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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, Fall 2015, Test 1</h3>
<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">$$\myv b=\myc{1,1,0};\ \ \ \myv c=(0,0,\sqrt 2).$$

<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 ${\bf a}=(u+v){\uv i}+2{\uv j}+v{\uv k}$, and
${\bf b}=3{\uv i}+(4u+v+1){\uv j} + 4{\uv k}$.  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 $\myv a \cdot \myv b=0$:
$$\begineq
0&=&a_xb_x+a_yb_y+a_zb_z=(u+v)3+2(4u+v+1)+(v)4\\
&=&3u+3v+8u+2v+2+4v\\
0&=&11u+9v+2
\endeq
$$
Many solutions are possible.  One is $u=-1;\ v=1$.


<li style="margin-top:25ex;">
Find an equation of the plane through the point $(0,-3,3)$ with normal vector ${\bf n} = \myc{2,-2, 1}$.  <!--(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 $2x-2y+1z=c$.  Substituting in the coordinates of the one point we know...
$$c=2(0)-2(-3)+1(3)=6+3=9$$
So the equation is:
$$2x-2y+z=9.$$

<li style="margin-top:25ex;">
Consider two vectors, ${\bf a}=\langle 4,2,0 \rangle$ and
${\bf b}=\langle 1,-2,1 \rangle$.
<ol type="a">
<li>What's the angle between ${\bf a}$ and ${\bf b}$? (Give your answer in 
radians.)
<p class="answer">$\myv a \cdot \myv b=4*1+2*(-2)+0*1=0$.  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">
$$\myv v=\myv a \times \myv b =\det \begincv
 \uv i& \uv j &\uv k\\
4 & 2 & 0\\
1 & -2 & 1
\endcv=\langle 2, -4, -10\rangle.$$

<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">
$$\myv r(t)=\myv r_0 +t\myv v=\langle 2t,3-4t,5-10t\rangle.$$
</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
${\bf r} (t) =  \langle e^t, e^{−t}, \sqrt{2t} \rangle$.
<ol type="a">
<li>What is the change in Oscar’s altitude ($z$-component) from t = 0 to t = 10?
<p class="answer">$\Delta z= z(10)-z(0)=\sqrt(2*10)-0=4.47$
<li>What is Oscar’s speed in the $x$-direction when $t$ = 4?
<p class="answer">$x'(t)=e^t;\ \ x'(4)=e^4=54.6$
<li>What is Oscar’s speed in the $y$-direction when $t$ = 4?
<p class="answer">$y'(t)=-e^{-t};\ \ y'(4)=-e^{-4}=-0.018.$
<li>Find a formula (but don't try to solve it) for the total distance that
Oscar travels from $t=0$ to $t=4$.
<p class="answer">$$\begineq L&=&\int_{t_o}^{t_f}|\myv r'(t)\,dt\\
&=&\int_0^4\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}\,dt\\
&=&\int_0^4\sqrt{(e^x)^2+(-e^{-x})^2+(1/\sqrt{2t})^2}\,dt\\
&=&\int_0^4\sqrt{e^{2x} - e^{-2x}+1/(2t)}\,dt\\
\endeq$$

</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">
A point moves along a circle with position vector given by ${\bf r} (t) =  \langle \cos t^2, \sin t^2 \rangle $. 
<ol type="a">
<li> Find ${\bf a}(t)$.
<li> Find the unit tangent vector $\uv T (t)$ for $t \neq 0$.
</ol>

<p class="answer">
$$\myv r'(t)=\langle -\sin(t^2)*2t,+\cos(t^2)*2t\rangle.$$
$$\begineq |\myv r'(t)|&=&\sqrt{(x'(t))^2+(y'(t))^2}\\
&=&\sqrt{4t^2*(\sin^2(t^2)+\cos^2(t^2))}=2t\sqrt 1=2t.
\endeq$$
So, the unit tangent vector 
$$\begineq \uv T(t)&=&\frac{\myv r'(t)}{|\myv r'(t)|}\\
&=& \langle
-\sin(t^2), \ \cos(t^2)
\rangle
\endeq $$


$$\begineq \myv a(t)&=&\frac{d}{dt}\myv r'(t)\\
&=&\langle x''(t),y''(t)\rangle.\endeq $$
where...
$$x''(t)=\frac{d}{dt}x'(t)=
-4t^2\cos(t^2)-2\sin(t^2)
$$

$$y''(t)=\frac{d}{dt}y'(t)= -4t^2sin(t^2)+2\cos(t^2)$$



</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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