Chapter 3

> 来源: OpenStax《Calculus Volume 3》| 原页: https://openstax.org/books/calculus-volume-3/pages/chapter-3

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Chapter 3

Calculus Volume 3Chapter 3

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Chapter 3

Checkpoint

3.1

$\mathbf{\text{r}}(0) = \mathbf{\text{j}},\mspace{2mu}\mathbf{\text{r}}(1) = -2\mspace{2mu}\mathbf{\text{i}} + 5\mspace{2mu}\mathbf{\text{j}},\mspace{2mu}\mathbf{\text{r}}(-4) = 28\mspace{2mu}\mathbf{\text{i}} - 15\mspace{2mu}\mathbf{\text{j}}$

The domain of $\mathbf{\text{r}}(t) = \left( {t^{2} - 3t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {4t + 1} \right)\mspace{2mu}\mathbf{\text{j}}$ is all real numbers.

3.2 3.3

$\underset{t\rightarrow-2}{\text{lim}}\mspace{2mu}\mathbf{\text{r}}(t) = 3\mspace{2mu}\mathbf{\text{i}} - 5\mspace{2mu}\mathbf{\text{j}} - \mathbf{\text{k}}$

3.4

$\mathbf{r^{\prime}}(t) = 4t\mspace{2mu}\mathbf{\text{i}} + 5\mspace{2mu}\mathbf{\text{j}}$

3.5

$\mathbf{r^{\prime}}(t) = \left( {1 + \text{ln}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{i}} + 5e^{t}\mspace{2mu}\mathbf{\text{j}} - \left( {\text{sin}\mspace{2mu} t + \text{cos}\mspace{2mu} t} \right)\mspace{2mu}\mathbf{\text{k}}$

3.6

$\frac{d}{dt}\left\lbrack {\mspace{2mu}\mathbf{\text{r}}(t) \cdot \mspace{2mu}\mathbf{r^{\prime}}(t)} \right\rbrack = 8e^{4t}$

$\begin{array}{l}

{\frac{d}{dt}\left\lbrack {\mspace{2mu}\mathbf{\text{u}}(t)\ \times \ \mathbf{\text{r}}(t)} \right\rbrack} \\

{= - \left( {e^{2t}\left( {\text{cos}\mspace{2mu} t + 2\mspace{2mu}\text{sin}\mspace{2mu} t} \right) + \text{cos}\mspace{2mu} 2t} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {e^{2t}\left( {2t + 1} \right) - \text{sin}\mspace{2mu} 2t} \right)\mspace{2mu}\mathbf{\text{j}} + \left( {t\mspace{2mu}\text{cos}\mspace{2mu} t + \text{sin}\mspace{2mu} t - \text{cos}\mspace{2mu} 2t} \right)\mspace{2mu}\mathbf{\text{k}}}

\end{array}$

3.7

$\mathbf{\text{T}}(t) = \frac{2t}{\sqrt{4t^{2} + 5}}\mspace{2mu}\mathbf{\text{i}} + \frac{2}{\sqrt{4t^{2} + 5}}\mspace{2mu}\mathbf{\text{j}} + \frac{1}{\sqrt{4t^{2} + 5}}\mspace{2mu}\mathbf{\text{k}}$

3.8

${\int_{1}^{3}{\left\lbrack {\left( {2t + 4} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {3t^{2} - 4t} \right)\mspace{2mu}\mathbf{\text{j}}} \right\rbrack dt}} = 16\mspace{2mu}\mathbf{\text{i}} + 10\mspace{2mu}\mathbf{\text{j}}$

3.9

$\mathbf{r^{\prime}}(t) = \left\langle {4t,4t,3t^{2}} \right\rangle,$ so $s = \frac{1}{27}\left( {113^{3\text{/}2} - 32^{3\text{/}2}} \right) \approx 37.785$

3.10

$s = 5t,$ or $t = {s\text{/}5}.$ Substituting this into $\mathbf{\text{r}}(t) = \left\langle {3\mspace{2mu}\text{cos}\mspace{2mu} t,3\mspace{2mu}\text{sin}\mspace{2mu} t,4t} \right\rangle$ gives

$\mathbf{\text{r}}(s) = \left\langle {3\mspace{2mu}\text{cos}\left( \frac{s}{5} \right),\ 3\mspace{2mu}\text{sin}\left( \frac{s}{5} \right),\frac{4s}{5}} \right\rangle,s \geq 0.$

3.11

$\kappa = \frac{6}{101^{3\text{/}2}} \approx 0.0059$

3.12

$\mathbf{\text{N}}(2) = \frac{\sqrt{2}}{2}\left( {\mathbf{\text{i}} - \mathbf{\text{j}}} \right)$

3.13

$\kappa = \frac{4}{\left\lbrack {1 + \left( {4x - 4} \right)^{2}} \right\rbrack^{3\text{/}2}}$

At the point $x = 1,$ the curvature is equal to 4. Therefore, the radius of the osculating circle is $\frac{1}{4}.$

A graph of this function appears next:

The vertex of this parabola is located at the point $\left( {1,3} \right).$ Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are $\left( {1,\frac{13}{4}} \right).$ The equation of the osculating circle is

$\left( {x - 1} \right)^{2} + \left( {y - \frac{13}{4}} \right)^{2} = \frac{1}{16}.$

3.14

$\begin{array}{l}

{\mathbf{\text{v}}(t) = \mathbf{r^{\prime}}(t) = \left( {2t - 3} \right)\mspace{2mu}\mathbf{\text{i}} + 2\mathbf{\text{j}} + \mathbf{\text{k}}} \\

{\mathbf{\text{a}}(t) = \mathbf{v^{\prime}}(t) = 2\mathbf{\text{i}}} \\

{v(t) = \left\| {\mathbf{r^{\prime}}(t)} \right\| = \sqrt{\left( {2t - 3} \right)^{2} + 2^{2} + 1^{2}} = \sqrt{4t^{2} - 12t + 14}}

\end{array}$

The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.

3.15

1. $\begin{matrix}

{\textbf{v}(t)} & = & {\mathbf{r}^{'}(t) = 4\textbf{i} + 2t\mspace{2mu}\textbf{j}} \\

{\textbf{a}(t)} & = & {\mathbf{v}^{'}(t) = 2\textbf{j}} \\

a_{\mathbf{\text{T}}} & = & {\frac{2t}{\sqrt{t^{2} + 4}},a_{\mathbf{\text{N}}} = \frac{4}{\sqrt{4 + t^{2}}}}

\end{matrix}$

2. $a_{\mathbf{\text{T}}}(-3) = - \frac{6\sqrt{13}}{13},a_{\mathbf{\text{N}}}(-3) = \frac{2\sqrt{13}}{13}$

3.16

967.15 m

3.17

$a = 1.224\ \times \ 10^{9}\text{m} \approx 1,224,000\ \text{km}$

Section 3.1 Exercises

1.

$f(t) = 3\mspace{2mu}\text{sec}\mspace{2mu} t,g(t) = 2\mspace{2mu}\text{tan}\mspace{2mu} t$

3.

5.

a\. $\left\langle {\frac{1}{2},\frac{\sqrt{3}}{2}} \right\rangle,$ b. $\left\langle {\frac{1}{2},\frac{\sqrt{3}}{2}} \right\rangle,$ c. Yes, the limit as *t* approaches $\pi\text{/}3$ is equal to $\mathbf{\text{r}}\left( {\pi\text{/}3} \right),$ d.

7.

a\. $\left\langle {e^{\pi\text{/}4},\frac{\sqrt{2}}{2},\text{ln}\left( \frac{\pi}{4} \right)} \right\rangle;$ b. $\left\langle {e^{\pi\text{/}4},\frac{\sqrt{2}}{2},\text{ln}\left( \frac{\pi}{4} \right)} \right\rangle;$ c. Yes

9.

$\left\langle {e^{\pi\text{/}2},1,\text{ln}\left( \frac{\pi}{2} \right)} \right\rangle$

11.

$2e^{2}\mspace{2mu}\mathbf{\text{i}} + \frac{2}{e^{4}}\mspace{2mu}\mathbf{\text{j}} + 2\mspace{2mu}\mathbf{\text{k}}$

13.

The limit does not exist because the limit of $\text{ln}(t - 1)$ as *t* approaches infinity does not exist.

15.

$t > 0,t \neq (2k + 1)\frac{\pi}{2},$ where k is an integer

17.

$t > 3,t \neq n\pi,$ where *n* is an integer

19.

21.

All *t* such that $t \in \left( {1,\infty} \right)$

23.

$y = 2\sqrt[3]{x},$ a variation of the cube-root function

25.

$x^{2} + y^{2} = 9,$ a circle centered at $\left( {0,0} \right)$ with radius 3, and a counterclockwise orientation

27.

29.

Find a vector-valued function that traces out the given curve in the indicated direction.

31.

For left to right, $y = x^{2},$ where t increases

33.

$(50,0,0)$

35.

37.

39.

One possibility is $r(t) = \text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} + \text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + \text{sin}(4t)\mspace{2mu}\mathbf{\text{k}}.$ By increasing the coefficient of *t* in the third component, the number of turning points will increase.

Section 3.2 Exercises

41.

$\left\langle {3t^{2},6t,\frac{1}{2}t^{2}} \right\rangle$

43.

$\left\langle {\text{−}e^{\text{−}t},3\mspace{2mu}\text{cos}(3t),\frac{5}{\sqrt{t}}} \right\rangle$

45.

$\left\langle {0,0,0} \right\rangle$

47.

$\left\langle {\frac{-1}{\left( {t + 1} \right)^{2}},\frac{1}{1 + t^{2}},\frac{3}{t}} \right\rangle$

49.

$\left\langle {0,12\mspace{2mu}\text{cos}(3t),\text{cos}\mspace{2mu} t - t\mspace{2mu}\text{sin}\mspace{2mu} t} \right\rangle$

51.

$\left\langle 1,-1,0 \right\rangle$

53.

$\left\langle \left. 6, - \frac{3}{4},32 \right\rangle \right.$

55.

$\frac{1}{\sqrt{9\mspace{2mu}\text{sin}^{2}(3t) + 144\mspace{2mu}\text{cos}^{2}(4t)}}\left\langle {0,-3\mspace{2mu}\text{sin}(3t),12\mspace{2mu}\text{cos}(4t)} \right\rangle$

57.

$\mathbf{\text{T}}(t) = \frac{-12}{13}\text{sin}(4t)\mspace{2mu}\mathbf{\text{i}} + \frac{12}{13}\text{cos}(4t)\mspace{2mu}\mathbf{\text{j}} + \frac{5}{13}\mspace{2mu}\mathbf{\text{k}}$

59.

$\left\langle {2t,4t^{3},-8t^{7}} \right\rangle$

61.

$\text{sin}(t) + 2te^{t} - 4t^{3}\text{cos}(t) + t\mspace{2mu}\text{cos}(t) + t^{2}e^{t} + t^{4}\text{sin}(t)$

63.

$900t^{7} + 16t$

65.

1.

2. Undefined or infinite

67.

$\mathbf{\text{r}}'(t) = - b\omega\mspace{2mu}\text{sin}(\omega t)\mspace{2mu}\mathbf{\text{i}} + b\omega\mspace{2mu}\text{cos}(\omega t)\mspace{2mu}\mathbf{\text{j}}.$ To show orthogonality, note that $\mathbf{\text{r}}'(t) \cdot \mspace{2mu}\mathbf{\text{r}}(t) = 0.$

69.

$0\mspace{2mu}\mathbf{\text{i}} + 2\mspace{2mu}\mathbf{\text{j}} + 4t\mspace{2mu}\mathbf{\text{k}}$

71.

$\frac{1}{3}\left( {10^{3\text{/}2} - 1} \right)$

73.

$\begin{array}{rll}

{\text{‖}{\mspace{2mu}\mathbf{\text{v}}(t)}\text{‖}} & = & k \\

{\mathbf{\text{v}}(t) \cdot \mspace{2mu}\mathbf{\text{v}}(t)} & = & k \\

{\frac{d}{dt}\left( {\mspace{2mu}\mathbf{\text{v}}(t) \cdot \mspace{2mu}\mathbf{\text{v}}(t)} \right)} & = & {\frac{d}{dt}k = 0} \\

{\mathbf{\text{v}}(t) \cdot \mspace{2mu}\mathbf{\text{v}}'(t) + \mspace{2mu}\mathbf{\text{v}}'(t) \cdot \mspace{2mu}\mathbf{\text{v}}(t)} & = & 0 \\

{2\mspace{2mu}\mathbf{\text{v}}(t) \cdot \mspace{2mu}\mathbf{\text{v}}'(t)} & = & 0 \\

{\mathbf{\text{v}}(t) \cdot \mspace{2mu}\mathbf{\text{v}}'(t)} & = & {0.}

\end{array}$

The last statement implies that the velocity and acceleration are perpendicular or orthogonal.

75.

$\mathbf{\text{v}}(t) = \left\langle {1 - \text{sin}\mspace{2mu} t,1 - \text{cos}\mspace{2mu} t} \right\rangle,$ ${\text{speed}\ = \text{‖}{\mspace{2mu}\mathbf{\text{v}}(t)}\text{‖}} = \sqrt{3 - 2(\text{sin}\mspace{2mu} t + \text{cos}\mspace{2mu} t)}$

77.

$x = t + 1,y = - t + 1,z = 0$

79.

$\mathbf{\text{r}}(t) = \left\langle {18,9} \right\rangle$ at $t = 3$

81.

$\sqrt{161}$

83.

$\mathbf{\text{v}}(t) = \left\langle {\text{−}\text{sin}\mspace{2mu} t,\text{cos}\mspace{2mu} t,1} \right\rangle$

85.

$\mathbf{\text{a}}(t) = - \text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{i}} - \text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} + 0\mspace{2mu}\mathbf{\text{j}}$

87.

$\mathbf{\text{v}}(t) = \left\langle {\text{−}\text{sin}\mspace{2mu} t,2\mspace{2mu}\text{cos}\mspace{2mu} t,0} \right\rangle$

89.

$\mathbf{\text{a}}(t) = \left\langle {- \frac{\sqrt{2}}{2}, - \sqrt{2},0} \right\rangle$

91.

${\text{‖}{\mspace{2mu}\mathbf{\text{v}}(t)}\text{‖}} = \sqrt{\text{sec}^{4}t + \text{sec}^{2}t\mspace{2mu}\text{tan}^{2}t} = \sqrt{\text{sec}^{2}t(\text{sec}^{2}t + \text{tan}^{2}t)}$

93.

$\sqrt{3 - 2\sqrt{2}}$

95.

$\left\langle {0,2\mspace{2mu}\text{sin}\mspace{2mu} t\left( {t - \frac{1}{t}} \right) - 2\mspace{2mu}\text{cos}\mspace{2mu} t\left( {1 + \frac{1}{t^{2}}} \right),2\mspace{2mu}\text{sin}\mspace{2mu} t\left( {1 + \frac{1}{t^{2}}} \right) + 2\mspace{2mu}\text{cos}\mspace{2mu} t\left( {t - \frac{2}{t}} \right)} \right\rangle$

97.

$\mathbf{\text{T}}(t) = \left\langle {\frac{t^{2}}{\sqrt{t^{4} + 1}},\frac{-1}{\sqrt{t^{4} + 1}}} \right\rangle$

99.

$\mathbf{\text{T}}(t) = \frac{1}{3}\left\langle {1,2,2} \right\rangle$

101.

$\frac{3}{4}\mspace{2mu}\textbf{i} + ln(2)\mspace{2mu}\textbf{j} + \left( 1 - \frac{1}{e} \right)\mspace{2mu}\textbf{k}$

Section 3.3 Exercises

103.

$8\sqrt{5}$

105.

$\frac{1}{54}\left( {37^{3\text{/}2} - 1} \right)$

107.

Length $= 2\pi$

109.

$6\pi$

111.

$e - \frac{1}{e}$

113.

$\mathbf{\text{T}}(0) = \mathbf{\text{j}},$ $\mathbf{\text{N}}(0) = - \mathbf{\text{i}}$

115.

$\mathbf{\text{T}}(t) = \left\langle {\frac{2}{\sqrt{6}},\frac{\text{cos}\mspace{2mu} t - \text{sin}\mspace{2mu} t}{\sqrt{6}},\frac{\text{cos}\mspace{2mu} t + \text{sin}\mspace{2mu} t}{\sqrt{6}}} \right\rangle$

117.

$\textbf{N}(0) = \left\langle 0, - \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right\rangle$

119.

$\mathbf{\text{T}}(t) = \frac{1}{\sqrt{4t^{2} + 2}} < 1,2t,1 >$

121.

$\mathbf{\text{T}}(t) = \frac{1}{\sqrt{100t^{2} + 13}}\left( {3\mathbf{\text{i}} + 10t\mathbf{\text{j}} + 2\mathbf{\text{k}}} \right)$

123.

$\mathbf{\text{T}}(t) = \frac{1}{\sqrt{9t^{4} + 76t^{2} + 16}}\left( {\left\lbrack {3t^{2} - 4} \right\rbrack\mathbf{\text{i}} + 10t\mathbf{\text{j}}} \right)$

125.

$\mathbf{\text{N}}(t) = \left\langle {\text{−}\text{sin}\mspace{2mu} t,0, - \text{cos}\mspace{2mu} t} \right\rangle$

127.

Arc-length function: $s(t) = 5t;$ r as a parameter of *s*: $\mathbf{\text{r}}(s) = \left( {3 - \frac{3s}{5}} \right)\mspace{2mu}\mathbf{\text{i}} + \frac{4s}{5}\mathbf{\text{j}}$

129.

$\mathbf{\text{r}}(s) = \left( {1 + \frac{s}{\sqrt{2}}} \right)\ \text{sin}\left( {\text{ln}(1 + \frac{s}{\sqrt{2}})} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {1 + \frac{s}{\sqrt{2}}} \right)\ \text{cos}\left\lbrack {\text{ln}\left( {1 + \frac{s}{\sqrt{2}}} \right)} \right\rbrack\mathbf{\text{j}}$

131.

The maximum value of the curvature occurs at $x = 1.$

133.

$\frac{1}{2}$

135.

$\kappa \approx \frac{49.477}{\left( {17 + 144t^{2}} \right)^{3\text{/}2}}$

137.

$\frac{1}{2\sqrt{2}}$

139.

The curvature approaches zero.

141.

$y = 6x + \pi$ and $x + 6y = 6\pi$

143.

$x + 2z = \frac{\pi}{2}$

145.

$\frac{a^{4}b^{4}}{\left( {b^{4}x^{2} + a^{4}y^{2}} \right)^{3\text{/}2}}$

147.

$\frac{10\sqrt{10}}{3}$

149.

$\frac{1}{4}\ln\left( {\sqrt{17} + 4} \right) + \sqrt{17} \approx 4.647$

151.

The curvature is decreasing over this interval.

153.

$\kappa = \frac{6}{x^{2\text{/}5}\left( {25 + 4x^{6\text{/}5}} \right)}$

Section 3.4 Exercises

155.

$\mathbf{\text{v}}(t) = (6t)\mathbf{\text{i}} + (2 - \text{cos}(t))\mathbf{\text{j}}$

157.

$\mathbf{\text{v}}(t) = \left\langle {-3\mspace{2mu}\text{sin}\mspace{2mu} t,3\mspace{2mu}\text{cos}\mspace{2mu} t,2t} \right\rangle,$ $\mathbf{\text{a}}(t) = \left\langle {-3\mspace{2mu}\text{cos}\mspace{2mu} t,-3\mspace{2mu}\text{sin}\mspace{2mu} t,2} \right\rangle,$ $\text{speed} = \sqrt{9 + 4t^{2}}$

159.

$\mathbf{v}(t) = -2\sin\ t\mspace{2mu}\mathbf{j} + 3\mspace{2mu}\cos\ t\ \mathbf{k},$ $\mathbf{\text{a}}(t) = -2\mspace{2mu}\text{cos}\mspace{2mu} t\mspace{2mu}\mathbf{\text{j}} - 3\mspace{2mu}\text{sin}\mspace{2mu} t\mspace{2mu}\mathbf{\text{k}},$ $\text{speed} = \sqrt{4\mspace{2mu}\text{sin}^{2}t + 9\text{cos}^{2}t}$

161.

$\mathbf{\text{v}}(t) = e^{t}\mathbf{\text{i}} - e^{\text{−}t}\mathbf{\text{j}},$ $\mathbf{\text{a}}(t) = e^{t}\mathbf{\text{i}} + e^{\text{−}t}\mathbf{\text{j}},$ speed = $\left\| {\mathbf{\text{v}}(t)} \right\| = \sqrt{e^{2t} + e^{-2t}}$

163.

$t = 4$

165.

$\mathbf{\text{v}}(t) = \left( {\omega - \omega\mspace{2mu}\text{cos}(\omega t)} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {\omega\mspace{2mu}\text{sin}(\omega t)} \right)\mspace{2mu}\mathbf{\text{j}},$

$\mathbf{\text{a}}(t) = \left( {\omega^{2}\text{sin}(\omega t)} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {\omega^{2}\text{cos}(\omega t)} \right)\mspace{2mu}\mathbf{\text{j}},$

$\text{speed} = \sqrt{\omega^{2} - 2\omega^{2}\text{cos}(\omega t) + \omega^{2}\text{cos}^{2}(\omega t) + \omega^{2}\text{sin}^{2}(\omega t)}\ \text{=}\ \sqrt{2\omega^{2}(1 - \text{cos}(\omega t))}$

167.

$\left\| {\mathbf{\text{v}}(t)} \right\| = \sqrt{9 + 4t^{2}}$

169.

$\mathbf{\text{v}}(t) = \left\langle {e^{-5t}(\text{cos}\mspace{2mu} t - 5\mspace{2mu}\text{sin}\mspace{2mu} t),\text{−}e^{-5t}(\text{sin}\mspace{2mu} t + 5\mspace{2mu}\text{cos}\mspace{2mu} t),-20e^{-5t}} \right\rangle$

171.

$\mathbf{\text{a}}(t) = \left\langle e \right.^{-5t}\left( {\text{−}\text{sin}\mspace{2mu} t - 5\mspace{2mu}\text{cos}\mspace{2mu} t} \right) - 5e^{-5t}\left( {\text{cos}\mspace{2mu} t - 5\mspace{2mu}\text{sin}\mspace{2mu} t} \right),$ $\text{−}e^{-5t}\left( {\text{cos}\mspace{2mu} t - 5\mspace{2mu}\text{sin}\mspace{2mu} t} \right) + 5e^{-5t}\left( {\text{sin}\mspace{2mu} t + 5\mspace{2mu}\text{cos}\mspace{2mu} t} \right),100\left. e^{-5t} \right\rangle$

173.

44.185 sec

175.

$t = 88.37$ sec

177.

88.37 sec

179.

The range is approximately 886.29 m.

181.

$\mathbf{\text{v}} = 42.16$ m/sec

183.

$\mathbf{\text{r}}(t) = 0\mathbf{\text{i}} + \left( {\frac{1}{6}t^{3} + 4.5t - \frac{14}{3}} \right)\mspace{2mu}\mathbf{\text{j}} + \left( {\frac{t^{3}}{6} - \frac{1}{2}t + \frac{1}{3}} \right)\mspace{2mu}\mathbf{\text{k}}$

185.

$a_{T} = 0,$ $a_{N} = a\omega^{2}$

187.

$a_{T} = \sqrt{3}e^{t},$ $a_{N} = \sqrt{2}e^{t}$

189.

$a_{T} = 2t,$ $a_{N} = 2$

191.

$a_{T}\frac{6t + 12t^{3}}{\sqrt{1 + t^{4} + t^{2}}},$ $a_{N} = 6\sqrt{\frac{1 + 4t^{2} + t^{4}}{1 + t^{2} + t^{4}}}$

193.

$a_{T} = 0,$ $a_{N} = 12\pi^{2}$

195.

$\mathbf{\text{r}}(t) = \left( {\frac{-1}{m}\text{cos}\mspace{2mu} t + c + \frac{1}{m}} \right)\mspace{2mu}\mathbf{\text{i}} + \left( {\frac{\text{−}\text{sin}\mspace{2mu} t}{m} + \left( {v_{0} + \frac{1}{m}} \right)t} \right)\mspace{2mu}\mathbf{\text{j}}$

197.

10.94 km/sec

201.

$a_{T} = 0.43\ \text{m/sec}^{2},$

$a_{N} = 2.46\ \text{m/sec}^{2}$

Review Exercises

203.

False, $\frac{d}{dt}\left\lbrack {\mathbf{\text{u}}(t)\ \times \ \mathbf{\text{u}}(t)} \right\rbrack = 0$

205.

False, it is $\left| {\mathbf{\text{r}^{\prime}}(t)} \right|$

207.

$t < 4,$ $t \neq \frac{n\pi}{2}$

209.

211.

$\mathbf{\text{r}}(t) = \left\langle {t,2 - \frac{t^{2}}{8},-2 - \frac{t^{2}}{8}} \right\rangle$

213.

$\mathbf{\text{u}^{\prime}}(t) = \left\langle {2t,2,20t^{4}} \right\rangle,$ $\mathbf{\text{u}\text{''}}(t) = \left\langle {2,0,80t^{3}} \right\rangle,$ $\frac{d}{dt}\left\lbrack {\mathbf{\text{u}^{\prime}}(t)\ \times \ \mathbf{\text{u}}(t)} \right\rbrack = \left\langle {-480t^{3} - 160t^{4},24 + 75t^{2},12 + 4t} \right\rangle,$ $\frac{d}{dt}\left\lbrack {\mathbf{\text{u}}(t)\ \times \ \mathbf{\text{u}^{\prime}}(t)} \right\rbrack = \left\langle {480t^{3} + 160t^{4},-24 - 75t^{2},-12 - 4t} \right\rangle,$ $\frac{d}{dt}\left\lbrack {\mathbf{\text{u}}(t) \cdot \mathbf{\text{u}^{\prime}}(t)} \right\rbrack = 720t^{8} - 9600t^{3} + 6t^{2} + 4,$ unit tangent vector: $\mathbf{\text{T}}(t) = \frac{2t}{\sqrt{400t^{8} + 4t^{2} + 4}}\mspace{2mu}\mathbf{\text{i}} + \frac{2}{\sqrt{400t^{8} + 4t^{2} + 4}}\mspace{2mu}\mathbf{\text{j}} + \frac{20t^{4}}{\sqrt{400t^{8} + 4t^{2} + 4}}\mspace{2mu}\mathbf{\text{k}}$

215.

$\frac{\text{ln}{(4)}^{2}}{2}\mathbf{\text{i}} + 2\mathbf{\text{j}} + \frac{2\left( {2 + \sqrt{2}} \right)}{\pi}\mathbf{\text{k}}$

217.

$\frac{\sqrt{37}}{2} + \frac{1}{12}\text{sinh}^{-1}(6)$

219.

$\mathbf{\text{r}}(t(s)) = \text{cos}\left( \frac{2s}{\sqrt{68}} \right)\mathbf{\text{i}} + \frac{8s}{\sqrt{68}}\mathbf{\text{j}} - \text{sin}\left( \frac{2s}{\sqrt{68}} \right)\mathbf{\text{k}}$

221.

$\frac{e^{2t}}{\left( {e^{2t} + 1} \right)^{2}}$

223.

$a_{T} = \frac{e^{2t}}{\sqrt{1 + e^{2t}}},$ $a_{N} = \frac{\sqrt{2e^{2t} + 1}}{\sqrt{1 + e^{2t}}}$

225.

$\mathbf{\text{v}}(t) = \left\langle {2t,\frac{1}{t},\pi\text{cos}\left( {\pi t} \right)} \right\rangle$ m/sec, $\mathbf{\text{a}}(t) = \left\langle {2, - \frac{1}{t^{2}},\text{−}\pi^{2}\text{sin}\left( {\pi t} \right)} \right\rangle{\ \text{m/sec}}^{2},$ $\text{speed} = \sqrt{4t^{2} + \frac{1}{t^{2}} + {\pi^{2}\text{cos}}^{2}\left( {\pi t} \right)}$ m/sec; at $t = 1,$ $\mathbf{\text{r}}(1) = \left\langle {1,0,0} \right\rangle$ m, $\mathbf{\text{v}}(1) = \left\langle {2,1,{–\pi}} \right\rangle$ m/sec, $\mathbf{\text{a}}(1) = \left\langle {2,-1,0} \right\rangle$ m/sec², and $\text{speed} = \sqrt{5 + \pi^{2}}$ m/sec

227.

$\mathbf{\text{r}}(t) = \textbf{v}_{0}t - \frac{g}{2}t^{2}\mspace{2mu}\mathbf{\text{j}},$ $\mathbf{\text{r}}(t) = \left\langle {\textbf{v}_{0}(\text{cos}\mspace{2mu}\theta)t,\textbf{v}_{0}(\text{sin}\mspace{2mu}\theta)t, - \frac{g}{2}t^{2}} \right\rangle$

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• Location: Houston, Texas
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• Section URL: https://openstax.org/books/calculus-volume-3/pages/chapter-3

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