Chapter 2

parametric equations

displacement

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$\parallel \overrightarrow{r}\left(t\right)\parallel =\sqrt{25{\cos }^{2}t+9{\sin }^{2}t}$.

$\parallel \overrightarrow{r}\left(t\right)\parallel =\sqrt{{\cos }^{2}t+t^2+t^4}$.

Answers may vary; three solutions are

$\overrightarrow{r}\left(t\right)=⟨3\sin t+5,3\cos t+5⟩$,

$\overrightarrow{r}\left(t\right)=⟨-3\cos t+5,3\sin t+5⟩$ and

$\overrightarrow{r}\left(t\right)=⟨3\cos t+5,-3\sin t+5⟩$.

Answers may vary, though most direct solutions are

$\overrightarrow{r}\left(t\right)=⟨-3\cos t+3,2\sin t-2⟩$,

$\overrightarrow{r}\left(t\right)=⟨3\cos t+3,-2\sin t-2⟩$ and

$\overrightarrow{r}\left(t\right)=⟨3\sin t+3,2\cos t-2⟩$.

Answers may vary, though most direct solutions are

$\overrightarrow{r}\left(t\right)=⟨t,-1/2\left(t-1\right)+5⟩$,

$\overrightarrow{r}\left(t\right)=⟨t+1,-1/2t+5⟩$,

$\overrightarrow{r}\left(t\right)=⟨-2t+1,t+5⟩$ and

$\overrightarrow{r}\left(t\right)=⟨2t+1,-t+5⟩$.

Answers may vary, though most direct solution is

$\overrightarrow{r}\left(t\right)=⟨3\cos \left(4\pi t\right),3\sin \left(4\pi t\right),3t⟩$.

Answers may vary, though most direct solution is $\overrightarrow{r}\left(t\right)=⟨\sqrt{5}\cos \left(2\pi t\right),2,\sqrt{5}\sin \left(2\pi t\right)⟩$.

$⟨1,1⟩$

$⟨1,2,7⟩$

component

It is difficult to identify the points on the graphs of $\overrightarrow{r}\left(t\right)$ and ${\overrightarrow{r}}^{\prime }\left(t\right)$ that correspond to each other.

$⟨e^3,0⟩$

$⟨2t,1,0⟩$

$(0,\mathrm{\infty })$

${\overrightarrow{r}}^{\prime }\left(t\right)=⟨-1/t^2,5/{\left(3t+1\right)}^2,{\sec }^{2}t⟩$

$r^{\prime }\left(t\right)=⟨2t,1⟩\cdot ⟨\sin t,2t+5⟩+⟨t^2+1,t-1⟩\cdot ⟨\cos t,2⟩=$

$\left(t^2+1\right)\cos t+2t\sin t+4t+3$

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${\overrightarrow{r}}^{\prime }\left(t\right)=⟨2t+1,2t-1⟩$

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${\overrightarrow{r}}^{\prime }\left(t\right)=⟨2t,3t^2-1⟩$

$\mathrm{\ell }\left(t\right)=⟨2,0⟩+t⟨3,1⟩$

$\mathrm{\ell }\left(t\right)=⟨-3,0,\pi ⟩+t⟨0,-3,1⟩$

$t=2n\pi$, where $n$ is an integer;

so $t=\mathrm{\dots }-4\pi ,-2\pi ,0,2\pi ,4\pi ,\mathrm{\dots }$

$\overrightarrow{r}\left(t\right)$ is not smooth at $t=3\pi /4+n\pi$, where $n$ is an integer

Both derivatives return $⟨5t^4,4t^3-3t^2,3t^2⟩$.

Both derivatives return

$⟨2t-e^t-1,\cos t-3t^2,\left(t^2+2t\right)e^t-\left(t-1\right)\cos t-\sin t⟩$.

$⟨{\tan }^{-1}t,\tan t⟩+\overrightarrow{C}$

$⟨4,-4⟩$

$\overrightarrow{r}\left(t\right)=⟨\ln \left|t+1\right|+1,-\ln \left|\cos t\right|+2⟩$

$\overrightarrow{r}\left(t\right)=⟨-\cos t+1,t-\sin t,e^t-t-1⟩$

$10\pi$

$\sqrt{2}\left(1-e^{-1}\right)$

$2\pi \sqrt{2+16{\pi }^2}+\ln \left(2\sqrt{2}\pi +\sqrt{1+8{\pi }^2}\right)$

$\sqrt{h^2+4{\pi }^2{R}^2{N}^2}$

Velocity is a vector, indicating an objects direction of travel and its rate of distance change (i.e., its speed). Speed is a scalar.

The average velocity is found by dividing the displacement by the time traveled – it is a vector. The average speed is found by dividing the distance traveled by the time traveled – it is a scalar.

One example is traveling at a constant speed $s$ in a circle, ending at the starting position. Since the displacement is $\overrightarrow{0}$, the average velocity is $\overrightarrow{0}$, hence $\parallel \overrightarrow{0}\parallel =0$. But traveling at constant speed $s$ means the average speed is also $s>0$.

$\overrightarrow{v}\left(t\right)=⟨2,5,0⟩$, $\overrightarrow{a}\left(t\right)=⟨0,0,0⟩$

$\overrightarrow{v}\left(t\right)=⟨-\sin t,\cos t⟩$, $\overrightarrow{a}\left(t\right)=⟨-\cos t,-\sin t⟩$

$\overrightarrow{v}\left(t\right)=⟨1,\cos t⟩$, $\overrightarrow{a}\left(t\right)=⟨0,-\sin t⟩$

$\overrightarrow{v}\left(t\right)=⟨2t+1,-2t+2⟩$, $\overrightarrow{a}\left(t\right)=⟨2,-2⟩$

$\parallel \overrightarrow{v}\left(t\right)\parallel =\sqrt{4t^2+1}$.

Min at $t=0$; Max at $t=\pm 1$.

$\parallel \overrightarrow{v}\left(t\right)\parallel =5$.

Speed is constant, so there is no difference between min/max

$\parallel \overrightarrow{v}\left(t\right)\parallel =\left|\sec t\right|\sqrt{{\tan }^{2}t+{\sec }^{2}t}$.

min: $t=0$; max: $t=\pi /4$

$\parallel \overrightarrow{v}\left(t\right)\parallel =13$.

speed is constant, so there is no difference between min/max

$\parallel \overrightarrow{v}\left(t\right)\parallel =\sqrt{4t^2+1+t^2/\left(1-t^2\right)}$.

min: $t=0$; max: there is no max; speed approaches $\mathrm{\infty }$ as $t\to \pm 1$

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1. (a)

   ${\overrightarrow{r}}_1\left(1\right)=⟨1,1⟩$; ${\overrightarrow{r}}_2\left(1\right)=⟨1,1⟩$

2. (b)

   ${\overrightarrow{v}}_1\left(1\right)=⟨1,2⟩$; $\parallel {\overrightarrow{v}}_1\left(1\right)\parallel =\sqrt{5}$; ${\overrightarrow{a}}_1\left(1\right)=⟨0,2⟩$

   ${\overrightarrow{v}}_2\left(1\right)=⟨2,4⟩$; $\parallel {\overrightarrow{v}}_2\left(1\right)\parallel =2\sqrt{5}$; ${\overrightarrow{a}}_2\left(1\right)=⟨2,12⟩$

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1. (a)

   ${\overrightarrow{r}}_1\left(2\right)=⟨6,4⟩$; ${\overrightarrow{r}}_2\left(2\right)=⟨6,4⟩$

2. (b)

   ${\overrightarrow{v}}_1\left(2\right)=⟨3,2⟩$; $\parallel {\overrightarrow{v}}_1\left(2\right)\parallel =\sqrt{13}$; ${\overrightarrow{a}}_1\left(2\right)=⟨0,0⟩$

   ${\overrightarrow{v}}_2\left(2\right)=⟨6,4⟩$; $\parallel {\overrightarrow{v}}_2\left(2\right)\parallel =2\sqrt{13}$; ${\overrightarrow{a}}_2\left(2\right)=⟨0,0⟩$

$\overrightarrow{v}\left(t\right)=⟨2t+1,3t+2⟩$,

$\overrightarrow{r}\left(t\right)=⟨t^2+t+5,3t^2/2+2t-2⟩$

$\overrightarrow{v}\left(t\right)=⟨\sin t,\cos t⟩$, $\overrightarrow{r}\left(t\right)=⟨1-\cos t,\sin t⟩$

Displacement: $⟨0,0,6\pi ⟩$; distance traveled: $2\sqrt{13}\pi \approx 22.65$ft; average velocity: $⟨0,0,3⟩$; average speed: $\sqrt{13}\approx 3.61$ft/s

Displacement: $⟨0,0⟩$; distance traveled: $2\pi \approx 6.28$ft; average velocity: $⟨0,0⟩$; average speed: $1$ft/s

At $t$-values of ${\sin }^{-1}\left(9/30\right)/\left(4\pi \right)+n/2\approx 0.024+n/2$ seconds, where $n$ is an integer.

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1. (a)

   Holding the crossbow at an angle of $0.013$ radians, $\approx {0.745}^{\circ }$ will hit the target $0.4$s later. (Another solution exists, with an angle of ${89}^{\circ }$, landing $18.75$s later, but this is impractical.)

2. (b)

   In the .4 seconds the arrow travels, a deer, traveling at 20mph or 29.33ft/s, can travel 11.7ft. So she needs to lead the deer by 11.7ft.

The position function is $\overrightarrow{r}\left(t\right)=⟨220t,-16t^2+1000⟩$. The $y$-component is 0 when $t=7.9$; $\overrightarrow{r}\left(7.9\right)=⟨1739.25,0⟩$, meaning the box will travel about 1740ft horizontally before it lands.

1

$\overrightarrow{T}\left(t\right)$ and $\overrightarrow{N}\left(t\right)$.

$\overrightarrow{T}\left(t\right)=⟨\frac{4t}{\sqrt{20t^2-4t+1}},\frac{2t-1}{\sqrt{20t^2-4t+1}}⟩$;

$\overrightarrow{T}\left(1\right)=⟨4/\sqrt{17},1/\sqrt{17}⟩$

$\overrightarrow{T}\left(t\right)=\frac{\cos t\sin t}{\sqrt{{\cos }^{2}t{\sin }^{2}t}}⟨-\cos t,\sin t⟩$. (Be careful; this cannot be simplified as just $⟨-\cos t,\sin t⟩$ as $\sqrt{{\cos }^{2}t{\sin }^{2}t}\ne \cos t\sin t$, but rather $\left|\cos t\sin t\right|$.) $\overrightarrow{T}\left(\pi /4\right)=⟨-\sqrt{2}/2,\sqrt{2}/2⟩$

$\mathrm{\ell }\left(t\right)=⟨2,0⟩+t⟨4/\sqrt{17},1/\sqrt{17}⟩$; in parametric form,

$\mathrm{\ell }\left(t\right)=\{\begin{array}{cc}x=2+4t/\sqrt{17}\hfill & \\ y=t/\sqrt{17}\hfill & \end{array}$

$\mathrm{\ell }\left(t\right)=⟨\sqrt{2}/4,\sqrt{2}/4⟩+t⟨-\sqrt{2}/2,\sqrt{2}/2⟩$; in parametric form,

$\mathrm{\ell }\left(t\right)=\{\begin{array}{cc}x=\sqrt{2}/4-\sqrt{2}t/2\hfill & \\ y=\sqrt{2}/4+\sqrt{2}t/2\hfill & \end{array}$

$\overrightarrow{T}\left(t\right)=⟨-\sin t,\cos t⟩$; $\overrightarrow{N}\left(t\right)=⟨-\cos t,-\sin t⟩$

$\overrightarrow{T}\left(t\right)=⟨-\frac{\sin t}{\sqrt{4{\cos }^{2}t+{\sin }^{2}t}},\frac{2\cos t}{\sqrt{4{\cos }^{2}t+{\sin }^{2}t}}⟩$;

$\overrightarrow{N}\left(t\right)=⟨-\frac{2\cos t}{\sqrt{4{\cos }^{2}t+{\sin }^{2}t}},-\frac{\sin t}{\sqrt{4{\cos }^{2}t+{\sin }^{2}t}}⟩$

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1. (a)

   Be sure to show work

2. (b)

   $\overrightarrow{N}\left(\pi /4\right)=⟨-5/\sqrt{34},-3/\sqrt{34}⟩$

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1. (a)

   Be sure to show work

2. (b)

   $\overrightarrow{N}\left(0\right)=⟨-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}⟩$

$\overrightarrow{T}\left(t\right)=\frac{1}{\sqrt{5}}⟨2,\cos t,-\sin t⟩$; $\overrightarrow{N}\left(t\right)=⟨0,-\sin t,-\cos t⟩$

$\overrightarrow{T}\left(t\right)=\frac{1}{\sqrt{a^2+b^2}}⟨-a\sin t,a\cos t,b⟩$;

$\overrightarrow{N}\left(t\right)=⟨-\cos t,-\sin t,0⟩$

$a_{\text{T}}=\frac{4t}{\sqrt{1+4t^2}}$ and $a_{\text{N}}=\sqrt{4-\frac{16t^2}{1+4t^2}}$

At $t=0$, $a_{\text{T}}=0$ and $a_{\text{N}}=2$;

At $t=1$, $a_{\text{T}}=4/\sqrt{5}$ and $a_{\text{N}}=2/\sqrt{5}$.

At $t=0$, all acceleration comes in the form of changing the direction of velocity and not the speed; at $t=1$, more acceleration comes in changing the speed than in changing direction.

$a_{\text{T}}=0$ and $a_{\text{N}}=2$

At $t=0$, $a_{\text{T}}=0$ and $a_{\text{N}}=2$;

At $t=\pi /2$, $a_{\text{T}}=0$ and $a_{\text{N}}=2$.

The object moves at constant speed, so all acceleration comes from changing direction, hence $a_{\text{T}}=0$. $\overrightarrow{a}\left(t\right)$ is always parallel to $\overrightarrow{N}\left(t\right)$, but twice as long, hence $a_{\text{N}}=2$.

$a_{\text{T}}=0$ and $a_{\text{N}}=a$

At $t=0$, $a_{\text{T}}=0$ and $a_{\text{N}}=a$;

At $t=\pi /2$, $a_{\text{T}}=0$ and $a_{\text{N}}=a$.

The object moves at constant speed, meaning that $a_{\text{T}}$ is always 0. The object “rises” along the $z$-axis at a constant rate, so all acceleration comes in the form of changing direction circling the $z$-axis. The greater the radius of this circle the greater the acceleration, hence $a_{\text{N}}=a$.

time and/or distance

Answers may include lines, circles, helixes

$\kappa$

$s=3t$, so $\overrightarrow{r}\left(s\right)=⟨2s/3,s/3,-2s/3⟩$

$s=\sqrt{13}t$, so

$\overrightarrow{r}\left(s\right)=⟨3\cos \left(s/\sqrt{13}\right),3\sin \left(s/\sqrt{13}\right),2s/\sqrt{13}⟩$

$\kappa =\frac{\left|6x\right|}{{\left(1+{\left(3x^2-1\right)}^2\right)}^{3/2}}$;

$\kappa \left(0\right)=0$, $\kappa \left(1/2\right)=\frac{192}{17\sqrt{17}}\approx 2.74$.

$\kappa =\frac{\left|\cos x\right|}{{\left(1+{\sin }^{2}x\right)}^{3/2}}$;

$\kappa \left(0\right)=1$, $\kappa \left(\pi /2\right)=0$

$\kappa =\frac{\left|2\cos t\cos \left(2t\right)+4\sin t\sin \left(2t\right)\right|}{{\left(4{\cos }^2\left(2t\right)+{\sin }^{2}t\right)}^{3/2}}$;

$\kappa \left(0\right)=1/4$, $\kappa \left(\pi /4\right)=8$

$\kappa =\frac{\left|6t^2+2\right|}{{\left(4t^2+{\left(3t^2-1\right)}^2\right)}^{3/2}}$;

$\kappa \left(0\right)=2$, $\kappa \left(5\right)=\frac{19}{1394\sqrt{1394}}\approx 0.0004$

$\kappa =0$;

$\kappa \left(0\right)=0$, $\kappa \left(1\right)=0$

$\kappa =\frac{3}{13}$;

$\kappa \left(0\right)=3/13$, $\kappa \left(\pi /2\right)=3/13$

maximized at $x=\pm \frac{\sqrt{2}}{\sqrt[4]{5}}$

maximized at $t=1/4$

radius of curvature is $5\sqrt{5}/4$.

radius of curvature is $9$.

$x^2+{\left(y-1/2\right)}^2=1/4$, or

$\overrightarrow{c}\left(t\right)=⟨1/2\cos t,1/2\sin t+1/2⟩$

$x^2+{\left(y+8\right)}^2=81$, or $\overrightarrow{c}\left(t\right)=⟨9\cos t,9\sin t-8⟩$

Let $\overrightarrow{r}\left(t\right)=⟨x\left(t\right),y\left(t\right),0⟩$ and apply the second formula of part 3.