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\newcommand{\ra}{\rangle}

\newcommand{\force}{{$\mbox{\;}$}}

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\begin{document}

%\pagestyle{empty}

\noindent
\parbox{2in}{\bf Math 2130-101}
\hfill {\Large \bf Test \#1} \hfill
\parbox{2in}{\bf \hfill Feb. $19^\mathrm{th}$, 2021}

\vspace{0.15in}

\noindent {\large\bf Name:} \underline{\Large\sc\color{red}\quad Answer Key \quad}
\hfill {\bf Be sure to show your work!}

\vspace{0.1in}

\noindent {\bf \large 1. (22 points)} Vector Basics: Let $\B u = \la 2,-2,1 \ra$, $\B v =\la -1,3,1 \ra$, and $\B w = \la 1,1,0 \ra$.
  
\begin{enumerate}[(a)]
   \item Find two unit vectors that are perpendicular to both $\B u$ and $\B v$.
   
$\displaystyle \B u \times \B v = \begin{vmatrix} \B i & \B j & \B k \\ 2 & -2 & 1 \\ -1 & 3 & 1 \end{vmatrix} = \la (-2)1-3(1), -(2(1)-(-1)1), 2(3)-(-1)(-2) \ra = \la -5, -3, 4 \ra$ is perpendicular to both $\B u$ and $\B v$. To find unit vectors, we need to normalize this cross product. Consider $|\B u \times \B v| = \sqrt{(-5)^2+(-3)^2+4^2} = \sqrt{25+9+16}=\sqrt{50}=5\sqrt{2}$. Thus \fbox{$\displaystyle \pm\dfrac{\la -5,-3,4\ra}{5\sqrt{2}}$} are both unit vectors that are perpendicular to $\B u$ and $\B v$.

  \item Find the volume of the parallelepiped spanned by $\B u$, $\B v$, and $\B w$.

Either we can compute a $3 \times 3$ determinant: $\displaystyle \begin{vmatrix} \B u \\ \B v \\ \B w \end{vmatrix} = \begin{vmatrix} 2 & -2 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 0 \end{vmatrix}$ or (since we've already computed $\B u \times \B v$) we can compute: $(\B u \times \B v) \bdot \B w = \la -5,-3,4 \ra \bdot \la 1,1,0 \ra = (-5)1+(-3)1+(4)0 = -8$. Thus, the volume of the parallelepiped is $|-8|=$ \fbox{8}.

   \item Find the angle between $\B u$ and $\B v$ (don't worry about evaluating inverse trig. functions). 

We use the formula $\B u \bdot \B v = |\B u| \,|\B v| \cos(\theta)$. We have $\B u \bdot \B v = 2(-1)+(-2)(3)+1(1) = -7$, $|\B u|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3$, and $|\B v|=\sqrt{(-1)^2+3^2+1^2}=\sqrt{11}$. Therefore, the angle between $\B u$ and $\B v$ is $\theta=$ \fbox{$\mathrm{arccos}\left(\dfrac{-7}{3\sqrt{11}}\right)$}. In particular, notice that $\B u \bdot \B v = -7<0$ so this angle is obtuse.

\vspace{0.05in}
      
      Is this angle... {\bf \quad right, \quad acute, \quad} or {\bf \;\; \fbox{obtuse}} \;\;? (Circle your answer.)

%\vspace{0.05in}

  \item Match the statement on the left to the corresponding statement on the right\dots

\parbox{3.5in}{
\noindent \hspace{-0.15in}
   \begin{tabular}{lcl}
  \fbox{{\bf\large B}} $\B a \bdot \B b = 0$  & \myspace & {\bf\large A)}  $\B a$ and $\B b$ are parallel \\
  \fbox{{\bf\large A}} $\B a \times \B b = \B 0$   & \myspace &  {\bf\large B)}  $\B a$ and $\B b$ are orthogonal\\
  \fbox{{\bf\large D}} $(\B a \bdot \B b) \times (\B a \bdot \B b) = \B 0$ & \myspace & {\bf\large C)} is always true\\
  \fbox{{\bf\large C}} $(\B a \times \B b) \bdot \B b= 0$  & \myspace & {\bf\large D)}   is nonsense
  \end{tabular}
} \hfill
\parbox{3.25in}{
Recall that the dot product vanishing means we have perpendicular vectors, the cross product being zero means we have parallel vectors, and since $\B a \times \B b$ is perpendicular to both $\B a$ and $\B b$ we have that $(\B a \times \B b) \bdot \B b=0$ is always true. Finally, $\B a \bdot \B b$ is a scalar (not a vector) and we only take cross products of vectors, so the remaining (third) option is nonsense.
}

   \item The vectors $\B a$ and $\B b$ are shown to the right. 
   
            They are based at the point $X$. Sketch the 
            
            vector $-\B a + \B b$ based at the point $P$ and sketch 
            
            the vector $2\B a - \B b$ based at the point $Q$.
   
\vspace{-0.85in}

\force \hfill
\begin{tikzpicture}[scale=0.6]

\colorlet{grayLines}{gray!50}

% parallel to b
\foreach \i in {1,...,7}
{
    \draw[ultra thick,color=grayLines] (-6,0.9*\i)--(6,0.9*\i);
}

% parallel to a
\draw[ultra thick,color=grayLines] (-6,5)--(-4.7,6.5);   % 3.4 horizontal step to next diagonal, slope of diagonals = 15/13
\draw[ultra thick,color=grayLines] (-6,6.5-15/13*4.7)--(-1.3,6.5);
\draw[ultra thick,color=grayLines] (2.1-6.1*13/15,0.4)--(2.1,6.5);
\draw[ultra thick,color=grayLines] (5.5-6.1*13/15,0.4)--(5.5,6.5);
\draw[ultra thick,color=grayLines] (8.9-6.1*13/15,0.4)--(6,3.153846154);

\draw[-,ultra thick,blue!70] (-2+0.9*13/15,2.7)--(-2,1.8);
\node[scale=1.5] (vecC) at (-2+0.45*13/15-0.65,2.25) {\scriptsize{\color{blue!70}$-\B a$}};
\draw[-,ultra thick,blue!70] (-2.4,1.8)--(1.4,1.8);
\node[scale=1.5] (vecB) at (-0.5,1.4) {{\scriptsize\color{blue!70}$\B b$}};

%\draw[-,ultra thick,blue!70] (-2+0.9*13/15,2.7)--(-2,1.8);
%\node[scale=1.5] (vecC) at (-2+0.45*13/15-0.65,2.25) {\scriptsize{\color{blue!70}$-\B a$}};
\draw[-,ultra thick,blue!70] (-2+2.7*13/15,4.5)--(-2+2.7*13/15-3.4,4.5)--(-2+4.5*13/15-3.4,6.3);
\node[scale=1.5] at (-2+2.7*13/15-1.7,4.1) {{\scriptsize\color{blue!70}$-\B b$}};
\node[scale=1.5] at (-2+3.15*13/15-3.8,5) {{\scriptsize\color{blue!70}$\B a$}};
\node[scale=1.5] at (-2+4.05*13/15-3.8,5.9) {{\scriptsize\color{blue!70}$\B a$}};


% vector a
\draw[->,ultra thick] (-5.4,1.8)--(-5.4+13/15*0.9,2.7);
\node[scale=1.5] (vecA) at (-5.4,2.3) {{\scriptsize$\B a$}};

% vector b
\draw[->,ultra thick] (-5.4,1.8)--(-2,1.8);
\node[scale=1.5] (vecB) at (-3.9,1.4) {{\scriptsize$\B b$}};

% point X (in red)
\filldraw[color=red] (-5.4,1.8) circle (0.1);
\node[scale=1.25] (ptX) at (-5.3,1.45) {{\color{red}\scriptsize$X$}};

% over by 0.9*13/15 and up by 0.9

% point P
\filldraw (-2+0.9*13/15,2.7) circle (0.1); 
\node[scale=1] (ptP) at (-2+0.9*13/15 +0.1,2.7 -0.35) {\scriptsize$P$};

% point Q
\filldraw (-2+2.7*13/15,4.5) circle (0.1);
\node[scale=1] (ptP) at (-2+2.7*13/15 +0.1,4.5 -0.35) {\scriptsize$Q$};

% vector -a+b
\draw[->,ultra thick,red] (-2+0.9*13/15,2.7)--(-2+3.4,1.8); 
\node[scale=1.5] (vecC) at (-2+0.45*13/15+2.35,2.45) {\scriptsize{$-\B a+\B b$}};


% vector 2a-b
\draw[->,ultra thick,red] (-2+2.7*13/15,4.5)--(-2+4.5*13/15-3.4,6.3);
\node[scale=1.5] (vecD) at  (-2+3.5*13/15-0.55,5.7) {\scriptsize{$2\B a-\B b$}};


\end{tikzpicture}
\hspace{1in}
\force

\end{enumerate}

\noindent {\bf \large 2. (8 points)} Let $\ell_1$ be parametrized by $\B r_1(t) = \la t, -t+1, 3t+2 \ra$ and let $\ell_2$ be the line which passes through the points $P=(-1,2,-1)$ and $Q=(2,1,0)$. Determine if $\ell_1$ and $\ell_2$ are\dots  (circle the correct answer)

\vspace{0.05in}

\force \hfill {\Large the same, \quad parallel} (but not the same){\Large, \quad \fbox{intersecting}, \quad } or {\Large \quad skew}. \hfill \force

\vspace{0.05in}

The line $\ell_2$ is parameterized by $\B r_2(t) = P+\overrightarrow{PQ}\,t = P+(Q-P)t = \la -1,2,-1\ra + \la 3,-1,1 \ra t = \la -1+3t,2-t,-1+t \ra$. Notice that $\ell_1$ has direction vector $\B r_1'(t) = \la 1,-1,3 \ra$ and $\ell_2$ has direction vector $\B r_2'(t) = \la 3,-1,1 \ra$. Clearly these vectors are not multiples of each other, thus our lines to not run in parallel directions (i.e., they must be intersecting or skew). 

Let's see if they intersect: $\B r_1(s)= \B r_2(t)$ so that $s=-1+3t$, $-s+1=2-t$, and $3s+2=-1+t$. Plugging the first equation into the second gives us $-(-1+3t)+1=2-t$ so $-3t+2=-t+2$ and so $0=2t$. Thus $t=0$. Plugging this into the first equation yields $s=-1$. Now let's see: $\B r_1(-1)=\la -1,2,-1\ra=\B r_2(0)$. Therefore, these lines intersect at $(-1,2,-1)$.


\newpage

\noindent {\bf \large 3. (12 points)} Plane old geometry.

\begin{enumerate}[(a)]
\item Find a (scalar) equation for the plane containing the points $A=(2,1,-1)$, $B=(3,2,1)$, and $C=(2,3,2)$.

\begin{tikzpicture}[scale=0.5]

\colorlet{grayLines}{gray!50}

\draw[ultra thick, color=grayLines] (2.1-6.1*13/15,0.4)--(2.1,6.5)--(5.5+6,6.5)--(5.5+6-6.1*13/15,0.4)--(2.1-6.1*13/15,0.4);

\draw[-,ultra thick,color=black] (5-2+2.7*13/15,4.5-2)--(5-2+4.5*13/15-3.4,6.3-2);

\filldraw (5-2+2.7*13/15-3.4,4.5-2) circle (0.15); 
\node[scale=1.5] (ptP) at (5-2+2.7*13/15-4.1,4.5-2.2) {$A$};

\filldraw (5-2+2.7*13/15,4.5-2) circle (0.15); 
\node[scale=1.5] (ptP) at (5-2+2.7*13/15+0.6,4.5-2) {$B$};

\filldraw (5-2+4.5*13/15-3.4,6.3-2) circle (0.15); 
\node[scale=1.5] (ptP) at (5-2+4.5*13/15-3.4,6.3-2+0.65) {$C$};

\draw[<-,ultra thick,red] (5-2+2.7*13/15,4.5-2)--(5-2+2.7*13/15-3.4,4.5-2);
\draw[->,ultra thick,red] (5-2+2.7*13/15-3.4,4.5-2)--(5-2+4.5*13/15-3.4,6.3-2);
\node[scale=1.5] at (5-2+2.7*13/15-1.7,3.9-2) {{\scriptsize\color{red}$\overrightarrow{AB}$}};
\node[scale=1.5] at (5-2+4.05*13/15-4.6,5.9-2.3) {{\scriptsize\color{red}$\overrightarrow{AC}$}};
\end{tikzpicture}

\vspace{-1.3in}

\hfill
\begin{minipage}{4in}
The vectors $\overrightarrow{AB}=B-A=\langle 1,1,2 \rangle$ and $\overrightarrow{AC}=C-A = \langle 0,2,3 \rangle$ are parallel to the plane containing $A$, $B$, and $C$. Thus $$\overrightarrow{AB} \times \overrightarrow{AC} = \langle 1,1,2 \rangle \times \langle 0,2,3 \rangle = \begin{vmatrix} \B i & \B j & \B k \\ 1 & 1 & 2 \\ 0 & 2 & 3 \end{vmatrix}$$ $$= \langle 1(3)-2(2), -(1(3)-0(2)), 1(2)-0(1) \rangle = \langle -1,-3,2 \rangle$$
\end{minipage}

\noindent
is perpendicular to the plane. Using this vector and one of our points (say $A$), we get a scalar equation for this plane. Therefore, \fbox{$-1(x-2)-3(y-1)+2(z+1)=0$} or equivalently $-x-3y+2z+7=0$ is an equation for this plane.

\vspace{0.05in}

\item Find the area of the triangle $\Delta ABC$ where $A$, $B$, and $C$ are the same points as in part (a).

Obviously, the area of $\Delta ABC$ is half of the area of the parallelogram spanned by $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Since $|\overrightarrow{AB} \times \overrightarrow{AC}|$ is the area of that parallelogram, the area of $\Delta ABC$ is $\dfrac{|\overrightarrow{AB} \times \overrightarrow{AC}|}{2} = \dfrac{|\langle -1,-3,2 \rangle |}{2} = \dfrac{\sqrt{(-1)^2+(-3)^2+2^2}}{2}=$ \fbox{$\dfrac{\sqrt{14}}{2}$}.

\end{enumerate}

\noindent {\bf \large 4. (10 points)} A strange object is observed to have velocity function $\B v(t) = 3t^2\B i-6\B j+e^t\B k$. In addition, this object's initial position was known to be $\B r_0 = 5\B i + 100\B k$. \qquad {}[For what it's worth\dots{}measurements are made in meters and seconds.]

\begin{enumerate}[(a)]

\item This object's acceleration function is $\B a(t)=$ \quad \underline{\qquad $\B v'(t) = 6t\B i+0\B j+e^t\B k =$ \fbox{$6t\B i+e^t\B k$} \qquad}

\item Find this object's position function $\B r(t)$.

Since $\B v(t) = \B r'(t)$, we need to integrate: $\B r(t) = \int \B v(t)\,dt = \int (3t^2\B i-6\B j+e^t\B k)\, dt = t^3\B i-6t\B j+e^t\B k + \B C$. We need to use the initial position to figure out what our constant $\B C$ is. We have $5\B i+100\B k =\B r_0 = \B r(0) = 0\B i+0\B j+e^0\B k+\B C = \B k+\B C$. Thus $\B C=5\B i+99\B k$. Therefore, \fbox{$\B r(t) = (t^3+5)\B i -6t\B j+(e^t+99)\B k$}.

Finally, initial speed is the magnitude of the initial velocity: $|\B v(0)| = |0\B i-6\B j+e^0\B k| = |-6\B j+\B k| = \sqrt{(-6)^2+1^2}=\sqrt{37}$.

Its initial speed was \underline{\quad \fbox{$\sqrt{37}$} \quad} meters per second.
\end{enumerate}

\vspace{-0.15in}

\noindent {\bf \large 5. (10 points)} Parameterize and set up an integral that computes the arc length of the ellipse $\dfrac{(x-1)^2}{4}+\dfrac{(y+2)^2}{9}=1$.

We can build a parameterization from cosine and sine using a kind of modified polar coordinates: 

\force \hfill
\fbox{$\B r(t) = \langle 2\cos(t)+1, 3\sin(t)-2 \rangle$ where $0 \leq t \leq 2\pi$}
\hfill \force

\noindent
noticing that this ellipse is centered at $(1,-2)$ and has radii $2$ and $3$.

For arc length we need the length of the derivative: $|\B r'(t)| = |\langle -2\sin(t),3\cos(t) \rangle |=\sqrt{4\sin^2(t)+9\cos^2(t)}$. 

\force \hfill
Arc Length $\displaystyle = \int_0^{2\pi} |\B r'(t)|\,dt=$ \fbox{$\displaystyle \int_0^{2\pi} \sqrt{4\sin^2(t)+9\cos^2(t)}\,dt$}
\hfill \force

\vspace{0.05in}

\noindent {\bf \large 6. (16 points)} Let $C$ be parameterized by $\B r(t) = \langle \sqrt{2} \sin(t), 2\cos(t), \sqrt{2}\sin(t) \rangle$.

\begin{enumerate}[(a)]
\item Find a parameterization, $\B \ell(t)$, for the line tangent to $C$ at $t=0$.

\vspace{0.05in}

We need a point and a direction vector. This tangent goes through the point $\B r(0)$ and heads in the $\B r'(0)$ direction. We have $\B r'(t) = \langle \sqrt{2}\cos(t),-2\sin(t),\sqrt{2}\cos(t) \rangle$. Thus $\B r(0) = \langle \sqrt{2}\sin(0),2\cos(0),\sqrt{2}\sin(0) \rangle = \langle 0,2,0 \rangle$ and $\B r'(0) = \langle \sqrt{2}\cos(0),-2\sin(0),\sqrt{2}\cos(0) \rangle = \langle \sqrt{2},0,\sqrt{2} \rangle$. Therefore, \fbox{$\ell(t) = \langle 0,2,0 \rangle + t\langle \sqrt{2},0,\sqrt{2} \rangle$}.

\vspace{-0.05in}

\item Find the TNB-frame for $C$.

\vspace{0.05in}

Notice that $|\B r'(t)| = |\langle \sqrt{2}\cos(t),-2\sin(t),\sqrt{2}\cos(t) \rangle | = \sqrt{2\cos^2(t)+4\sin^2(t)+2\cos^2(t)} = \sqrt{4\cos^2(t)+4\sin^2(t)} = \sqrt{4}=2$. Thus \fbox{$\displaystyle \B T(t) = \dfrac{\B r'(t)}{|\B r'(t)|} = \dfrac{1}{2}\langle \sqrt{2}\cos(t),-2\sin(t),\sqrt{2}\cos(t) \rangle$}. Next, $\B T'(t) = \dfrac{1}{2} \langle -\sqrt{2}\sin(t),-2\cos(t),-\sqrt{2}\sin(t) \rangle$ and so $|\B T'(t)| = \dfrac{1}{2}\sqrt{2\sin^2(t)+4\cos^2(t)+2\sin^2{t}} = \dfrac{1}{2}\sqrt{4\sin^2{t}+4\cos^2{t}} = 1$. Therefore,\\
\fbox{$\displaystyle \B N(t) = \dfrac{\B T'(t)}{|\B T'(t)|} = \dfrac{1}{2} \langle -\sqrt{2}\sin(t),-2\cos(t),-\sqrt{2}\sin(t) \rangle$}. Finally, $\B B(t) = \B T(t) \times \B N(t) = \dfrac{1}{2}\langle \sqrt{2}\cos(t),-2\sin(t),\sqrt{2}\cos(t) \rangle \times \dfrac{1}{2} \langle -\sqrt{2}\sin(t),-2\cos(t),-\sqrt{2}\sin(t) \rangle$ $=\dfrac{1}{4} \langle 2\sqrt{2}\sin^2(t)-(-2\sqrt{2}\cos^2(t)),$ $-(-2\sqrt{2}\cos(t)\sin(t)-(2\sqrt{2}\sin(t)\cos(t))),$\\ $-2\sqrt{2}\cos^2(t)-(2\sqrt{2}\sin^2(t))\rangle$ $=\dfrac{1}{4}\langle 2\sqrt{2},0,-2\sqrt{2}\rangle$. Thus \fbox{$\displaystyle \B B(t) = \left\langle \dfrac{\sqrt{2}}{2},0,-\dfrac{\sqrt{2}}{2} \right\rangle$}.

\vspace{-0.15in}

\item Compute the curvature of $C$.

\vspace{-0.25in}

\force \hfill
$\kappa = \dfrac{|\B T'(t)|}{|\B r'(t)|} =$ \fbox{$\dfrac{1}{2}$}
\hfill \hfill \force

%\vspace{-0.05in}

\end{enumerate}

It may be interesting to note that since our binormal, $\B B(t)$, is a constant vector, $C$ is a planar curve. Moreover, our curvature is constant: $\kappa = 1/2$. It turns out that a planar curve with (non-zero) constant curvature must be a circle (or a piece of a circle) of radius $1/\kappa$. Thus, $C$ is actually a circle of radius $2$ with center $(0,0,0)$ and lying in the plane perpendicular to $\B B(t)$. Specifically, this plane is $x=z$.

\vspace{0.15in}

\noindent {\bf \large 7. (22 points)} Consider the curve $C$ parameterized by $\B r(t) = \la t^2, 3t, e^t \ra$, $-3 \leq t \leq 10$. 

\vspace{-0.1in}

$$\B r'(t) = \langle 2t,3,e^t \rangle \qquad \qquad \B r''(t) = \langle 2,0,e^t \rangle \qquad \qquad \B r'''(t) = \langle 0,0,e^t \rangle$$

\vspace{-0.15in}

\begin{enumerate}[(a)]
\item Compute the curvature of $\B r(t)$.

\vspace{-0.05in}

$$\B r'(t) \times \B r''(t) =  \langle 2t,3,e^t \rangle \times  \langle 2,0,e^t \rangle = \langle 3e^t-0(e^t), -(2te^t-2e^t), 0-6 \rangle = \langle 3e^t,2(1-t)e^t,-6\rangle$$

Thus $|\B r'(t) \times \B r''(t)| = \sqrt{9e^{2t}+4(t-1)^2e^{2t}+36\strut}$ and also $|\B r'(t)| = \sqrt{4t^2+9+e^{2t}\strut}$. Therefore,

\force\hfill $\displaystyle \kappa(t) = \dfrac{|\B r'(t) \times \B r''(t)|}{|\B r'(t)|^3} =$ \fbox{$\dfrac{\sqrt{9e^{2t}+4(t-1)^2e^{2t}+36\strut}}{\left(4t^2+9+e^{2t}\strut\right)^{3/2}}$} \hfill \force

\vspace{-0.15in}

\item Compute the torsion of $\B r(t)$.

\vspace{0.05in}

Note that $(\B r'(t) \times \B r''(t)) \bdot \B r'''(t) =  \langle 3e^t,2(1-t)e^t,-6\rangle \bdot \langle 0,0,e^t \rangle = -6e^t$. Therefore,

\force\hfill $\displaystyle \tau(t) = \dfrac{(\B r'(t) \times \B r''(t)) \bdot \B r'''(t)}{|\B r'(t) \times \B r''(t)|^2}=$ \fbox{$\displaystyle \dfrac{-6e^t}{9e^{2t}+4(t-1)^2e^{2t}+36}$} \hfill \force

\vspace{-0.1in}

\item Set up the integral $\displaystyle\int_C (z+x\sin(y))\,ds$  \quad {}[Obviously we cannot hope to evaluate this by hand -- please don't try.]

\vspace{0.05in}

From our parameterization, $\B r(t)$, we have $x=t^2$, $y=3t$, and $z=e^t$. Also, $ds = |\B r'(t)|\,dt = \sqrt{4t^2+9+e^{2t}\strut}\,dt$. Therefore,

\force\hfill $\displaystyle\int_C (z+x\sin(y))\,ds =$ \fbox{$\displaystyle\int_{-3}^{10} (e^t+t^2\sin(3t))\sqrt{4t^2+9+e^{2t}\strut}\,dt$} \hfill\force

\vspace{0.05in}

\item Compute the tangential and normal components of acceleration of $\B r(t)$.

\vspace{0.05in}

Note that $\B r'(t) \bdot \B r''(t) = \langle 2t,3,e^t \rangle \bdot \langle 2,0,e^t \rangle = 2t(2)+3(0)+e^t(e^t) = 4t+e^{2t}$.

\force\hfill $a_T=\dfrac{\B r'(t) \bdot \B r''(t)}{|\B r'(t)|} =$ \fbox{$\dfrac{4t+e^{2t}}{\sqrt{4t^2+9+e^{2t}\strut}}$} \hfill
$a_N=\dfrac{|\B r'(t) \times \B r''(t)|}{|\B r'(t)|} =$ \fbox{$\dfrac{\sqrt{9e^{2t}+4(t-1)^2e^{2t}+36\strut}}{\sqrt{4t^2+9+e^{2t}\strut}}$} 
\hfill\force

\vspace{0.05in}

\item Does this curve lie in a plane? Why or why not?

\vspace{0.05in}

\fbox{No.} When a curve is planar, its binormal is constant and its torsion is zero. Notice that the torsion $\tau(t) \not=0$, so our curve is not planar.

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\end{enumerate}

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