Chapter 3

$\text{r}\left(0\right)=\text{j},\text{r}\left(1\right)=\mathrm{-2}\text{i}+5\text{j},\text{r}\left(\mathrm{-4}\right)=28\text{i}-15\text{j}$

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

$\underset{t\to \mathrm{-2}}{\text{lim}}\text{r}\left(t\right)=3\text{i}-5\text{j}-\text{k}$

$r^{\prime }\left(t\right)=4t\text{i}+5\text{j}$

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

$\frac{d}{dt}\left[\text{r}\left(t\right)·r^{\prime }\left(t\right)\right]=8e^{4t}$

$\begin{array}{l}\frac{d}{dt}\left[\text{u}\left(t\right)\times \text{r}\left(t\right)\right]\\ =-\left(e^{2t}\left(\text{cos}t+2\text{sin}t\right)+\text{cos}2t\right)\text{i}+\left(e^{2t}\left(2t+1\right)-\text{sin}2t\right)\text{j}+\left(t\text{cos}t+\text{sin}t-\text{cos}2t\right)\text{k}\end{array}$

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

${\displaystyle {\int }_1^3\left[\left(2t+4\right)\text{i}+\left(3t^2-4t\right)\text{j}\right]dt}=16\text{i}+10\text{j}$

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

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

$\text{r}\left(s\right)=\langle 3\text{cos}\left(\frac{s}{5}\right),3\text{sin}\left(\frac{s}{5}\right),\frac{4s}{5}\rangle ,s\ge 0.$

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

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

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

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

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

$\begin{array}{c}\text{v}\left(t\right)=r^{\prime }\left(t\right)=\left(2t-3\right)\text{i}+2\text{j}+\text{k}\hfill \\ \text{a}\left(t\right)=v^{\prime }\left(t\right)=2\text{i}\hfill \\ v\left(t\right)=\Vert r^{\prime }\left(t\right)\Vert =\sqrt{{\left(2t-3\right)}^2+2^2+1^2}=\sqrt{4t^2-12t+14}\hfill \end{array}$

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

1. $\begin{array}{ccc}\mathbf{\text{v}}\left(t\right)& =& {\mathbf{r}}^{\mathbf{\text{'}}}\left(t\right)=4\mathbf{\text{i}}+2t\mathbf{\text{j}}\\ \mathbf{\text{a}}\left(t\right)& =& {\mathbf{v}}^{\mathbf{\text{'}}}\left(t\right)=2\mathbf{\text{j}}\\ a_{\text{T}}& =& \frac{2t}{\sqrt{t^2+4}},a_{\text{N}}=\frac{4}{\sqrt{4+t^2}}\end{array}$
2. $a_{\text{T}}\left(\mathrm{-3}\right)=-\frac{6\sqrt{13}}{13},a_{\text{N}}\left(\mathrm{-3}\right)=\frac{2\sqrt{13}}{13}$

967.15 m

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$(50,0,0)$

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

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

$\langle \text{−}{e}^{\text{−}t},3\text{cos}(3t),\frac{5}{\sqrt{t}}\rangle$

$\langle 0,0,0\rangle$

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

$\langle 0,12\text{cos}(3t),\text{cos}t-t\text{sin}t\rangle$

$\frac{1}{\sqrt{2}}\langle 1,\mathrm{-1},0\rangle$

$\frac{1}{\sqrt{1060.5625}}\langle 6,-\frac{3}{4},32\rangle$

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

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

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

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

$900t^7+16t$

1. 
   

   
2. Undefined or infinite

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

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

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

$\begin{array}{ccc}\hfill \text{‖}\text{v}(t)\text{‖}& =\hfill & k\hfill \\ \hfill \text{v}(t)·\text{v}(t)& =\hfill & k\hfill \\ \hfill \frac{d}{dt}\left(\text{v}(t)·\text{v}(t)\right)& =\hfill & \frac{d}{dt}k=0\hfill \\ \hfill \text{v}(t)·\text{v}\text{'}(t)+\text{v}\text{'}(t)·\text{v}(t)& =\hfill & 0\hfill \\ \hfill 2\text{v}(t)·\text{v}\text{'}(t)& =\hfill & 0\hfill \\ \hfill \text{v}(t)·\text{v}\text{'}(t)& =\hfill & 0.\hfill \end{array}$
The last statement implies that the velocity and acceleration are perpendicular or orthogonal.

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

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

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

$\sqrt{161}$

$\text{v}(t)=\langle \text{−}\text{sin}t,\text{cos}t,1\rangle$

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

$\text{v}(t)=\langle \text{−}\text{sin}t,2\text{cos}t,0\rangle$

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

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

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

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

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

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

$\frac{3}{4}\text{i}+\text{ln}(2)\text{j}+\left(1-\frac{1}{e}\right)\text{j}$

$8\sqrt{5}$

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

Length $=2\pi$

$6\pi$

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

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

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

$\mathbf{\text{N}}\left(0\right)=⟨0,-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}⟩$

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

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

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

$\text{N}(t)=\langle \text{−}\text{sin}t,0,-\text{cos}t\rangle$

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

$\text{r}(s)=\left(1+\frac{s}{\sqrt{2}}\right)\text{sin}\left(\text{ln}(1+\frac{s}{\sqrt{2}})\right)\text{i}+\left(1+\frac{s}{\sqrt{2}}\right)\text{cos}\left[\text{ln}\left(1+\frac{s}{\sqrt{2}}\right)\right]\text{j}$

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

$\frac{1}{2}$

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

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

The curvature approaches zero.

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

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

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

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

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

The curvature is decreasing over this interval.

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

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

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

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

$\text{v}(t)=e^t\text{i}-e^{\text{−}t}\text{j},$ $\text{a}(t)=e^t\text{i}+e^{\text{−}t}\text{j},$ speed = $\Vert \text{v}(t)\Vert =\sqrt{e^{2t}+e^{\mathrm{-2}t}}$

$t=4$

$\text{v}(t)=\left(\omega -\omega \text{cos}(\omega t)\right)\text{i}+\left(\omega \text{sin}(\omega t)\right)\text{j},$
$\text{a}(t)=\left({\omega }^2\text{sin}(\omega t)\right)\text{i}+\left({\omega }^2\text{cos}(\omega t)\right)\text{j},$
$\text{speed}=\sqrt{{\omega }^2-2{\omega }^2\text{cos}(\omega t)+{\omega }^2{\text{cos}}^2(\omega t)+{\omega }^2{\text{sin}}^2(\omega t)}\text{=}\sqrt{2{\omega }^2(1-\text{cos}(\omega t))}$

$\Vert \text{v}(t)\Vert =\sqrt{9+4t^2}$

$\text{v}(t)=\langle e^{\mathrm{-5}t}(\text{cos}t-5\text{sin}t),\text{−}{e}^{\mathrm{-5}t}(\text{sin}t+5\text{cos}t),\mathrm{-20}{e}^{\mathrm{-5}t}\rangle$

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

44.185 sec

$t=88.37$ sec

88.37 sec

The range is approximately 886.29 m.

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

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

$a_T=0,$ $a_N=a{\omega }^2$

$a_T=\sqrt{3}{e}^t,$ $a_N=\sqrt{2}{e}^t$

$a_T=2t,$ $a_N=2$

$a_T\frac{6t+12t^3}{\sqrt{1+t^4+t^2}},$ $a_N=6\sqrt{\frac{1+4t^2+t^4}{1+t^2+t^4}}$