\(\frac{-5\!\left(x-4\right)^{3}}{4^{4}}\)

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

Answer 1.
\(\frac{1}{\sqrt{2}}\)
Answer 2.
\(\frac{-1\!\left(x-\frac{\pi }{4}\right)}{\sqrt{2}}\)
Answer 3.
\(\frac{-1\!\left(x-\frac{\pi }{4}\right)^{2}}{2\sqrt{2}}\)
Answer 4.
\(\frac{1\!\left(x-\frac{\pi }{4}\right)^{3}}{6\sqrt{2}}\)

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

Answer 1.
\(\ln\!\left(3\right)\)
Answer 2.
\(\frac{1}{3}\!\left(x-3\right)\)
Answer 3.
\(\frac{-\left(\frac{1}{3}\right)^{2}}{2}\!\left(x-3\right)^{2}\)
Answer 4.
\(\frac{\left(\frac{1}{3}\right)^{3}}{3}\!\left(x-3\right)^{3}\)
Answer 5.
\(\frac{-\left(\frac{1}{3}\right)^{4}}{4}\!\left(x-3\right)^{4}\)

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.
\(\frac{e\!\left(-1\right)^{k}\!\left(x--1\right)^{k}}{k!}\)

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. \(P_3(x) = -1 + 3x - \frac{4}{2!}x^2 + \frac{6}{3!}x^3 \text{,}\) which is the same polynomial as \(f(x)\text{.}\)
  2. For \(n \ge 3\text{,}\) \(P_n(x) = f(x)\text{.}\)
  3. For \(n \ge m\text{,}\) \(P_n(x) = f(x)\text{.}\)

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.
\(\frac{5x^{3}}{3}-\frac{5^{3}x^{7}}{42}\)
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. \(\frac{dT}{dt}\vert_{T=105} = -2\text{;}\) when \(T = 105\text{,}\) the coffee’s temperature is decreasing at an instantaneous rate of \(-2\) degrees F per minute.
  2. \(T\) decreasing at \(t=0\text{.}\)
  3. \(T(1) \approx 103 \) degrees F.
  4. For \(T \lt 75\text{,}\) \(T\) increases. For \(T \gt 75\text{,}\) \(T\) decreases.
  5. Room temperature is \(75\) degrees F.
  6. Substitute \(T(t) = 75 + 30e^{-t/15}\) in for \(T\) in the differential equation \(\frac{dT}{dt}= -\frac1{15}T+5\) and verify the equality holds; \(T(0) = 75 + 30e^0 = 75 + 30 = 105\text{;}\) \(T(t) = 75 + 30e^{-t/15} \to 75\) as \(t \to \infty\text{.}\)

8.1.3.5. Population growth.

Answer.
  1. \(1 \lt P \lt 3\text{.}\)
  2. \(P \lt 1\) and \(3 \lt P \lt 4\text{.}\)
  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 \(3\text{.}\)
  6. The population will decrease toward \(P = 3\) with \(P\) always being above \(3\text{.}\)
  7. There’s a maximum threshold of \(P = 3\text{.}\)

8.1.3.6. A look at solutions to differential equations.

Answer.
    1. \(y(t) = t + 1 + 2e^t\) 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.
  1. \(k = 9\text{.}\)

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.
\(\text{positive}\)
Answer 2.
\(\text{negative}\)
Answer 3.
\(\text{negative}\)
Answer 4.
\(\text{negative}\)

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 \(\frac{dy}{dt} = t-y \).

Answer.
  1. Sketch curves through appropriate points in the slope field above.
  2. \(y(t) = t-1\text{.}\)
  3. \(t\) and \(y\) are equal.

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

Answer.
  1. Any solution curve that starts with \(P(0) \gt 3\) will decrease to \(P(t) = 3\) as \(t \to \infty\text{;}\) any curve that starts with \(1 \lt P(0) \lt 3\) will increase to \(P(t) = 3\text{;}\) any curve that starts with \(0 \lt P(0) \lt 1\) will decrease to \(P(t) = 0\text{.}\)
  2. \(P = 0\text{,}\) \(P = 1\text{,}\) and \(P = 3\text{.}\) \(P = 1\) is unstable; \(P = 0\) and \(P = 3\) are stable.
  3. The population will stabilize either at the value \(P = 3\) or at \(P = 0\text{.}\)
  4. \(P(t) = 1\) 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. \(\frac{dP}{dt} = g(P) = P(6-P)-1 \) ; the equilibrium at \(P\approx 0.172\) is unstable; the equilibrium at \(P \approx 5.83\) is stable.
  3. If \(P \lt \frac{6-\sqrt{32}}{2}\text{,}\) then the fish population will die out. If \(\frac{6-\sqrt{32}}{2} \lt \text{,}\) then the fish population will approach \(\frac{6+\sqrt{32}}{2}\) thousand fish.
  4. \(\frac{dP}{dt} = g(P) = P(6-P)-h \text{;}\) equilibrium solutions \(P = \frac{6+\sqrt{36-4h}}{2}, \ \frac{6-\sqrt{36-4h}}{2} \text{.}\)
  5. \(9000\) fish; harvesting at that rate will maintain the number of fish we start with, provided it’s at least \(3000\text{.}\)

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

Answer.
  1. \(\frac{dy}{dt} = 20y\text{.}\)
  2. \(\displaystyle \frac{dy}{dt} = 20y - C\frac{y}{2+y} \)
  3. For positive \(y\) near \(0\text{,}\) \(M(y) = \frac{y}{2+y} \approx 0\text{;}\) for large values of \(y\text{,}\) \(M(y) = \frac{y}{2+y} \approx 1\text{.}\)
  4. The only equilibrium solution is \(y = 0\text{,}\) 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: \(\frac{dT_A}{dt} \vert_{T = 100} = -0.5(30) = -15\) degrees per minute; Bob’s coffee: \(\frac{dT_B}{dt} \vert_{T = 100} = -0.1(30) = -3\) degrees per minute.
  2. Consider the insulation of the containers.
  3. Alice’s coffee:
    \begin{equation*} \frac{dT_A}{dt} = -0.5(T_A-(70+10\sin t))\text{,} \end{equation*}
    with the inital condition \(T_A(0) = 100\text{.}\)
    \(t\) \(T_A(t)\)
    \(0.0\) \(100\)
    \(0.1\) \(98.5\)
    \(0.2\) \(97.12492\)
    \(0.3\) \(95.86801\)
    \(0.4\) \(94.72237\)
    \(\vdots\) \(\vdots\)
    \(49.6\) \(65.56715\)
    \(49.7\) \(65.48008\)
    \(49.8\) \(65.43816\)
    \(49.9\) \(65.44183\)
    \(50\) \(65.49103\)
  4. \(t\) \(T_A(t)\)
    \(0.0\) \(100\)
    \(0.1\) \(99.7\)
    \(0.2\) \(99.41298\)
    \(0.3\) \(99.13872\)
    \(0.4\) \(98.87689\)
    \(\vdots\) \(\vdots\)
    \(49.6\) \(69.39515\)
    \(49.7\) \(69.33946\)
    \(49.8\) \(69.29248\)
    \(49.9\) \(69.25467\)
    \(50\) \(69.22638\)
  5. Compare the rate of initial decrease and amplitude of oscillation.

8.3.4.5. Accelerated convergence.

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

8.3.4.6. The Improved Euler’s Method.

Answer.
  1. \(y(1) \approx y_5 = 2.7027\text{.}\)
  2. \(y(1) \approx y_{10} = 2.7141\text{.}\)
  3. The square of \(\Delta t\text{.}\)

8.4 Separable differential equations
8.4.3 Exercises

8.4.3.1. Initial value problem for \(dy/dx=x^8 y\).

Answer.
\(y = \frac{3}{e^{\frac{1}{5}}}e^{\frac{x^{5}}{5}}\)

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

Answer.
\(-245e^{0.5t}+300\)

8.4.3.3. Initial value problem for \(dy/dt=y^2(8+t)\).

Answer.
\(\frac{12}{2+\left(1-t\right)\!\left(13+t\right)\cdot 6}\)

8.4.3.4. Initial value problem for \(du/dt=e^{6u+10t}\).

Answer.
\(\frac{-\ln\!\left(e^{-8\cdot 3}+\frac{3}{8}-\frac{3}{8}e^{8t}\right)}{3}\)

8.4.3.5. Initial value problem for \(dy/dx=170yx^{16}\).

Answer.
\(6\exp\!\left(9x^{16}\right)\)

8.4.3.6. Radioactive decay.

Answer.
  1. \(\frac{dM}{dt} = kM \text{.}\)
  2. \(M(t) = M_0e^{kt} \text{.}\)
  3. \(\displaystyle M(t) = M_0e^{-\frac{\ln(2)}{5730}t} \approx M_0e^{-0.000121t} \)
  4. \(t = \frac{5730\ln(4)}{\ln(2)} \approx 11460\) years.
  5. \(t = -\frac{5730\ln(0.3)}{\ln(2)} \approx 9952.8\) years.

8.4.3.7. Initial value problem for \(dy/dt= -\frac{t}{y} \).

Answer.
  1. \(y = \sqrt{64 - t^2} \text{.}\)
  2. \(-8 \le t \le 8\text{.}\)
  3. \(y(8) = 0\text{.}\)
  4. \(\frac{dy}{dt} = -\frac ty\) is not defined when \(y = 0\text{.}\)

8.4.3.8. Torricelli’s Law.

Answer.
  1. \(\frac{dh}{dt} = k \sqrt{h} \text{.}\)
  2. The tank with \(k = -10\) has water leaving the tank much more rapidly.
  3. \(k = -5\text{.}\)
  4. \(h(t) = \left( 5 - 2.5t \right)^2 \text{.}\)
  5. \(2.5\) minutes.
  6. No.

8.4.3.9. The Gompertz equation.

Answer.
  1. \(P = 3\) is stable.
  2. \(P(t) = 3e^{\ln \left(\frac{1}{3} \right) e^{-t}} \text{.}\)
  3. \(P(t) = 3e^{\ln \left( 2 \right) e^{-t}} \text{.}\)
  4. Yes.

8.5 Modeling with Differential Equations
8.5.4 Exercises

8.5.4.1. Mixing problem.

Answer 1.
\(0\ {\rm kg}\)
Answer 2.
\(0.24-4\frac{S}{2960}\)
Answer 3.
\(177.6\!\left(1-e^{-0.0013514t}\right)\)
Answer 4.
\(9.1176\ {\rm kg}\)

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.
\(260\cdot 2.71828^{0.298627t}\)
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.
\(\frac{k}{0.06}\!\left(\exp\!\left(0.06t\right)-1\right)\)
Answer 2.
\(28236.5578786596\)

8.5.4.6. Comparing lottery investment options.

Answer.
  1. \(\displaystyle \frac{dA}{dt} = 1 + 0.05A \)
  2. \(A(25) = 49.80686\) million dollars.
  3. \(A(25) = 34.90343\) million dollars.
  4. The first.
  5. \(t = 20 \ln(2) \approx 13.86 \ \text{years} \text{.}\)

8.5.4.7. Velocity of a skydiver.

Answer.
  1. \(\displaystyle \frac{dv}{dt} = 9.8 - kv \)
  2. \(v = \frac{9.8}{k}\) is a stable equilibrium.
  3. \(\displaystyle v(t) = \frac{9.8 - 9.8e^{-kt}}{k} \)
  4. \(k = 9.8/54 \approx 0.181481\text{.}\)
  5. \(t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894\) seconds.

8.5.4.8. Weight gain.

Answer.
  1. \(\frac{dw}{dt} = \frac{k}{w} \text{.}\)
  2. \(w(t) = \sqrt{7t+64} \text{;}\) \(w(12) = \sqrt{148} \approx 12.17\) 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. \(\displaystyle \frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}} \)
  4. \(\frac{3S(t)}{100} \frac{\text{grams}}{\text{minute}} \text{.}\)
  5. \(\frac{dS}{dt} = 60 - \frac{3}{100} S \text{.}\)
  6. \(S = 2000\) is a stable equilibrium solution.
  7. \(S(t) = 2000 - 2000e^{-\frac{3}{100}t} \text{.}\)
  8. \(S(t) \to 2000\text{.}\)

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
\begin{align*} y\amp = a_0\left( 1+\sum_{k=1}^{\infty} \frac{x^{3k}}{(2)(3)(5)(6) \cdots (3k-1)(3k)} \right)\\ \amp\phantom{={}}+ a_1 \left( x + \sum_{k=1}^{\infty} \frac{x^{3k+1}}{(3)(4)(6)(7) \cdots (3k)(3k+1)} \right)\text{.} \end{align*}

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.
\(\left(0,1.5\right)\)
Answer 6.
\(\left(1.5,\infty \right)\)
Answer 7.
\(1.5\)
Answer 8.
\(0.75\)
Answer 9.
\(\left(0,1.5\right)\)
Answer 10.
\(\left(1.5,\infty \right)\)
Answer 11.
\(0, 1.5\)
Answer 12.
\(\frac{1.5}{2}\)

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.
\(\frac{4750}{1+\frac{\left(4750-20\right)e^{-1.6t}}{20}}\)
Answer 2.
\(\frac{4750}{2}\)

8.6.4.4. Analyzing a population growth model.

Answer 1.
\(\left(0,1.74796\right)\)
Answer 2.
\(\left(1.74796,\infty \right)\)
Answer 3.
\(\infty \)
Answer 4.
\(\frac{b}{a}\)
Answer 5.
\(0\)

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

Answer.
  1. \(p(t) \to 1\) as \(t \to \infty\) provided \(p(0) \gt 0\text{.}\)
  2. \(p(t) = \frac{1}{9e^{-0.2t} + 1} \text{.}\)
  3. \(t = -5 \ln(1/9) \approx 10.986\) days.
  4. \(t = -5 \ln(0.2/9) \approx 19.033\) days.

8.6.4.6. Per capita growth rate of bacteria.

Answer.
  1. \(\displaystyle \frac{db}{dt} = \frac{1}{3000} b(15000 - b) \)
  2. \(b = 15000\text{.}\)
  3. When \(b = 7500\text{.}\)
  4. \(t = -\frac{1}{5} \ln(1/70) \approx 0.8497\) days.

8.6.4.7. A logistic equation modeling a fish population.

Answer.
  1. 10000 fish.
  2. \(\frac{dP}{dt} = 0.1P(10 - P) - 0.2P \text{.}\)
  3. \(8000\) fish.
  4. \(P(1) \approx 8.7899 \) thousand fish.
  5. \(t = -1.2 \ln(5/11) \approx 0.9461\) years.