AC Answers to Activities

Appendix B Answers to Activities

This appendix contains answers to all activities in the text. Answers for preview activities are not included.

1 Understanding the Derivative
1.1 How do we measure velocity?
1.1.2 Position and average velocity

Activity 1.1.2.

1.1.2.a

Answer.

$AV_{[0.4,0.8]} = 12.8$ ft/sec; $AV_{[0.7,0.8]} = 8$ ft/sec; the other average velocities are, respectively, 6.56, 6.416, 0, 4.8, 6.24, 6.384, all in ft/sec.

1.1.2.b

Answer.

$m = 12.8$ is the average velocity of the ball between $t = 0.4$ and $t = 0.8\text{.}$

1.1.2.c

Answer.

Like a straight line with slope about 6.4.

1.1.3 Instantaneous Velocity

Activity 1.1.3.

1.1.3.a

Answer.

$AV_{[1.5,2]} = -24$ ft/sec, which is negative.

1.1.3.b

Answer.

The instantaneous velocity at $t = 1.5$ is approximately $-16$ ft/sec; at $t = 2\text{,}$ the instantaneous velocity is about $-32$ ft/sec, and $-16>-32\text{.}$

1.1.3.c

Answer.

When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.

1.2 The notion of limit
1.2.2 The Notion of Limit

Activity 1.2.2.

1.2.3 Instantaneous Velocity

Activity 1.2.3.

Activity 1.2.4.

1.2.4.a

Answer.

$AV_{[0.5,1]} = \frac{1-1}{1-0.5} = 0\text{,}$ $AV_{[1.5,2.5]} = \frac{3-1}{2.5-1.5} = 2\text{,}$ and $AV_{[0,5]} = \frac{5-0}{5-0} = 1\text{.}$

Take shorter and shorter time intervals and draw the lines whose slopes represent average velocity. If those lines’ slopes are approaching a single number, that number represents the instantaneous velocity.

1.2.4.c

Answer.

The instantaneous velocity at $t = 2$ is greater than the average velocity on $[1.5,2.5]\text{.}$

1.3 The derivative of a function at a point
1.3.2 The Derivative of a Function at a Point

Activity 1.3.2.

1.3.2.a

Answer.

$f$ is linear.

1.3.2.b

Answer.

The average rate of change on $[1,4]\text{,}$ $[3,7]\text{,}$ and $[5,5+h]$ is $-2\text{.}$

1.3.2.c

Answer.

$f'(1)=-2\text{.}$

1.3.2.d

Answer.

$f'(2)=-2\text{,}$ $f'(\pi)=-2\text{,}$ and $f'(-\sqrt{2})=-2\text{,}$ since the slope of a linear function is the same at every point.

Activity 1.3.3.

1.3.3.a

Answer.

The vertex is $(\frac{1}{2},36)\text{.}$

1.3.3.b

Answer.

$\frac{s(2)-s(1)}{2-1} = -32$ feet per second.

1.3.3.c

Answer.

$s'(1) = -16\text{.}$

1.3.3.d

Answer.

A plot of the quadratic function $s(t) = -16t^2 + 16t + 32\text{,}$ the secant line through $(1,s(1))$ and $(2,s(2))\text{,}$ and the tangent line through $(1,s(1))$ with slope $s'(1)=-16$

The point $(1,32)$ is of most interest, and the curve, secant line, and tangent line all pass through this point. The function $s(t)$ is a parabola that opens down with vertex $(\frac12,36)$ and a $t$-intercept at $(2,0)\text{.}$ The secant line passes through $(1,32)$ and $(2,0)\text{,}$ while the tangent line only passes through $(1,32)\text{.}$

1.3.3.e

Answer.

$s'(a)$ is positive whenever $0 \le a \lt \frac{1}{2}\text{;}$ $s'(a)$ to be negative whenever $\frac{1}{2} \lt a \lt 2\text{;}$ $s'(\frac{1}{2}) = 0\text{.}$

Activity 1.3.4.

1.3.4.a

Answer.

A plot of $P(t) = 25000 e^{t/5}$ on the interval $0 \lt t \lt 5$ (in decades). The values of $P$ on the vertical axis range from $0$ to $100$ (in thousands). A secant line and a tangent line are also shown.

The function $P(t) = 25000 e^{t/5}$ is exponential and increases as we move from left to right, beginning at the $y$-intercept $(0,25)\text{.}$ The points $(2,P(2))$ and $(4,P(4)$ are also shown, along with the secant line that joins those points and the tangent line to the curve at $(2,P(2))\text{.}$

1.3.4.b

Answer.

$AV_{[2,4]} \approx 9171$ people per decade is expected to be the average rate of change of the city’s population over the two decades from 2030 to 2050.

1.3.4.c

Answer.

\begin{equation*} P'(2) = \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} = \lim_{h \to 0} 25000e^{2/5}\left( \frac{e^{h/5} - 1}{h}\right) \end{equation*}

Because there is no way to remove a factor of $h$ from the numerator, we cannot eliminate the $h$ that is making the denominator go to zero.

1.3.4.d

Answer.

\begin{equation*} P'(2) = \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} \approx 7458.5 \end{equation*}

which is measured in people per decade.

1.3.4.e

Answer.

See the graph provided in (a). The magenta line has slope equal to the average rate of change of $P$ on $[2,4]\text{,}$ while the green line is the tangent line at $(2,P(2))$ with slope $P'(2)\text{.}$

1.3.4.f

Answer.

It appears that the tangent line’s slope at the point $(a,P(a))$ will increase as $a$ increases.

1.4 The derivative function
1.4.2 How the derivative is itself a function

Activity 1.4.2.

Answer.

ADD ALT TEXT TO THIS IMAGE

Activity 1.4.3.

1.4.3.a

Answer.

$f'(x) = 0\text{.}$

1.4.3.b

Answer.

$g'(t) = 1\text{.}$

1.4.3.c

Answer.

$p'(z) = 2z\text{.}$

1.4.3.d

Answer.

$q'(s) = 3s^2\text{.}$

1.4.3.e

Answer.

$F'(t) = \dfrac{-1}{t^2}\text{.}$

1.4.3.f

Answer.

$G'(y) = \frac{1}{2\sqrt{y}}\text{.}$

1.5 Interpreting, estimating, and using the derivative
1.5.2 Units of the derivative function

Activity 1.5.2.

1.5.2.a

Answer.

$V'(10250) = -0.89$ means that the instantaneous rate of change of the car’s value when the car has been driven $10250$ miles is $-0.89$ dollars per mile. In addition, when the car has been driven $10250$ miles, if the car is driven one more mile, we expect that the value of the car will decrease by about $0.89$ dollars.

1.5.2.b

Answer.

$W'(0.75) = 3.43$ means that the instantaneous rate of change of volume of water in the tank when the water is $0.75$ meters deep is $3.43$ liters per meter. This tells us that when the water is $0.75$ meters deep, if the water rises one more meter, we expect that the volume of water in the tank will increase by about $3.43$ liters.

1.5.2.c

Answer.

$S'(20) = -0.527$ means that the instantaneous rate of change of the soda’s temperature at the instant $t = 20$ minutes is $-0.527$ degrees Celsius per minute. This tells us that after $20$ minutes have elapsed, if one more minute passes, we expect that the soda’s temperature will drop by about $0.527$ degrees Celsius.

1.5.2.d

Answer.

$C'(19) = 52.1$ means that the instantaneous rate of change of the rate at which the biker is burning calories when traveling at a speed of $19$ kilometers per hour is $52.1$ calories per hour per kilometer per hour. This tells us that when the person is riding at $19$ kilometers per hour, if they increase their speed by $1$ kilometer per hour, we expect that they will burn about $52.1$ additional calories over the next hour.

1.5.3 Toward more accurate derivative estimates

Activity 1.5.3.

1.5.3.a

Answer.

\begin{equation*} F'(20) \approx \frac{F(30)-F(10)}{30-10} = \frac{167.6-94.7}{20} = 3.645 \end{equation*}

degrees per minute.

1.5.3.b

Answer.

\begin{equation*} F'(40) \approx \frac{F(50)-F(30)}{50-30} = \frac{197.9-167.6}{20} = 1.515 \end{equation*}

degrees per minute.

1.5.3.c

Answer.

We expect $F'(50) \gt F'(60)\text{.}$

1.5.3.d

Answer.

Because at time $t = 46$ the potato’s temperature is increasing at 1.3941 degrees per minute, we expect that at $t = 47\text{,}$ the temperature will be about 1.3941 degrees greater than at $t = 46\text{.}$

1.5.3.e

Answer.

We might say that on the interval $[10,40]$“the temperature of the potato is increasing, but at a decreasing rate.”

Activity 1.5.4.

1.5.4.a

Answer.

$f'(90) \approx 0.0006$ liters per kilometer per kilometer per hour.

1.5.4.b

Answer.

At 80 kilometers per hour, the car is using fuel at a rate of 0.015 liters per kilometer.

1.5.4.c

Answer.

When the car is traveling at 90 kilometers per hour, its rate of fuel consumption per kilometer is increasing at a rate of 0.0006 liters per kilometer per kilometer per hour.

1.6 The second derivative
1.6.4 Concavity

Activity 1.6.2.

1.6.2.a

Answer.

Increasing: $0\lt t\lt 2\text{,}$ $3\lt t\lt 5\text{,}$ $7\lt t\lt 9\text{,}$ and $10\lt t\lt 12\text{.}$ Decreasing: never.

1.6.2.b

Answer.

Velocity is increasing on $0\lt t\lt 1\text{,}$ $3\lt t\lt 4\text{,}$ $7\lt t\lt 8\text{,}$ and $10\lt t\lt 11\text{;}$ $y = v(t)$ is decreasing on $1\lt t\lt 2\text{,}$ $4\lt t\lt 5\text{,}$ $8\lt t\lt 9\text{,}$ and $11\lt t\lt 12\text{.}$ Velocity is constant on $2\lt t\lt 3\text{,}$ $5\lt t\lt 7\text{,}$ and $9\lt t\lt 10\text{.}$

1.6.2.c

Answer.

$a(t) = v'(t)$ and $a(t) = s''(t)\text{.}$

1.6.2.d

Answer.

$s''(t)$ is positive since $s'(t)$ is increasing.

1.6.2.e

Answer.

increasing.
decreasing.
constant.
increasing.
decreasing.
constant.
concave up.
concave down.
linear.

Activity 1.6.3.

1.6.3.a

Answer.

Degrees Fahrenheit per minute.

1.6.3.b

Answer.

$F''(30) \approx -0.0516\text{.}$

1.6.3.c

Answer.

At the moment $t = 30\text{,}$ the temperature of the potato is $167.6$ degrees; its temperature is rising at an instaneous rate of $2.605$ degrees Fahrenheit per minute; and the rate at which the temperature is rising is falling at a rate of $0.0516$ degrees Fahrenheit per minute per minute. Over the minute from $t = 30$ to $t = 31\text{,}$ we expect the temperature of the potato to rise about $2.605$ degrees Fahrenheit and for the rate at which its temperature is increasing to drop by about $0.0516$ degrees Fahrenheit per minute. We expect that $F(31) \approx 169.7$ degrees Fahrenheit, and $F'(31) \approx 2.01343$ degrees Fahrenheit per minute.

1.6.3.d

Answer.

Increasing at a decreasing rate.

Activity 1.6.4.

Answer.

ADD ALT TEXT TO THIS IMAGE

1.7 Limits, continuity, and differentiability
1.7.2 Having a limit at a point

Activity 1.7.2.

1.7.2.a

Answer.

$f(-2) = 1\text{;}$ $f(-1)$ is not defined; $f(0) = \frac{7}{3}\text{;}$ $f(1) = 2\text{;}$ $f(2) = 2\text{.}$

1.7.2.b

Answer.

\begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1 \end{equation*}

\begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3} \end{equation*}

\begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3} \end{equation*}

\begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3 \end{equation*}

\begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2 \end{equation*}

1.7.2.c

Answer.

$\lim_{x \to -2} f(x)$ does not exist. The values of the limits as $x \to a$ for $a = -1, 0, 1, 2$ are $\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}$

1.7.2.d

Answer.

$a = -2\text{,}$ $a = -1\text{,}$ and $a = 1\text{.}$

1.7.2.e

Answer.

ADD ALT TEXT TO THIS IMAGE

1.7.3 Being continuous at a point

Activity 1.7.3.

1.7.3.a

Answer.

$a = -2\text{;}$ $a = +2\text{.}$

1.7.3.b

Answer.

$a = 3\text{.}$

1.7.3.c

Answer.

$a = -1\text{;}$ $a = 3\text{.}$

1.7.3.d

Answer.

$a=-2\text{;}$ $a = 2\text{;}$ $a = 3\text{;}$ $a = -1\text{.}$

1.7.3.e

Answer.

“If $f$is continuous at $x = a\text{,}$ then $f$has a limit at $x = a\text{.}$”

1.7.4 Being differentiable at a point

Activity 1.7.4.

1.7.4.a

Answer.

$g$ is piecewise linear.

1.7.4.b

Answer.

\begin{align*} g'(0) =\mathstrut \amp \lim_{h \to 0} \frac{g(0+h)-g(0)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|0+h|-|0|}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|h|}{h} \end{align*}

1.7.4.c

Answer.

$\lim_{h \to 0^+} \frac{|h|}{h} = 1 \text{,}$ but $\lim_{h \to 0^-} \frac{|h|}{h} = -1 \text{.}$

1.7.4.d

Answer.

$a = -3, -2, -1, 1, 2, 3\text{.}$

1.7.4.e

Answer.

True.

1.8 The tangent line approximation
1.8.3 The local linearization

Activity 1.8.2.

1.8.2.a

Answer.

$L(-1) = -2\text{;}$ $L'(-1) = 3\text{.}$

1.8.2.b

Answer.

$g(-1) = -2\text{;}$ $g'(-1) = 3\text{.}$

1.8.2.c

Answer.

Less.

1.8.2.d

Answer.

$g(-1.03) \approx L(-1.03) = -2.09\text{.}$

1.8.2.e

Answer.

Concave up.

1.8.2.f

Answer.

The illustration below shows a possible graph of $y = g(x)$ near $x = -1\text{,}$ along with the tangent line $y = L(x)$ through $(-1, g(-1))\text{.}$

ADD ALT TEXT TO THIS IMAGE

Activity 1.8.3.

1.8.3.a

Answer.

$L(x) = -1 + 2(x-2)\text{.}$

1.8.3.b

Answer.

$f(2.07) \approx L(2.07) = -0.86\text{.}$

1.8.3.c

Answer.

See the image in part 1.8.3.e.

1.8.3.d

Answer.

Neither.

1.8.3.e

Answer.

See the image below, which shows, at left, a possible graph of $y = f(x)$ near $x = 2\text{,}$ along with the tangent line $y = L(x)$ through $(2, f(2))\text{.}$

ADD ALT TEXT TO THIS IMAGE

1.8.3.f

Answer.

Too large.

Activity 1.8.4.

1.8.4.a

Answer.

$f(3) = 1\text{,}$ $f'(3) = -0.5\text{,}$ $g(3) = 1\text{,}$ and $g'(3) = -0.5\text{;}$ this means $f$ and $g$ will have the same tangent line approximation at $x = 3\text{.}$

1.8.4.b

Answer.

$L(x) = 1 - 0.5(x-3)$ is the tangent line approximation for both $f$ and $g$ at $x = 3\text{.}$

1.8.4.c

Answer.

$x$	$2.9$	$2.99$	$2.999$	$3$	$3.001$	$3.01$	$3.1$
$f(x)$	$1.051$	$1.00501$	$1.0005001$	$1$	$0.9995001$	$0.99501$	$0.951$
$g(x)$	$1.045$	$1.00495$	$1.0004995$	$1$	$0.9994995$	$0.99495$	$0.945$
$L(x)$	$1.05$	$1.005$	$1.0005$	$1$	$0.9995$	$0.995$	$0.95$

Near $x = 3\text{,}$ $L(x)$ always underestimates the value of $f(x)\text{,}$ and $L(x)$ always overestimates the value of $g(x)\text{.}$

1.8.4.d

Answer.

ADD ALT TEXT TO THIS IMAGE

Since $f''(3) = 0.2$ ande $g''(3)=-1\text{,}$ $g$ has “more” concavity at $x = 3$ than $f\text{,}$ so $L(x)$ is a closer approximation of the value of $f(x)$ than of the value of $g(x)$ near $x = 3\text{.}$

2 Computing Derivatives
2.1 Elementary derivative rules
2.1.3 Constant, Power, and Exponential Functions

Activity 2.1.2.

2.1.2.a

Answer.

$f'(t) = 0\text{.}$

2.1.2.b

Answer.

$g'(z) = 7^z \ln(7)\text{.}$

2.1.2.c

Answer.

$h'(w) = \frac{3}{4} w^{-1/4}\text{.}$

2.1.2.d

Answer.

$\frac{dp}{dx} = 0\text{.}$

2.1.2.e

Answer.

$r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}$

2.1.2.f

Answer.

$\frac{d}{dq}[q^{-1}] = -q^{-2}\text{.}$

2.1.2.g

Answer.

$\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}$

2.1.4 Constant Multiples and Sums of Functions

Activity 2.1.3.

2.1.3.a

Answer.

$f'(x) = \frac{5}{3}x^{2/3} - 4 x^3 + 2^x \ln(2)\text{.}$

2.1.3.b

Answer.

$g'(x) = 14e^x + 3 \cdot 5x^4 - 1\text{.}$

2.1.3.c

Answer.

$h'(z) = \frac{1}{2}z^{-1/2} - 4z^{-5} + 5^z \ln(5)\text{.}$

2.1.3.d

Answer.

$\frac{dr}{dt} = \sqrt{53} \cdot 7 t^6 - \pi e^t\text{.}$

2.1.3.e

Answer.

$\frac{ds}{dy} = 4y^3\text{.}$

2.1.3.f

Answer.

$q'(x) = 2x - 2x^{-2}\text{.}$

2.1.3.g

Answer.

$p'(a) = 12a^3 - 6 a^2 + 14a - 1\text{.}$

Activity 2.1.4.

2.1.4.a

Answer.

$h'(4) = \frac{3}{16}\text{.}$

2.1.4.b

Answer.

(i.)$P'(4) = 2(1.37)^4 \ln(1.37) \approx 2.218$ million cells per day; (ii.) the population is growing at an increasing rate.

2.1.4.c

Answer.

$y - 25 = -33(a+1)\text{.}$

2.1.4.d

Answer.

The slope is a number, while the equation is, well, an equation.

2.2 The sine and cosine functions
2.2.2 The sine and cosine functions

Activity 2.2.2.

2.2.2.a

Answer.

ADD ALT TEXT TO THIS IMAGE

2.2.2.b

Answer.

$1,0,-1,0,1,0,-1,0,1\text{.}$

2.2.2.c

Answer.

$f'(0) = f'(-2\pi) = f'(2\pi) = 1\text{.}$

2.2.2.d

Answer.

ADD ALT TEXT TO THIS IMAGE

2.2.2.e

Answer.

$\frac{d}{dx}[\sin(x)] = \cos(x)\text{.}$

Activity 2.2.3.

2.2.3.a

Answer.

ADD ALT TEXT TO THIS IMAGE

2.2.3.b

Answer.

$0,-1,0,1,0,-1,0,1,0\text{.}$

2.2.3.c

Answer.

$g'(\frac{\pi}{2})=g'(-\frac{3\pi}{2})=-1\text{.}$

2.2.3.d

Answer.

ADD ALT TEXT TO THIS IMAGE

2.2.3.e

Answer.

$\frac{d}{dx}[\cos(x)] = -\sin(x)\text{.}$

Activity 2.2.4.

2.2.4.a

Answer.

$\frac{dh}{dt} = -3\sin(t) - 4\cos(t)\text{.}$

2.2.4.b

Answer.

$f'(\frac{\pi}{6}) = 2 + \frac{\sqrt{3}}{4}\text{.}$

2.2.4.c

Answer.

$y - \frac{\pi^2}{4} = (\pi-2)(x-\frac{\pi}{2})\text{.}$

2.2.4.d

Answer.

$p'(z) = 4z^3 + 4^z \ln(4) - 4\sin(z)\text{.}$

2.2.4.e

Answer.

$P'(2) = 8\cos(2) \approx -3.329$ hundred animals per decade.

2.3 The product and quotient rules
2.3.2 The product rule

Activity 2.3.2.

2.3.2.a

Answer.

$m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.}$

2.3.2.b

Answer.

$h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\text{.}$

2.3.2.c

Answer.

$f'(1) = e(\cos(1) + \sin(1)) \approx 3.756\text{.}$

2.3.2.d

Answer.

$L(x) = -\frac{1}{2}(x+1)\text{.}$

2.3.3 The quotient rule

Activity 2.3.3.

2.3.3.a

Answer.

$r'(z)=\frac{(z^4+1) 3^z \ln(3) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.}$

2.3.3.b

Answer.

$v'(t) = \frac{(\cos(t) + t^2)\cos(t) - \sin(t)(-\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.}$

2.3.3.c

Answer.

$R'(0) = \frac{2}{9}\text{.}$

2.3.3.d

Answer.

$I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$ $I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$ and $I'(5) = \frac{-400}{e^5} \approx -2.695\text{,}$ each in candles per millisecond.

2.3.4 Combining rules

Activity 2.3.4.

2.3.4.a

Answer.

$f'(r) = (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)]\text{.}$

2.3.4.b

Answer.

$p'(t) = \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2}\text{.}$

2.3.4.c

Answer.

$g'(z) = 3 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 1 - z(2z)}{(z^2 + 1)^2}\text{.}$

2.3.4.d

Answer.

$s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414$ feet per second.

2.3.4.e

Answer.

$p'(3) = 30$ and $q'(3) = \frac{13}{8}\text{.}$

2.4 Derivatives of other trigonometric functions
2.4.2 Derivatives of the cotangent, secant, and cosecant functions

Activity 2.4.2.

2.4.2.a

Answer.

All real numbers $x$ such that $x \ne \frac{\pi}{2} + k\pi\text{,}$ where $k = \pm 1, \pm 2, \ldots\text{.}$

2.4.2.b

Answer.

$h'(x) = \frac{\sin(x)}{\cos^2(x)}\text{.}$

2.4.2.c

Answer.

$h'(x) = \sec(x) \tan(x)\text{.}$

2.4.2.d

Answer.

$h$ and $h'$ have the same domain: all real numbers $x$ such that $x \ne \frac{\pi}{2}+k\pi\text{,}$ where $k = 0, \pm 1, \pm 2, \ldots\text{.}$

Activity 2.4.3.

2.4.3.a

Answer.

All real numbers $x$ such that $x \ne k\pi\text{,}$ where $k = 0, \pm 1, \pm 2, \ldots\text{.}$

2.4.3.b

Answer.

$h'(x) = -\frac{\cos(x)}{\sin^2(x)}\text{.}$

2.4.3.c

Answer.

$h'(x) = -\csc(x) \cot(x)\text{.}$

2.4.3.d

Answer.

$p$ and $p'$ have the same domain: all real numbers $x$ such that $x \ne k\pi\text{,}$ where $k = 0, \pm 1, \pm 2, \ldots\text{.}$

Activity 2.4.4.

2.4.4.a

Answer.

$m = f'(\frac{\pi}{3}) =10\sqrt{3} + \frac{4}{3}\text{.}$

2.4.4.b

Answer.

$p'(\frac{\pi}{4}) = \frac{\pi^2}{16} \sqrt{2} + \frac{\sqrt{2}\pi}{2} + \frac{\pi}{2} - 1\text{.}$

2.4.4.c

Answer.

$h'(t) = \frac{(t^2+1) \sec^2(t) - 2t \tan(t)}{(t^2 + 1)^2} + 2e^t \sin(t) - 2 e^t\cos(t)\text{.}$

2.4.4.d

Answer.

$g'(r) = \frac{r \sec(r) \tan(r) + \sec(r) - r ln(5) \sec(r)}{5^r}\text{.}$

2.4.4.e

Answer.

$s'(2) = \frac{15\cos(2) - 15\sin(2)}{e^2} \approx -2.69$ inches per second.

2.5 The chain rule
2.5.2 The chain rule

Activity 2.5.2.

2.5.2.a

Answer.

$h'(x) = -4x^3\sin(x^4)\text{.}$

2.5.2.b

Answer.

$h'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.}$

2.5.2.c

Answer.

$h'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.}$

2.5.2.d

Answer.

$h'(x) = -5\cot^4(x) \csc^2(x)\text{.}$

2.5.2.e

Answer.

$h'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.}$

2.5.3 Using multiple rules simultaneously

Activity 2.5.3.

2.5.3.a

Answer.

$p'(r) = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.}$

2.5.3.b

Answer.

$m'(v) = -3v^2 \sin(v^2)\sin(v^3) + 2v \cos(v^3)\cos(v^2)\text{.}$

2.5.3.c

Answer.

$h'(y) = \frac{(e^{4y}+1) [-10\sin(10y)] - \cos(10y) [4e^{4y}]}{(e^{4y}+1)^2}\text{.}$

2.5.3.d

Answer.

$s'(z) = 2^{z^2\sec(z)} \ln(2) [z^2 \sec(z)\tan(z) + \sec(z) \cdot 2z]\text{.}$

2.5.3.e

Answer.

$c'(x) = \cos(e^{x^2}) [e^{x^2}\cdot 2x]\text{.}$

Activity 2.5.4.

2.5.4.a

Answer.

$y - 2 = \frac{1}{4}(x-0)\text{.}$

2.5.4.b

Answer.

$v(1) = s'(1) = -\frac{3}{8}$ inches per second; the particle is moving left at the instant $t = 1\text{.}$

2.5.4.c

Answer.

$P'(1000) = 30 e^{-0.0323} (-0.0000323) \approx -0.000938$ inches of mercury per foot.

2.5.4.d

Answer.

$C'(2) = -10 \text{;}$ $D'(-1) = -20\text{.}$

2.6 Derivatives of inverse functions
2.6.3 The derivative of the natural logarithm function

Activity 2.6.2.

2.6.2.a

Answer.

$h'(x) = x + 2x\ln(x)\text{.}$

2.6.2.b

Answer.

$p'(t) = \frac{(e^t + 1) \frac{1}{t} - \ln(t) \cdot e^t}{(e^t + 1)^2}\text{.}$

2.6.2.c

Answer.

$s'(y) = \frac{1}{\cos(y) + 2} \cdot (-\sin(y))\text{.}$

2.6.2.d

Answer.

$z'(x) = \sec^2(\ln(x)) \cdot \frac{1}{x}\text{.}$

2.6.2.e

Answer.

$m'(z) = \frac{1}{\ln(z)} \cdot \frac{1}{z}\text{.}$

2.6.4 Inverse trigonometric functions and their derivatives

Activity 2.6.3.

2.6.3.a

Answer.

$\tan(r(x)) = x\text{.}$

2.6.3.b

Answer.

$r'(x) = \cos^2(r(x))\text{.}$

2.6.3.c

Answer.

$r'(x) = \cos^2(\arctan(x))\text{.}$

2.6.3.d

Answer.

With $\theta = \arctan(x)\text{,}$

2.6.3.e

Answer.

$\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.}$

2.6.3.f

Answer.

$r'(x) = \frac{1}{1+x^2}\text{.}$

Activity 2.6.4.

2.6.4.a

Answer.

$f'(x) = \left[x^3 \cdot \frac{1}{1+x^2} + \arctan(x) \cdot 3x^2 \right] + \left[e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x\right]\text{.}$

2.6.4.b

Answer.

$p'(t) = 2^{t\arcsin(t)} \ln(2) [t \cdot \frac{1}{\sqrt{1-t^2}} + \arcsin(t) \cdot 1]\text{.}$

2.6.4.c

Answer.

$h'(z) = 27(\arcsin(5z) + \arctan(4-z))^{26} \left[\frac{1}{\sqrt{1-(5z)^2}} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right]\text{.}$

2.6.4.d

Answer.

$s'(y) = -\frac{1}{y^2}\text{.}$

2.6.4.e

Answer.

$m'(v) = \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right]\text{.}$

2.6.4.f

Answer.

$g'(w) = \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{(1+w^2) \frac{1}{w} - \ln(w) \cdot 2w}{(1+w^2)^2} \right] $

2.7 Derivatives of functions given implicitly
2.7.2 Implicit Differentiation

Activity 2.7.2.

2.7.2.a

Answer.

The graph of the curve fails the vertical line test.

2.7.2.b

Answer.

$\frac{dy}{dx} = \frac{1}{5y^4 - 15y^2 + 4}\text{.}$

2.7.2.c

Answer.

$y = -\frac{1}{6}x + 1\text{.}$

2.7.2.d

Answer.

$(1.418697,0.543912)\text{,}$ $(-1.418697,-0.543912)\text{,}$ $(-3.63143, 1.64443)\text{,}$ and $(3.63143, -1.64443)\text{.}$

Activity 2.7.3.

2.7.3.a

Answer.

Horizontal at $x \approx 0.42265\text{,}$ thus $(0.42265, -1.05782); (0.42265, 0.229478); (0.42265, 0.770522); (0.42265, 2.05782)\text{.}$ There are four more points where $x \approx 1.57735\text{.}$

2.7.3.b

Answer.

When $y = \frac{1}{2}, \frac{1 \pm \sqrt{5}}{2}\text{,}$ so one point is $(2.21028, \frac{1}{2})\text{.}$

2.7.3.c

Answer.

$y - 1 = \frac{1}{2}(x-1)\text{.}$

Activity 2.7.4.

2.7.4.a

Answer.

$\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2 $ and the tangent line has equation $y - 3 = 1(x+3)\text{.}$

2.7.4.b

Answer.

$\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}$ and the tangent line has equation $y = \frac{1}{2}x\text{.}$

2.7.4.c

Answer.

$\frac{dy}{dx} = \frac{3e^{-xy} - 3xye^{-xy}}{3x^2e^{-xy}+2y}$ and the tangent line is $y - 1 = 0.234950(x - 0.619061)\text{.}$

3 Using Derivatives
3.1 Related rates
3.1.2 Related Rates Problems

Activity 3.1.2.

3.1.2.a

Answer.

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3.1.2.b

Answer.

$r = \frac{3}{4}h\text{.}$

3.1.2.c

Answer.

$V = \frac{3}{16} \pi h^3\text{.}$

3.1.2.d

Answer.

$\frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt} \text{.}$

3.1.2.e

Answer.

$\left. \frac{dh}{dt} \right|_{h=3} = \frac{64}{81\pi} \approx 0.2515$ feet per minute.

3.1.2.f

Answer.

Most rapidly when $h = 3\text{.}$

Activity 3.1.3.

3.1.3.a

Answer.

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3.1.3.b

Answer.

$\frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.}$

3.1.3.c

Answer.

$h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}$

3.1.3.d

Answer.

$\left. \frac{dz}{dt} \right|_{h=3000} = 360 \ \text{feet/sec}; $$\left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{12}{125} $ radians per second.

3.1.3.e

Answer.

greater.

Activity 3.1.4.

3.1.4.a

Answer.

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3.1.4.b

Answer.

$3s = 2x\text{.}$

3.1.4.c

Answer.

$3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.}$

3.1.4.d

Answer.

$\left. \frac{ds}{dt} \right|_{x=8} = 2$ feet per second.

3.1.4.e

Answer.

at a constant rate.

3.1.4.f

Answer.

Let $y$ represent the location of the tip of the shadow; $\frac{dy}{dt} = 5$ feet/sec.

Activity 3.1.5.

Answer.

Let $x$ denote the position of the ball at time $t$ and $z$ the distance from the ball to first base, as pictured below.

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$\left. \frac{dz}{dt} \right|_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 \ \text{feet/sec} \text{.}$

Let $r$ be the runner’s position at time $t$ and let $s$ be the distance between the runner and the ball, as pictured.

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$\left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{feet/sec} \text{.}$

3.2 Using derivatives to evaluate limits
3.2.2 Using derivatives to evaluate indeterminate limits of the form $\frac{0}{0}\text{.}$

Activity 3.2.2.

3.2.2.a

Answer.

$\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1\text{.}$

3.2.2.b

Answer.

$\lim_{x \to \pi} \frac{\cos(x)}{x} = -\frac{1}{\pi}\text{.}$

3.2.2.c

Answer.

$\lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}} = -2\text{.}$

3.2.2.d

Answer.

$\lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = 0\text{.}$

Activity 3.2.3.

3.2.3.a

Answer.

$\lim_{x \to 2} \frac{f(x)}{g(x)} = \frac{1}{8}\text{.}$

3.2.3.b

Answer.

$\lim_{x \to 2} \frac{p(x)}{q(x)} = 1\text{.}$

3.2.3.c

Answer.

$\lim_{x \to 2} \frac{r(x)}{s(x)} \lt 0\text{.}$

3.2.3 Limits involving $\infty$

Activity 3.2.4.

3.2.4.a

Answer.

$\lim_{x \to \infty} \frac{x}{\ln(x)} = \infty\text{.}$

3.2.4.b

Answer.

$\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \frac{1}{2}\text{.}$

3.2.4.c

Answer.

$\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = 0\text{.}$

3.2.4.d

Answer.

$\lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.}$

3.2.4.e

Answer.

$\lim_{x \to \infty} xe^{-x} = 0\text{.}$

3.3 Using derivatives to identify extreme values
3.3.2 Critical numbers and the first derivative test

Activity 3.3.2.

3.3.2.a

Answer.

$x = -4$ and $x = 2$ are the only two critical numbers of $g\text{.}$

3.3.2.b

Answer.

For $x \lt -4\text{,}$ $g'(x) \gt 0\text{;}$ for $-4 \lt x \lt 2\text{,}$ $g'(x) \lt 0\text{;}$ and for $x \gt 2\text{,}$ $g'(x) \gt 0\text{.}$

3.3.2.c

Answer.

$g$ has a local maximum at $x = -4\text{;}$ $g$ has a local minimum at $x = 2\text{.}$

3.3.2.d

Answer.

3.3.3 The second derivative test

Activity 3.3.3.

3.3.3.a

Answer.

$g$ is concave up for $x \lt -1\text{,}$ concave down for $-1 \lt x \lt 2\text{,}$ and concave down for $x \gt 2\text{.}$

3.3.3.b

Answer.

$x = -1$ is an inflection point of $g\text{.}$

3.3.3.c

Answer.

$g$ has a local minimum at $x = -1.67857351\text{.}$

3.3.3.d

Answer.

$g$ is a degree 5 polynomial.

Activity 3.3.4.

3.3.4.a

Answer.

In the graph below, $h(x) = x^2 + \cos(3x)$ is given in dark blue, while $h(x) = x^2 + \cos(1.6x)$ is shown in light blue.

3.3.4.b

Answer.

If $\frac{2}{k^2} \gt 1\text{,}$ then the equation $\cos(kx) = \frac{2}{k^2}$ has no solution. Hence, whenever $k^2 \lt 2\text{,}$ or $k \lt \sqrt{2} \approx 1.414\text{,}$ it follows that the equation $\cos(kx) = \frac{2}{k^2}$ has no solutions $x\text{,}$ which means that $h''(x)$ is never zero (indeed, for these $k$-values, $h''(x)$ is always positive so that $h$ is always concave up). On the other hand, if $k \ge \sqrt{2}\text{,}$ then $\frac{2}{k^2} \le 1\text{,}$ which guarantees that $\cos(kx) = \frac{2}{k^2}$ has infinitely many solutions, due to the periodicity of the cosine function. At each such point, $h''(x) = 2 - k^2 \cos(kx)$ changes sign, and therefore $h$ has infinitely many inflection points whenever $k \ge \sqrt{2}\text{.}$

3.3.4.c

Answer.

To see why $h$ can only have a finite number of critical numbers regardless of the value of $k\text{,}$ consider the equation

\begin{equation*} 0 = h'(x) = 2x - k\sin(kx)\text{,} \end{equation*}

which implies that $2x = k\sin(kx)\text{.}$ Since $-1 \le \sin(kx) \le 1\text{,}$ we know that $-k \le k\sin(kx) \le k\text{.}$ Once $|x|$ is sufficiently large, we are guaranteed that $|2x| \gt k\text{,}$ which means that for large $x\text{,}$ $2x$ and $k\sin(kx)$ cannot intersect. Moreover, for relatively small values of $x\text{,}$ the functions $2x$ and $k\sin(kx)$ can only intersect finitely many times since $k\sin(kx)$ oscillates a finite number of times. This is why $h$ can only have a finite number of critical numbers, regardless of the value of $k\text{.}$

3.4 Using derivatives to describe families of functions
3.4.2 Describing families of functions in terms of parameters

Activity 3.4.2.

3.4.2.a

Answer.

$p$ has two critical numbers ($x = \pm \sqrt{\frac{a}{3}}$) whenever $a \gt 0$ and no critical numbers when $a \lt 0\text{.}$

3.4.2.b

Answer.

When $a \lt 0\text{,}$ $p$ is always increasing and has no relative extreme values. When $a\gt 0\text{,}$ $p$ has a relative maximum at $x = -\sqrt{\frac{a}{3}}$ and a relative minimum at $x = +\sqrt{\frac{a}{3}}\text{.}$

3.4.2.c

Answer.

$p$ is CCD for $x \lt 0$ and $p$ is CCU for $x\gt 0\text{,}$ making $x = 0$ an inflection point.

3.4.2.d

Answer.

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Activity 3.4.3.

3.4.3.a

Answer.

$h$ is an always increasing function.

3.4.3.b

Answer.

$h$ is always concave down.

3.4.3.c

Answer.

$\lim_{x \to \infty} a(1-e^{-bx}) = a\text{,}$ and $\lim_{x \to \infty} a(1-e^{-bx}) = -\infty\text{.}$

3.4.3.d

Answer.

If $b$ is large and $x$ is close to zero, $h'(x)$ is relatively large near $x = 0\text{,}$ and the curve’s slope will quickly approach zero as $x$ increases. If $b$ is small, the graph is less steep near $x = 0$ and its slope goes to zero less quickly as $x$ increases.

3.4.3.e

Answer.

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Activity 3.4.4.

3.4.4.a

Answer.

$L$ is an always increasing function.

3.4.4.b

Answer.

$L$ is concave up for all $t \lt -\frac{1}{k} \ln \left(\frac{1}{c}\right)$ and concave up for all other values of $t\text{.}$

3.4.4.c

Answer.

$\lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = A\text{,}$ and

\begin{equation*} \lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = 0\text{.} \end{equation*}

3.4.4.d

Answer.

The inflection point on the graph of $L$ is $( -\frac{1}{k} \ln \left(\frac{1}{c}\right), \frac{A}{2})\text{.}$

3.4.4.e

Answer.

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3.5 Global optimization
3.5.2 Global Optimization

Activity 3.5.2.

3.5.2.a

Answer.

$x = \pm \sqrt{2} \approx \pm 1.414\text{.}$

3.5.2.b

Answer.

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3.5.2.c

Answer.

On $[-2,3]\text{,}$ $g$ has a global maximum at $x = 3$ and a global minimum at $x = \sqrt{2}\text{.}$

3.5.2.d

Answer.

On $[-2,2]\text{,}$ $g$ has a global maximum at $x = -\sqrt{2}$ and a global minimum at $x = \sqrt{2}\text{.}$

3.5.2.e

Answer.

On $[-2,1]\text{,}$ $g$ has a global maximum at $x = -\sqrt{2}$ and a global minimum at $x = 1\text{.}$

Activity 3.5.3.

3.5.3.a

Answer.

Absolute maximum: $e^{-1}\text{;}$ absolute minimum: $0\text{.}$

3.5.3.b

Answer.

Absolute maximum: $\sqrt{2}\text{;}$ absolute minimum: $-1\text{.}$

3.5.3.c

Answer.

Absolute maximum: 9.8; absolute minimum: 8.

3.5.3.d

Answer.

Absolute minimum 3; no absolute maximum.

3.5.3.e

Answer.

Absolute minimum $0\text{;}$ absolute maximum $\frac{1}{a}e^{-1}\text{.}$

3.5.3.f

Answer.

Absolute minimum $b-1\text{;}$ no absolute maximum.

3.5.3 Moving toward applications

Activity 3.5.4.

3.5.4.a

Answer.

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3.5.4.b

Answer.

$V(x) = x (10-2x) (15-2x) = 4x^3 - 50x^2 + 150x\text{.}$

3.5.4.c

Answer.

$1 \le x \le 3\text{.}$

3.5.4.d

Answer.

$x = \frac{25 \pm 5\sqrt{7}}{6} \approx 6.371459426, 1.961873908\text{.}$

3.5.4.e

Answer.

$\displaystyle V(1.961873908) = 132.0382370$
$\displaystyle V(1) = 104$
$\displaystyle V(3) = 108$

3.5.4.f

Answer.

Absolute maximum: 132.0382370; absolute minimum: 104.

3.6 Applied optimization
3.6.2 More applied optimization problems

Activity 3.6.2.

3.6.2.a

Answer.

Let the can have radius $r$ and height $h\text{.}$

3.6.2.b

Answer.

$V = \pi r^2 h\text{;}$ $S = 2 \pi r^2 + 2 \pi r h\text{;}$ $C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.}$

3.6.2.c

Answer.

$C(r) = 0.054 \pi r^2 + 0.48 \frac{1}{r}\text{,}$ $r \gt 0\text{.}$

3.6.2.d

Answer.

$r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.12259\text{;}$ $h \approx 4.041337\text{;}$ minimum cost $C(1.12259) \approx 0.64137\text{.}$

Activity 3.6.3.

Answer.

The absolute minimum time the hiker can achieve is $0.99302$ hours, which is attained by hiking about 2.2 km from $P$ to $Q$ and then turning into the woods for the remainder of the trip.

Activity 3.6.4.

Answer.

$A(1.19606) \approx 2.2018$ is the absolute maximum cross-sectional area, which leads to the absolute maximum volume.

Activity 3.6.5.

Answer.

Maximum area: $A(\frac{5}{\sqrt{3}}) = \frac{500}{9}\sqrt{3} \approx 96.225\text{.}$ Maximum perimeter: $P(1) = 52\text{.}$

4 The Definite Integral
4.1 Determining distance traveled from velocity
4.1.2 Area under the graph of the velocity function

Activity 4.1.2.

4.1.2.a

Answer.

\begin{align*} A =\mathstrut \amp v(0.0) \cdot 0.5 + v(0.5) \cdot 0.5 + v(1.0) \cdot 0.5 + v(1.5) \cdot 0.5\\ =\mathstrut \amp 1.500 \cdot 0.5 + 1.9375 \cdot 0.5 + 2.000 \cdot 0.5 + 2.0625 \cdot 0.5\\ =\mathstrut \amp 3.75 \end{align*}

Thus, $D \approx 3.75$ miles.

4.1.2.b

Answer.

Using 8 rectangles of width $0.25\text{,}$ $D \approx 3.875\text{.}$

4.1.2.c

Answer.

$s(t) = \frac{1}{8}t^4 - \frac{1}{2} t^3 + \frac{3}{4} t^2 + \frac{3}{2}t\text{.}$

4.1.2.d

Answer.

$s(2) - s(0) = \frac{1}{8}2^4 - \frac{1}{2}2^3 + \frac{3}{4}2^2 + \frac{3}{2} 2 = 4\text{.}$

4.1.3 Two approaches: area and antidifferentiation

Activity 4.1.3.

4.1.3.a

Answer.

On $(0,1)\text{,}$ $s$ is increasing because velocity is positive.

4.1.3.b

Answer.

$s(t) = 32t - 16t^2\text{.}$

4.1.3.c

Answer.

$s(1) - s(\frac{1}{2}) = 4\text{.}$

4.1.3.d

Answer.

$A = 4$ feet is the total distance the ball traveled vertically on $[\frac{1}{2},1]\text{.}$

4.1.3.e

Answer.

$s(1) - s(0) = 16$ is the vertical distance the ball traveled on the interval $[0,1]\text{.}$ Equivalently, the area between the velocity curve and the $t$-axis on $[0,1]$ is $A = 16$ feet.

4.1.3.f

Answer.

$s(2) - s(0) = 0\text{,}$ so the ball has zero change in position on the interval $[0,2]\text{.}$

4.1.4 When velocity is negative

Activity 4.1.4.

4.1.4.a

Answer.

Total distance traveled is $2\text{;}$ change in position is $0\text{.}$

4.1.4.b

Answer.

$0 \lt t \lt 1$ and $4 \lt t \lt 8 \text{.}$

4.1.4.c

Answer.

$s(8) - s(0) = 5 \ \mbox{m} \text{,}$ while the distance traveled on $[0,8]$ is $D = 13\text{,}$ and thus these two quantities are different.

4.1.4.d

Answer.

See the figure below.

4.2 Riemann sums
4.2.2 Sigma Notation

Activity 4.2.2.

4.2.2.a

Answer.

$65 $

4.2.2.b

Answer.

$32 $

4.2.2.c

Answer.

\begin{equation*} 3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.} \end{equation*}

4.2.2.d

Answer.

\begin{equation*} 4 + 8 + 16 + 32 + \cdots + 256 = \sum_{i=2}^{8} 2^i\text{.} \end{equation*}

4.2.2.e

Answer.

\begin{equation*} \sum_{i=1}^{6} \frac{1}{2^i} = \frac{63}{64}\text{.} \end{equation*}

4.2.3 Riemann Sums

Activity 4.2.3.

4.2.3.a

Answer.

ADD ALT TEXT TO THIS IMAGE

4.2.3.b

Answer.

$L_4 = \frac{311}{48} \approx 6.47917\text{,}$ $R_4 = \frac{335}{48} \approx 6.97917\text{,}$ and $M_4 = \frac{637}{96} \approx 6.63542\text{.}$

4.2.3.c

Answer.

\begin{equation*} \frac{L_4 + M_4}{2} = \frac{646}{96} \ne \frac{637}{96} = M_4\text{.} \end{equation*}

4.2.3.d

Answer.

$L_n$ is an under-estimate; $R_n$ is an over-estimate.

4.2.4 When the function is sometimes negative

Activity 4.2.4.

4.2.4.a

Answer.

$M_5 = -\frac{36}{25} = -1.44$

4.2.4.b

Answer.

The change in position is approximately $-1.44$ feet.

4.2.4.c

Answer.

$D \approx 2.336\text{.}$

4.2.4.d

Answer.

$-\frac{4}{3}$ is the object’s total change in position on $[1,5]\text{.}$

4.3 The definite integral
4.3.2 The definition of the definite integral

Activity 4.3.2.

4.3.2.a

Answer.

$\int_0^1 3x \, dx = \frac{3}{2}\text{.}$

4.3.2.b

Answer.

$\int_{-1}^4 (2-2x) \, dx = -5\text{.}$

4.3.2.c

Answer.

$\int_{-1}^1 \sqrt{1-x^2} \, dx = \frac{\pi}{2}\text{.}$

4.3.2.d

Answer.

$\int_{-3}^4 g(x) \, dx = \frac{3\pi}{4} - \frac{3}{2}\text{.}$

4.3.3 Some properties of the definite integral

Activity 4.3.3.

4.3.3.a

Answer.

$\int_5^2 f(x) \,dx = -2\text{.}$

4.3.3.b

Answer.

$\int_0^5 g(x) \,dx = 3\text{.}$

4.3.3.c

Answer.

$\int_0^5 (f(x) + g(x))\, dx = 2\text{.}$

4.3.3.d

Answer.

$\int_2^5 (3x^2 - 4x^3) \, dx = -492\text{.}$

4.3.3.e

Answer.

$\int_5^0 (2x^3 - 7g(x)) \, dx = -\frac{583}{2}\text{.}$

4.3.4 How the definite integral is connected to a function’s average value

Activity 4.3.4.

4.3.4.a

Answer.

$y = v(t) = \sqrt{4-(t-2)^2}$ is the top half of the circle $(t-2)^2 + y^2 = 4\text{,}$ which has radius 2 and is centered at $(2,0)\text{.}$

4.3.4.b

Answer.

$\int_0^4 v(t) \, dt = 2\pi\text{.}$

4.3.4.c

Answer.

The object moved $2 \pi$ meters in 4 minutes.

4.3.4.d

Answer.

$v_{\text{AVG} }[0,4] = \frac{\pi}{2}\text{,}$ meters per minute,.

4.3.4.e

Answer.

The height of the rectangle is the average value of $v\text{,}$ $v_{\text{AVG} }[0,4] = \frac{\pi}{2} \approx 1.57\text{.}$

4.3.4.f

Answer.

$D = 2\pi\text{.}$

4.4 The Fundamental Theorem of Calculus
4.4.2 The Fundamental Theorem of Calculus

Activity 4.4.2.

4.4.2.a

Answer.

$\int_{-1}^4 (2-2x) \, dx = -5\text{.}$

4.4.2.b

Answer.

$\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = 1\text{.}$

4.4.2.c

Answer.

$\int_0^1 e^x \, dx = e-1\text{.}$

4.4.2.d

Answer.

$\int_{-1}^{1} x^5 \, dx = 0\text{.}$

4.4.2.e

Answer.

$\int_0^2 (3x^3 - 2x^2 - e^x) \, dx = \frac{23}{3} - e^2\text{.}$

4.4.3 Basic antiderivatives

Activity 4.4.3.

4.4.3.a

Answer.

given function, $f(x)$	antiderivative, $F(x)$
$k\text{,}$ ($k \ne 0$)	$kx$
$x^n\text{,}$ $n \ne -1$	$\frac{1}{n+1}x^{n+1}$
$\frac{1}{x}\text{,}$ $x \gt 0$	$\ln(x)$
$\sin(x)$	$-\cos(x)$
$\cos(x)$	$\sin(x)$
$\sec(x) \tan(x)$	$\sec(x)$
$\csc(x) \cot(x)$	$-\csc(x)$
$\sec^2 (x)$	$\tan(x)$
$\csc^2 (x)$	$-\cot(x)$
$e^x$	$e^x$
$a^x$ $(a \gt 1)$	$\frac{1}{\ln(a)} a^x$
$\frac{1}{1+x^2}$	$\arctan(x)$
$\frac{1}{\sqrt{1-x^2}}$	$\arcsin(x)$

$\int_0^1 \left(x^3 - x - e^x + 2\right) \,dx = \frac{11}{4} - e\text{.}$

4.4.3.b

Answer.

$\int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt = 1 - \sqrt{3} - \frac{\pi^2}{3}\text{.}$

4.4.3.c

Answer.

$\int_0^1 (\sqrt{x} - x^2) \, dx = \frac{1}{3}\text{.}$

4.4.4 The total change theorem

Activity 4.4.4.

4.4.4.a

Answer.

The person burned exactly $\frac{400}{3}$ calories in the first 10 minutes of the workout.

4.4.4.b

Answer.

$C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt$ is the total calories burned on $[0,40]\text{.}$

4.4.4.c

Answer.

The exact average rate at which the person burned calories on $0 \le t \le 40$ is

\begin{equation*} c_{\operatorname{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt = \frac{1}{40} \cdot \frac{1700}{3} = \frac{1700}{120} \approx 14.17 \ \text{cal/min}\text{.} \end{equation*}

4.4.4.d

Answer.

One time at which the instantaneous rate at which calories are burned equals the average rate on $[0,40]$ is $t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}$

5 Evaluating Integrals
5.1 Constructing accurate graphs of antiderivatives
5.1.2 Constructing the graph of an antiderivative

Activity 5.1.2.

5.1.2.a

Answer.

$F$ is increasing on $(0,2)$ and $(5,7)\text{;}$ $F$ is decreasing on $(2,5)\text{.}$

5.1.2.b

Answer.

$F$ is concave up on $(0,1)\text{,}$ $(4,6)\text{;}$ concave down on $(1,3)\text{,}$ $(6,7)\text{;}$ neither on $(3,4)\text{.}$

5.1.2.c

Answer.

A relative maximum at $x = 2\text{;}$ a relative minimum at $x = 5\text{.}$

5.1.2.d

Answer.

$F(1) = -\frac{1}{2}\text{;}$ $F(2) = \frac{\pi}{4} - \frac{1}{2}\text{;}$ $F(3) = \frac{\pi}{4} - 1\text{;}$ $F(4) = \frac{\pi}{4}-2\text{;}$ $F(5) = \frac{\pi}{4} - \frac{5}{2}\text{;}$ $F(6) = \frac{\pi}{2} - \frac{5}{2}\text{;}$ $F(7) = \frac{3\pi}{4} - \frac{5}{2}\text{;}$ $F(8) = \frac{3\pi}{4} - \frac{5}{2}\text{;}$ and $F(-1) = -1\text{.}$

5.1.2.e

Answer.

Use the function values found in (d) and the earlier information regarding the shape of $F\text{.}$

5.1.2.f

Answer.

$G(x) = F(x) + 1\text{.}$

5.1.3 Multiple antiderivatives of a single function

Activity 5.1.3.

5.1.3.a

Answer.

This image shows three possible antiderivatives of $g(x)$ on the interval $[-2,2]\text{.}$

The first, $G(x)\text{,}$ looks like two quadratic functions joined together at $x = 0\text{:}$ one that’s concave down on the left, and another that’s concave up on the right. Three important points on the graph of $G(x)$ are $G(-1) = 0\text{,}$ $G(0) = -\frac12\text{,}$ and $G(1) = -1\text{.}$

The other two antiderivatives, $H(x)$ and $F(x)\text{,}$ are vertical shifts of $G(x)\text{.}$ $H(x)$ satisfies $H(-2)=0$ and $F(x)$ satisfies $F(-2)=2\text{.}$

5.1.3.b

Answer.

5.1.3.c

Answer.

This image shows the described antiderivative $P(x)$ on the interval $[-1,3]\text{.}$

On the intervals $[-1,0]$ and $[2,3]\text{,}$ $P(x)=0\text{.}$ On the interval $(0,1)\text{,}$ $P(x)$ is increasing and concave up. There’s a corner point on the graph of $P(x)$ at the point $(1,4/3)\text{,}$ and there $P(x)$ changes to become decreasing and concave up on the interval $(1,2)\text{.}$

5.1.4 Functions defined by integrals

Activity 5.1.4.

5.1.4.a

Answer.

$A$ is increasing on $(0,1.5)\text{,}$ $(4,6)\text{;}$ $A$ is decreasing on $(1.5,4)\text{.}$

5.1.4.b

Answer.

$A$ is concave up on $(0,1)$ and $(3,5)\text{;}$ $A$ is concave down on $(1,3)$ and $(5,6)\text{.}$

5.1.4.c

Answer.

At $x = 1.5\text{,}$ $A$ has a relative maximum; $A$ has a relative minimum at $x = 4\text{.}$

5.1.4.d

Answer.

$A(0) = -\frac{1}{2}\text{;}$ $A(1) = 0\text{;}$ $A(2) = 0\text{;}$ $A(3) = -2\text{;}$ $A(4) = -3.5\text{,}$ $A(5) = -2\text{,}$ $A(6) = -0.5\text{.}$

5.1.4.e

Answer.

Use your work in (a)-(d) appropriately.

5.1.4.f

Answer.

$B(x) = A(x) + \frac{1}{2}\text{.}$

5.2 The Second Fundamental Theorem of Calculus
5.2.2 The Second Fundamental Theorem of Calculus

Activity 5.2.2.

5.2.2.a

Answer.

$A'(x) = f(x)\text{.}$

5.2.2.b

Answer.

$A(1) = -\frac{\pi}{4}\text{,}$ $A(3) = -\frac{1}{2}\text{.}$

5.2.2.c

Answer.

$A$ is increasing wherever $f$ is positive; $A$ is CCU wherever $f$ is increasing. $A(2) = 0\text{,}$ $A(3) = -0.5\text{,}$ $A(4) = -1.5\text{,}$ $A(5) = -2\text{,}$ $A(6) = -2 + \frac{\pi}{4}\text{,}$ and $A(7) = -2 + \frac{\pi}{2}\text{.}$

5.2.2.d

Answer.

$F$ and $A$ differ by the constant $\frac{\pi}{4} - \frac{1}{2}\text{.}$

5.2.2.e

Answer.

$B$ and $C$ have the same shape as $A$ and $F\text{,}$ and differ from $A$ by a constant. Observe that $B(3) = 0$ and $C(1) = 0\text{.}$

5.2.3 Understanding Integral Functions

Activity 5.2.3.

5.2.3.a

Answer.

$F' = f\text{.}$

5.2.3.b

Answer.

$F$ is increasing for all $x \gt 0\text{;}$ $F$ is decreasing for $x \lt 0$

5.2.3.c

Answer.

$F$ is CCU on $-1 \lt x \lt 1$ and CCD for $x \lt -1$ and $x \gt 1\text{.}$

5.2.3.d

Answer.

$x$	$-10$	$-5$	$0$	$5$	$10$
$F(x)$	2.35973	1.64038	0	1.64038	2.35973

5.2.3.e

Answer.

A plot of $y = F(x)\text{.}$ The $x$ values on the coordinate axes range horizontally from $-10.5$ to $-10.5\text{,}$ and the $y$ values range vertically from $-5.5$ to $-5.5\text{.}$ The function $F$ is decreasing for $x \lt 0$ and increasing for $x \gt 0$ with a global minimum at $x = 0\text{.}$ The function $F$ is concave up for approximately $-1 \lt x \lt 1$ and concave down otherwise.

5.2.3.f

Answer.

$g'(x)=\frac{x}{1+x^2}$ and $g(0)=0\text{.}$ $F=g$

5.2.4 Differentiating an Integral Function

Activity 5.2.4.

5.2.4.a

Answer.

$\displaystyle \frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}$

5.2.4.b

Answer.

$\displaystyle \int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \frac{x^4}{1+x^4} - \frac{16}{17}\text{.}$

5.2.4.c

Answer.

$\displaystyle \frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}$

5.2.4.d

Answer.

$\displaystyle \int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \ln(1+x^2)-\ln(10)\text{.}$

5.2.4.e

Answer.

$\displaystyle \frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \sin(x^6) \cdot 3x^2\text{.}$

5.3 Integration by substitution
5.3.2 Reversing the Chain Rule: First Steps

Activity 5.3.2.

5.3.2.a

Answer.

$\displaystyle \int \sin(8-3x) \, dx = -\frac{1}{3} (-\cos(8-3x)) + C\text{.}$

5.3.2.b

Answer.

$\displaystyle \int \sec^2 (4x) \, dx = \frac{1}{4} \tan(4x) + C\text{.}$

5.3.2.c

Answer.

$\displaystyle \int \frac{1}{11x - 9} \, dx = \frac{1}{11} \ln|11x - 9| + C\text{.}$

5.3.2.d

Answer.

$\displaystyle \int \csc(2x+1) \cot(2x+1) \, dx = -\frac{1}{2}\csc(2x+1) + C\text{.}$

5.3.2.e

Answer.

$\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx = \frac{1}{4} \arcsin(4x) + C$

5.3.2.f

Answer.

$\displaystyle \int 5^{-x}\, dx = -\frac{1}{\ln(5)}5^{-x} + C\text{.}$

5.3.3 Reversing the Chain Rule: $u$-substitution

Activity 5.3.3.

5.3.3.a

Answer.

$\displaystyle \int \frac{x^2}{5x^3+1} \, dx = \frac{1}{15} \ln(5x^3 + 1) + C\text{.}$

5.3.3.b

Answer.

$\displaystyle \int e^x \sin(e^x) \, dx = -\cos(e^x) + C\text{.}$

5.3.3.c

Answer.

$\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx = 2\sin(\sqrt{x}) + C\text{.}$

5.3.4 Evaluating Definite Integrals via $u$-substitution

Activity 5.3.4.

5.3.4.a

Answer.

$\displaystyle \int_{x=1}^{x=2} \frac{x}{1 + 4x^2} \, dx = \frac{1}{8} (\ln(17) - \ln(5))\text{.}$

5.3.4.b

Answer.

$\displaystyle \int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx = -\frac{1}{20}(2e^{-1}+3)^{10} + \frac{1}{20}(2e^{0}+3)^{10}\text{.}$

5.3.4.c

Answer.

$\displaystyle \int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx = 1 - \frac{\sqrt{2}}{2}\text{.}$

5.4 Integration by parts
5.4.2 Reversing the Product Rule: Integration by Parts

Activity 5.4.2.

5.4.2.a

Answer.

$\displaystyle \int t e^{-t} dt = -te^{-t} - e^{-t} + c\text{.}$

5.4.2.b

Answer.

$\displaystyle \int 4x \sin(3x) dx = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + c \text{.}$

5.4.2.c

Answer.

$\displaystyle \int z \sec^2(z) dz = z \tan(z) + \ln |\cos(z)| + c \text{.}$

5.4.2.d

Answer.

$\displaystyle \int x\ln(x) dx = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + c \text{.}$

5.4.3 Some Subtleties with Integration by Parts

Activity 5.4.3.

5.4.3.a

Answer.

$\displaystyle \int{\arctan(x) dx} = x\arctan(x) - \frac{1}{2} \ln \left( | 1 + x^2 | \right) + c \text{.}$

5.4.3.b

Answer.

$\displaystyle \int \ln(z) dz = z \ln(z) - z + c \text{.}$

5.4.3.c

Answer.

$\displaystyle \int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) + C \text{.}$

5.4.3.d

Answer.

$\int s^5 e^{s^3} ds = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + c \text{.}$

5.4.3.e

Answer.

$\int e^{2t} \cos\left( e^t \right) dt = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + c \text{.}$

5.4.4 Using Integration by Parts Multiple Times

Activity 5.4.4.

5.4.4.a

Answer.

$\displaystyle \int x^2 \sin(x) dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + c \text{.}$

5.4.4.b

Answer.

$\displaystyle \int t^3 \ln(t) dt = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + c \text{.}$

5.4.4.c

Answer.

$\displaystyle \int e^z \sin(z) dz = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + c \text{.}$

5.4.4.d

Answer.

$\displaystyle \int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + c \text{.}$

5.4.4.e

Answer.

$\displaystyle \int t \arctan(t) dt = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t + \frac{1}{2} \arctan(t) + c \text{.}$

5.5 Other options for finding algebraic antiderivatives
5.5.2 The Method of Partial Fractions

Activity 5.5.2.

5.5.2.a

Answer.

$\displaystyle \int \frac{1}{x^2 - 2x - 3} \, dx = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| + C\text{.}$

5.5.2.b

Answer.

$\displaystyle \int \frac{x^2+1}{x^3 - x^2} \, dx = -\ln|x| + x^{-1} + 2\ln|x-1| + C\text{.}$

5.5.2.c

Answer.

$\displaystyle \int \frac{x-2}{x^4 + x^2}\, dx = \ln|x| + 2x^{-1} - \frac{1}{2} \ln|1+x^2| + 2\arctan(x) + C\text{.}$

5.5.3 Using an Integral Table

Activity 5.5.3.

5.5.3.a

Answer.

$\displaystyle \int \sqrt{x^2 + 4} \, dx = \frac{x}{2} \sqrt{x^2+4} + 2 \ln | x + \sqrt{x^2+4}| + C\text{.}$

5.5.3.b

Answer.

$\displaystyle \int \frac{x}{\sqrt{x^2 +4}} \, dx = \sqrt{x^2 + 4} + C\text{.}$

5.5.3.c

Answer.

$\displaystyle \int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln| 5x + \sqrt{16+25x^2} | + C\text{.}$

5.5.3.d

Answer.

$\displaystyle \int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = - \frac{\sqrt{49-36x^2}}{49x} + C\text{.}$

5.6 Numerical integration
5.6.2 The Trapezoid Rule

Activity 5.6.2.

5.6.2.a

Answer.

$\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}$

5.6.2.b

Answer.

The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.

$n$	$T_n$	$E_{T,n}$
$4$	$0.50899$	$0.00899$
$8$	$0.50227$	$0.00227$
$16$	$0.50057$	$0.00057$

5.6.2.c

Answer.

The table below gives values of the midpoint rule and corresponding errors for different $n$-values.

$n$	$M_n$	$E_{M,n}$
$4$	$0.49555$	$-0.00445$
$8$	$0.49887$	$-0.00113$
$16$	$0.49972$	$-0.00028$

5.6.2.d

Answer.

The trapezoid rule overestimates; the midpoint rule underestimates.

5.6.2.e

Answer.

$f(x) = \dfrac{1}{x^2}$ is concave up on $[1, 2]\text{.}$

5.6.4 Simpson’s Rule

Activity 5.6.3.

5.6.3.a

Answer.

Plot the data.

5.6.3.b

Answer.

$\int_0^{1.8} v(t) dt\text{.}$

5.6.3.c

Answer.

\begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft}\text{.} \end{align*}

$R_3$ and $T_3$ are underestimates.

5.6.3.d

Answer.

$M_3 = 143.4 \text{ ft}\text{;}$ overestimate.

5.6.3.e

Answer.

$S_6 = 140.8 \text{ ft}\text{.}$

5.6.3.f

Answer.

Simpson’s rule gives the best approximation of the distance traveled, $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft}\text{,}$ which leads to $AV_{[0,1.8]} \approx \frac{140.8}{1.8} \approx 78.22 \text{ ft/sec}\text{.}$

5.6.5 Overall observations regarding $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ $M_n\text{,}$ and $S_{2n}\text{.}$

Activity 5.6.4.

5.6.4.a

Answer.

$L_1 = 2$ for each of the three functions and $R_1 = 1$ for each.

The values of $L_1$ and $R_1$ are the same for all three.

5.6.4.b

Answer.

For $f\text{,}$ $g\text{,}$ and $h\text{,}$ respectively, $M_1 = 7/4\text{,}$ $15/8\text{,}$ and $31/16\text{.}$

5.6.4.c

Answer.

For $f\text{,}$ $S_2 = (2M_1 + T_1)/3 = 5/3$
For $g\text{,}$ $S_2 = (2M_1 + T_1)/3 = 7/4$
For $h\text{,}$ $S_2 = (2M_1 + T_1)/3 = 43/24$

5.6.4.d

Answer.

\begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5} \end{align*}

5.6.4.e

Answer.

Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson’s rule is exact for both $f$ and $g\text{,}$ while a slight underestimate of $\int_0^1 h(x) dx\text{.}$

6 Using Definite Integrals
6.1 Using definite integrals to find area and length
6.1.2 The Area Between Two Curves

Activity 6.1.2.

6.1.2.a

Answer.

$A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx = \frac{32}{3}\text{.}$

6.1.2.b

Answer.

$A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-2x^2)-(x^2-8)) \, dx \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}$

6.1.2.c

Answer.

$A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.}$

6.1.2.d

Answer.

The left-hand region has area

\begin{equation*} A_1 = \int_{\frac{1 - \sqrt{5}}{2}}^0 \left( \left(x^3 - x \right) - x^2\right) dx = \dfrac{13 - 5\sqrt{5}}{24} \approx 0.075819\text{.} \end{equation*}

The right-hand region has area

\begin{equation*} A_2 = \int_0^{\frac{1 + \sqrt{5}}{2}} \left( x^2 - \left(x^3 - x \right) \right) dx = \dfrac{13 + 5\sqrt{5}}{24} \approx 1.007514 \text{.} \end{equation*}

The total area is $A_1 + A_2 \approx 1.083333\text{.}$

6.1.3 Finding Area with Horizontal Slices

Activity 6.1.3.

6.1.3.a

Answer.

$A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.}$

6.1.3.b

Answer.

$A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = \frac{4}{3}\text{.}$

6.1.3.c

Answer.

$A = \int_{y=0}^{y=1} \left(2-y - \sqrt{y} \right) \, dy = \frac{5}{6} $

6.1.3.d

Answer.

$A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \frac{9}{2}\text{.}$

6.1.4 Finding the length of a curve

Activity 6.1.4.

6.1.4.a

Answer.

$L \approx 2.95789\text{.}$

6.1.4.b

Answer.

$L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.}$

6.1.4.c

Answer.

$L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.}$

6.1.4.d

Answer.

We will usually have to estimate the value of $\int_a^b \sqrt{1+f'(x)^2} \, dx$ using computational technology.

6.1.4.e

Answer.

Approximately $(14.9165,f(14.9165)) = (14.9165, 23.2502)\text{.}$

6.2 Using definite integrals to find volume
6.2.2 The Volume of a Solid of Revolution

Activity 6.2.2.

6.2.2.a

Answer.

$V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.}$

6.2.2.b

Answer.

$V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.}$

6.2.2.c

Answer.

$V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.}$

6.2.2.d

Answer.

$V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.}$

6.2.2.e

Answer.

$V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.}$

6.2.3 Revolving about the $y$-axis

Activity 6.2.3.

6.2.3.a

Answer.

$V = \int_0^2 \pi y^4 \Delta \, dy\text{.}$

6.2.3.b

Answer.

$V = \int_0^2 \pi (16 - y^4) \, dy\text{.}$

6.2.3.c

Answer.

$V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.}$

6.2.3.d

Answer.

$V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.}$

6.2.3.e

Answer.

$V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.}$

6.2.4 Revolving about horizontal and vertical lines other than the coordinate axes

Activity 6.2.4.

6.2.4.a

Answer.

\begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (2x+2)^2 - (x^3 + 2)^2 ) \, dx = \frac{4}{21}(21+8\sqrt{2}) \pi \approx 19.336\text{.} \end{equation*}

6.2.4.b

Answer.

\begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (4 - x^3)^2 - (4-2x)^2 ) \, dx = \left( 8-\frac{32\sqrt{2}}{21} \right)\pi \approx 18.3626\text{.} \end{equation*}

6.2.4.c

Answer.

\begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (y^{1/3} + 1)^2 - (\frac{1}{2}y + 1)^2 ) \, dy = \frac{2}{15}(15 + 8\sqrt{2}) \pi \approx 11.022\text{.} \end{equation*}

6.2.4.d

Answer.

\begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (5 - \frac{1}{2}y)^2 - (5 - y^{1/3})^2 ) \, dy = \frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677\text{.} \end{equation*}

6.3 Density, mass, and center of mass
6.3.2 Density

Activity 6.3.2.

6.3.2.a

Answer.

$M = 10 - 10e^{-2} \approx 8.64665$ grams.

6.3.2.b

Answer.

$V = \int_{0}^{5} \pi (4 - \frac{4}{5}x)^2 \, dx = \frac{80\pi}{3} \approx 83.7758 \mbox{m}^3\text{.}$
$M = \frac{64000\pi}{3} \approx 67020.6433 \mbox{kg} \text{.}$
$\displaystyle M = \int_{0}^{5} (400 + \frac{200}{1+x^2}) \cdot \pi (4-\frac{4}{5}x)^2 \, dx = 128 \pi (\frac{265}{3} + 24 \arctan(5) - 5 \ln(26)) \approx 42224.8024 \mbox{kg}$

6.3.2.c

Answer.

$b \approx 3.0652\text{.}$

6.3.3 Weighted Averages

Activity 6.3.3.

6.3.3.a

Answer.

$\overline{x} = \frac{x_1 + x_2}{2} = 3\text{.}$

6.3.3.b

Answer.

$\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 3\text{.}$

6.3.3.c

Answer.

$\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 2.75\text{.}$

6.3.3.d

Answer.

$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{16}{7}\text{.}$

6.3.3.e

Answer.

$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{17}{7}\text{.}$

6.3.3.f

Answer.

$\overline{x} = \frac{2x_1 + 3x_2 + 2x_3 + 1x_4}{8} = \frac{20}{8}\text{.}$

6.3.3.g

Answer.

Answers will vary.

6.3.3.h

Answer.

If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

6.3.4 Center of Mass

Activity 6.3.4.

6.3.4.a

Answer.

$M = \int_{0}^{20} 4 + 0.1x \, dx = 100$ g.

6.3.4.b

Answer.

Greater than 10.

6.3.4.c

Answer.

$\overline{x} = \frac{\int_{0}^{20} x (4 + 0.1x)) \, dx}{\int_{0}^{20} 4 + 0.1x \, dx} = \frac{32}{3}\text{.}$

6.3.4.d

Answer.

5 g/cm.

6.3.4.e

Answer.

Slightly to the right of the center of mass for $\rho(x)\text{.}$

6.3.4.f

Answer.

$\overline{x} = \frac{\int_{0}^{20} x 4e^{0.020732x} \, dx}{\int_{0}^{20} 4e^{0.020732x} \, dx} \approx 10.6891\text{,}$

6.4 Physics applications: work, force, and pressure
6.4.2 Work

Activity 6.4.2.

6.4.2.a

Answer.

$W = \int_0^{200} 0.3(200-h) \, dh = 6000 \text{ foot-pounds}\text{.}$

6.4.2.b

Answer.

$W = \int_0^{100} (40-0.1h) \, dh = 3500 \text{foot-pounds}\text{.}$

6.4.2.c

Answer.

$B_{\text{AVG} [0,100]} \approx 27.98 \text{ pounds}\text{.}$

6.4.2.d

Answer.

For the given spring,

$k = 15\text{.}$
$W = \int_0^1 15x \, dx = \frac{15}{2} \text{ foot-pounds}\text{.}$
$W = \int_1^{1.5} 15x \, dx = 9.375 \text{ foot-pounds}\text{.}$

6.4.3 Work: Pumping Liquid from a Tank

Activity 6.4.3.

6.4.3.a

Answer.

\begin{equation*} W = \int_{2}^{3} 9.81 \cdot 4000\pi \cdot x \, dx = 308~190 \, \text{newton-meters}\text{.} \end{equation*}

6.4.3.b

Answer.

\begin{equation*} W = \int_{3}^{8} 62.4 \pi (100-x^2)(x+5) \, dx \approx 673593 \, \text{foot-pounds}\text{.} \end{equation*}

6.4.3.c

Answer.

\begin{equation*} W = \int_{1}^{3} 62.4 (50 - \frac{25}{2}x) x \, dx = 5720 \, \text{foot-pounds}\text{.} \end{equation*}

6.4.4 Force due to Hydrostatic Pressure

Activity 6.4.4.

6.4.4.a

Answer.

$F = \int_{x = 0}^{x=50} (6240 x) dx = 7~800~000 \text{ pounds } \text{.}$

6.4.4.b

Answer.

$F = \int_{x=10}^{x=30} 124.8 (x - 10)\sqrt{900 - x^2} dx = 800~244 \text{ pounds } \text{.}$

6.4.4.c

Answer.

$F = \int_{x=1}^{x=4} 62.4 (x - 1)(5 - 1.25x) dx = 351 \text{ pounds } \text{.}$

6.5 Improper integrals
6.5.2 Improper Integrals Involving Unbounded Intervals

Activity 6.5.2.

6.5.2.a

Answer.

$\int_1^{10} \frac{1}{x} dx = \ln(10)$ $\int_1^{1000} \frac{1}{x} dx = \ln(1000)$ $\int_1^{100000} \frac{1}{x} dx = \ln(100000)$
$\int_1^b \frac{1}{x} dx = \ln(b)\text{.}$
$\displaystyle \lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \ln(b) = \infty$

6.5.2.b

Answer.

$\int_1^{10} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{10}}$ $\int_1^{1000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{1000}}$ $\int_1^{100000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{100000}}$
$\int_1^b \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{b}}\text{.}$
$\displaystyle \lim_{b \to \infty} \int_1^b \frac{1}{x^{3/2}} dx = \lim_{b \to \infty} \left( 2 - \frac{2}{\sqrt{b}} \right) = 2$

6.5.2.c

Answer.

Both graphs have a vertical asymptote at $x = 0$ and for both graphs, the $x$-axis is a horizontal asymptote. However, the graph of $y = \frac{1}{x^{3/2}}$ will ’’approach the $x$-axis faster’’ than the graph of $y = \frac{1}{x}\text{.}$

6.5.2.d

Answer.

The area bounded by the graph of $y = \frac{1}{x}\text{,}$ the $x$-axis, and the vertical line $x = 1$ is infinite or unbounded. However, The area bounded by the graph of $y = \frac{1}{x^{3/2}}\text{,}$ the $x$-axis, and the vertical line $x = 1$ is equal to 2.

6.5.3 Convergence and Divergence

Activity 6.5.3.

6.5.3.a

Answer.

$\int_1^\infty \frac{1}{x^2} dx = 1 $

6.5.3.b

Answer.

$\int_0^\infty e^{-x/4} dx = 4 $

6.5.3.c

Answer.

$\int_2^\infty \frac{9}{(x+5)^{2/3}} dx = \infty $

6.5.3.d

Answer.

$\int_4^\infty \frac{3}{(x+2)^{5/4}} dx = \frac{12}{6^{1/4}} $

6.5.3.e

Answer.

$\int_0^\infty x e^{-x/4} dx = 16 $

6.5.3.f

Answer.

If $0 \lt p \lt 1\text{,}$ $\int_1^\infty \frac{1}{x^p} dx$ diverges, while if $p \gt 1\text{,}$ the integral converges.

6.5.4 Improper Integrals Involving Unbounded Integrands

Activity 6.5.4.

6.5.4.a

Answer.

$\int_0^1 \frac{1}{x^{1/3}}dx = \frac{3}{2} $

6.5.4.b

Answer.

$\int_0^2 e^{-x} dx = 1 - e^{-2} $

6.5.4.c

Answer.

$\int_1^4 \frac{1}{\sqrt{4-x}} dx = 2\sqrt{3} $

6.5.4.d

Answer.

$\int_{-2}^2 \frac{1}{x^2} \, dx$ diverges.

6.5.4.e

Answer.

$\int_0^{\pi/2} \tan(x) dx = \infty $

6.5.4.f

Answer.

$\int_0^1 \frac{1}{\sqrt{1-x^2}} dx = \frac{\pi}{2} $

7 Differential Equations
7.1 An introduction to differential equations
7.1.2 What is a differential equation?

Activity 7.1.2.

7.1.2.a

Answer.

Let $P$ be the population $t$ the time in years; $\frac{dP}{dt} = 0.0125P\text{.}$

7.1.2.b

Answer.

Let $m$ be the mass $t$ the time in days; $\frac{dm}{dt} = -0.056m\text{.}$

7.1.2.c

Answer.

Let $B$ be the balance $t$ be time in years; $\frac{dB}{dt} = 0.04B - 1000\text{.}$

7.1.2.d

Answer.

Let $t$ be time in minutes $H$ the temperature of the hot chocolate; $\frac{dH}{dt} = -0.1(H - 70)\text{.}$

7.1.2.e

Answer.

Let $t$ be time in minutes and $H$ the temperature of the soda;

\begin{equation*} \frac{dH}{dt} = 0.1(70 - H) = -0.1(H - 70)\text{.} \end{equation*}

7.1.3 Differential equations in the world around us

Activity 7.1.3.

7.1.3.a

Answer.

For the skydiver:

\begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp = 1.25 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5)} \amp = 0.75 \\ \left. \frac{dv}{dt}\right|_{(v = 2 )} \amp = 0.5 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp = 0.25 \end{align*}

7.1.3.b

Answer.

For the meteorite:

\begin{align*} \left. \frac{dv}{dt}\right|_{(v = 3.5)} \amp = -0.25 \amp \left. \frac{dv}{dt}\right|_{(v = 4)} \amp = -0.5\\ \left. \frac{dv}{dt}\right|_{(v = 4.5)} \amp = -0.75 \amp \left. \frac{dv}{dt}\right|_{(v = 5.5)} \amp \approx -1.25 \end{align*}

A graph of the points from parts (a) and (b) is shown in the following diagram:

A plot of points of the form $(v, \frac{dv}{dt})$ on the grid provided in part (a), using the data generated from the graphs in parts (a) and (b). The data points appear to lie on a line that passes through the point $(3,0)$ and has slope $m = -0.5\text{.}$

7.1.3.c

Answer.

$\frac{dv}{dt} = -0.6v + 1.8\text{.}$

7.1.3.d

Answer.

The rate of change of velocity with respect to time is a linear function of velocity.

7.1.3.e

Answer.

$0 \lt v \lt 3\text{.}$

7.1.3.f

Answer.

$3 \lt v \lt 5\text{.}$

7.1.3.g

Answer.

$v = 3\text{,}$ which is the terminal velocity of any falling object.

7.1.4 Solving a differential equation

Activity 7.1.4.

7.1.4.a

Answer.

$v(t) = 1.5t - 0.25t^2$ is not a solution to the given DE.

7.1.4.b

Answer.

$v(t) = 3 + 2e^{-0.5t}$ is a solution to the given DE.

7.1.4.c

Answer.

$v(t) = 3$ is a solution to the given DE.

7.1.4.d

Answer.

$v(t) = 3 + Ce^{-0.5t}$ is a solution to the given DE for any choice of $C\text{.}$

7.2 Qualitative behavior of solutions to differential equations
7.2.2 Slope fields

Activity 7.2.2.

7.2.2.a

Answer.

When $y \lt 4\text{,}$ $y$ is an increasing function of $t\text{.}$ When $y \gt 4\text{,}$ $y$ is a decreasing function of $t\text{.}$

Plot of $\frac{dy}{dt}$ as a function of $y\text{.}$ The graph is a straight line with slope $m=-\frac{1}{2}$ that passes through the point $(4,0)\text{.}$

7.2.2.b

Answer.

The slope field shows a constant solution of $y = 4$ and that solutions that satisfy $y(0) \gt 4$ are decreasing while solutions that satisfy $y(0) \lt 4$ are increasing

For a fixed value of $y\text{,}$ the line segments in the slope field all have the same slope, regardless of $t\text{.}$ When $y=4\text{,}$ the slopes are all $0\text{.}$ When $y \gt 4\text{,}$ the slopes are all negative, and the larger the value of $y$ the more negative the slope is. When $y \lt 4\text{,}$ the slopes are all positive, and the smaller the value of $y$ the more positive the slope is.

7.2.2.c

Answer.

The graph shows the plots of four different solutions to the differential equation with the graphs superimposed on the slope field.

The solution that satisfies $y(0) = 6$ is a decreasing, concave up function that approaches $y = 4$ as $t$ increases without bound. The solution that satisfies $y(0) = 4$ is constant (that is, a horizontal line). The solutions that satisfy $y(0) = 0$ and $y(0) = 2$ are increasing, concave down functions that approach $y = 4$ as $t$ increases without bound.

7.2.2.d

Answer.

\begin{equation*} \frac{dy}{dt} = 2 \left( -\frac{1}{2} e^{-t/2} \right) = -e^{-t/2} \end{equation*}

and

\begin{equation*} -\frac{1}{2}( y - 4 ) = -\frac{1}{2} \left( 4 + 2e^{-t/2} \right) = -e^{-t/2} \end{equation*}

In addition, $y(0) = 4 + 2e^0 = 6\text{.}$

7.2.2.e

Answer.

A constant function.

7.2.3 Equilibrium solutions and stability

Activity 7.2.3.

7.2.3.a

Answer.

When $y \lt 0$ and when $y \gt 4\text{,}$ $y$ is a decreasing function of $t\text{.}$ When $0 \lt y \lt 4\text{,}$ $y$ is a increasing function of $t\text{.}$

Plot of $\frac{dy}{dt}$ as a function of $y\text{.}$ The graph is a concave down parabola that has zeros at $y = 0$ and $y=4$ with a maximum at the point $(2,2)\text{.}$

7.2.3.b

Answer.

$y = 0$ and $y = 4\text{.}$

7.2.3.c

Answer.

The slope field shows constant solutions of $y = 0$ and $y = 4\text{.}$ Solutions that satisfy $y(0) \gt 4$ or $y(0) \lt 0$ are decreasing, while solutions that satisfy $0 \lt y(0) \lt 4$ are increasing

For a fixed value of $y\text{,}$ the line segments in the slope field all have the same slope, regardless of $t\text{.}$ When $y \gt 4\text{,}$ the slopes are all negative, and the larger the value of $y$ the more negative the slope is. When $y=4\text{,}$ the slopes are all $0\text{.}$ When $0 \lt y \lt 4\text{,}$ the slopes are all positive, and the closer $y$ is to $0$ or $4$ the closer the slope is to $0\text{;}$ the largest positive slope occurs when $y=2\text{.}$ Finally, when $y \lt 0\text{,}$ the slopes are all negative, and the more negative the value of $y$ the more negative the slope is.

7.2.3.d

Answer.

The graph shows the plots of the different solutions to the differential equation with the graphs superimposed on the slope field.

The solution that satisfies $y(0) = 5$ is a decreasing, concave up function that approaches $y = 4$ as $t$ increases without bound. The solution that satisfies $y(0) = 4$ is constant (that is, a horizontal line). The solutions that satisfy $y(0) = 1\text{,}$ $y(0) = 2\text{,}$ and $y(0) = 3$ are increasing functions that approach $y = 4$ as $t$ increases without bound. Each of these three functions is concave down for all values of $y$ such that $y \gt 2\text{.}$ The solution that satisfies $y(0) = 0$ is constant (that is, a horizontal line). Finally, the solution that satisfies $y(0) = -1$is a decreasing concave down function that decreases without bound.

7.2.3.e

Answer.

$y = 4$ is stable; $y = 0$ is unstable.

7.2.3.f

Answer.

Tend to 4.

7.2.3.g

Answer.

The left image is for an ustable equilibrium; the right image is for a stable equilibrium.

7.3 Euler’s method
7.3.2 Euler’s Method

Activity 7.3.2.

7.3.2.a

Answer.

$t_i$	$y_i$	$dy/dt$	$\Delta y$
$0$	$0$	$-1$	$-0.2$
$0.2$	$-0.2$	$-0.6$	$-0.12$
$0.4$	$-0.32$	$-0.2$	$-0.04$
$0.6$	$-0.36$	$0.2$	$0.04$
$0.8$	$-0.32$	$0.6$	$0.12$
$1.0$	$-0.2$	$1$	$0.2$

This image is a plot of the $5$ points that result from applying Euler’s method to the given differential equation and the initial condition.

After the initial point $(0,0)\text{,}$ the plotted points are $(0.2,-0.2)\text{,}$ $(0.4,-0.32)\text{,}$ $(0.6,-0.36)\text{,}$ $(0.8,-0.32)\text{,}$ $(1,-0.2)\text{.}$

7.3.2.b

Answer.

$y = t^2 - t\text{,}$ with errors $e_1 = 0.04\text{,}$ $e_2 = 0.08\text{,}$ $e_3 = 0.12\text{,}$ $e_4 = 0.16\text{,}$ $e_5 = 0.2\text{.}$

7.3.2.c

Answer.

Solutions to this differential equation all differ by only a constant.

7.3.2.d

Answer.

If we first think about how $y_1$ is generated for the initial value problem $\frac{dy}{dt} = f(t) = 2t-1, \ y(0) = 0\text{,}$ we see that $y_1 = y_0 + \Delta t \cdot f(t_0)\text{.}$ Since $y_0 = 0\text{,}$ we have $y_1 = \Delta t \cdot f(t_0)\text{.}$ From there, we know that $y_2$ is given by $y_2 = y_1 + \Delta t f(t_1)\text{.}$ Substituting our earlier result for $y_1\text{,}$ we see that $y_2 = \Delta t \cdot f(t_0) + \Delta t f(t_1)\text{.}$ Continuing this process up to $y_5\text{,}$ we get

\begin{equation*} y_5 = \Delta t \cdot f(t_0) + \Delta t f(t_1) + \Delta t f(t_2) + \Delta t f(t_3) + \Delta t f(t_4) \end{equation*}

This is precisely the left Riemann sum with five subintervals for the definite integral $\int_0^1 (2t-1)~dt\text{.}$

Activity 7.3.3.

7.3.3.a

Answer.

The slope field shows constant solutions of $y = 0$ and $y = 6\text{.}$ Solutions that satisfy $y(0) \gt 6$ are decreasing, while solutions that satisfy $0 \lt y(0) \lt 6$ are increasing

For a fixed value of $y\text{,}$ the line segments in the slope field all have the same slope, regardless of $t\text{.}$ When $y \gt 6\text{,}$ the slopes are all negative, and the larger the value of $y$ the more negative the slope is. When $y=6\text{,}$ the slopes are all $0\text{.}$ When $0 \lt y \lt 6\text{,}$ the slopes are all positive, and the closer $y$ is to $0$ or $6$ the closer the slope is to $0\text{;}$ the largest positive slope occurs when $y=3\text{.}$ Finally, when $y = 0\text{,}$ the slopes are all zero.

7.3.3.b

Answer.

$y = 0$ or $y = 6\text{;}$ $y = 0$ is unstable, $y = 6$ is stable.

7.3.3.c

Answer.

The solution will tend to $y = 6\text{.}$

7.3.3.d

Answer.

$t_i$	$y_i$	$dy/dt$	$\Delta y$
$0.0$	$1.0000$	$5.0000$	$1.0000$
$0.2$	$2.0000$	$8.0000$	$1.6000$
$0.4$	$3.6000$	$8.6400$	$1.7280$
$0.6$	$5.3280$	$3.5804$	$0.7161$
$0.8$	$6.0441$	$-0.2664$	$-0.0533$
$1.0$	$5.9908$	$0.0551$	$0.0110$

This image is a plot of the $5$ points that result from applying Euler’s method to the given differential equation and the initial condition.

After the initial point $(0,1)\text{,}$ the plotted points are $(0.2,2)\text{,}$ $(0.4,3.6)\text{,}$ $(0.6,5.328)\text{,}$ $(0.8,6.0441)\text{,}$ $(1,5.9908)\text{.}$

7.3.3.e

Answer.

The value of $y_i = 6$ for every value of $i\text{.}$

7.4 Separable differential equations
7.4.2 Solving separable differential equations

Activity 7.4.2.

7.4.2.a

Answer.

$\frac{dP}{dt} = 0.03 P$

7.4.2.b

Answer.

$P = Ce^{0.03t}\text{.}$

7.4.2.c

Answer.

$P = 10000 e^{0.03t}\text{.}$

7.4.2.d

Answer.

The doubling time is $t = \frac{\ln(2)}{0.03} \approx 23.105$ years.

7.4.2.e

Answer.

The doubling time is $t = \frac{1}{k} \ln(2)\text{.}$

Activity 7.4.3.

7.4.3.a

Answer.

$k = \frac{1}{30}$

7.4.3.b

Answer.

$T = 75 + Ce^{-t/30}$

7.4.3.c

Answer.

The temperature of the coffee tends to 75 degrees.

7.4.3.d

Answer.

$T(20) = 75 + 30e^{-2/3} \approx 90.4^\circ$F.

7.4.3.e

Answer.

$t = -30 \ln \left( \frac{1}{6} \right) \approx 53.75$ minutes.

Activity 7.4.4.

7.4.4.a

Answer.

$y = -1 + C e^{\left(2t - \frac{t^2}{2} \right)}\text{.}$

7.4.4.b

Answer.

$y = \frac{1}{2} \ln \left( e^{t^2} + C \right)\text{.}$

7.4.4.c

Answer.

$y = -1 + 3 e^{2t}\text{.}$

7.4.4.d

Answer.

$y = -\frac{1}{2t + \frac{3}{2}} = -\frac{2}{4t + 3}\text{.}$

7.4.4.e

Answer.

$y = \frac{4}{t^2 + 1}\text{.}$

7.5 Modeling with differential equations
7.5.2 Developing a differential equation

Activity 7.5.2.

7.5.2.a

Answer.

$\frac{dA}{dt} = 0.05A\text{.}$

7.5.2.b

Answer.

$\frac{dA}{dt} = 0.05A - 10000\text{.}$

7.5.2.c

Answer.

The only equilibrium solution is $A = 200000\text{.}$

7.5.2.d

Answer.

$t = 20 \ln(2) \approx 13.86$years.

7.5.2.e

Answer.

At least $200000.

7.5.2.f

Answer.

Up to $15000 every year.

Activity 7.5.3.

7.5.3.a

Answer.

$\frac{dM}{dt} = -kM\text{,}$ where $k$ is a positive constant.

7.5.3.b

Answer.

$k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}$

7.5.3.c

Answer.

$\frac{dM}{dt} = 3 - kM\text{,}$ where $k$ is a positive constant.

7.5.3.d

Answer.

The equilibrium solution $mM = \frac{3}{k}$ is stable.

7.5.3.e

Answer.

$M = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}$

7.5.3.f

Answer.

About 2.426 milligrams per hour.

7.6 Population growth and the logistic equation
7.6.2 The earth’s population

Activity 7.6.2.

7.6.2.a

Answer.

$P'(0) \approx 0.0755\text{.}$

7.6.2.b

Answer.

$P(0) = 6.084\text{.}$

7.6.2.c

Answer.

$k \approx 0.012041\text{.}$

7.6.2.d

Answer.

$P(t) = 6.084 e^{0.012041t}\text{.}$

7.6.2.e

Answer.

$P(10) \approx 6.8878\text{.}$

7.6.2.f

Answer.

$t = \frac{1}{0.012041} \ln \left( \frac{12}{6.084}\right) \approx 56.41\text{,}$ or in the year 2056.

7.6.2.g

Answer.

$P(500) \approx 3012.3$ billion.

7.6.3 Solving the logistic differential equation

Activity 7.6.3.

7.6.3.a

Answer.

When $P = \frac{N}{2}\text{.}$

7.6.3.b

Answer.

When the population is 6.125 billion.

7.6.3.c

Answer.

$P = \frac{12.5}{1.0546e^{-0.025t} + 1} \text{;}$ $P(100) = 11.504$ billion.

7.6.3.d

Answer.

$t = \frac{1}{-0.025} \ln \left( \frac{\left( \frac{12.5}{9} - 1 \right)}{1.0546}\right) \approx 39.9049$ (so in about year $2040$).

7.6.3.e

Answer.

$\lim_{t \to \infty} P(t) = N\text{.}$

8 Taylor Polynomials and Taylor Series
8.1 Extending local linearization
8.1.2 Finding a quadratic approximation

Activity 8.1.2.

8.1.2.a

Answer.

$T_2''(x) = 2b_2\text{.}$

8.1.2.b

Answer.

$f'(x) = f''(x)= e^x\text{.}$

8.1.2.c

Answer.

Table 8.1.1. Formulas and values for $f(x)$ and $T_2(x)\text{.}$

$f(x)=$	$e^x$	$T_2(x)=$	$b_0 + b_1 x + b_2 x^2$
$f'(x)=$	$e^x$	$T_2'(x)=$	$b_1 + 2b_2 x$
$f''(x)=$	$e^x$	$T_2''(x)=$	$2b_2$

$f(0)=$	$1$	$T_2(0)=$	$b_0$
$f'(0)=$	$1$	$T_2'(0)=$	$b_1$
$f''(0)=$	$1$	$T_2''(0)=$	$2b_2$

8.1.2.d

Answer.

See the bottom half of the table above.

8.1.2.e

Answer.

$b_0 = 1\text{;}$ $b_1 = 1\text{;}$ $b_2 = \frac{1}{2}\text{.}$ So, $T_2(x) = 1 + x + \frac{1}{2} x^2\text{.}$

8.1.2.f

Answer.

$T_2(x)$ is a better approximation to $f(x) = e^x$ near $a = 0$ than the tangent line; $|f(x)-T_2(x)| \lt 0.1$ for approximately $-0.9 \lt x \lt 0.8\text{,}$ and for any $x$-value in that interval, $|f(x)-T_2(x)| \lt |f(x)-T_1(x)|\text{.}$

8.1.3 Over and over again

Activity 8.1.3.

8.1.3.a

Answer.

Table 8.1.5. Formulas and values for $f(x)$ and $T_3(x)\text{.}$

$f(x)=$	$e^x$	$T_3(x)=$	$c_0 + c_1 x + c_2 x^2 + c_3 x^3$
$f'(x)=$	$e^x$	$T_3'(x)=$	$c_1 + 2 c_2 x + 3 c_3 x^2$
$f''(x)=$	$e^x$	$T_3''(x)=$	$2 c_2 + 3 \cdot 2 c_3 x$
$f'''(x)=$	$e^x$	$T_3'''(x)=$	$3 \cdot 2 c_3$

$f(0)=$	$1$	$T_3(0)=$	$c_0$
$f'(0)=$	$1$	$T_3'(0)=$	$c_1$
$f''(0)=$	$1$	$T_3''(0)=$	$2c_2$
$f'''(0)=$	$1$	$T_3'''(0)=$	$6c_3$

8.1.3.b

Answer.

$c_0 = 1\text{;}$ $c_1 = 1\text{;}$ $c_2 = \frac{1}{2}\text{;}$ $c_3 = \frac{1}{6}\text{.}$

8.1.3.c

Answer.

$T_3(x)$ appears to be an even better approximation than $T_2(x)$ near $a = 0$ and that the quality of the approximation extends further; $|f(x)-T_3(x)| \lt 0.1$ for approximately $-1.3 \lt x \lt 1.1\text{.}$

8.1.3.d

Answer.

$T_4(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \frac{1}{24}x^4\text{;}$ $T_4(x)$ is an even better approximation to $f(x) = e^x$ and on a still wider interval.

8.1.4 As the degree of the approximation increases

Activity 8.1.4.

8.1.4.a

Answer.

The first seven columns and eleven rows of the spreadsheet are:

Table 8.1.11. Comparing $f(x) = e^x$ and its degree $1\text{,}$ $2\text{,}$ $3\text{,}$ $4$ and approximations near $a = 0\text{.}$

$\Delta x$	$x$	$f(x)$	$T_1(x)$	$T_2(x)$	$T_3(x)$	$T_4(x)$
$0.1$	$-1.0$	$0.36787$	$0.00000$	$0.50000$	$0.33333$	$0.37500$
$0.1$	$-0.9$	$0.40657$	$0.10000$	$0.50500$	$0.38350$	$0.41083$
$0.1$	$-0.8$	$0.44933$	$0.20000$	$0.52000$	$0.43467$	$0.45173$
$0.1$	$-0.7$	$0.49659$	$0.30000$	$0.54500$	$0.48783$	$0.49784$
$0.1$	$-0.6$	$0.54881$	$0.40000$	$0.58000$	$0.54400$	$0.54940$
$0.1$	$-0.5$	$0.60653$	$0.50000$	$0.62500$	$0.60417$	$0.60677$
$0.1$	$-0.4$	$0.67032$	$0.60000$	$0.68000$	$0.66933$	$0.67040$
$0.1$	$-0.3$	$0.74082 $	$0.70000 $	$0.74500 $	$0.74050 $	$0.74084$
$0.1$	$-0.2$	$0.81873 $	$0.80000 $	$0.82000 $	$0.81867 $	$0.81873$
$0.1$	$-0.1$	$0.90484 $	$0.90000 $	$0.90500 $	$0.90483 $	$0.90484$
$0.1$	$0.0$	$1.00000$	$1.00000$	$1.00000$	$1.00000$	$1.00000$

The next four columns and four rows of the spreadsheet are:

Table 8.1.12. The absolute error between $f(x) = e^x$ and its degree $1\text{,}$ $2\text{,}$ $3\text{,}$ and $4$ approximations at $x=-1$ and $x=-0.9\text{.}$

$\|f(x)-T_1(x)\|$	$\|f(x)-T_2(x)\|$	$\|f(x)-T_3(x)\|$	$\|f(x)-T_4(x)\|$
$0.36787$	$0.13212$	$0.03454$	$0.00712$
$0.30657$	$0.09843$	$0.02307$	$0.00426$
$0.24933 $	$0.07067 $	$0.01466 $	$0.00240$
$0.19659 $	$0.04841 $	$0.00875 $	$0.00125$

8.1.4.b

Answer.

$|f(-1) - T_2(-1)| \approx 0.13212\text{;}$ $|f(1) - T_2(1)| \approx 0.21828\text{.}$

8.1.4.c

Answer.

$|f(-1) - T_3(-1)| \approx 0.03455\text{;}$ $|f(1) - T_3(1)| \approx 0.05162\text{.}$

8.1.4.d

Answer.

$|f(-1) - T_4(-1)| \approx 0.00712\text{;}$ $|f(1) - T_4(1)| \approx 0.00995\text{.}$

8.1.4.e

Answer.

As the degree of the approximation increases, at each fixed $x$-value, the approximation gets better, and in addition the interval of values on which the approximation is within a certain tolerance gets wider.

8.1.4.f

Answer.

Answers will vary. But, as we widen the interval of $x$-values, the errors of each polynomial approximation increase near the endpoints of the interval.

8.2 Taylor polynomials
8.2.2 Taylor polynomials

Activity 8.2.2.

8.2.2.a

Answer.

Table 8.2.7. Finding the derivatives of $f(x) = \cos(x)$ at $a = 0\text{.}$

$f(x) =$	$\cos(x)$	$f(0) =$	$\cos(0) = 1$
$f'(x) = $	$-\sin(x)$	$f'(0) = $	$0$
$f''(x) = $	$-\cos(x)$	$f''(0) = $	$-1$
$f'''(x) = $	$\sin(x)$	$f'''(0) = $	$0$
$f^{(4)}(x) = $	$\cos(x)$	$f^{(4)}(0) = $	$1$
$f^{(5)}(x) = $	$-\sin(x)$	$f^{(5)}(0) = $	$0$
$f^{(6)}(x) = $	$-\cos(x)$	$f^{(6)}(0) = $	$-1$
$f^{(7)}(x) = $	$\sin(x)$	$f^{(7)}(0) = $	$0$
$f^{(8)}(x) = $	$\cos(x)$	$f^{(8)}(0) = $	$1$

8.2.2.b

Answer.

\begin{align*} T_2(x) \amp= 1 + 0x - \frac{1}{2!}x^2 \\ T_4(x) \amp= 1 + 0x - \frac{1}{2!}x^2 + 0x^3 + \frac{1}{4!}x^4 \\ T_6(x) \amp= 1 + 0x - \frac{1}{2!}x^2 + 0x^3 + \frac{1}{4!}x^4 + 0x^5 - \frac{1}{6!}x^6\\ T_8(x) \amp= 1 + 0x - \frac{1}{2!}x^2 + 0x^3 + \frac{1}{4!}x^4 + 0x^5 - \frac{1}{6!}x^6 + 0x^7 + \frac{1}{8!}x^8 \end{align*}

8.2.2.c

Answer.

$T_{10}(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + \frac{1}{8!}x^8 - \frac{1}{10!}x^{10}\text{.}$

8.2.2.d

Answer.

As the degree of the approximation increases, the accuracy of the approximation improves at each fixed $x$-value and in how large the interval is on which the approximation is accurate.

8.2.2.e

Answer.

Table 8.2.14. Comparing $f(x) = \cos(x)$ and its degree $2\text{,}$ $4\text{,}$ and $6$ approximations near $a = 0\text{.}$

$\Delta x$	$x$	$f(x)$	$T_2(x)$	$T_4(x)$	$T_6(x)$
$0.2$	$-2.0$	$-0.41615$	$-1.00000$	$-0.33333$	$-0.42222$
$0.2$	$-1.8$	$-0.22720$	$-0.62000$	$-0.18260$	$-0.22984$
$0.2$	$-1.6$	$-0.02920$	$-0.28000$	$-0.00693$	$-0.03024$
$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$	$\cdots$
$0.2$	$1.6$	$-0.02920$	$-0.28000$	$-0.00693$	$-0.03024$
$0.2$	$1.8$	$-0.22720$	$-0.62000$	$-0.18260$	$-0.22984$
$0.2$	$2.0$	$-0.41615$	$-1.00000$	$-0.33333$	$-0.42222$

Table 8.2.15. The absolute error between $f(x) = \cos(x)$ and its degree $2\text{,}$ $4\text{,}$ and $6$ approximations.

$\|f(x)-T_2(x)\|$	$\|f(x)-T_4(x)\|$	$\|f(x)-T_6(x)\|$
$0.58385$	$0.08281$	$0.00608$
$0.39280$	$0.04460$	$0.00263$
$0.25080$	$0.02227$	$0.00104$
$\cdots$	$\cdots$	$\cdots$
$0.25080$	$0.02227$	$0.00104$
$0.39280$	$0.04460$	$0.00263$
$0.58385$	$0.08281$	$0.00608$

$|f(x) - T_2(x)| \lt 0.1$ for roughly $-1.2 \lt x \lt 1.2\text{;}$ $|f(x) - T_4(x)| \lt 0.1$ for $-2 \lt x \lt 2\text{;}$ $|f(x) - T_6(x)| \lt 0.1$ for approximately $-2.8 \lt x \lt 2.8\text{.}$

8.2.3 Taylor polynomial approximations centered at an arbitrary value $a$

Activity 8.2.3.

8.2.3.a

Answer.

$f(x) =$	$\ln(x)$	$f(1) =$	$0$
$f'(x) = $	$x^{-1}$	$f'(1) = $	$1$
$f''(x) = $	$-1 \cdot x^{-2}$	$f''(1) = $	$-1$
$f'''(x) = $	$(-2)(-1)x^{-3}$	$f'''(1) = $	$(-2)(-1)$
$f^{(4)}(x) = $	$(-3)(-2)(-1)x^{-4}$	$f^{(4)}(1) = $	$(-3)(-2)(-1)$

8.2.3.b

Answer.

$T_4(x) = 1 (x-1) - \frac{1}{2} (x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 \text{.}$

8.2.3.c

Answer.

The function $f(x)=\ln(x)$ graphed together with its degree $1$ and $4$ Taylor polynomial approximations near the point $(1,f(1))\text{.}$

The graph shows that the function and the two Taylor approximations all intersect at the point $(1,f(1))$ and have the same slope at that point.

The three functions are very close together on the interval $(0.5,1.5)\text{,}$ but outside of that interval there start to be visual differences as the tangent line continues straight, but the function $\ln(x)$ is curved, and the degree $4$ approximation stays close to the function for longer.

The degree $4$ Taylor polynomial approximation looks very close to the function $\ln(x)$ on the interval $(0.3,1.75)\text{.}$

$T_4(x)$ provides a much more accurate approximation of $f(x)$ than $T_1(x)\text{.}$

8.2.3.d

Answer.

$|f(x) - T_4(x)| \lt 0.1$ for approximately $0.29 \lt x \lt 1.97\text{.}$

8.2.3.e

Answer.

$T_5(x) = 1 (x-1) - \frac{1}{2} (x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5\text{;}$ $T_6(x) = 1 (x-1) - \frac{1}{2} (x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 - \frac{1}{6}(x-1)^6\text{.}$

$|f(x) - T_5(x)| \lt 0.1$ for about $0.24 \lt x \lt 1.999\text{;}$ $|f(x) - T_6(x)| \lt 0.1$ for about $0.21 \lt x \lt 2\text{.}$ While the interval of accuracy gets wider as the degree increases, it seems not to extend past $x = 2$ and doesn’t move much to the left.

Activity 8.2.4.

8.2.4.a

Answer.

$f(x) =$	$\ln(x)$	$f(2) =$	$\ln(2)$
$f'(x) = $	$x^{-1}$	$f'(2) = $	$\frac{1}{2}$
$f''(x) = $	$-1 \cdot x^{-2}$	$f''(2) = $	$-\frac{1}{2^2}$
$f'''(x) = $	$(-2)(-1)x^{-3}$	$f'''(2) = $	$\frac{(-2)(-1)}{2^3}$
$f^{(4)}(x) = $	$(-3)(-2)(-1)x^{-4}$	$f^{(4)}(2) = $	$\frac{(-3)(-2)(-1)}{2^4}$

8.2.4.b

Answer.

$T_4(x) = \ln(2) + \frac{1}{2}(x-2) - \frac{1}{2 \cdot 2^2} (x-2)^2 + \frac{1}{3 \cdot 2^3}(x-2)^3 - \frac{1}{4 \cdot 2^4}(x-2)^4\text{.}$

8.2.4.c

Answer.

The function $f(x)=\ln(x)$ graphed together with its degree $1$ and $4$ Taylor polynomial approximations near the point $(2,f(2))\text{.}$

The graph shows that the function and the two Taylor approximations all intersect at the point $(2,f(2))$ and have the same slope at that point.

The three functions are very close together on the interval $(1.5,2.5)\text{,}$ but outside of that interval there start to be visual differences as the tangent line continues straight, but the function $\ln(x)$ is curved, and the degree $4$ approximation stays close to the function for longer.

The degree $4$ Taylor polynomial approximation looks very close to the function $\ln(x)$ on the interval $(0.7,3.5)\text{.}$

$T_4(x)$ provides a much better approximation of $f(x)$ near $a = 2$ and on a wider interval.

8.2.4.d

Answer.

$|f(x) - T_4(x)| \lt 0.1$ for approximately $0.58 \lt x \lt 3.95\text{.}$

8.2.4.e

Answer.

$T_5(x) = \ln(2) + \frac{1}{2}(x-2) - \frac{1}{2 \cdot 2^2} (x-2)^2 + \frac{1}{3 \cdot 2^3}(x-2)^3 - \frac{1}{4 \cdot 2^4}(x-2)^4 + \frac{1}{5 \cdot 2^5} (x-2)^5 \text{;}$ $T_6(x) = \ln(2) + \frac{1}{2}(x-2) - \frac{1}{2 \cdot 2^2} (x-2)^2 + \frac{1}{3 \cdot 2^3}(x-2)^3 - \frac{1}{4 \cdot 2^4}(x-2)^4 + \frac{1}{5 \cdot 2^5} (x-2)^5 - \frac{1}{6 \cdot 2^6} (x-2)^6 \text{.}$

$|f(x) - T_5(x)| \lt 0.1$ for roughly $0.48 \lt x \lt 4\text{;}$ $|f(x) - T_6(x)| \lt 0.1$ for about $0.42 \lt x lt 4\text{.}$ By moving to $a = 2\text{,}$ which is further away from the asymptote at $x = 0\text{,}$ we can get approximations of $ln(x)$ that seem to be good all the way up to $x = 4\text{.}$

8.3 Geometric sums
8.3.2 Finite Geometric Series

Activity 8.3.2.

8.3.2.a

Answer.

$\frac{2}{5} \cdot S_n = \frac{2}{5} + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots + \left(\frac{2}{5}\right)^n \text{.}$

8.3.2.b

Answer.

The key observation is that $1 + \frac{2}{5} + \left(\frac{2}{5}\right)^2 + \cdots + \left(\frac{2}{5}\right)^{n-1} - \frac{2}{5} - \left(\frac{2}{5}\right)^2 - \left(\frac{2}{5}\right)^3 - \cdots - \left(\frac{2}{5}\right)^n = 1 - \left(\frac{2}{5}\right)^n\text{.}$

8.3.2.c

Answer.

$S_n = \frac{1 - \left(\frac{2}{5}\right)^n}{\frac{3}{5}}$

8.3.2.d

Answer.

$S_n = \frac{7 - 7 \cdot \left(\frac{2}{5}\right)^n}{\frac{3}{5}}$

8.3.3 Infinite Geometric Series

Activity 8.3.3.

8.3.3.a

Answer.

$S_5 = \frac{a - ar^5}{1-r} = \frac{1 - (1/3)^5}{1-(1/3)} = \frac{121}{81} \approx 1.4938 \text{;}$ $S_{10} = \frac{a - ar^{10}}{1-r} = \frac{1 - (1/3)^{10}}{1-(1/3)} = \frac{29524}{19683} \approx 1.49997 \text{;}$ $S = \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \text{.}$

8.3.3.b

Answer.

$S_5 = \frac{a - ar^5}{1-r} = \frac{4 - 4\cdot(-1/2)^5}{1-(-1/2)} = \frac{11}{4} \text{;}$ $S_{10} = \frac{a - ar^{10}}{1-r} = \frac{4 - 4\cdot(-1/2)^{10}}{1-(-1/2)} = \frac{341}{128} = 2.6640625 \text{;}$ $S = \frac{a}{1-r} = \frac{4}{1+\frac{1}{2}} = \frac{4}{\frac{3}{2}} = \frac{8}{3} \text{.}$

8.3.3.c

Answer.

$S_5 = \frac{a - ar^5}{1-r} = \frac{2 - 2\cdot(4/3)^5}{1-(4/3)} = \frac{1562}{81} \approx 19.284 \text{;}$ $S_{10} = \frac{a - ar^{10}}{1-r} = \frac{2 - 2\cdot(4/3)^{10}}{1-(4/3)} = \frac{1979054}{19683} \approx 100.546 \text{;}$ the infinite series diverges.

8.3.3.d

Answer.

$S_5 = \frac{a - ar^5}{1-r} = \frac{5 - 5\cdot(3/4)^5}{1-(3/4)} = \frac{3905}{256} \approx 15.254 \text{;}$ $S_{10} = \frac{a - ar^{10}}{1-r} = \frac{5 - 5\cdot(3/4)^{10}}{1-(3/4)} = \frac{4947635}{262144} \approx 18.874 \text{;}$ $S = \frac{a}{1-r} = \frac{5}{1-\frac{3}{4}} = 20 \text{.}$

8.3.3.e

Answer.

$S_5 = \frac{a - ar^5}{1-r} = \frac{\frac{4}{3} - \frac{4}{3} \cdot (-2/3)^5}{1-(-2/3)} = \frac{220}{243} \approx 0.905 \text{;}$ $S_{10} = \frac{a - ar^{10}}{1-r} = \frac{\frac{4}{3} - \frac{4}{3} \cdot (-2/3)^{10}}{1-(-2/3)} = \frac{46420}{59049} \approx 0.786 \text{;}$ $S = \frac{a}{1-r} = \frac{\frac{4}{3}}{1+\frac{2}{3}} = \frac{4}{5} = 0.8 \text{.}$

8.3.4 How geometric series naturally connect to Taylor polynomials

Activity 8.3.4.

8.3.4.a

Answer.

$f(x) =$	$\frac{1}{1-x} = (1-x)^{-1}$
$f'(x) =$	$(-1)(1-x)^{-2}(-1)$
$f''(x) = $	$(-2)(-1)(1-x)^{-3}(-1)(-1)$
$f'''(x) = $	$(-3)(-2)(-1)(1-x)^{-4}(-1)(-1)(-1)$
$f^{(4)}(x) = $	$(-4)(-3)(-2)(-1)(1-x)^{-5}(-1)(-1)(-1)(-1)$
$f^{(5)}(x) = $	$(-5)(-4)(-3)(-2)(-1)(1-x)^{-6}(-1)(-1)(-1)(-1)(-1)$

8.3.4.b

Answer.

$f(0) =$	$\frac{1}{1-0} = 1$	$c_0 =$	$f(0) = 1$
$f'(0) =$	$(-1)(1-0)^{-2}(-1) = 1$	$c_1 =$	$\frac{f'(0)}{1!} = \frac{1}{1!} = 1$
$f''(0) = $	$(-2)(-1)(1)^{-3}(-1)(-1) = 2!$	$c_2 =$	$\frac{f''(0)}{2!} = \frac{2!}{2!} = 1 $
$f'''(0) = $	$(-3)(-2)(-1)(1)^{-4}(-1)(-1)(-1) = 3!$	$c_3 = $	$\frac{f'''(0)}{3!} = \frac{3!}{3!} = 1 $
$f^{(4)}(0) = $	$(-4)(-3)(-2)(-1)(1)^{-5}(-1)(-1)(-1)(-1) = 4!$	$c_4 =$	$\frac{f^{(4)}(0)}{4!} = \frac{4!}{4!} = 1 $
$f^{(5)}(0) = $	$(-5)(-4)(-3)(-2)(-1)(1)^{-6}(-1)(-1)(-1)(-1)(-1) = 5!$	$c_5 =$	$\frac{f^{(5)}(0)}{5!} = \frac{5!}{5!} = 1 $

8.3.4.c

Answer.

$T_5(x) = 1 + x + x^2 + \cdots + x^5$ and $T_n(x) = 1 + x + x^2 + \cdots + x^n \text{.}$

8.3.4.d

Answer.

$T_n(x)$ is a finite geometric sum with $r = x$ and $a = 1\text{.}$

8.3.4.e

Answer.

$T(x) = 1 + x + x^2 + \cdots + x^n + \cdots$ which is an infinite geometric sum with $a = 1$ and $r = x$ whose sum (for $|r| = |x| \lt 1$) is $\frac{a}{1-r} = \frac{1}{1-x} = f(x)\text{.}$

8.4 Taylor series
8.4.2 Taylor series and the Ratio Test

Activity 8.4.2.

8.4.2.a

Answer.

$r_n(x) = \frac{n}{n+1} \cdot \frac{1}{2} \cdot (x-1)$

8.4.2.b

Answer.

$r(x) = \frac{1}{2} \cdot (x-1)$

8.4.2.c

Answer.

$T(x)$ is guaranteed to converge whenever $-1 \lt x \lt 3\text{.}$

8.4.2.d

Answer.

On $1 \lt x \lt 3\text{,}$ $T_{10}$ and $f$ are nearly indistinguishable for most of the interval. This suggests that $T(x)$ is the Taylor series of $f(x) = \ln(2) - \ln(3-x)\text{.}$

8.4.3 Taylor series of several important functions

Activity 8.4.3.

8.4.3.a

Answer.

$f^{(k)}(0) = 1$ for every natural number $k$ because $f^{(k)}(x) = e^x$ for every natural number $k\text{.}$

8.4.3.b

Answer.

$T_f(x) = \sum_{n=0}^{\infty} \frac{1}{k!} x^k = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots + \frac{1}{n!}x^n + \cdots$

8.4.3.c

Answer.

$r_n(x) = \frac{1}{n+1} \cdot x \text{.}$

8.4.3.d

Answer.

$r(x) = \lim_{n \to \infty} r_n(x) = 0 \text{;}$ $|r(x)| \lt 1$ for every value of $x\text{;}$ $T_f(x)$ converges for every real number $x\text{.}$

8.4.3.e

Answer.

$f(x)$ and $T_{10}(x)$ are almost indistinguishable on the interval $-4 \lt x \lt 4\text{;}$ $f(x)$ and $T_{20}(x)$ are almost indistinguishable on $-8 \lt x \lt 8\text{.}$

8.5 Finding and using Taylor series
8.5.2 Using substitution and algebra to find new Taylor series expressions

Activity 8.5.2.

8.5.2.a

Answer.

$g(x) = x^3 \sin(x^2) = x^5 - \frac{1}{3!}x^9 + \frac{1}{5!}x^{13} - \cdots \text{,}$ which converge for all real numbers $x\text{.}$

8.5.2.b

Answer.

$h(x) = 1 - x^4 + \frac{1}{2!}x^8 - \frac{1}{3!}x^{12} + \cdots\text{,}$ which converges for all real numbers $x\text{.}$

8.5.2.c

Answer.

$p(x) = 1 - 5x + (-5x)^2 + (-5x)^3 + \cdots = 1 - 5x + 5^2x^2 - 5^3x^3 + \cdots\text{,}$ which converges whenever $|-5x| \lt 1\text{,}$ or for $-\frac{1}{5} \lt x \lt \frac{1}{5}\text{.}$

8.5.2.d

Answer.

$q(x) = x^6 - \frac{1}{2}x^{10} + \frac{1}{3}x^{14} - \cdots\text{,}$ which converges for $|x| \lt 1\text{.}$

8.5.2.e

Answer.

$r(x) = 1 + \frac{1}{2!}(3x) + \frac{1}{3!}(3x)^2 + \frac{1}{4!}(3x)^3 + \cdots\text{,}$ which converges for all real numbers $x\text{.}$

8.5.3 Differentiating and integrating Taylor series

Activity 8.5.3.

8.5.3.a

8.5.3.a.i

Answer.

$f(x) = \cos(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + \cdots$

8.5.3.a.ii

Answer.

$f'(x) = -\sin(x) = - x + \frac{1}{3!}\cdot x^3 + \frac{1}{5!} \cdot x^5 + \cdots$

8.5.3.a.iii

Answer.

$\frac{d}{dx}\left[ (-1)^{k} \frac{x^{2k}}{(2k)!} \right] = (-1)^{k} \frac{1}{(2k-1)!} \cdot x^{2k-1}$

8.5.3.a.iv

Answer.

$f'(x) = -\sin(x) = \sum_{k=1}^{\infty} (-1)^{k} \frac{x^{2k-1}}{(2k-1)!}$

8.5.3.b

8.5.3.b.i

Answer.

$g'(x) = \frac{1}{1+x^2}$

8.5.3.b.ii

Answer.

$f(x) = g'(x) = \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots$

8.5.3.b.iii

Answer.

By the Second FTC we have $g(x) = \int_0^x f(u) \, du \text{;}$ integrating, it follows that $g(x) = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 + \cdots \text{.}$

8.5.3.b.iv

Answer.

The series for $g(x)$ is certain to converge on $|x| \lt 1\text{.}$

Activity 8.5.4.

8.5.4.a

Answer.

$e^{-t^2} = 1 - t^2 + \frac{1}{2!}t^4 - \frac{1}{3!}t^6 + \cdots$

8.5.4.b

Answer.

$\int_0^x e^{-t^2} \, dt = x - \frac{1}{3}x^3 + \frac{1}{5 \cdot 2!}x^5 - \frac{1}{7 \cdot 3!} x^7 + \cdots $

8.5.4.c

Answer.

$\erf(x) = \frac{2}{\sqrt{\pi}} \left( x - \frac{1}{3}x^3 + \frac{1}{5 \cdot 2!}x^5 - \frac{1}{7 \cdot 3!} x^7 + \cdots \right)$

8.5.4.d

Answer.

for all real numbers.

8.5.4.e

Answer.

$\erf(0.5) \approx \frac{2}{\sqrt{\pi}} \left( \frac{1}{2} - \frac{1}{3 \cdot 2^3} + \frac{1}{5 \cdot 2! \cdot 2^5} - \frac{1}{7 \cdot 3! \cdot 2^7}\right) = 0.52049 \ldots$

8.6 Quantifying the accuracy of approximations
8.6.2 Alternating series of real numbers

Activity 8.6.2.

8.6.2.a

Answer.

$\sin(1) \approx 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} = \frac{4241}{5040} = 0.8414682539682539\ldots \text{,}$ and this approximation has error at most $\frac{1}{9!} \approx 0.00000276\text{.}$

8.6.2.b

Answer.

$\int_0^1 e^{-x^2} \, dx \approx 1 - \frac{1}{3} + \frac{1}{2! \cdot 5} - \frac{1}{3! \cdot 7} + \frac{1}{4! \cdot 9} - \frac{1}{5! \cdot 11} + \frac{1}{6! \cdot 13} = \frac{1614779}{2162160} = 0.7468360343\ldots$ and this approximation has error at most $\frac{1}{7! \cdot 15} = \frac{1}{756600} \approx 0.0000132$

8.6.2.c

Answer.

$\int_0^1 \cos(x^2) \, dx \approx 1 - \frac{1}{5 \cdot 2!} + \frac{1}{9 \cdot 4!} - \frac{1}{13 \cdot 6!} = \frac{25399}{28080} = 0.904522792022792022792\ldots $ and this approximation has error at most $\frac{1}{17 \cdot 8!} = \frac{1}{685440} = 0.0000014589 \ldots \text{.}$

8.6.2.d

Answer.

The alternating series

\begin{equation*} 1 - \frac{1}{2} \cdot 1^2 + \frac{1}{3} \cdot 1^3 - \cdots + (-1)^{n-1} \frac{1}{n} \cdot 1^n + \cdots \end{equation*}

converges by the Alternating Series Theorem, and its exact sum is $\ln(2)\text{.}$

Using the Alternating Series Estimation Theorem to approximate within $0.01$ results in

\begin{equation*} \ln(2) \approx 1 - \frac{1}{2} + \frac{1}{3} - \cdots + \frac{1}{99} = 0.69817217931 \ldots\text{.} \end{equation*}

8.6.3 Error Approximations for Taylor Polynomials

Activity 8.6.3.

8.6.3.a

Answer.

For $f(x) = e^2\text{,}$ the Lagrange Error Bound gives $\left| f(2) - T_{10}(2) \right| \leq e^2 \cdot \frac{|2-0|^{11}}{(11)!} = \frac{e^2 \cdot 2^{11}}{(11)!} \approx 0.00037910821 \text{.}$ The actual error is $\left| f(2) - T_{10}(2) \right| \approx 0.00006138994 $

8.6.3.b

Answer.

Using $M = 1$ as the bound on the $(n+1)^{\text{st}}$ derivative of $\cos(x)\text{,}$ the Lagrange error bound tells us that we need to use $n = 11$ ($n+1 = 12$) to achieve the desired accuracy, and that $\cos(1) \approx T_{10}(1) = 1 - \frac{1^2}{2!} + \frac{1^4}{4!} - \cdots - \frac{1^{10}}{10!} \approx 0.54030230379189\text{.}$

8.6.3.c

Answer.

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\(x\)	\(2.9\)	\(2.99\)	\(2.999\)	\(3\)	\(3.001\)	\(3.01\)	\(3.1\)
\(f(x)\)	\(1.051\)	\(1.00501\)	\(1.0005001\)	\(1\)	\(0.9995001\)	\(0.99501\)	\(0.951\)
\(g(x)\)	\(1.045\)	\(1.00495\)	\(1.0004995\)	\(1\)	\(0.9994995\)	\(0.99495\)	\(0.945\)
\(L(x)\)	\(1.05\)	\(1.005\)	\(1.0005\)	\(1\)	\(0.9995\)	\(0.995\)	\(0.95\)

given function, \(f(x)\)	antiderivative, \(F(x)\)
\(k\text{,}\) (\(k \ne 0\))	\(kx\)
\(x^n\text{,}\) \(n \ne -1\)	\(\frac{1}{n+1}x^{n+1}\)
\(\frac{1}{x}\text{,}\) \(x \gt 0\)	\(\ln(x)\)
\(\sin(x)\)	\(-\cos(x)\)
\(\cos(x)\)	\(\sin(x)\)
\(\sec(x) \tan(x)\)	\(\sec(x)\)
\(\csc(x) \cot(x)\)	\(-\csc(x)\)
\(\sec^2 (x)\)	\(\tan(x)\)
\(\csc^2 (x)\)	\(-\cot(x)\)
\(e^x\)	\(e^x\)
\(a^x\) \((a \gt 1)\)	\(\frac{1}{\ln(a)} a^x\)
\(\frac{1}{1+x^2}\)	\(\arctan(x)\)
\(\frac{1}{\sqrt{1-x^2}}\)	\(\arcsin(x)\)

\(n\)	\(T_n\)	\(E_{T,n}\)
\(4\)	\(0.50899\)	\(0.00899\)
\(8\)	\(0.50227\)	\(0.00227\)
\(16\)	\(0.50057\)	\(0.00057\)

\(n\)	\(M_n\)	\(E_{M,n}\)
\(4\)	\(0.49555\)	\(-0.00445\)
\(8\)	\(0.49887\)	\(-0.00113\)
\(16\)	\(0.49972\)	\(-0.00028\)

\(t_i\)	\(y_i\)	\(dy/dt\)	\(\Delta y\)
\(0\)	\(0\)	\(-1\)	\(-0.2\)
\(0.2\)	\(-0.2\)	\(-0.6\)	\(-0.12\)
\(0.4\)	\(-0.32\)	\(-0.2\)	\(-0.04\)
\(0.6\)	\(-0.36\)	\(0.2\)	\(0.04\)
\(0.8\)	\(-0.32\)	\(0.6\)	\(0.12\)
\(1.0\)	\(-0.2\)	\(1\)	\(0.2\)

\(f(x)=\)	\(e^x\)	\(T_2(x)=\)	\(b_0 + b_1 x + b_2 x^2\)
\(f'(x)=\)	\(e^x\)	\(T_2'(x)=\)	\(b_1 + 2b_2 x\)
\(f''(x)=\)	\(e^x\)	\(T_2''(x)=\)	\(2b_2\)

\(f(0)=\)	\(1\)	\(T_2(0)=\)	\(b_0\)
\(f'(0)=\)	\(1\)	\(T_2'(0)=\)	\(b_1\)
\(f''(0)=\)	\(1\)	\(T_2''(0)=\)	\(2b_2\)

\(f(x)=\)	\(e^x\)	\(T_3(x)=\)	\(c_0 + c_1 x + c_2 x^2 + c_3 x^3\)
\(f'(x)=\)	\(e^x\)	\(T_3'(x)=\)	\(c_1 + 2 c_2 x + 3 c_3 x^2\)
\(f''(x)=\)	\(e^x\)	\(T_3''(x)=\)	\(2 c_2 + 3 \cdot 2 c_3 x\)
\(f'''(x)=\)	\(e^x\)	\(T_3'''(x)=\)	\(3 \cdot 2 c_3\)

\(f(0)=\)	\(1\)	\(T_3(0)=\)	\(c_0\)
\(f'(0)=\)	\(1\)	\(T_3'(0)=\)	\(c_1\)
\(f''(0)=\)	\(1\)	\(T_3''(0)=\)	\(2c_2\)
\(f'''(0)=\)	\(1\)	\(T_3'''(0)=\)	\(6c_3\)

\(\Delta x\)	\(x\)	\(f(x)\)	\(T_1(x)\)	\(T_2(x)\)	\(T_3(x)\)	\(T_4(x)\)
\(0.1\)	\(-1.0\)	\(0.36787\)	\(0.00000\)	\(0.50000\)	\(0.33333\)	\(0.37500\)
\(0.1\)	\(-0.9\)	\(0.40657\)	\(0.10000\)	\(0.50500\)	\(0.38350\)	\(0.41083\)
\(0.1\)	\(-0.8\)	\(0.44933\)	\(0.20000\)	\(0.52000\)	\(0.43467\)	\(0.45173\)
\(0.1\)	\(-0.7\)	\(0.49659\)	\(0.30000\)	\(0.54500\)	\(0.48783\)	\(0.49784\)
\(0.1\)	\(-0.6\)	\(0.54881\)	\(0.40000\)	\(0.58000\)	\(0.54400\)	\(0.54940\)
\(0.1\)	\(-0.5\)	\(0.60653\)	\(0.50000\)	\(0.62500\)	\(0.60417\)	\(0.60677\)
\(0.1\)	\(-0.4\)	\(0.67032\)	\(0.60000\)	\(0.68000\)	\(0.66933\)	\(0.67040\)
\(0.1\)	\(-0.3\)	\(0.74082 \)	\(0.70000 \)	\(0.74500 \)	\(0.74050 \)	\(0.74084\)
\(0.1\)	\(-0.2\)	\(0.81873 \)	\(0.80000 \)	\(0.82000 \)	\(0.81867 \)	\(0.81873\)
\(0.1\)	\(-0.1\)	\(0.90484 \)	\(0.90000 \)	\(0.90500 \)	\(0.90483 \)	\(0.90484\)
\(0.1\)	\(0.0\)	\(1.00000\)	\(1.00000\)	\(1.00000\)	\(1.00000\)	\(1.00000\)

\(\|f(x)-T_1(x)\|\)	\(\|f(x)-T_2(x)\|\)	\(\|f(x)-T_3(x)\|\)	\(\|f(x)-T_4(x)\|\)
\(0.36787\)	\(0.13212\)	\(0.03454\)	\(0.00712\)
\(0.30657\)	\(0.09843\)	\(0.02307\)	\(0.00426\)
\(0.24933 \)	\(0.07067 \)	\(0.01466 \)	\(0.00240\)
\(0.19659 \)	\(0.04841 \)	\(0.00875 \)	\(0.00125\)

\(f(x) =\)	\(\cos(x)\)	\(f(0) =\)	\(\cos(0) = 1\)
\(f'(x) = \)	\(-\sin(x)\)	\(f'(0) = \)	\(0\)
\(f''(x) = \)	\(-\cos(x)\)	\(f''(0) = \)	\(-1\)
\(f'''(x) = \)	\(\sin(x)\)	\(f'''(0) = \)	\(0\)
\(f^{(4)}(x) = \)	\(\cos(x)\)	\(f^{(4)}(0) = \)	\(1\)
\(f^{(5)}(x) = \)	\(-\sin(x)\)	\(f^{(5)}(0) = \)	\(0\)
\(f^{(6)}(x) = \)	\(-\cos(x)\)	\(f^{(6)}(0) = \)	\(-1\)
\(f^{(7)}(x) = \)	\(\sin(x)\)	\(f^{(7)}(0) = \)	\(0\)
\(f^{(8)}(x) = \)	\(\cos(x)\)	\(f^{(8)}(0) = \)	\(1\)

\(\|f(x)-T_2(x)\|\)	\(\|f(x)-T_4(x)\|\)	\(\|f(x)-T_6(x)\|\)
\(0.58385\)	\(0.08281\)	\(0.00608\)
\(0.39280\)	\(0.04460\)	\(0.00263\)
\(0.25080\)	\(0.02227\)	\(0.00104\)
\(\cdots\)	\(\cdots\)	\(\cdots\)
\(0.25080\)	\(0.02227\)	\(0.00104\)
\(0.39280\)	\(0.04460\)	\(0.00263\)
\(0.58385\)	\(0.08281\)	\(0.00608\)

\(f(x) =\)	\(\ln(x)\)	\(f(1) =\)	\(0\)
\(f'(x) = \)	\(x^{-1}\)	\(f'(1) = \)	\(1\)
\(f''(x) = \)	\(-1 \cdot x^{-2}\)	\(f''(1) = \)	\(-1\)
\(f'''(x) = \)	\((-2)(-1)x^{-3}\)	\(f'''(1) = \)	\((-2)(-1)\)
\(f^{(4)}(x) = \)	\((-3)(-2)(-1)x^{-4}\)	\(f^{(4)}(1) = \)	\((-3)(-2)(-1)\)

\(f(x) =\)	\(\ln(x)\)	\(f(2) =\)	\(\ln(2)\)
\(f'(x) = \)	\(x^{-1}\)	\(f'(2) = \)	\(\frac{1}{2}\)
\(f''(x) = \)	\(-1 \cdot x^{-2}\)	\(f''(2) = \)	\(-\frac{1}{2^2}\)
\(f'''(x) = \)	\((-2)(-1)x^{-3}\)	\(f'''(2) = \)	\(\frac{(-2)(-1)}{2^3}\)
\(f^{(4)}(x) = \)	\((-3)(-2)(-1)x^{-4}\)	\(f^{(4)}(2) = \)	\(\frac{(-3)(-2)(-1)}{2^4}\)

\(f(x) =\)	\(\frac{1}{1-x} = (1-x)^{-1}\)
\(f'(x) =\)	\((-1)(1-x)^{-2}(-1)\)
\(f''(x) = \)	\((-2)(-1)(1-x)^{-3}(-1)(-1)\)
\(f'''(x) = \)	\((-3)(-2)(-1)(1-x)^{-4}(-1)(-1)(-1)\)
\(f^{(4)}(x) = \)	\((-4)(-3)(-2)(-1)(1-x)^{-5}(-1)(-1)(-1)(-1)\)
\(f^{(5)}(x) = \)	\((-5)(-4)(-3)(-2)(-1)(1-x)^{-6}(-1)(-1)(-1)(-1)(-1)\)

\(f(0) =\)	\(\frac{1}{1-0} = 1\)	\(c_0 =\)	\(f(0) = 1\)
\(f'(0) =\)	\((-1)(1-0)^{-2}(-1) = 1\)	\(c_1 =\)	\(\frac{f'(0)}{1!} = \frac{1}{1!} = 1\)
\(f''(0) = \)	\((-2)(-1)(1)^{-3}(-1)(-1) = 2!\)	\(c_2 =\)	\(\frac{f''(0)}{2!} = \frac{2!}{2!} = 1 \)
\(f'''(0) = \)	\((-3)(-2)(-1)(1)^{-4}(-1)(-1)(-1) = 3!\)	\(c_3 = \)	\(\frac{f'''(0)}{3!} = \frac{3!}{3!} = 1 \)
\(f^{(4)}(0) = \)	\((-4)(-3)(-2)(-1)(1)^{-5}(-1)(-1)(-1)(-1) = 4!\)	\(c_4 =\)	\(\frac{f^{(4)}(0)}{4!} = \frac{4!}{4!} = 1 \)
\(f^{(5)}(0) = \)	\((-5)(-4)(-3)(-2)(-1)(1)^{-6}(-1)(-1)(-1)(-1)(-1) = 5!\)	\(c_5 =\)	\(\frac{f^{(5)}(0)}{5!} = \frac{5!}{5!} = 1 \)

\(t_i\)	\(y_i\)	\(dy/dt\)	\(\Delta y\)
\(0.0\)	\(1.0000\)	\(5.0000\)	\(1.0000\)
\(0.2\)	\(2.0000\)	\(8.0000\)	\(1.6000\)
\(0.4\)	\(3.6000\)	\(8.6400\)	\(1.7280\)
\(0.6\)	\(5.3280\)	\(3.5804\)	\(0.7161\)
\(0.8\)	\(6.0441\)	\(-0.2664\)	\(-0.0533\)
\(1.0\)	\(5.9908\)	\(0.0551\)	\(0.0110\)

\(\Delta x\)	\(x\)	\(f(x)\)	\(T_2(x)\)	\(T_4(x)\)	\(T_6(x)\)
\(0.2\)	\(-2.0\)	\(-0.41615\)	\(-1.00000\)	\(-0.33333\)	\(-0.42222\)
\(0.2\)	\(-1.8\)	\(-0.22720\)	\(-0.62000\)	\(-0.18260\)	\(-0.22984\)
\(0.2\)	\(-1.6\)	\(-0.02920\)	\(-0.28000\)	\(-0.00693\)	\(-0.03024\)
\(\cdots\)	\(\cdots\)	\(\cdots\)	\(\cdots\)	\(\cdots\)	\(\cdots\)
\(0.2\)	\(1.6\)	\(-0.02920\)	\(-0.28000\)	\(-0.00693\)	\(-0.03024\)
\(0.2\)	\(1.8\)	\(-0.22720\)	\(-0.62000\)	\(-0.18260\)	\(-0.22984\)
\(0.2\)	\(2.0\)	\(-0.41615\)	\(-1.00000\)	\(-0.33333\)	\(-0.42222\)

Appendix B Answers to Activities

1 Understanding the Derivative1.1 How do we measure velocity?1.1.2 Position and average velocity

1.1.3 Instantaneous Velocity

1.2 The notion of limit1.2.2 The Notion of Limit

1.2.3 Instantaneous Velocity

1.3 The derivative of a function at a point1.3.2 The Derivative of a Function at a Point

1.4 The derivative function1.4.2 How the derivative is itself a function

1.5 Interpreting, estimating, and using the derivative1.5.2 Units of the derivative function

1.5.3 Toward more accurate derivative estimates

1.6 The second derivative1.6.4 Concavity

1.7 Limits, continuity, and differentiability1.7.2 Having a limit at a point

1.7.3 Being continuous at a point

1.7.4 Being differentiable at a point

1.8 The tangent line approximation1.8.3 The local linearization

1 Understanding the Derivative
1.1 How do we measure velocity?
1.1.2 Position and average velocity

1.2 The notion of limit
1.2.2 The Notion of Limit

1.3 The derivative of a function at a point
1.3.2 The Derivative of a Function at a Point

1.4 The derivative function
1.4.2 How the derivative is itself a function

1.5 Interpreting, estimating, and using the derivative
1.5.2 Units of the derivative function

1.6 The second derivative
1.6.4 Concavity

1.7 Limits, continuity, and differentiability
1.7.2 Having a limit at a point

1.8 The tangent line approximation
1.8.3 The local linearization