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Test 1
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<h2 class="justme">There was no cross product on this one</h2>
<h3>Name-<br>
Math 213, May 2018, Test 1</h3>
<p class="justme answer">Results:<br>
10|&nbsp;23
&nbsp;9|&nbsp;44<br>
&nbsp;8|&nbsp;6<br>
&nbsp;7|&nbsp;18<br>
6|

<p>On this exam, I'll write vector quantities in a bold-face, non-italic font  like: $\bf v$, $\bf a_T$. Scalar quantities will be in a non-bold, italic font, like this: $s$, $m_x$.  I'll write unit vectors as vectors with a caret overtop, like this: $\uv k$, $\uv T$.
<ol><li>
Let $\bf a=\overrightarrow{OP}$ where $P$ is the point $(2,2,\sqrt 2)$.<br>
<img src="0g/T1.2.jpg"><br>
<ol type="a">
<li>Find the components of vectors $\bf b$ and $\bf c$. (The tail of ${\bf b}$ is at the origin.)
<p class="answer">$$\myv b=\myc{2,2,0};\ \ \ \myv c=(0,0,\sqrt 2).$$

<li>
Find the distance of the point $P$ from the origin.
<p class="answer">$$\sqrt{2^2+2^2+\sqrt 2^2}=\sqrt{10}=3.16$$
<li>
Find the distance of the point $P$ from the $z$-axis.
<p class="answer">$$\sqrt{2^2+2^2}=\sqrt 8=2.83$$
<li>
What are the coordinates $(\rho, \theta, \phi)$ of $P$ in spherical coordinates?
<p class="answer">$\rho = \sqrt{10}=3.16$;
<br>$\theta = \arctan(1/1)=\pi/4=0.785$<br>
$\phi = \arcsin(\sqrt 8/\sqrt{10})=\arccos(\sqrt 2/\sqrt{10})=1.11$ [R]=$63.4^o$.
</ol>



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<li style="page-break-before:always">
Let ${\bf a}=(u+v){\uv i}+{\uv j}+3{\uv k}$, and
${\bf b}=2{\uv i}+(2u+3v+1){\uv j} + {\uv k}$.  Find a value for $u$ and a value for $v$ such 
that ${\bf a} \perp {\bf b}$.

<p class="answer">If $\myv a \perp \myv b$, then $\myv a \cdot \myv b=0$:
$$\begineq
0&=&a_xb_x+a_yb_y+a_zb_z=(u+v)*2+1*(2u+3v+1)+3*1\\
0&=&4u+5v+4
\endeq
$$
Many solutions are possible, as long as $4u+5v=-4$.  One is $u=9;\ v= -8$.


<li style="margin-top:25ex;">
Find an equation of the plane through the point $(0,-2,3)$ with normal vector ${\bf n} = \myc{3,-2, 1}$.  <!--(Second part:  Find an equation for the intersection of this plane with the x-y plane?-->

<p class="answer">Using the components of the normal vector, the plane could have the equation $3x-2y+1z=c$.  Substituting in the coordinates of the one point we know...
$$c=3(0)-2(-2)+1(3)=4+3=7$$
So the equation is:
$$3x-2y+z=7.$$

<li style="margin-top:25ex;">
Consider two vectors, ${\bf a}=\langle -5,1,3 \rangle$ and
${\bf b}=\langle 2,4,2 \rangle$.
<ol type="a">
<li>What's the angle between ${\bf a}$ and ${\bf b}$? (Give your answer in 
radians.)
<p class="answer">$\myv a \cdot \myv b=-5*2+1*4+3*2=0$.  So the two vectors are at right angles ($90^o$) to each other.  $90^o *(\pi \text{ [Rad] }/180^o)=\frac{\pi}{2}.$

<li>Find a vector,  ${\bf v}$, which is perpendicular to both $\bf a$ and $\bf b$.
<p class="answer">
$$\myv v=\myv a \times \myv b =\det \begincv
 \uv i& \uv j &\uv k\\
-5 & 1 & 3\\
2 & 4 & 2
\endcv=\langle -10, 16, -22\rangle.$$
But $\myc{10,-16,+22}$, $\myc{5,-8,11}$ and any scalar multiple of these also work.

<li>Find an equation for the line which passes through the point $\myc{0,3,5}$ and is parallel to  $\bf v$.
 <p class="answer">
$$\myv r(t)=\myv r_0 +t\myv v=\langle -10t,3+16t,5-22t\rangle.$$
</ol>
 
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<li style="page-break-before:always">
Which curve below is the path traced out by ${\bf r}(t)=
\langle \sin \pi t, \cos \pi t, \frac14t^2\rangle$, $0\leq t \leq 2$?  Justify your answer.<br>
<img src="0g/T1curves.jpg"><br>

<p class="answer">Some considerations:
<ul class="answer">
<li>$\left.\myv r(t)\right|^{t=0}=\langle 0,1,0 \rangle$, eliminating #3.
<li>As $t$ increases, $z(t)=\frac14t^2$ will always increase, eliminating #2. 
<li>$\left.\myv r(t)\right|^{t=2}=\langle 0,1,1 \rangle$, eliminating #2 and #3.
</ul>
<p class="answer">#1 is the correct graph.

<li style="margin-top: 8ex">
A crazed ostrich ("Oscar") runs along a mountain path with
coordinates
given by
${\bf r} (t) =  \langle  3e^{−t}, 4(\cos t+t), t^2/18 \rangle$.
<ol type="a">
<li>What is the change in Oscar’s altitude ($z$-component) from t = 0 to t = 6?
<p class="answer">$\Delta z= z(6)-z(0)=36/18=2.$
<li>What is the $x$ component of Oscar’s <strike>speed</strike> <span style="color: red">velocity</span> ($v_x$) when $t$ = 2?
<p class="answer">$x'(t)=-3e^{-t};\ \ x'(2)=-0.406$
<li>What is Oscar’s initial speed (when $t$ = 0)?
<p class="answer">$\myv r'(t)
=-3e^{-t}\uv i+4(-\sin(t)+1)\uv j$, so $\myv r'(0)= \myc{-3,4,0}$, and 
speed = $|\myv r'(0)|=\sqrt{(-3)^2+4^2}=5$.
<li>Find a formula (but don't try to solve the integral) for the total distance that
Oscar travels from $t=0$ to $t=6$.
<p class="answer">$$\begineq L&=&\int_{t_o}^{t_f}|\myv r'(u)|\,du\\
&=&\int_{t=0}^6\sqrt{(x'(u))^2+(y'(u))^2+(z'(u))^2}\,du\\
&=&\int_0^6\sqrt{(-3e^{-u})^2+(4-4\sin u)^2+(u/9)^2}\,du\\
\endeq$$

</ol>



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<li style="page-break-before: always;">
Consider the vector function ${\bf r} (t)$
 describing the curve shown below. Put the curvatures of ${\bf r} (t)$ at $A$, $B$, and $C$ in order from smallest to largest. 
<br>
<p><img src="0g/T1.curvatures.jpg" style="width:400px; padding-right:6ex;float:left;">
<span class="answer">B</span><-- smallest<br><br>
<span class="answer">A</span><--<br><br>
<span class="answer">C</span><--largest
</p>

<br clear="both">

<li>

Match the equations with their graphs.  Give a reason for your choices.<br>
<img src="0g/1.matchmtext.jpg"><br>
<img src="0g/1.matchmgraphs.jpg">
<p class="answer">
a-II - only equation of a plane.<br>
b-I - hold $x$ constant, $z=\cos y$; hold $y$ constant, $z=\sin x.$<br>
c-IV - at (0,0), $z=1$, which is the maximum value for the function $\sin$("something").<br>
d-III - No $x$ dependence, so surface should be curve $z=e^y$ dragged back and forth in the $x$ direction.<br>


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<li style="page-break-before:always">
Let $y = x^2$ be the parabolic trajectory of a particle in the $xy$ plane parameterized by ${\bf r}(t)=
\myc{t,t^2,0}$.  This is shown below, as well as an arrow indicating the direction in which the particle traverses the trajectory.
<br><img src="0g/T1.xyparabolaArrows.jpg"></br>
<ol type="a">
<li> 
Compute ${\bf r}'(t)$ and ${\bf r}''(t)$
<p class="answer">
$$\myv r'(t)=\myc{ 1, 2t, 0};\ \ \ \myv r''(t)=\myc {0,2,0}.$$

<li style="margin-top:20ex;">
What are the vectors $\uv T$ and $\uv N$ (the unit tangent, and the unit normal) at the origin (0,0,0)? (No complicated computations are needed.)

<p class="answer"><b>Easy way</b>: The tangent vector is <i>tangent</i> to the curve.  At (0,0,0) the curve is tangent to the $x$ axis.  
$$\uv T = \myc{1,0,0}.$$
The normal vector is perpendicular to the tangent vector, and points towards the center of the osculating circle.  At (0,0,0), this would be in the $y$ direction:
$$\uv N = \myc{0,1,0}.$$
<p class="answer"><b>Hard way</b>: At the origin, $\myv r(t)=(0,0,0)$ so this implies that $t=0$.<br><br>
The tangent vector is
$$\uv T(t)=\frac{\myv r'}{|\myv r'|}=\frac{\myc{1,2t,0}}{\sqrt{1+4t^2}}.$$
Evaluating this at $t=0$:
$$\uv T(0)=\frac{\myc{1,0,0}}{\sqrt 1}=\myc{1,0,0}.$$
A normal vector is $\myv N = \frac{d}{dt}\uv T$.  Evaluating the components of this vector:
$$ N_x=-\frac12 (1+4t^2)^{-3/2} 8t;
\ \ \left.N_x\right|^{t=0}=0.$$
$$N_y=2(1+4t^2)^{-1/2}+2t(1+4t^2)^{-3/2}8t;\ \ \left.N_x\right|^{t=0}=2.$$
$$N_z=0.$$
So $\myv N=\myc{0,2,0}$ and a unit vector is 
$$\uv N=\myc{0,1,0}.$$
</ol>





</ol>
<p style="page-break-before:always">
<i>This page is left blank for you to use as scratch paper.  If you would like
any work here to count for partial credit, label it with the question number.</i>
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