Vector Functions

11.1 Assess Your Understanding
Printed Page 765

765

Concepts and Vocabulary

The domain of $\mathbf{r}(t) =( 4t+1) \mathbf{i}+\sqrt{4-t}\mathbf{j}+\mathbf{k}$ is _______.

$\{t|t \leq 4\}$

True or False A vector function is continuous at a real number $t_{0}$ if at least one of its component functions is continuous at $t_{0}$ .

False

True or False To find the derivative of a vector function $\mathbf{r=r}(t),$ find the derivative of each of its component functions.

True

True or False The derivative of a dot product is a scalar function.

True

True or False If $y=f(t)$ is a differentiable real (scalar) function and $\mathbf{r=r}(t)$ is a differentiable vector function, then $\dfrac{d}{dt}\left[ f(t) \mathbf{r}(t) \right] =f^{\prime} (t) \dfrac{d}{dt}\mathbf{r}(t) .$

False

If $\mathbf{u=u}(t)$ and $\mathbf{v=v}(t)$ are vector functions in space that are differentiable, then $\dfrac{d}{dt}\left( \mathbf{u} \times \mathbf{v}\right)=$ ______.

$\mathbf{u}' \times \mathbf{v}\,{+}\,\mathbf{u} \times \mathbf{v}'$

Skill Building

In Problems 7–14, find the value of each vector function at t.

$\mathbf{r}(t)=t^{2}\mathbf{i}-2t\mathbf{j}$ at $t=1$

$\mathbf{i} - 2\mathbf{j}$

$\mathbf{r}(t)=t^{3}\mathbf{i}+2t\mathbf{j}$ at $t=2$

$\mathbf{v}(t)=\sin t\mathbf{i}-\cos t\mathbf{j}$ at $t=\dfrac{\pi }{4}$

$\dfrac{\sqrt{2}}{2} \mathbf{i} - \dfrac{\sqrt{2}}{2} \mathbf{j}$

$\mathbf{v}(t)=\tan t\mathbf{i}+\cos ( 2t) \mathbf{j}$ at $t=0$

$\mathbf{u}(t)=e^{2t}\mathbf{i}+t^{2} \mathbf{j}-2\mathbf{k}$ at $t=0$

$\mathbf{i}\,{-}\,2\mathbf{k}$

$\mathbf{u}(t)=\ln t \mathbf{i}-\mathbf{j}+t^{3}\mathbf{k}$ at $t=1$

$\mathbf{g}(t)=t\mathbf{i}-\cos \left( \dfrac{\pi t}{4}\right) \mathbf{j}+2t\mathbf{k}$ at $t=4$

$4\mathbf{i}\,{+}\,\mathbf{j}\,{+}\,8\mathbf{k}$

$\mathbf{f}(t)=\sin \left( \dfrac{3\pi t}{4}\right) \mathbf{i}+3\mathbf{j}-3t^{2}\mathbf{k}$ at $t=1$

In Problems 15–22, find the domain of each vector function.

$\mathbf{r}(t) =\cos t\mathbf{i}+\sin t\mathbf{j}+2\mathbf{k}$

All real numbers

$\mathbf{r}(t) =\cos (2t) \mathbf{i}+\sin t\mathbf{j}+2t\mathbf{k}$

$\mathbf{f}(t) =\sqrt{t}\mathbf{i}+t\mathbf{j-k}$

$\left\{t | t \geq 0\right\}$

$\mathbf{r}(t) =t^{2}\mathbf{i}+\dfrac{1}{\sqrt{t}}\mathbf{j}-t\mathbf{k}$

$\mathbf{u}(t) =\dfrac{( t^{2}+1) }{t}\mathbf{i}+3t\mathbf{j}-\dfrac{2}{t}\mathbf{k}$

$\left\{ t | t \neq 0\right\}$

$\mathbf{v}(t) =e^{t}\mathbf{i}+( 5+3t) \mathbf{j}-\dfrac{2}{t}\mathbf{k}$

$\mathbf{v}(t) =\ln t\mathbf{i}+( t+1)\mathbf{j}+4t\mathbf{k}$

$\left\{ t | t > 0\right\}$

$\mathbf{v}(t) =\sqrt{t}\kern.7pt\mathbf{i}+\ln t\mathbf{j}+t\mathbf{k}$

In Problems 23–36, graph the curve $C$ traced out by each vector function and show its orientation.

$\mathbf{r}(t)=2t\mathbf{i}+t^{2}\mathbf{j},\quad t\geq 0$

$\mathbf{r}(t)=t^{2}\mathbf{i}-4t\mathbf{j},\quad t\leq 0$

$\mathbf{r}(t)=t\mathbf{i},\quad -1\leq t\leq 1$

$\mathbf{r}(t)=t\mathbf{j},\quad -1\leq t\leq 1$

$\mathbf{r}(t)=t\mathbf{i}+t\mathbf{j}$

$\mathbf{r}(t)=3t\mathbf{i}+2t \mathbf{j}$

$\mathbf{r}(t)=3t\mathbf{i}-2t\mathbf{j}$

$\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}$

$\mathbf{r}(t)=\cos t\mathbf{i}-\sin t\mathbf{j},\quad 0\leq t\leq \dfrac{\pi }{2}$

$\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j},\quad 0\leq t\leq \dfrac{\pi }{2}$

$\mathbf{r}(t)=\sin ^{2}t\mathbf{i}+\cos ^{2}t\mathbf{j},\quad 0\leq t\leq \dfrac{\pi }{2}$

$\mathbf{r}(t)=\sin ^{2}t\mathbf{i}-\cos ^{2}t\mathbf{j},\quad 0\leq t\leq \dfrac{\pi }{2}$

$\mathbf{r}(t)=\sin t\mathbf{i}+t^{2}\mathbf{j},\quad 0\leq t\leq 4\pi$

$\mathbf{r}(t)=\sin ^{3}t\mathbf{i}+\cos ^{3}t\mathbf{j},\quad 0\leq t\leq 2\pi$

In Problems 37–40, match each vector function to its plane curve.

$\mathbf{r}(t)=t^{2}\mathbf{i}+\cos t\mathbf{j},\quad 0\leq t\leq 2\pi$

$\mathbf{r}(t)=\cos t\mathbf{i}+t^{2}\mathbf{j},\quad 0\leq t\leq 2\pi$

$\mathbf{r}(t)=\sqrt{t}\kern.7pt\mathbf{i}+\cos t\mathbf{j},\quad 0\leq t\leq 2\pi$

$\mathbf{r}(t)=\cos t\mathbf{i}-t\mathbf{j},\quad 0\leq t\leq 2\pi$

In Problems 41–48, graph the curve $C$ traced out by each vector function.

$\mathbf{r}(t)=( 2t+1) \mathbf{i}+\dfrac{t}{3}\mathbf{j}-( t+1) \mathbf{k}$

$\mathbf{r}(t)=t\mathbf{i}+t\mathbf{j}-( t+1) \mathbf{k}$

$\mathbf{r}(t)=2\cos t\mathbf{i}+t\mathbf{j}+2\sin t\mathbf{k},\quad -2\pi \leq t\leq 2\pi$

$\mathbf{r}(t)=\dfrac{t}{2}\mathbf{i}+3\sin t\mathbf{j}+2\cos t\mathbf{k},\quad -2\pi \leq t\leq 2\pi$

$\mathbf{r}(t)=4\cos t\mathbf{i}+\sin t\mathbf{j}+e^{t}\mathbf{k},\quad -\pi \leq t\leq 2\pi$

$\mathbf{r}(t)=t\mathbf{i}+\sin t\mathbf{j}+e^{t}\mathbf{k},\quad -5\leq t\leq 5$

$\mathbf{r}(t)=3\cos t\mathbf{i}+3\sin t\mathbf{j} +\sin ( 2t) \mathbf{k},\quad 0\leq t\leq 2\pi$

$\mathbf{r}(t)=t\mathbf{i}+t\cos \left(3t\right) \mathbf{j}\ +t\sin ( 3t) \mathbf{k},\quad 0\leq t\leq 2\pi$

In Problems 49–56, determine where each vector function is continuous.

$\mathbf{r}(t)=t^{2}\mathbf{i}+\dfrac{2}{1-t^{2}}\mathbf{j}+\sin t\mathbf{k}$

$\left\{t|t \neq -1, 1\right\}$

$\mathbf{r} (t) =\ln t\;\mathbf{i}+e^{-t}\mathbf{j}$

766

$\mathbf{r}(t) =\sec t\mathbf{i}-\cos t\mathbf{j}$

$\left\{ t|t \neq \dfrac{(2k+1) \pi}{2}\right\}$

$\mathbf{r} (t) =\sqrt{t+1}\kern.7pt\mathbf{i}+\tan t\mathbf{j}-\dfrac{\sin t-1}{t}\mathbf{k}$

$\mathbf{r}(t) =\dfrac{t}{t+1}\mathbf{i}-t \mathbf{j}$

$\left\{t|t \neq -1\right\}$

$\mathbf{r}(t) =\dfrac{t}{t^{2}+1}\mathbf{i}+t\mathbf{j}$

$\mathbf{r}(t) =\sin t\mathbf{i}-\tan t\mathbf{j+k}$

$\left\{t | t \neq \dfrac{(2k+1) \pi}{2}\right\}$

$\mathbf{r}(t) =\sec t\mathbf{i}+\csc t\mathbf{j+k}$

In Problems 57–72, find $\mathbf{r}^{\prime} (t)$ and $\mathbf{r}^{\prime \prime} (t)$ .

$\mathbf{r}(t)=4t^{2}\mathbf{i}-2t^{3}\mathbf{j}$

$\mathbf{r} \, '(t) = 8t \mathbf{i} - 6t^2 \mathbf{j}$ and $\mathbf{r}\,''(t) = 8 \mathbf{i} - 12t \mathbf{j}$

$\mathbf{r}(t)=8t\mathbf{i}+4t^{3}\mathbf{j}$

$\mathbf{r}(t)=4\sqrt{t}\kern.7pt\mathbf{i}+2e^{t}\mathbf{j}$

$\mathbf{r}\,'(t) = \dfrac{2\sqrt{t}}{t} \mathbf{i} + 2e^t \mathbf{j}$ and $\mathbf{r}\,''(t) = - \dfrac{1}{t^{3/2}} \mathbf{i} + 2e^t \mathbf{j}$

$\mathbf{r}(t)=e^{3t}\mathbf{i}+\sqrt[3]{t}\mathbf{j}$

$\mathbf{r}(t)=\mathbf{i}+t\mathbf{j}+t^{2}\mathbf{k}$

$\mathbf{r}\,'(t) = \mathbf{j} + 2t \mathbf{k}$ and $\mathbf{r}\,''(t) = 2\mathbf{k}$

$\mathbf{r}(t)=\mathbf{i}-\mathbf{j}+t\mathbf{k}$

$\mathbf{r}(t)=t^{2}\mathbf{i}+t^{3}\mathbf{j}-t\mathbf{k}$

$\mathbf{r}\,'(t) = 2t \mathbf{i} + 3t ^2 \mathbf{j} - \mathbf{k}$ and $\mathbf{r}\,''(t) = 2 \mathbf{i} + 6t \mathbf{j}$

$\mathbf{r}(t)=(1+t)\mathbf{i}-3t^{2}\mathbf{j}+t\mathbf{k}$

$\mathbf{r}(t)=\sin ^{2}t\mathbf{i}-\cos ^{2}t\mathbf{j}$

$\mathbf{r}\,'(t) = \sin (2t) \mathbf{i} + \sin (2t) \mathbf{j}$ and $\mathbf{r}\,''(t) = 2 \cos (2t) \mathbf{i} + 2 \cos (2t) \mathbf{j}$

$\mathbf{r}(t)=\sin ( 2t) \mathbf{i}-\cos (2t) \mathbf{j}$

$\mathbf{r}(t)=5e^{t^{2}+t}\mathbf{i}-\ln e^{t^{2}+t}\mathbf{j}$

$\mathbf{r}\,'(t) = 5e^{t^2+t}(2t+1) \mathbf{i}-(2t+1) \mathbf{j}$ and $\mathbf{r}\,''(t) = 5e^{t^2+t}(4t^2+4t+3) \mathbf{i} - 2 \mathbf{j}$

$\mathbf{r}(t)= \sin (3t^{3}-t)\mathbf{i} -\cos ^{2}(3t^{3}-t)\mathbf{j}$

$\mathbf{r}(t)=e^{t}\cos t\mathbf{i}+e^{t}\sin t\mathbf{j}+t\mathbf{k}$

$\mathbf{r}\,'(t) = (e^t \cos t - e^t \sin t) \mathbf{i} + (e^t \sin t + e^t \cos t) \mathbf{j} + \mathbf{k}$ and $\mathbf{r}\,''(t) = -2e^t \sin t \mathbf{i} + 2e^t \cos t \mathbf{j}$

$\mathbf{r}(t)=e^{-t}\cos t\mathbf{i}+e^{-t}\sin t\mathbf{j}-t\mathbf{k}$

$\mathbf{r}(t)=(t-t^{3})\mathbf{i}+(t+t^{3})\mathbf{j}-t\mathbf{k}$

$\mathbf{r}\,'(t) = (1-3t^2) \mathbf{i} + (1+3t^2)\mathbf{j} - \mathbf{k}$ and $\mathbf{r}\,''(t) = -6t\mathbf{i} + 6t \mathbf{j}$

$\mathbf{r}(t)=(t^{2}-t)\mathbf{i}+(t^{2}+t)\mathbf{j}+t\mathbf{k}$

In Problems 73–78, find $\frac {d}{dt}\left[ \mathbf{u}(t)\,{\cdot}\, \mathbf{v}(t)\right]$ .

$\mathbf{u}(t)=t^{2}\mathbf{i}-t\mathbf{j}$ $\quad$ and $\quad \mathbf{v}(t)=t\mathbf{i}+t^{2}\mathbf{j}$

$\dfrac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)]=0$

$\mathbf{u}(t)=t^{3}\mathbf{i}+t\mathbf{j}$ $\quad$ and $\quad \mathbf{v}(t)=t\mathbf{i}-2t\mathbf{j}$

$\mathbf{u}(t)=e^{t}\mathbf{i}+e^{-t}\mathbf{j}$ $\quad$ and $\quad \mathbf{v}(t)=t\mathbf{i}-t^{2}\mathbf{j}$

$\dfrac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)]= te^t+e^t+t^2e^{-t} - 2te^{-t}$

$\mathbf{u}(t)=t\mathbf{i}+4\sqrt{t}\mathbf{j}$ $\quad$ and $\quad \mathbf{v}(t)=2t\mathbf{i}-t^{2}\mathbf{j}$

$\mathbf{u}(t)=\sin ( \omega t) \mathbf{i}+\cos( \omega t) \mathbf{j}$ $\quad$ and $\quad \mathbf{v}(t)=\mathbf{i}+\mathbf{j}$

$\dfrac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)]= \omega \cos (\omega t) - \omega \sin (\omega t)$

$\mathbf{u}(t)=\sin ^{2}t\mathbf{i}-\cos^{2}t\mathbf{j}$ $\quad$ and $\quad \mathbf{v}(t)=\mathbf{i}-\mathbf{j}$

In Problems 79–82, find $\dfrac{d}{dt}[\mathbf{u}(t)\,{\cdot}\,\mathbf{v}(t)]$ and $\dfrac{d}{dt}[\mathbf{u}(t)\times \mathbf{v}(t)]$ .

$\mathbf{u}(t)=2t\mathbf{i}+t^{2}\mathbf{j}-5\mathbf{k}$ $\quad$ and $\quad \mathbf{v}(t)=t^{2}\mathbf{i}+2t\mathbf{j}+\mathbf{k}$

$\dfrac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)]= 12t^2$ and $\dfrac{d}{dt}[\mathbf{u}(t)\times\mathbf{v}(t)]= (2t+10) \mathbf{i} -(10t+2) \mathbf{j} + (8t-4t^3) \mathbf{k}$

$\mathbf{u}(t)=t^{3}\mathbf{i}-t^{2}\mathbf{j}+t\mathbf{k}$ $\quad$ and $\quad \mathbf{v}(t)=t\mathbf{i}-t^{2}\mathbf{j}+t^{3}\mathbf{k}$

$\mathbf{u}(t)=\cos ( 2t) \mathbf{i}+\sin (2t) \mathbf{j}+\mathbf{k}$ $\quad$ and $\quad \mathbf{v}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+\mathbf{k}$

$\dfrac{d}{dt}[\mathbf{u}(t)\cdot\mathbf{v}(t)]=\sin t \cos(2t)-\sin(2t)\cos t$ and $\dfrac{d}{dt}[\mathbf{u}(t)\times\mathbf{v}(t)]= (2\cos(2t)-\cos t) \mathbf{i} + (2 \sin(2t) - \sin t) \mathbf{j}-(\sin(2t)\sin t+\cos(2t)\cos t) \mathbf{k}$

$\mathbf{u}(t)=e^{2t}\mathbf{i}+e^{-2t}\mathbf{j}+t\mathbf{k}$ $\quad$ and $\quad \mathbf{v}(t)=e^{-t}\mathbf{i}+e^{-2t}\mathbf{j}-t\mathbf{k}$

Applications and Extensions

Given the vector function $\mathbf{u}(t) =\cos( \omega t) \mathbf{i}+\sin ( \omega t) \mathbf{j}$ , find $\dfrac{d\mathbf{u}}{dt}$ and $\left\Vert \dfrac{d\mathbf{u}}{dt}\right\Vert$ .

$\dfrac{d\mathbf{u}}{dt} = -\omega \sin(\omega t) \mathbf{i} + \omega \cos(\omega t) \mathbf{j}$ and $\left\| \dfrac{d\mathbf{u}}{dt} \right\|\;=\;|\omega|$

Given the vector function $\mathbf{v}(t) =t\mathbf{i}+t^{2}\mathbf{j}+t^{3}\mathbf{k}$ , find $\dfrac{d^{2}\mathbf{v}}{dt^{2}}$ and $\left\Vert \dfrac{d^{2}\mathbf{v}}{dt^{2}}\right\Vert$ .

For $\mathbf{f}(t)=\sin t\mathbf{i}-\cos t\mathbf{j}$ , show that $\mathbf{f}(t)$ and $\mathbf{f}^{\,\prime \prime} (t)$ are parallel.

See Student Solutions Manual.

For $\mathbf{f}(t)=e^{3t}\mathbf{i}+e^{-3t}\mathbf{j}$ , show that $\mathbf{f}(t)$ and $\mathbf{f}^{\,\prime \prime} (t)$ are parallel.

In Problems 87–94, a vector function $\mathbf{r}=\mathbf{r}(t)$ , defining a curve in the plane is given. Graph each curve, indicating its orientation. Include in the graph the vector $\mathbf{r}(0)$ and $\mathbf{r}^{\prime} (0)$ . Draw $\mathbf{r}^{\prime} (0)$ so its initial point is at the terminal point of $\mathbf{r}(0)$ .

$\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}$

$\mathbf{r}(t)=2t^{2}\mathbf{i}-t\mathbf{j}$

$\mathbf{r}(t)=t\mathbf{i}+e^{t}\mathbf{j}$

$\mathbf{r}(t)=t\mathbf{i}+\ln (1+t)\mathbf{j}$

$\mathbf{r}(t)=3\sin t\mathbf{i}-3\cos t\mathbf{j}$

$\mathbf{r}(t)=4\sin t\mathbf{i}+4\cos t \mathbf{j}$

$\mathbf{r}(t)=2\cos t\mathbf{i}-3\sin t\mathbf{j}$

$\mathbf{r}(t)=-\cos t\mathbf{i}+2\sin t\mathbf{j}$

In Problems 95–98, the position of an object at time $t$ it is given.

(a) Eliminate the parameter $t$ to find $y$ as a function of $x$ .

(b) Graph $\mathbf{r}=\mathbf{r}(t)$ and indicate the orientation.

$\mathbf{r}(t) =(1-t^{2})\mathbf{i}+t^{2}\mathbf{j}$

$y = 1-x$

$\mathbf{r}(t) =t^{2}\mathbf{i}+(4t^{2}-t^{4})\mathbf{j}$

$\mathbf{r}(t) =\sin ^{2}t\mathbf{i}+\tan t\mathbf{j,}\quad 0<t<\dfrac{\pi }{2}$

$y = \sqrt{\dfrac{x}{1-x}}$ for $0 < x < 1$
(b)

$\mathbf{r}(t) =e^{t}\mathbf{i}+e^{-t}\mathbf{j}$

Proof of the Sum Formula If the vector functions $\mathbf{u}=\mathbf{u}(t)$ and $\mathbf{v}=\mathbf{v}(t)$ are differentiable, show that $\begin{equation*} \left[ \mathbf{u}(t)+\mathbf{v}(t)\right] ^{\prime} =\mathbf{u}^{\prime} (t)+ \mathbf{v}^{\prime} (t) \end{equation*}$

See Student Solutions Manual.

Proof of the Dot Product Formula If the vector functions $\mathbf{u}=\mathbf{u}(t)$ and $\mathbf{v}=\mathbf{v}(t)$ are differentiable, show that $\begin{equation*} \left[ \mathbf{u}(t)\,{\cdot}\, \mathbf{v}(t)\right] ^{\prime} =\mathbf{u}^{\prime} (t)\,{\cdot}\, \mathbf{v}(t)+\mathbf{u}(t)\,{\cdot}\, \mathbf{v}^{\prime} (t) \end{equation*}$

Proof of the Cross Product Formula If the vector functions $\mathbf{u}=\mathbf{u}(t)$ and $\mathbf{v}=\mathbf{v}(t)$ in space are differentiable, show that $\begin{equation*} \left[ \mathbf{u}(t)\times \mathbf{v}(t)\right] ^{\prime} =\mathbf{u}^{\prime} (t)\times \mathbf{v}(t)+\mathbf{u}(t)\times \mathbf{v}^{\prime} (t) \end{equation*}$

See Student Solutions Manual.

Proof of the Chain Rule If the vector function $\mathbf{v}=\mathbf{v}(t)$ is differentiable and the real function $y=f(t)$ is differentiable, show that $\begin{equation*} \dfrac{d}{dt}\mathbf{v}\left( f(t) \right) =\mathbf{v}^{\prime} (f(t) )f^{\prime} (t) \end{equation*}$

The derivative of a cross product is not commutative. Give an example of two vector functions that demonstrate this fact.

Answers will vary.

Suppose the vector function $\mathbf{r}=\mathbf{r}(t)$ is twice differentiable; show that $[ \mathbf{r}(t)\times \mathbf{r}^{\prime}(t)] ^{\prime} =\mathbf{r}(t)\times \mathbf{r}^{\prime \prime} (t).$

Show that $\lim\limits_{t\rightarrow t_{\hbox{0}}}\mathbf{r}(t)=\mathbf{L}$ if and only if $\lim\limits_{t\rightarrow t_{\hbox{0}}}\left\Vert \mathbf{r}(t)-\mathbf{L}\right\Vert =0.$

See Student Solutions Manual.

If the vector function $\mathbf{r=r}(t)$ is differentiable, show that $\dfrac{d}{dt}\left\Vert \mathbf{r}(t)\right\Vert\; =\dfrac{\mathbf{r}(t)\,{\cdot}\, \mathbf{r}^{\prime} (t)}{\left\Vert \mathbf{r}(t)\right\Vert }.$ Use this result to show that $\left\Vert \mathbf{r}(t)\right\Vert$ is constant if and only if $\mathbf{r}(t)\,{\cdot}\, \mathbf{r}^{\prime} (t)=0$ for all $t$ .

Challenge Problems

If the vector function $\mathbf{r}$ is differentiable, show that $\begin{equation*} \frac{d}{dt}\frac{\mathbf{r}(t)}{\left\Vert \mathbf{r}(t)\right\Vert }=\frac{ \mathbf{r}^{\prime} (t)}{\left\Vert \mathbf{r}(t)\right\Vert }-\frac{\mathbf{r} (t)\,{\cdot}\, \mathbf{r}^{\prime} (t)}{\left\Vert \mathbf{r}(t)\right\Vert ^{3}} \mathbf{r}(t) \end{equation*}$

See Student Solutions Manual.

1. Discuss the curve $C$ traced out by the vector function $\mathbf{r}(t)=e^{t}\sin t\mathbf{i}+e^{t}\cos t\mathbf{j}+e^{t}\mathbf{k}$ .
2. Name the surface on which the curve $C$ lies.

11.1 Assess Your UnderstandingPrinted Page 765

11.1 Assess Your Understanding
Printed Page 765