# Active Calculus

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

### 1Understanding the Derivative1.1How do we measure velocity?1.1.1Position and average velocity

#### Activity1.1.2.

1. $$AV_{[0.4,0.8]} = 12.8$$ 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.
2. $$m = 12.8$$ is the average velocity of the ball between $$t = 0.4$$ and $$t = 0.8\text{.}$$
3. Like a straight line with slope about 6.4.
4. About 6.4 feet per second.

### 1.1.2Instantaneous Velocity

#### Activity1.1.3.

1. $$AV_{[1.5,2]} = -24$$ ft/sec, which is negative.
2. 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 $$-16>-32\text{.}$$
3. When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.
4. Zero.

#### Activity1.1.4.

$$AV_{[2, 2+h]} = -32 - 16h$$

### 1.2The notion of limit1.2.1The Notion of Limit

#### Activity1.2.2.

1. $$2\text{.}$$
2. $$12\text{.}$$
3. $$\frac{1}{2}\text{.}$$

### 1.2.2Instantaneous Velocity

#### Activity1.2.3.

1. $$6 + h\text{.}$$
2. $$6.2$$ meters/min.
3. $$6$$ meters per minute.

#### Activity1.2.4.

1. $$AV_{[0.5,1]} = \frac{1-1}{1-0.5} = 0\text{,}$$ $$AV_{[1.5,2.5]} = \frac{3-1}{2.5-1.5} = 2\text{,}$$ and $$AV_{[0,5]} = \frac{5-0}{5-0} = 1\text{.}$$
2. 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.
3. The instantaneous velocity at $$t = 2$$ is greater than the average velocity on $$[1.5,2.5]\text{.}$$

### 1.3The derivative of a function at a point1.3.1The Derivative of a Function at a Point

#### Activity1.3.2.

1. $$f$$ is linear.
2. The average rate of change on $$[1,4]\text{,}$$ $$[3,7]\text{,}$$ and $$[5,5+h]$$ is $$-2\text{.}$$
3. $$f'(1)=-2\text{.}$$
4. $$f'(2)=-2\text{,}$$ $$f'(\pi)=-2\text{,}$$ and $$f'(-\sqrt{2})=-2\text{,}$$ since the slope of a linear function is the same at every point.

#### Activity1.3.3.

1. The vertex is $$(\frac{1}{2},36)\text{.}$$
2. $$\frac{s(2)-s(1)}{2-1} = -32$$ feet per second.
3. $$s'(1) = -16\text{.}$$
4. $$s'(a)$$ is positive whenever $$0 \le a \lt \frac{1}{2}\text{;}$$ $$s'(a)$$ to be negative whenever $$\frac{1}{2} \lt a \lt 2\text{;}$$ $$s'(\frac{1}{2}) = 0\text{.}$$

#### Activity1.3.4.

1. $$AV_{[2,4]} \approx 9171$$ 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.
2. \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.
3. \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.
4. See the graph provided in (a) above. 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 $$P'(2)\text{.}$$
5. It appears that the tangent line’s slope at the point $$(a,P(a))$$ will increase as $$a$$ increases.

### 1.4The derivative function1.4.1How the derivative is itself a function

#### Activity1.4.3.

1. $$f'(x) = 0\text{.}$$
2. $$g'(t) = 1\text{.}$$
3. $$p'(z) = 2z\text{.}$$
4. $$q'(s) = 3s^2\text{.}$$
5. $$F'(t) = \frac{-1}{t^2}\text{.}$$
6. $$G'(y) = \frac{1}{2\sqrt{y}}\text{.}$$

### 1.5Interpreting, estimating, and using the derivative1.5.2Toward more accurate derivative estimates

#### Activity1.5.2.

1. $$F'(30) \approx = 3.85$$ degrees per minute.
2. $$F'(60) \approx = 1.56$$ degrees per minute.
3. $$F'(75) \gt F'(90)\text{.}$$
4. The value $$F(64) = 330.28$$ is the temperature of the potato in degrees Fahrenheit at time 64, while $$F'(64) = 1.341$$ measures the instantaneous rate of change of the potato’s temperature with respect to time at the instant $$t = 64\text{,}$$ and its units are degrees per minute. Because at time $$t = 64$$ the potato’s temperature is increasing at 1.341 degrees per minute, we expect that at $$t = 65\text{,}$$ the temperature will be about 1.341 degrees greater than at $$t = 64\text{,}$$ or in other words $$F(65) \approx 330.28 + 1.341 = 331.621\text{.}$$ Similarly, at $$t = 66\text{,}$$ two minutes have elapsed from $$t = 64\text{,}$$ so we expect an increase of $$2 \cdot 1.341$$ degrees: $$F(66) \approx 330.28 + 2 \cdot 1.341 = 332.962\text{.}$$
5. Throughout the time interval $$[0,90]\text{,}$$ the temperature $$F$$ of the potato is increasing. But as time goes on, the rate at which the temperature is rising appears to be decreasing. That is, while the values of $$F$$ continue to get larger as time progresses, the values of $$F'$$ are getting smaller (while still remaining positive). We thus might say that “the temperature of the potato is increasing, but at a decreasing rate.”

1. It costs 800 to make 2000 feet of rope. 2. “dollars per foot.” 3. $$C(2100) \approx = 835\text{,.}$$ 4. Either $$C'(2000) = C'(3000)$$ or $$C'(2000) > C'(3000)\text{.}$$ 5. Impossible. The total cost function $$C(r)$$ can never decrease. #### Activity1.5.4. Answer. 1. $$f'(90) \approx 0.0006$$ liters per kilometer per kilometer per hour. 2. At 80 kilometers per hour, the car is using fuel at a rate of 0.015 liters per kilometer. 3. 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. ### 1.6The second derivative1.6.3Concavity #### Activity1.6.2. Answer. 1. 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. 2. 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 $$9\lt t\lt 10\text{.}$$ 3. $$a(t) = v'(t)$$ and $$a(t) = s''(t)\text{.}$$ 4. $$s''(t)$$ is positive since $$s'(t)$$ is increasing. • increasing. • decreasing. • constant. • increasing. • decreasing. • constant. • concave up. • concave down. • linear. #### Activity1.6.3. Answer. 1. Degrees Fahrenheit per minute. 2. $$F''(30) \approx -0.119\text{.}$$ 3. At the moment $$t = 30\text{,}$$ the temperature of the potato is 251 degrees; its temperature is rising at a rate of 3.85 degrees per minute; and the rate at which the temperature is rising is falling at a rate of 0.119 degrees per minute per minute. 4. Increasing at a decreasing rate. #### Activity1.6.4. Answer. ### 1.7Limits, Continuity, and Differentiability1.7.1Having a limit at a point #### Activity1.7.2. Answer. 1. $$f(-2) = 1\text{;}$$ $$f(-1)$$ is not defined; $$f(0) = \frac{7}{3}\text{;}$$ $$f(1) = 2\text{;}$$ $$f(2) = 2\text{.}$$ 2. \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*} 3. $$\lim_{x \to -2} f(x)$$ does not exist. The values of the limits as $$x \to a$$ for $$a = -1, 0, 1, 2$$ are $$\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}$$ 4. $$a = -2\text{,}$$ $$a = -1\text{,}$$ and $$a = 1\text{.}$$ ### 1.7.2Being continuous at a point #### Activity1.7.3. Answer. 1. $$a = -2\text{;}$$ $$a = +2\text{.}$$ 2. $$a = 3\text{.}$$ 3. $$a = -1\text{;}$$ $$a = 3\text{.}$$ 4. $$a=-2\text{;}$$ $$a = 2\text{;}$$ $$a = 3\text{;}$$ $$a = -1\text{.}$$ 5. “If $$f$$ is continuous at $$x = a\text{,}$$ then $$f$$ has a limit at $$x = a\text{.}$$ ### 1.7.3Being differentiable at a point #### Activity1.7.4. Answer. 1. $$g$$ is piecewise linear. 2. \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*} 3. $$\lim_{h \to 0^+} \frac{|h|}{h} = 1 \text{,}$$ but $$\lim_{h \to 0^-} \frac{|h|}{h} = -1 \text{.}$$ 4. $$a = -3, -2, -1, 1, 2, 3\text{.}$$ 5. True. ### 1.8The Tangent Line Approximation1.8.2The local linearization #### Activity1.8.2. Answer. 1. $$L(-1) = -2\text{;}$$ $$L'(-1) = 3\text{.}$$ 2. $$g(-1) = -2\text{;}$$ $$g'(-1) = 3\text{.}$$ 3. Less. 4. $$g(-1.03) \approx L(-1.03) = -2.09\text{.}$$ 5. Concave up. 6. The illustration below shows a possible graph of $$y = g(x)$$ near $$x = -1\text{,}$$ along with the tangent line $$y = L(x)$$ through $$(-1, g(-1))\text{.}$$ #### Activity1.8.3. Answer. 1. $$L(x) = -1 + 2(x-2)\text{.}$$ 2. $$f(2.07) \approx L(2.07) = -0.86\text{.}$$ 3. See the image in part e. 4. Neither. 5. 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 $$(2, f(2))\text{.}$$ 6. Too large. ### 2Computing Derivatives2.1Elementary derivative rules2.1.2Constant, Power, and Exponential Functions #### Activity2.1.2. Answer. 1. $$f'(t) = 0\text{.}$$ 2. $$g'(z) = 7^z \ln(7)\text{.}$$ 3. $$h'(w) = \frac{3}{4} w^{-1/4}\text{.}$$ 4. $$\frac{dp}{dx} = 0\text{.}$$ 5. $$r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}$$ 6. $$\frac{d}{dq}[q^{-1}] = -q^{-2}\text{.}$$ 7. $$\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}$$ ### 2.1.3Constant Multiples and Sums of Functions #### Activity2.1.3. Answer. 1. $$f'(x) = \frac{5}{3}x^{2/3} - 4 x^3 + 2^x \ln(2)\text{.}$$ 2. $$g'(x) = 14e^x + 3 \cdot 5x^4 - 1\text{.}$$ 3. $$h'(z) = \frac{1}{2}z^{-1/2} - 4z^{-5} + 5^z \ln(5)\text{.}$$ 4. $$\frac{dr}{dt} = \sqrt{53} \cdot 7 t^6 - \pi e^t\text{.}$$ 5. $$\frac{ds}{dy} = 4y^3\text{.}$$ 6. $$q'(x) = 2x - 2x^{-2}\text{.}$$ 7. $$p'(a) = 12a^3 - 6 a^2 + 14a - 1\text{.}$$ #### Activity2.1.4. Answer. 1. $$h'(4) = \frac{3}{16}\text{.}$$ 2. (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. 3. $$y - 25 = -33(a+1)\text{.}$$ 4. The slope is a number, while the equation is, well, an equation. ### 2.2The sine and cosine functions2.2.1The sine and cosine functions #### Activity2.2.2. Answer. 1. $$1,0,-1,0,1,0,-1,0,1\text{.}$$ 2. $$f'(0) = f'(-2\pi) = f'(2\pi) = 1\text{.}$$ 3. $$\frac{d}{dx}[\sin(x)] = \cos(x)\text{.}$$ #### Activity2.2.3. Answer. 1. $$0,-1,0,1,0,-1,0,1,0\text{.}$$ 2. $$g'(\frac{\pi}{2})=g'(-\frac{3\pi}{2})=-1\text{.}$$ 3. $$\frac{d}{dx}[\cos(x)] = -\sin(x)\text{.}$$ #### Activity2.2.4. Answer. 1. $$\frac{dh}{dt} = -3\sin(t) - 4\cos(t)\text{.}$$ 2. $$f'(\frac{\pi}{6}) = 2 + \frac{\sqrt{3}}{4}\text{.}$$ 3. $$y - \frac{\pi^2}{4} = (\pi-2)(x-\frac{\pi}{2})\text{.}$$ 4. $$p'(z) = 4z^3 + 4^z \ln(4) - 4\sin(z)\text{.}$$ 5. $$P'(2) = 8\cos(2) \approx -3.329$$ hundred animals per decade. ### 2.3The product and quotient rules2.3.1The product rule #### Activity2.3.2. Answer. 1. $$m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.}$$ 2. $$h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\text{.}$$ 3. $$f'(1) = e(\cos(1) + \sin(1)) \approx 3.756\text{.}$$ 4. $$L(x) = -\frac{1}{2}(x+1)\text{.}$$ ### 2.3.2The quotient rule #### Activity2.3.3. Answer. 1. $$r'(z)=\frac{(z^4+1) 3^z \ln(3) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.}$$ 2. $$v'(t) = \frac{(\cos(t) + t^2)\cos(t) - \sin(t)(-\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.}$$ 3. $$R'(0) = \frac{2}{9}\text{.}$$ 4. $$I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$$ $$I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$$ and $$I'(5) = \frac{-400}{e^5} \approx -2.695\text{,}$$ each in candles per millisecond. ### 2.3.3Combining rules #### Activity2.3.4. Answer. 1. $$f'(r) = (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)]\text{.}$$ 2. $$p'(t) = \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2}\text{.}$$ 3. $$g'(z) = 3 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 1 - z(2z)}{(z^2 + 1)^2}\text{.}$$ 4. $$s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414$$ feet per second. 5. $$p'(3) = 30$$ and $$q'(3) = \frac{13}{8}\text{.}$$ ### 2.4Derivatives of other trigonometric functions2.4.1Derivatives of the cotangent, secant, and cosecant functions #### Activity2.4.2. Answer. 1. All real numbers $$x$$ such that $$x \ne \frac{\pi}{2} + k\pi\text{,}$$ where $$k = \pm 1, \pm 2, \ldots\text{.}$$ 2. $$h'(x) = \frac{\sin(x)}{\cos^2(x)}\text{.}$$ 3. $$h'(x) = \sec(x) \tan(x)\text{.}$$ 4. $$h$$ and $$h'$$ have the same domain: all real numbers $$x$$ such that $$x \ne \frac{\pi}{2}+k\pi\text{,}$$ where $$k = 0, \pm 1, \pm 2, \ldots\text{.}$$ #### Activity2.4.3. Answer. 1. All real numbers $$x$$ such that $$x \ne k\pi\text{,}$$ where $$k = 0, \pm 1, \pm 2, \ldots\text{.}$$ 2. $$h'(x) = -\frac{\cos(x)}{\sin^2(x)}\text{.}$$ 3. $$h'(x) = -\csc(x) \cot(x)\text{.}$$ 4. $$p$$ and $$p'$$ have the same domain: all real numbers $$x$$ such that $$x \ne k\pi\text{,}$$ where $$k = 0, \pm 1, \pm 2, \ldots\text{.}$$ #### Activity2.4.4. Answer. 1. $$m = f'(\frac{\pi}{3}) =10\sqrt{3} + \frac{4}{3}\text{.}$$ 2. $$p'(\frac{\pi}{4}) = \frac{\pi^2}{16} \sqrt{2} + \frac{\sqrt{2}\pi}{2} + \frac{\pi}{2} - 1\text{.}$$ 3. $$h'(t) = \frac{(t^2+1) \sec^2(t) - 2t \tan(t)}{(t^2 + 1)^2} + 2e^t \sin(t) - 2 e^t\cos(t)\text{.}$$ 4. $$g'(r) = \frac{r \sec(r) \tan(r) + \sec(r) - r ln(5) \sec(r)}{5^r}\text{.}$$ 5. $$s'(2) = \frac{15\cos(2) - 15\sin(2)}{e^2} \approx -2.69$$ inches per second. ### 2.5The chain rule2.5.1The chain rule #### Activity2.5.2. Answer. 1. $$h'(x) = -4x^3\sin(x^4)\text{.}$$ 2. $$h'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.}$$ 3. $$h'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.}$$ 4. $$h'(x) = -5\cot^4(x) \csc^2(x)\text{.}$$ 5. $$h'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.}$$ ### 2.5.2Using multiple rules simultaneously #### Activity2.5.3. Answer. 1. $$p'(r) = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.}$$ 2. $$m'(v) = -3v^2 \sin(v^2)\sin(v^3) + 2v \cos(v^3)\cos(v^2)\text{.}$$ 3. $$h'(y) = \frac{(e^{4y}+1) [-10\sin(10y)] - \cos(10y) [4e^{4y}]}{(e^{4y}+1)^2}\text{.}$$ 4. $$s'(z) = 2^{z^2\sec(z)} \ln(2) [z^2 \sec(z)\tan(z) + \sec(z) \cdot 2z]\text{.}$$ 5. $$c'(x) = \cos(e^{x^2}) [e^{x^2}\cdot 2x]\text{.}$$ #### Activity2.5.4. Answer. 1. $$y - 2 = \frac{1}{4}(x-0)\text{.}$$ 2. $$v(1) = s'(1) = -\frac{3}{8}$$ inches per second; the particle is moving left at the instant $$t = 1\text{.}$$ 3. $$P'(1000) = 30 e^{-0.0323} (-0.0000323) \approx -0.000938$$ inches of mercury per foot. 4. $$C'(2) = -10 \text{;}$$ $$D'(-1) = -20\text{.}$$ ### 2.6Derivatives of Inverse Functions2.6.2The derivative of the natural logarithm function #### Activity2.6.2. Answer. 1. $$h'(x) = x + 2x\ln(x)\text{.}$$ 2. $$p'(t) = \frac{(e^t + 1) \frac{1}{t} - \ln(t) \cdot e^t}{(e^t + 1)^2}\text{.}$$ 3. $$s'(y) = \frac{1}{\cos(y) + 2} \cdot (-\sin(y))\text{.}$$ 4. $$z'(x) = \sec^2(\ln(x)) \cdot \frac{1}{x}\text{.}$$ 5. $$m'(z) = \frac{1}{\ln(z)} \cdot \frac{1}{z}\text{.}$$ ### 2.6.3Inverse trigonometric functions and their derivatives #### Activity2.6.3. Answer. 1. $$\tan(r(x)) = x\text{.}$$ 2. $$r'(x) = \cos^2(r(x))\text{.}$$ 3. $$r'(x) = \cos^2(\arctan(x))\text{.}$$ 4. With $$\theta = \arctan(x)\text{,}$$ 5. $$\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.}$$ 6. $$r'(x) = \frac{1}{1+x^2}\text{.}$$ #### Activity2.6.4. Answer. 1. $$f'(x) = \left[x^3 \cdot \frac{1}{1+x^2} + \arctan(x) \cdot 3x^2 \right] + \left[e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x\right]\text{.}$$ 2. $$p'(t) = 2^{t\arcsin(t)} \ln(2) [t \cdot \frac{1}{\sqrt{1-t^2}} + \arcsin(t) \cdot 1]\text{.}$$ 3. $$h'(z) = 27(\arcsin(5z) + \arctan(4-z))^{26} \left[\frac{1}{\sqrt{1-(5z)^2}} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right]\text{.}$$ 4. $$s'(y) = -\frac{1}{y^2}\text{.}$$ 5. $$m'(v) = \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right]\text{.}$$ 6. $$\displaystyle g'(w) = \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{(1+w^2) \frac{1}{w} - \ln(w) \cdot 2w}{(1+w^2)^2} \right]$$ ### 2.7Derivatives of Functions Given Implicitly2.7.1Implicit Differentiation #### Activity2.7.2. Answer. 1. The graph of the curve fails the vertical line test. 2. $$\frac{dy}{dx} = \frac{1}{5y^4 - 15y^2 + 4}\text{.}$$ 3. $$y = -\frac{1}{6}x + 1\text{.}$$ 4. $$(1.418697,0.543912)\text{,}$$ $$(-1.418697,-0.543912)\text{,}$$ $$(-3.63143, 1.64443)\text{,}$$ and $$(3.63143, -1.64443)\text{.}$$ #### Activity2.7.3. Answer. 1. 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. When $$y = \frac{1}{2}, \frac{1 \pm \sqrt{5}}{2}\text{,}$$ so one point is $$(2.21028, \frac{1}{2})\text{.}$$ 3. $$y - 1 = \frac{1}{2}(x-1)\text{.}$$ #### Activity2.7.4. Answer. 1. $$\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2$$ and the tangent line has equation $$y - 3 = 1(x+3)\text{.}$$ 2. $$\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}$$ and the tangent line has equation $$y = \frac{1}{2}x\text{.}$$ 3. $$\frac{dy}{dx} = \frac{3e^{-xy} - 3xye^{-xy}}{3x^2e^{-xy}+2y}$$ and the tangent line is $$y - 1 = 0.234950(x - 0.619061)\text{.}$$ ### 2.8Using Derivatives to Evaluate Limits2.8.1Using derivatives to evaluate indeterminate limits of the form $$\frac{0}{0}\text{.}$$ #### Activity2.8.2. Answer. 1. $$\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1\text{.}$$ 2. $$\lim_{x \to \pi} \frac{\cos(x)}{x} = -\frac{1}{\pi}\text{.}$$ 3. $$\lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}} = -2\text{.}$$ 4. $$\lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = 0\text{.}$$ #### Activity2.8.3. Answer. 1. $$\lim_{x \to 2} \frac{f(x)}{g(x)} = \frac{1}{8}\text{.}$$ 2. $$\lim_{x \to 2} \frac{p(x)}{q(x)} = 1\text{.}$$ 3. $$\lim_{x \to 2} \frac{r(x)}{s(x)} \lt 0\text{.}$$ ### 2.8.2Limits involving $$\infty$$ #### Activity2.8.4. Answer. 1. $$\lim_{x \to \infty} \frac{x}{\ln(x)} = \infty\text{.}$$ 2. $$\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \frac{1}{2}\text{.}$$ 3. $$\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = 0\text{.}$$ 4. $$\lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.}$$ 5. $$\lim_{x \to \infty} xe^{-x} = 0\text{.}$$ ### 3Using Derivatives3.1Using derivatives to identify extreme values3.1.1Critical numbers and the first derivative test #### Activity3.1.2. Answer. 1. $$x = -4$$ or $$x = 1\text{.}$$ 2. $$g$$ has a local maximum at $$x = -4$$ and neither a max nor min at $$x = 1\text{.}$$ 3. $$g$$ does not have a global minimum; it is unclear (at this point in our work) if $$g$$ increases without bound, so we can’t say for certain whether or not $$g$$ has a global maximum. 4. $$\lim_{x \to \infty} g'(x) = \infty\text{.}$$ 5. A possible graph of $$g$$ is the following. ### 3.1.2The second derivative test #### Activity3.1.3. Answer. 1. $$x = -1$$ is an inflection point of $$g\text{.}$$ 2. $$g$$ 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. $$g$$ has a local minimum at $$x = -1.67857351\text{.}$$ 4. $$g$$ is a degree 5 polynomial. #### Activity3.1.4. Answer. 1. 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. 2. 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. 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 $$k\text{.}$$ ### 3.2Using derivatives to describe families of functions3.2.1Describing families of functions in terms of parameters #### Activity3.2.2. Answer. 1. $$p$$ has two critical numbers ($$x = \pm \sqrt{\frac{a}{3}}$$) whenever $$a \gt 0$$ and no critical numbers when $$a \lt 0\text{.}$$ 2. 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. $$p$$ is CCD for $$x \lt 0$$ and $$p$$ is CCU for $$x\gt 0\text{,}$$ making $$x = 0$$ an inflection point. #### Activity3.2.3. Answer. 1. $$h$$ is an always increasing function. 2. $$h$$ is always concave down. 3. $$\lim_{x \to \infty} a(1-e^{-bx}) = a\text{,}$$ and $$\lim_{x \to \infty} a(1-e^{-bx}) = -\infty\text{.}$$ 4. 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. #### Activity3.2.4. Answer. 1. $$L$$ is an always increasing function. 2. $$L$$ 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. $$\lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = A\text{,}$$ and \begin{equation*} \lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = 0\text{.} \end{equation*} 4. The inflection point on the graph of $$L$$ is $$( -\frac{1}{k} \ln \left(\frac{1}{c}\right), \frac{A}{2})\text{.}$$ ### 3.3Global Optimization3.3.1Global Optimization #### Activity3.3.2. Answer. 1. $$x = \pm \sqrt{2} \approx \pm 1.414\text{.}$$ 2. On $$[-2,3]\text{,}$$ $$g$$ has a global maximum at $$x = 3$$ and a global minimum at $$x = \sqrt{2}\text{.}$$ 3. On $$[-2,2]\text{,}$$ $$g$$ has a global maximum at $$x = -\sqrt{2}$$ and a global minimum at $$x = \sqrt{2}\text{.}$$ 4. On $$[-2,3]\text{,}$$ $$g$$ has a global maximum at $$x = -\sqrt{2}$$ and a global minimum at $$x = 1\text{.}$$ #### Activity3.3.3. Answer. 1. Absolute maximum: $$e^{-1}\text{;}$$ absolute minimum: $$0\text{.}$$ 2. Absolute maximum: $$\sqrt{2}\text{;}$$ absolute minimum: $$-1\text{.}$$ 3. Absolute maximum: 9.8; absolute minimum: 8. 4. Absolute minimum at $$x = 2\text{;}$$ no absolute maximum. ### 3.3.2Moving toward applications #### Activity3.3.4. Answer. 1. $$V(x) = x (10-2x) (15-2x) = 4x^3 - 50x^2 + 150x\text{.}$$ 2. $$1 \le x \le 3\text{.}$$ 3. $$x = \frac{25 \pm 5\sqrt{7}}{6} \approx 6.371459426, 1.961873908\text{.}$$ • $$\displaystyle V(1.961873908) = 132.0382370$$ • $$\displaystyle V(1) = 104$$ • $$\displaystyle V(3) = 108$$ 4. Absolute maximum: 132.0382370; absolute minimum: 104. ### 3.4Applied Optimization3.4.1More applied optimization problems #### Activity3.4.2. Answer. 1. Let the can have radius $$r$$ and height $$h\text{.}$$ 2. $$V = \pi r^2 h\text{;}$$ $$S = 2 \pi r^2 + 2 \pi r h\text{;}$$ $$C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.}$$ 3. $$C(r) = 0.054 \pi r^2 + 0.48 \frac{1}{r}\text{,}$$ $$r \gt 0\text{.}$$ 4. $$r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.12259\text{;}$$ $$h \approx 4.041337\text{;}$$ minimum cost $$C(1.12259) \approx 0.64137\text{.}$$ #### Activity3.4.3. 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. #### Activity3.4.4. Answer. Maximum area: $$A(\frac{5}{\sqrt{3}}) = \frac{500}{9}\sqrt{3} \approx 96.225\text{.}$$ Maximum perimeter: $$P(1) = 52\text{.}$$ At $$x = \frac{\sqrt{82}-1}{3}$$ the absolute maximum of combined perimeter and area occurs. #### Activity3.4.5. Answer. $$A(1.19606) \approx 2.2018$$ is the absolute maximum cross-sectional area, which leads to the absolute maximum volume. ### 3.5Related Rates3.5.1Related Rates Problems #### Activity3.5.2. Answer. 1. $$r = \frac{3}{4}h\text{.}$$ 2. $$V = \frac{3}{16} \pi h^3\text{.}$$ 3. $$\frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt} \text{.}$$ 4. $$\left. \frac{dh}{dt} \right|_{h=3} = \frac{64}{81\pi} \approx 0.2515$$ feet per minute. 5. Most rapidly when $$h = 3\text{.}$$ #### Activity3.5.3. Answer. 1. $$\frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.}$$ 2. $$h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}$$ 3. $$\left. \frac{dz}{dt} \right|_{h=3000} = 360 \ \text{feet/sec};$$ $$\left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{12}{125}$$ radians per second. 4. greater. #### Activity3.5.4. Answer. 1. $$3s = 2x\text{.}$$ 2. $$3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.}$$ 3. $$\left. \frac{ds}{dt} \right|_{x=8} = 2$$ feet per second. 4. at a constant rate. 5. Let $$y$$ represent the location of the tip of the shadow; $$\frac{dy}{dt} = 5$$ feet/sec. #### Activity3.5.5. 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. $$\left. \frac{dz}{dt} \right|_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 \ \text{feet/sec} \text{.}$$ Let $$r$$ be the runner’s position at time $$t$$ and let $$s$$ be the distance between the runner and the ball, as pictured. $$\left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{feet/sec} \text{.}$$ ### 4The Definite Integral4.1Determining distance traveled from velocity4.1.1Area under the graph of the velocity function #### Activity4.1.2. Answer. 1. \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. 2. Using 8 rectangles of width $$0.25\text{,}$$ $$D \approx 3.875\text{.}$$ 3. $$s(t) = \frac{1}{8}t^4 - \frac{1}{2} t^3 + \frac{3}{4} t^2 + \frac{3}{2}t\text{.}$$ 4. $$s(2) - s(0) = \frac{1}{8}2^4 - \frac{1}{2}2^3 + \frac{3}{4}2^2 + \frac{3}{2} 2 = 4\text{.}$$ ### 4.1.2Two approaches: area and antidifferentiation #### Activity4.1.3. Answer. 1. On $$(0,1)\text{,}$$ $$s$$ is increasing because velocity is positive. 2. $$s(t) = 32t - 16t^2\text{.}$$ 3. $$s(1) - s(\frac{1}{2}) = 4\text{.}$$ 4. $$A = 4$$ feet is the total distance the ball traveled vertically on $$[\frac{1}{2},1]\text{.}$$ 5. $$s(1) - s(0) = 16$$ 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. 6. $$s(2) - s(0) = 0\text{,}$$ so the ball has zero change in position on the interval $$[0,2]\text{.}$$ ### 4.1.3When velocity is negative #### Activity4.1.4. Answer. 1. Total distance traveled is $$2\text{;}$$ change in position is $$0\text{.}$$ 2. $$0 \lt t \lt 1$$ and $$4 \lt t \lt 8 \text{.}$$ 3. $$s(8) - s(0) = 5 \ \mbox{m} \text{,}$$ while the distance traveled on $$[0,8]$$ is $$D = 13\text{,}$$ and thus these two quantities are different. 4. See the figure below. ### 4.2Riemann Sums4.2.1Sigma Notation #### Activity4.2.2. Answer. 1. $$\displaystyle 65$$ 2. $$\displaystyle 32$$ 3. \begin{equation*} 3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.} \end{equation*} 4. \begin{equation*} 4 + 8 + 16 + 32 + \cdots + 256 = \sum_{i=2}^{8} 2^i\text{.} \end{equation*} 5. \begin{equation*} \sum_{i=1}^{6} \frac{1}{2^i} = \frac{63}{64}\text{.} \end{equation*} ### 4.2.2Riemann Sums #### Activity4.2.3. Answer. 1. $$L_4 = \frac{311}{48} \approx 6.47917\text{,}$$ $$R_4 = \frac{335}{48} \approx 6.97917\text{,}$$ and $$M_4 = \frac{637}{96} \approx 6.63542\text{.}$$ 2. \begin{equation*} \frac{L_4 + M_4}{2} = \frac{646}{96} \ne \frac{637}{96} = M_4\text{.} \end{equation*} 3. $$L_n$$ is an under-estimate; $$R_n$$ is an over-estimate. ### 4.2.3When the function is sometimes negative #### Activity4.2.4. Answer. 1. $$\displaystyle M_5 = -\frac{36}{25} = -1.44$$ 2. The change in position is approximately $$-1.44$$ feet. 3. $$D \approx 2.336\text{.}$$ 4. $$-\frac{4}{3}$$ is the object’s total change in position on $$[1,5]\text{.}$$ ### 4.3The Definite Integral4.3.1The definition of the definite integral #### Activity4.3.2. Answer. 1. $$\int_0^1 3x \, dx = \frac{3}{2}\text{.}$$ 2. $$\int_{-1}^4 (2-2x) \, dx = -5\text{.}$$ 3. $$\int_{-1}^1 \sqrt{1-x^2} \, dx = \frac{\pi}{2}\text{.}$$ 4. $$\int_{-3}^4 g(x) \, dx = \frac{3\pi}{4} - \frac{3}{2}\text{.}$$ ### 4.3.2Some properties of the definite integral #### Activity4.3.3. Answer. 1. $$\int_5^2 f(x) \,dx = -2\text{.}$$ 2. $$\int_0^5 g(x) \,dx = 3\text{.}$$ 3. $$\int_0^5 (f(x) + g(x))\, dx = 2\text{.}$$ 4. $$\int_2^5 (3x^2 - 4x^3) \, dx = -492\text{.}$$ 5. $$\int_5^0 (2x^3 - 7g(x)) \, dx = -\frac{583}{2}\text{.}$$ ### 4.3.3How the definite integral is connected to a function’s average value #### Activity4.3.4. Answer. 1. $$y = v(t) = \sqrt{4-(t-2)^2}$$ is the top half of the circle $$(t-2)^2 + y^2 = 4\text{,}$$ which has radius 2 and is centered at $$(2,0)\text{.}$$ 2. $$\int_0^4 v(t) \, dt = 2\pi\text{.}$$ 3. The object moved $$2 \pi$$ meters in 4 minutes. 4. $$v_{\text{AVG} }[0,4] = \frac{\pi}{2}\text{,}$$ meters per minute,. 5. The height of the rectangle is the average value of $$v\text{,}$$ $$v_{\text{AVG} }[0,4] = \frac{\pi}{2} \approx 1.57\text{.}$$ 6. $$D = 2\pi\text{.}$$ ### 4.4The Fundamental Theorem of Calculus4.4.1The Fundamental Theorem of Calculus #### Activity4.4.2. Answer. 1. $$\int_{-1}^4 (2-2x) \, dx = -5\text{.}$$ 2. $$\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = 1\text{.}$$ 3. $$\int_0^1 e^x \, dx = e-1\text{.}$$ 4. $$\int_{-1}^{1} x^5 \, dx = 0\text{.}$$ 5. $$\int_0^2 (3x^3 - 2x^2 - e^x) \, dx = \frac{23}{3} - e^2\text{.}$$ ### 4.4.2Basic antiderivatives #### Activity4.4.3. Answer.  given function, $$f(x)$$ antiderivative, $$F(x)$$ $$k\text{,}$$ ($$k \ne 0$$) $$kx$$ $$x^n\text{,}$$ $$n \ne -1$$ $$\frac{1}{n+1}x^{n+1}$$ $$\frac{1}{x}\text{,}$$ $$x \gt 0$$ $$\ln(x)$$ $$\sin(x)$$ $$-\cos(x)$$ $$\cos(x)$$ $$\sin(x)$$ $$\sec(x) \tan(x)$$ $$\sec(x)$$ $$\csc(x) \cot(x)$$ $$-\csc(x)$$ $$\sec^2 (x)$$ $$\tan(x)$$ $$\csc^2 (x)$$ $$-\cot(x)$$ $$e^x$$ $$e^x$$ $$a^x$$ $$(a \gt 1)$$ $$\frac{1}{\ln(a)} a^x$$ $$\frac{1}{1+x^2}$$ $$\arctan(x)$$ $$\frac{1}{\sqrt{1-x^2}}$$ $$\arcsin(x)$$ 1. $$\int_0^1 \left(x^3 - x - e^x + 2\right) \,dx = \frac{11}{4} - e\text{.}$$ 2. $$\int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt = 1 - \sqrt{3} - \frac{\pi^2}{3}\text{.}$$ 3. $$\int_0^1 (\sqrt{x} - x^2) \, dx = \frac{1}{3}\text{.}$$ ### 4.4.3The total change theorem #### Activity4.4.4. Answer. 1. The person burned exactly $$\frac{400}{3}$$ calories in the first 10 minutes of the workout. 2. $$C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt$$ is the total calories burned on $$[0,40]\text{.}$$ 3. 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. One time at which the instantaneous rate at which calories are burned equals the average rate on $$[0,40]$$ is $$t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}$$ ### 5Evaluating Integrals5.1Constructing Accurate Graphs of Antiderivatives5.1.1Constructing the graph of an antiderivative #### Activity5.1.2. Answer. 1. $$F$$ is increasing on $$(0,2)$$ and $$(5,7)\text{;}$$ $$F$$ is decreasing on $$(2,5)\text{.}$$ 2. $$F$$ is concave up on $$(0,1)\text{,}$$ $$(4,6)\text{;}$$ concave down on $$(1,3)\text{,}$$ $$(6,7)\text{;}$$ neither on $$(3,4)\text{.}$$ 3. A relative maximum at $$x = 2\text{;}$$ a relative minimum at $$x = 5\text{.}$$ 4. $$F(1) = -\frac{1}{2}\text{;}$$ $$F(2) = \frac{\pi}{4} - \frac{1}{2}\text{;}$$ $$F(3) = \frac{\pi}{4} - 1\text{;}$$ $$F(4) = \frac{\pi}{4}-2\text{;}$$ $$F(5) = \frac{\pi}{4} - \frac{5}{2}\text{;}$$ $$F(6) = \frac{\pi}{2} - \frac{5}{2}\text{;}$$ $$F(7) = \frac{3\pi}{4} - \frac{5}{2}\text{;}$$ $$F(8) = \frac{3\pi}{4} - \frac{5}{2}\text{;}$$ and $$F(-1) = -1\text{.}$$ 5. Use the function values found in (d) and the earlier information regarding the shape of $$F\text{.}$$ 6. $$G(x) = F(x) + 1\text{.}$$ ### 5.1.2Multiple antiderivatives of a single function #### Activity5.1.3. Answer. 1. $$H(x) = -\cos(x) + 2\text{.}$$ ### 5.1.3Functions defined by integrals #### Activity5.1.4. Answer. 1. $$A$$ is increasing on $$(0,1.5)\text{,}$$ $$(4,6)\text{;}$$ $$A$$ is decreasing on $$(1.5,4)\text{.}$$ 2. $$A$$ is concave up on $$(0,1)$$ and $$(3,5)\text{;}$$ $$A$$ is concave down on $$(1,3)$$ and $$(5,6)\text{.}$$ 3. At $$x = 1.5\text{,}$$ $$A$$ has a relative maximum; $$A$$ has a relative minimum at $$x = 4\text{.}$$ 4. $$A(0) = -\frac{1}{2}\text{;}$$ $$A(1) = 0\text{;}$$ $$A(2) = 0\text{;}$$ $$A(3) = -2\text{;}$$ $$A(4) = -3.5\text{,}$$ $$A(5) = -2\text{,}$$ $$A(6) = -0.5\text{.}$$ 5. Use your work in (a)-(d) appropriately. 6. $$B(x) = A(x) + \frac{1}{2}\text{.}$$ ### 5.2The Second Fundamental Theorem of Calculus5.2.1The Second Fundamental Theorem of Calculus #### Activity5.2.2. Answer. 1. $$A'(x) = f(x)\text{.}$$ 2. $$A(1) = -\frac{\pi}{4}\text{.}$$ 3. $$A$$ 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 $$A(7) = -2 + \frac{\pi}{2}\text{.}$$ 4. $$F$$ and $$A$$ differ by the constant $$\frac{\pi}{4} - \frac{1}{2}\text{.}$$ 5. $$B$$ and $$C$$ have the same shape as $$A$$ and $$F\text{,}$$ and differ from $$A$$ by a constant. Observe that $$B(3) = 0$$ and $$C(1) = 0\text{.}$$ ### 5.2.2Understanding Integral Functions #### Activity5.2.3. Answer. 1. See the plot at below left. 2. $$F' = f\text{.}$$ 3. $$F$$ is increasing for all $$x \gt 0\text{;}$$ $$F$$ is decreasing for $$x \lt 0$$ 4. $$F$$ is CCU on $$-1 \lt x \lt 1$$ and CCD for $$x \lt -1$$ and $$x \gt 1\text{.}$$ 5. $$F(5) \approx 1.64038\text{;}$$ $$F(10) \approx 2.35973\text{.}$$ 6. See the graph at below right. ### 5.2.3Differentiating an Integral Function #### Activity5.2.4. Answer. 1. $$\frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}$$ 2. $$\int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \frac{x^4}{1+x^4} - \frac{16}{17}\text{.}$$ 3. $$\frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}$$ 4. $$\int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \ln(1+x^2)-\ln(10)\text{.}$$ 5. $$\frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \sin(x^6) \cdot 3x^2\text{.}$$ ### 5.3Integration by Substitution5.3.1Reversing the Chain Rule: First Steps #### Activity5.3.2. Answer. 1. $$\int \sin(8-3x) \, dx = -\frac{1}{3} (-\cos(8-3x)) + C\text{.}$$ 2. $$\int \sec^2 (4x) \, dx = \frac{1}{4} \tan(4x) + C\text{.}$$ 3. $$\int \frac{1}{11x - 9} \, dx = \frac{1}{11} \ln|11x - 9| + C\text{.}$$ 4. $$\int \csc(2x+1) \cot(2x+1) \, dx = -\frac{1}{2}\csc(2x+1) + C\text{.}$$ 5. $$\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx = \frac{1}{4} \arcsin(4x) + C$$ 6. $$\int 5^{-x}\, dx = -\frac{1}{\ln(5)}5^{-x} + C\text{.}$$ ### 5.3.2Reversing the Chain Rule: $$u$$-substitution #### Activity5.3.3. Answer. 1. $$\int \frac{x^2}{5x^3+1} \, dx = \frac{1}{15} \ln(5x^3 + 1) + C\text{.}$$ 2. $$\int e^x \sin(e^x) \, dx = -\cos(e^x) + C\text{.}$$ 3. $$\int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx = 2\sin(\sqrt{x}) + C\text{.}$$ ### 5.3.3Evaluating Definite Integrals via $$u$$-substitution #### Activity5.3.4. Answer. 1. $$\int_{x=1}^{x=2} \frac{x}{1 + 4x^2} \, dx = \frac{1}{8} (\ln(17) - \ln(5))\text{.}$$ 2. $$\int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx = -\frac{1}{20}(2e^{-1}+3)^{10} + \frac{1}{20}(2e^{0}+3)^{10}\text{.}$$ 3. $$\int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx = 1 - \frac{\sqrt{2}}{2}\text{.}$$ ### 5.4Integration by Parts5.4.1Reversing the Product Rule: Integration by Parts #### Activity5.4.2. Answer. 1. $$\int t e^{-t} dt = -te^{-t} - e^{-t}\text{.}$$ 2. $$\int 4x \sin(3x) dx = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + c \text{.}$$ 3. $$\int z \sec^2(z) dz = z \tan(z) + \ln |\cos(z)| + c \text{.}$$ 4. $$\int x\ln(x) dx = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + c \text{.}$$ ### 5.4.2Some Subtleties with Integration by Parts #### Activity5.4.3. Answer. 1. $$\int{\arctan(x) dx} = x\arctan(x) - \frac{1}{2} \ln \left( | 1 + x^2 | \right) + c \text{.}$$ 2. $$\int \ln(z) dz = z \ln(z) - z + c \text{.}$$ 3. $$\int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) \text{.}$$ 4. $$\int s^5 e^{s^3} ds = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + c \text{.}$$ 5. $$\int e^{2t} \cos\left( e^t \right) dt = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + c \text{.}$$ ### 5.4.3Using Integration by Parts Multiple Times #### Activity5.4.4. Answer. 1. $$\int x^2 \sin(x) dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + c \text{.}$$ 2. $$\int t^3 \ln(t) dt = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + c \text{.}$$ 3. $$\int e^z \sin(z) dz = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + c \text{.}$$ 4. $$\int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + c \text{.}$$ 5. $$\int t \arctan(t) dt = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t - \frac{1}{2} \arctan(t) + c \text{.}$$ ### 5.5Other Options for Finding Algebraic Antiderivatives5.5.1The Method of Partial Fractions #### Activity5.5.2. Answer. 1. $$\int \frac{1}{x^2 - 2x - 3} \, dx = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| + C\text{.}$$ 2. $$\int \frac{x^2+1}{x^3 - x^2} \, dx = -\ln|x| + x^{-1} + 2\ln|x-1| + C\text{.}$$ 3. $$\int \frac{x-2}{x^4 + x^2}\, dx = \ln|x| + 2x^{-1} - \frac{1}{2} \ln|1+x^2| + 2\arctan(x) + C\text{.}$$ ### 5.5.2Using an Integral Table #### Activity5.5.3. Answer. 1. $$\int \sqrt{x^2 + 4} \, dx = \frac{x}{2} \sqrt{x^2+4} + 2 \ln | x + \sqrt{x^2+4}| + C\text{.}$$ 2. $$\int \frac{x}{\sqrt{x^2 +4}} \, dx = \sqrt{x^2 + 4} + C\text{.}$$ 3. $$\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln| 5x + \sqrt{16+25x^2} | + C\text{.}$$ 4. $$\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = - \frac{\sqrt{49-36x^2}}{49x} + C\text{.}$$ ### 5.6Numerical Integration5.6.1The Trapezoid Rule #### Activity5.6.2. Answer. 1. $$\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}$$ 2. The table below gives values of the trapezoid rule and corresponding errors for different $$n$$-values.  $$n$$ $$T_n$$ $$E_{T,n}$$ $$4$$ $$0.50899$$ $$0.00899$$ $$8$$ $$0.50227$$ $$0.00227$$ $$16$$ $$0.50057$$ $$0.00057$$ 3. The table below gives values of the midpoint rule and corresponding errors for different $$n$$-values.  $$n$$ $$M_n$$ $$E_{M,n}$$ $$4$$ $$0.49555$$ $$-0.00445$$ $$8$$ $$0.49887$$ $$-0.00113$$ $$16$$ $$0.49972$$ $$-0.00028$$ 4. The trapezoid rule overestimates; the midpoint rule underestimates. 5. $$f(x) = \dfrac{1}{x^2}$$ is concave up on $$[1, 2]\text{.}$$ ### 5.6.3Simpson’s Rule #### Activity5.6.3. Answer. 1. Plot the data. 2. $$\int_0^{1.8} v(t) dt\text{.}$$ 3. \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. 4. $$M_3 = 143.4 \text{ ft }$$ ; overestimate. 5. $$S_6 = 140.8 \text{ ft } \text{.}$$ 6. Simpson’s rule gives the best approximation of the distance traveled, $$\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft }\text{.}$$ ### 5.6.4Overall observations regarding $$L_n\text{,}$$ $$R_n\text{,}$$ $$T_n\text{,}$$ $$M_n\text{,}$$ and $$S_{2n}\text{.}$$ #### Activity5.6.4. Answer. 1. For $$L_1$$ and $$T_1\text{:}$$ The values of $$L_1$$ and $$R_1$$ are the same for all three. 2. For the $$M_1\text{,}$$ 3. For $$T_1$$ and $$S_2\text{,}$$ 4. \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. 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 overestimate of $$\int_0^1 h(x) dx\text{.}$$ ### 6Using Definite Integrals6.1Using Definite Integrals to Find Area and Length6.1.1The Area Between Two Curves #### Activity6.1.2. Answer. 1. $$A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx = \frac{32}{3}\text{.}$$ 2. $$A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-2x^2)-(x^2-8)) \, dx \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}$$ 3. $$A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.}$$ 4. 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*} ### 6.1.2Finding Area with Horizontal Slices #### Activity6.1.3. Answer. 1. $$A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.}$$ 2. $$A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = \frac{4}{3}\text{.}$$ 3. $$\displaystyle A = \int_{y=0}^{y=1} \left(2-y - \sqrt{y} \right) \, dy = \frac{5}{6}$$ 4. $$A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \frac{9}{2}\text{.}$$ ### 6.1.3Finding the length of a curve #### Activity6.1.4. Answer. 1. $$L \approx 2.95789\text{.}$$ 2. $$L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.}$$ 3. $$L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.}$$ 4. We will usually have to estimate the value of $$\int_a^b \sqrt{1+f'(x)^2} \, dx$$ using computational technology. 5. Approximately $$(14.9165,f(14.9165)) = (14.9165, 23.2502)\text{.}$$ ### 6.2Using Definite Integrals to Find Volume6.2.1The Volume of a Solid of Revolution #### Activity6.2.2. Answer. 1. $$V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.}$$ 2. $$V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.}$$ 3. $$V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.}$$ 4. $$V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.}$$ 5. $$V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.}$$ ### 6.2.2Revolving about the $$y$$-axis #### Activity6.2.3. Answer. 1. $$V = \int_0^2 \pi y^4 \Delta \, dy\text{.}$$ 2. $$V = \int_0^2 \pi (16 - y^4) \, dy\text{.}$$ 3. $$V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.}$$ 4. $$V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.}$$ 5. $$V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.}$$ ### 6.2.3Revolving about horizontal and vertical lines other than the coordinate axes #### Activity6.2.4. Answer. 1. \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*} 2. \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*} 3. \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*} 4. \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*} ### 6.3Density, Mass, and Center of Mass6.3.1Density #### Activity6.3.2. Answer. 1. $$M = 10 - 10e^{-2} \approx 8.64665$$ grams. 1. $$V = \int_{0}^{5} \pi (4 - \frac{4}{5}x)^2 \, dx = \frac{80\pi}{3} \approx 83.7758 \mbox{m}^3\text{.}$$ 2. $$M = \frac{64000\pi}{3} \approx 67020.6433 \mbox{kg} \text{.}$$ 3. $$\displaystyle M = \int_{0}^{5} (400 + \frac{200}{1+x^2}) \cdot \pi (4-\frac{4}{5}x)^2 \, dx = 128 \pi (\frac{265}{3} + 24 \arctan(5) - 5 \ln(26)) \approx 42224.8024 \mbox{kg}$$ 2. $$b \approx 3.0652\text{.}$$ ### 6.3.2Weighted Averages #### Activity6.3.3. Answer. 1. $$\overline{x} = \frac{x_1 + x_2}{2} = 3\text{.}$$ 2. $$\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 3\text{.}$$ 3. $$\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 2.75\text{.}$$ 4. $$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{16}{7}\text{.}$$ 5. $$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{17}{7}\text{.}$$ 6. $$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 2x_4}{7} = \frac{22}{7}\text{.}$$ 7. Answers will vary. 8. 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. ### 6.3.3Center of Mass #### Activity6.3.4. Answer. 1. $$M = \int_{0}^{20} 4 + 0.1x \, dx = 100$$ g. 2. Greater than 10. 3. $$\overline{x} = \frac{\int_{0}^{20} x (4 + 0.1x)) \, dx}{\int_{0}^{20} 4 + 0.1x \, dx} = \frac{32}{3}\text{.}$$ 4. 5 g/cm. 5. Slightly to the right of the center of mass for $$\rho(x)\text{.}$$ 6. $$\overline{x} = \frac{\int_{0}^{20} x 4e^{0.020732x} \, dx}{\int_{0}^{20} 4e^{0.020732x} \, dx} \approx 10.6891\text{,}$$ ### 6.4Physics Applications: Work, Force, and Pressure6.4.1Work #### Activity6.4.2. Answer. 1. $$W = \int_0^{200} 0.3(200-h) \, dh = 6000 \text{ foot-pounds}\text{.}$$ 2. $$W = \int_0^{100} (40-0.1h) \, dh = 3500 \text{foot-pounds}\text{.}$$ 3. $$B_{\text{AVG} [0,100]} \approx 25.9798 \text{ pounds}\text{.}$$ 4. For the given spring, 1. $$k = 15\text{.}$$ 2. $$W = \int_0^1 15x \, dx = \frac{15}{2} \text{ foot-pounds}\text{.}$$ 3. $$W = \int_1^{1.5} 15x \, dx = 9.375 \text{ foot-pounds}\text{.}$$ ### 6.4.2Work: Pumping Liquid from a Tank #### Activity6.4.3. Answer. 1. \begin{equation*} W = \int_{2}^{3} 9.81 \cdot 4000\pi \cdot x \, dx = 308~190 \, \text{newton-meters}\text{.} \end{equation*} 2. \begin{equation*} W = \int_{3}^{8} 62.4 \pi (100-x^2)(x+5) \, dx \approx 673593 \, \text{foot-pounds}\text{.} \end{equation*} 3. \begin{equation*} W = \int_{1}^{3} 62.4 (50 - \frac{25}{2}x) x \, dx = 5720 \, \text{foot-pounds}\text{.} \end{equation*} ### 6.4.3Force due to Hydrostatic Pressure #### Activity6.4.4. Answer. 1. $$F = \int_{x = 0}^{x=50} (6240 x) dx = 7~800~000 \text{ pounds } \text{.}$$ 2. $$F = \int_{x=10}^{x=30} 124.8 (x - 10)\sqrt{900 - x^2} dx = 800~244 \text{ pounds } \text{.}$$ 3. $$F = \int_{x=1}^{x=4} 62.4 (x - 1)(5 - 1.25x) dx = 351 \text{ pounds } \text{.}$$ ### 6.5Improper Integrals6.5.1Improper Integrals Involving Unbounded Intervals #### Activity6.5.2. Answer. 1. $$\int_1^{10} \frac{1}{x} dx = \ln(10)$$ $$\int_1^{1000} \frac{1}{x} dx = \ln(1000)$$ $$\int_1^{100000} \frac{1}{x} dx = \ln(100000)$$ 2. $$\int_1^b \frac{1}{x} dx = \ln(b)\text{.}$$ 3. $$\displaystyle \lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \ln(b) = \infty$$ 1. $$\int_1^{10} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{10}}$$ $$\int_1^{1000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{1000}}$$ $$\int_1^{100000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{100000}}$$ 2. $$\int_1^b \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{b}}\text{.}$$ 3. $$\displaystyle \lim_{b \to \infty} \int_1^b \frac{1}{x^{3/2}} dx = \lim_{b \to \infty} \left( 2 - \frac{2}{\sqrt{b}} \right) = 2$$ 1. 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 $$y = \frac{1}{x}\text{.}$$ 2. 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. ### 6.5.2Convergence and Divergence #### Activity6.5.3. Answer. 1. $$\displaystyle \int_1^\infty \frac{1}{x^2} dx = 1$$ 2. $$\displaystyle \int_0^\infty e^{-x/4} dx = 4$$ 3. $$\displaystyle \int_2^\infty \frac{9}{(x+5)^{2/3}} dx = \infty$$ 4. $$\displaystyle \int_4^\infty \frac{3}{(x+2)^{5/4}} dx = \frac{12}{6^{1/4}}$$ 5. $$\displaystyle \int_0^\infty x e^{-x/4} dx = 16$$ 6. 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. ### 6.5.3Improper Integrals Involving Unbounded Integrands #### Activity6.5.4. Answer. 1. $$\displaystyle \int_0^1 \frac{1}{x^{1/3}}dx = \frac{3}{2}$$ 2. $$\displaystyle \int_0^2 e^{-x} dx = 1 - e^{-2}$$ 3. $$\displaystyle \int_1^4 \frac{1}{\sqrt{4-x}} dx = 2\sqrt{3}$$ 4. $$\int_{-2}^2 \frac{1}{x^2} \, dx$$ diverges. 5. $$\displaystyle \int_0^{\pi/2} \tan(x) dx = \infty$$ 6. $$\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}} dx = \frac{\pi}{2}$$ ### 7Differential Equations7.1An Introduction to Differential Equations7.1.1What is a differential equation? #### Activity7.1.2. Answer. 1. Let $$P$$ be the population $$t$$ the time in years; $$\frac{dP}{dt} = 0.0125P\text{.}$$ 2. Let $$m$$ be the mass $$t$$ the time in days; $$\frac{dm}{dt} = -0.056m\text{.}$$ 3. Let $$B$$ be the balance $$t$$ be time in years; $$\frac{dB}{dt} = 0.04B - 1000\text{.}$$ 4. Let $$t$$ be time in minutes $$H$$ the temperature of the hot chocolate; $$\frac{dH}{dt} = -0.1(H - 70)\text{.}$$ 5. 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*} ### 7.1.2Differential equations in the world around us #### Activity7.1.3. Answer. 1. For the skydiver: \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp \approx 1.5 \amp \left. \frac{dv}{dt}\right|_{(v = 1)} \amp \approx 1.2 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5 )} \amp \approx 0.9\\ \left. \frac{dv}{dt}\right|_{(v = 2)} \amp \approx 0.6 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp \approx 0.3 \end{align*} 2. For the meteorite: \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 3.5)} \amp \approx -0.3 \amp \left. \frac{dv}{dt}\right|_{(v = 4)} \amp \approx -0.6\\ \left. \frac{dv}{dt}\right|_{(v = 4.5)} \amp \approx -0.9 \amp \left. \frac{dv}{dt}\right|_{(v = 5)} \amp \approx -1.2 \end{align*} A graph of the points from parts (a) and (b) is shown in the following diagram: 3. $$\frac{dv}{dt} = -0.6v + 1.8\text{.}$$ 4. The rate of change of velocity with respect to time is a linear function of velocity. 5. $$0 \lt v \lt 3\text{.}$$ 6. $$3 \lt v \lt 5\text{.}$$ 7. $$v = 3\text{.}$$ ### 7.1.3Solving a differential equation #### Activity7.1.4. Answer. 1. $$v(t) = 1.5t - 0.25t^2$$ is not a solution to the given DE. 2. $$v(t) = 3 + 2e^{-0.5t}$$ is a solution to the given DE. 3. $$v(t) = 3$$ is a solution to the given DE. 4. $$v(t) = 3 + Ce^{-0.5t}$$ is a solution to the given DE for any choice of $$C\text{.}$$ ### 7.2Qualitative behavior of solutions to DEs7.2.1Slope fields #### Activity7.2.2. Answer. 1. When $$y \lt 4\text{,}$$ $$y$$ is an increasing function of $$t\text{.}$$ When $$y \gt 4\text{,}$$ $$y$$ is a decreasing function of $$t\text{.}$$ 2. \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, $$y(0) = 4 + 2e^0 = 6\text{.}$$ 3. A constant function. ### 7.2.2Equilibrium solutions and stability #### Activity7.2.3. Answer. 1. 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 $$t\text{.}$$ 2. $$y = 0$$ and $$y = 4\text{.}$$ 3. $$y = 4$$ is stable; $$y = 0$$ is unstable. 4. Tend to 4. 5. Figure 7.2.11 is for an ustable equilibrium; Figure 7.2.12 is for a stable equilibrium. ### 7.3Euler’s method7.3.1Euler’s Method #### Activity7.3.2. Answer. 1.  $$t_i$$ $$y_i$$ $$dy/dt$$ $$\Delta y$$ $$0$$ $$0$$ $$-1$$ $$-0.2$$ $$0.2$$ $$-0.2$$ $$-0.6$$ $$-0.12$$ $$0.4$$ $$-0.32$$ $$-0.2$$ $$-0.04$$ $$0.6$$ $$-0.36$$ $$0.2$$ $$0.04$$ $$0.8$$ $$-0.32$$ $$0.6$$ $$0.12$$ $$1.0$$ $$-0.2$$ $$1$$ $$0.2$$ 2. $$y = t^2 - t\text{,}$$ with errors $$e_1 = 0.04\text{,}$$ $$e_2 = 0.08\text{,}$$ $$e_3 = 0.12\text{,}$$ $$e_4 = 0.16\text{,}$$ $$e_5 = 0.2\text{.}$$ 3. 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 $$\int_0^1 (2t-1)~dt\text{.}$$ 4. Solutions to this differential equation all differ by only a constant. #### Activity7.3.3. Answer. 1. $$y = 0$$ or $$y = 6\text{;}$$ $$y = 0$$ is unstable, $$y = 6$$ is stable. 2. The solution will tend to $$y = 6\text{.}$$ 3.  $$t_i$$ $$y_i$$ $$dy/dt$$ $$\Delta y$$ $$0.0$$ $$1.0000$$ $$5.0000$$ $$1.0000$$ $$0.2$$ $$2.0000$$ $$8.0000$$ $$1.6000$$ $$0.4$$ $$3.6000$$ $$8.6400$$ $$1.7280$$ $$0.6$$ $$5.3280$$ $$3.5804$$ $$0.7161$$ $$0.8$$ $$6.0441$$ $$-0.2664$$ $$-0.0533$$ $$1.0$$ $$5.9908$$ $$0.0551$$ $$0.0110$$ 4. The value of $$y_i = 6$$ for every value of $$i\text{.}$$ ### 7.4Separable differential equations7.4.1Solving separable differential equations #### Activity7.4.2. Answer. 1. $$\displaystyle \frac{dP}{dt} = 0.03 P$$ 2. $$P = Ce^{0.03t}\text{.}$$ 3. $$P = 10000 e^{0.03t}\text{.}$$ 4. The doubling time is $$t = \frac{\ln(2)}{0.03} \approx 23.105$$ years. 5. The doubling time is $$t = \frac{1}{k} \ln(2)\text{.}$$ #### Activity7.4.3. Answer. 1. $$\displaystyle k = \frac{1}{30}$$ 2. $$\displaystyle T = 75 + Ce^{-t/30}$$ 3. The temperature of the coffee tends to 75 degrees. 4. $$T(20) = 75 + 30e^{-2/3} \approx 90.4^\circ$$F. 5. $$t = -30 \ln \left( \frac{1}{6} \right) \approx 53.75$$ minutes. #### Activity7.4.4. Answer. 1. $$y = -1 + C e^{\left(2t - \frac{t^2}{2} \right)}\text{.}$$ 2. $$y = \frac{1}{2} \ln \left( e^{t^2} + C \right)\text{.}$$ 3. $$y = -1 + 3 e^{2t}\text{.}$$ 4. $$y = -\frac{1}{2t + \frac{3}{2}} = -\frac{2}{4t + 3}\text{.}$$ 5. $$y = \frac{4}{t^2 + 1}\text{.}$$ ### 7.5Modeling with differential equations7.5.1Developing a differential equation #### Activity7.5.2. Answer. 1. $$\frac{dA}{dt} = 0.05A\text{.}$$ 2. $$\frac{dA}{dt} = 0.05A - 10000\text{.}$$ 3. The only equilibrium solution is $$A = 200000\text{.}$$ 4. $$t = 20 \ln(2) \approx 13.86$$years. 5. At least200000.
6. Up to \$15000 every year.

#### Activity7.5.3.

1. $$\frac{dM}{dt} = -kM\text{,}$$ where $$k$$ is a positive constant.
2. $$k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}$$
3. $$\frac{dM}{dt} = 3 - kM\text{,}$$ where $$k$$ is a positive constant.
4. The equilibrium solution $$mM = \frac{3}{k}$$ is stable.
5. $$M = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}$$
6. About 2.426 milligrams per hour.

### 7.6Population Growth and the Logistic Equation7.6.1The earth’s population

#### Activity7.6.2.

1. $$P'(0) \approx 0.0755\text{.}$$
2. $$P(0) = 6.084\text{.}$$
3. $$k \approx 0.012041\text{.}$$
4. $$P(t) = 6.084 e^{0.012041t}\text{.}$$
5. $$P(10) \approx 6.8878\text{.}$$
6. $$t = \frac{1}{0.012041} \ln \left( \frac{12}{6.084}\right) \approx 56.41\text{,}$$ or in the year 2056.
7. $$P(500) \approx 3012.3$$ billion.

### 7.6.2Solving the logistic differential equation

#### Activity7.6.3.

1. When $$P = \frac{N}{2}\text{.}$$
2. When the population is 6.125 billion.
3. $$P = \frac{12.5}{1.0546e^{-0.025t} + 1} \text{;}$$ $$P(100) = 11.504$$ billion.
4. $$t = \frac{1}{-0.025} \ln \left( \frac{\left( \frac{12.5}{9} - 1 \right)}{1.0546}\right) \approx 39.9049$$ (so in about year $$2040$$).
5. $$\lim_{t \to \infty} P(t) = N\text{.}$$