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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 June $5^\mathrm{th}$, 2019}

\vspace{0.05in}

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

\vspace{0.1in}

%\noindent
%$\displaystyle{\mbox{proj}_\B w(\B v) = \frac{\B v \bdot \B w}{\,|\B w|^2}\B w }$ \hfill 
%$\displaystyle{\mathbf{r}''(t) = \left(\frac{\mathbf{r}'(t) \bdot \mathbf{r}''(t)}{|\mathbf{r}'(t)|} \right) \mathbf{T}(t) + 
%                            \left(\frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|} \right) \mathbf{N}(t)}$ \hfill
%$\displaystyle{\kappa = \left|\frac{d\mathbf{T}}{ds}\right| = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}}$
%
%\vspace{0.1in}
%
%\noindent
%$\displaystyle m = \int_C \delta(x,y,z)\,ds \qquad \left(\bar{x},\bar{y},\bar{z}\right) = \frac{1}{m}\left( \int_C x\delta(x,y,z)\,ds, \int_C y\delta(x,y,z)\,ds, \int_C z\delta(x,y,z)\,ds \right)$ \hfill \hfill
%$\displaystyle \tau = \dfrac{(\B r'(t) \times \B r''(t)) \bdot \B r'''(t)}{|\B r'(t) \times \B r''(t)|^2}$
%
%\vspace{0.1in}

\noindent {\bf \large 1. (24 points)} Vector Basics: Let $\B v = \la -2,0,1 \ra$ and $\B w = \la 1,2,-2 \ra$.
 
\vspace{-0.1in} 
  
\begin{enumerate}[(a)]
  \item Compute $\mathrm{proj}_{\B w}(\B v) = \dfrac{\B v \bdot \B w}{|\B w|^2}\B w = \dfrac{(-2)1+0(2)+1(-2)}{1^2+2^2+(-2)^2}\langle 1,2,-2\rangle =$ \fbox{$\dfrac{-4}{9}\langle 1,2,-2\rangle$}.

   \item Find two {\it unit} vectors which are {\it orthogonal} (i.e., perpendicular) to both $\B v$ and $\B w$.

\vspace{-0.3in}

$$\B v \times \B w = \begin{vmatrix} \B i & \B j & \B k \\ -2 & 0 & 1 \\ 1 & 2 & -2 \end{vmatrix} = 
\begin{vmatrix} 0 & 1 \\  2 & -2 \end{vmatrix}\B i - 
\begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix}\B j +
\begin{vmatrix} -2 & 0 \\ 1 & 2 \end{vmatrix}\B k = (0(-2)-2(1))\B i-((-2)(-2)-1(1))\B j+((-2)2-1(0))\B k$$

\vspace{-0.1in}

$= -2\B i-3\B j -4\B k = \langle -2,-3,-4 \rangle$ is perpendicular to both $\B v$ and $\B w$. To get a unit vector we need to normalize: $\dfrac{\B v \times \B w}{|\B v \times \B w|} = \dfrac{1}{\sqrt{(-2)^2+(-3)^2+(-4)^2}}\langle -2,-3,-4 \rangle$. To get a second unit vector (perpendicular to $\B v$ and $\B w$) we negate. Thus \fbox{$\pm \dfrac{\B v \times \B w}{|\B v \times \B w|} = \dfrac{\pm 1}{\sqrt{29}}\langle 2,3,4 \rangle$} give us two unit vectors perpendicular to both $\B v$ and $\B w$.

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

Recall that $\B v \bdot \B w = |\B v|\,|\B w|\cos(\theta)$ where $\theta$ is the angle between $\B v$ and $\b w$. Solving for $\theta$ we get\dots

\vspace{-0.1in}

$$\theta = \mathrm{arccos}\left(\dfrac{\B v \bdot \B w}{|\B v|\,|\B w|}\right) = \mathrm{arccos}\left( \dfrac{-4}{\sqrt{5}\sqrt{9}}\right) = \mbox{\fbox{$\mathrm{arccos}\left(\dfrac{-4}{3\sqrt{5}}\right)$}}$$

\vspace{-0.1in}
      
      Is this angle... {\bf \quad right, \quad acute, \quad} or {\bf \;\; \fbox{obtuse}} \;\;? (Circle your answer.) since $\B v \bdot \B w = -4<0$ (negative).

\vspace{0.05in}

   \item Let $\B a$, $\B b$, and $\B c$ be vectors in $\mathbb{R}^3$. Fill in the blanks:

If $\B a \times \B b = \B 0$, then $\B a$ and $\B b$ are \underline{\quad parallel \quad}.

If $\B a \bdot \B b = 0$, then $\B a$ and $\B b$ are \underline{\quad perpendicular (or orthogonal) \ \ }.

If $(\B a \times \B b) \bdot \B c = 0$, then $\B a$, $\B b$, and $\B c$ are \underline{\quad coplanar \quad}.

%\vspace{0.1in}

   \item The vectors $\B a$ and $\B b$ are shown to the right. 
   
            They are based at the point $X$. Sketch the 
            
            vector $2\B a - \B b$ based at the point $P$ and sketch 
            
            the vector $-\B a + \B b$ based at the point $Q$.
   
\vspace{-1.65in}

\force \hfill
\begin{tikzpicture}[scale=0.7]

\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);

% 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) {{$\B a$}};

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

% point X (in red)
\filldraw[color=red] (-5.4,1.8) circle (0.1);
\node[scale=1.25] (ptX) at (-5.3,1.35) {{\color{red}$X$}};

% 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) {$P$};

% vector 2a-b
\draw[->,ultra thick,color=green] (-2+0.9*13/15,2.7)--(-2+0.9*13/15+13/15*0.9,2.7+0.9);
\draw[->,ultra thick,color=green] (-2+0.9*13/15+13/15*0.9,2.7+0.9)--(-2+2*0.9*13/15+13/15*0.9,2.7+2*0.9);
\draw[->,ultra thick,color=green] (-2+2*0.9*13/15+13/15*0.9,2.7+2*0.9)--(-2+2*0.9*13/15+13/15*0.9-3.4,2.7+2*0.9);
\draw[->,ultra thick,color=blue] (-2+0.9*13/15,2.7)--(-2+2*0.9*13/15+13/15*0.9-3.4,2.7+2*0.9);
\node[scale=1.5] (ansA) at (-2+0.9*13/15+13/15*0.9-2.7,2.7+0.5) {{\color{blue}$2\B a- \B b$}};


% 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) {$Q$};

% vector -a+b
\draw[->,ultra thick,color=red] (-2+2.7*13/15,4.5)--(-2+2.7*13/15 - 0.9*13/15, 4.5 - 0.9);
\draw[->,ultra thick,color=red] (-2+2.7*13/15 - 0.9*13/15, 4.5 - 0.9)--(-2+2.7*13/15 - 0.9*13/15 + 3.4, 4.5 - 0.9);
\draw[->,ultra thick,color=blue] (-2+2.7*13/15,4.5)--(-2+2.7*13/15 - 0.9*13/15 + 3.4, 4.5 - 0.9);
\node[scale=1.5] (ansB) at (-2+2.7*13/15 - 0.9*13/15 + 3.25, 4.5 - 0.2) {{\color{blue}$-\B a+ \B b$}};
\end{tikzpicture}

\end{enumerate}

\vspace{-0.15in}

\noindent {\bf \large 2. (14 points)} Give vector valued functions which parameterize the following curves.\\ \force \hfill {\bf Don't forget} to specify a domain for your parameter.

\vspace{-0.1in}

\begin{enumerate}[(a)]

\item The line segment from $A = (1,2,0)$ to $B = (4,0,1)$. \quad $\B r(t) = A+\overrightarrow{AB}\,t = A+(B-A)t =$ \fbox{$\langle 1,2,0 \rangle + \langle 3,-2,1 \rangle t$, \quad $0 \leq t \leq 1$}

\item The circle $(x+1)^2+(y-3)^2=25$. \qquad \fbox{$\B r(t) = \langle 5\cos(t)-1, 5\sin(t)+3 \rangle$, \quad $0 \leq t \leq 2\pi$}

Of course, there are infinitely many correct (and incorrect) answers for problem \#2. For example, if we exchanged sine and cosine, we would still parameterize our above circle (centered at $(-1,3)$ with radius $\sqrt{25}=5$). The given parameterization is the ``standard'' one. This traverses the circle once in a counter-clockwise direction beginning and ending at the rightmost point (i.e., $(x,y) = (-1,3)+(5,0) = (4,3)$). Interchanging sine and cosine would cause the circle to be traced out in a clockwise direction beginning and ending at its top.

 Also, neglecting the domain on part (a) would yield the entire line through $A$ and $B$. Neglecting the domain on part (b) would be ok (we would just trace out the circle over and over and over \dots).

\end{enumerate}

\noindent {\bf \large 3. (14 points)} Lines and Planes

\vspace{-0.1in}

\begin{enumerate}[(a)]
\item Find an equation for the plane which passes through the points $A=(1,0,-1)$, $B=(2,3,0)$, and $C=(3,-1,1)$.

We need a point and a normal vector. Any point ($A$, $B$, or $C$) will do. To get a normal vector we note that $\overrightarrow{AB}=B-A=\langle 1,3,1 \rangle$ and $\overrightarrow{AC} = C-A = \langle 2,-1,2 \rangle$ are parallel to the plane through $A,B,C$. Thus $\langle 1,3,1 \rangle \times \langle 2,-1,2\rangle = \langle 7,0,-7\rangle$ is perpendicular to the plane (it's our desired normal vector). Using point $A$ and our normal vector, we get \fbox{$7(x-1)+0(y-0)-7(z+1)=0$} (this simplifies to $x-z-2=0$).


\item Consider the line parameterized by $\B r(t) = \la 2t,-4t+1,6t-2\ra$  and the plane $-x+2y-3z=10$. 

 The line and plane are (circle {\bf all} that apply)\dots \hfill {\Large parallel \qquad \fbox{perpendicular} \qquad \fbox{intersecting}}.

The line is parallel to $\B r' = \langle 2,-4,6 \rangle$. The plane is perpendicular to $\B n = \langle -1,2,-3 \rangle$.  Notice that $\B r' = -2\B n$ (i.e., the line's direction is {\it parallel} to the plane's normal vector). Thus means that the line and plane are perpendicular (since $\B r'$ goes with the line and $\B n$ goes perpendicular to the plane) to each other (so they also necessarily intersect). 

\end{enumerate}

\noindent {\bf \large 4. (12 points)} Suppose that a particle's velocity is given by $\B v(t) = t^2\B i + 5e^t\B j$. In addition, we have that this particle's initial position is $\B r_0 = \B i +2\B j$. \hfill [For what it's worth, measurements are made in meters and seconds.]

\begin{enumerate}[(a)]

\item The particle's initial speed is \underline{\qquad $|\B v(0)| = |\langle 0^2,5e^0 \rangle| = |\langle 0,5 \rangle| = 5$ \qquad} meters per second.

\item Find the particle's acceleration $\B a(t)$. \qquad  Since $\B v(t)=\B r'(t)$. $\B a(t) = \B r''(t) = \B v'(t) =$ \fbox{$\langle 2t,5e^t \rangle$} (or $2t\B i+5e^t\B j$).

\item Find the particle's position function $\B r(t)$. \qquad $\B r(t) = \int \B r'(t)\,dt = \int \B v(t)\,dt = \int \langle t^2,5e^t \rangle\,dt = \left\langle \dfrac{t^3}{3}, 5e^t \right\rangle +\B C$. However, $\langle 1,2 \rangle = \B r_0 = \B r(0) = \langle 0^3/3, 5e^0 \rangle + \B C$ so $\B C = \langle 1,2 \rangle - \langle 0,5 \rangle = \langle 1,-3 \rangle$. Thus \fbox{$\B r(t) = \left\langle \dfrac{t^3}{3}+1, 5e^t-3 \right\rangle$}.

\end{enumerate}


\noindent {\bf \large 5. (18 points)} Let $C$ be the curve parameterized by $\B r(t) = \la 3t, 4\sin(t), 4\cos(t) \ra$, $-\pi \leq t \leq 3\pi$.

\begin{enumerate}[(a)]

\item Compute the $\B T \B N \B B$-frame of $C$. \quad
$\B r'(t) = \langle 3,4\cos(t),-4\sin(t) \rangle$ and $|\B r'(t)| = \sqrt{9+16\cos^2(t)+16\sin^2(t)} = \sqrt{9+16}=5$. Thus 

\vspace{-0.3in}

$$\B T(t) = \dfrac{\B r'(t)}{|\B r'(t)|} = \mbox{\fbox{$\dfrac{1}{5}\langle 3,4\cos(t),-4\sin(t) \rangle$}}$$

$\B T'(t) = \dfrac{1}{5}\langle 0, -4\sin(t), -4\cos(t) \rangle$ and $|\B T'(t)| = \dfrac{1}{5} \sqrt{0^2+16\sin^2(t)+16\cos^2(t)} = \dfrac{1}{5}\sqrt{16} = \dfrac{4}{5}$. Thus 

\vspace{-0.15in}

$$\B N(t) = \dfrac{\B T'(t)}{|\B T'(t)|} = \dfrac{\dfrac{1}{5}\langle 0,-4\sin(t),-4\cos(t) \rangle}{4/5} = \mbox{\fbox{$\langle 0,-\sin(t),-\cos(t) \rangle$}}$$

\vspace{-0.1in}

$$\B B(t) = \B T(t) \times \B N(t) = \dfrac{1}{5}\langle -4\cos^2(t)-4\sin^2(t), -(-3\cos(t)), -3\sin(t) \rangle = \mbox{\fbox{$\dfrac{1}{5}\langle -4,3\cos(t),-3\sin(t) \rangle$}}$$

\vspace{-0.15in}

\item Compute the curvature of $C$. 

\vspace{-0.05in}

If we use the {\it right formula} (taking into account part (a)'s calculations) this is easy: $\kappa = \dfrac{|\B T'|}{|\B r'|} = \dfrac{4/5}{5} =$ \fbox{$\dfrac{4}{25}$}.

\vspace{-0.05in}

\item Compute the arc length of $C$.

\vspace{-0.25in}

$$\qquad \qquad \mathrm{Total Arc Length} = \int_a^b |\B r'(t)|\,dt = \int_{-\pi}^{3\pi} 5\,dt = 5t\Big|_{-\pi}^{3\pi} = \mbox{\fbox{$20\pi$}}$$

\item Set up (but {\bf do not evaluate}) the line integral $\int_C x\,e^{y^2+z^2}\,ds$.

We plug in our parameterization: $x(t) = 3t$, $y(t) = 4\sin(t)$, $z(t) = 4\cos(t)$, and $ds = |\B r'(t)|\,dt = 5\,dt$. Thus\dots

\vspace{-0.05in}

$$\int_C x\,e^{y^2+z^2}\,ds = \int_{-\pi}^{3\pi} 3t\,e^{16\cos^2(t)+16\sin^2(t)}5\,dt = \mbox{\fbox{$\displaystyle \int_{-\pi}^{3\pi} 15e^{16}\,t\,dt $}}$$

\end{enumerate}


\noindent {\bf \large 6. (18 points)} Consider the curve $C$ parameterized by $\B r(t) = \la t, t^2, t^3 \ra$. 

\begin{enumerate}[(a)]
\item Find a parameterization for the line tangent to $C$ at $t=2$.

For future reference: $\B r'(t) = \langle 1,2t,3t^2 \rangle$, $\B r''(t) = \langle 0,2,6t \rangle$, and $\B r'''(t) = \langle 0,0,6 \rangle$.

Thus $\B r(2) = \langle 2,4,8 \rangle$ (point) and $\B r'(2) = \langle 1,4,12 \rangle$ (direction). 

Therefore, \fbox{$\ell(t) = \langle 2,4,8 \rangle +\langle 1,4,12 \rangle t$} parameterizes our tangent line.

\item Compute the curvature of $C$. 

Since we haven't computed this curve's TNB-frame (which would be fairly horrible), we are better off using the cross product formula for curvature.

$\B r' \times \B r'' = \langle 1,2t,3t^2 \rangle \times \langle 0,2,6t \rangle = \langle 2t(6t)-2(3t^2), -(1(6t)-0(3t^2)), 1(2)-0(2t) \rangle = \langle 6t^2,-6t,2 \rangle = 2\langle 3t^2,-3t,1\rangle$. 
$$\kappa(t) = \dfrac{|\B r' \times \B r''|}{|\B r'|^3} = \mbox{\fbox{$\dfrac{2\sqrt{9t^4+9t^2+1\strut}}{\left(1+4t^2+9t^4\strut\right)^{3/2}}$}}$$

\item Compute the torsion of $C$.

$$\tau(t) = \dfrac{(\B r' \times \B r'') \bdot \B r'''}{|\B r' \times \B r''|^2} = \dfrac{\langle 6t^2,-6t,2\rangle \bdot \langle 0,0,6 \rangle}{36t^4+36t^2+4} = \dfrac{12}{36t^4+36t^2+4}= \mbox{\fbox{$\dfrac{3}{9t^4+9t^2+1}$}}$$

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

$$a_{\B T} = \dfrac{\B r' \bdot \B r''}{|\B r'|} = \mbox{\fbox{$\dfrac{4t+18t^3}{\sqrt{9t^4+4t^2+1}}$}} 
\qquad \text{and} \qquad
a_{\B N} = \dfrac{|\B r' \times \B r''|}{|\B r'|} = \mbox{\fbox{$\dfrac{2\sqrt{9t^4+9t^2+1}}{\sqrt{9t^4+4t^2+1}}$}}$$

\item Does this curve lie in a plane? Why or why not? \qquad \fbox{No, since $\tau \not=0$.}

A smooth curve is planar if and only if its torsion is identically zero. Alternatively, if we can compute a  TNB-frame, a curve is planar if and only if the binormal is constant.


\end{enumerate}


\end{document}