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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>
<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 $\bf c$.
<p class="answer">$$\myv b=\myc{1,1,0};\ \ \ \myv c=(0,0,\sqrt 2).$$

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

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<li style="page-break-before:always">
Draw a picture of the $xyz$ coordinate axes as they would appear if viewed from point whose coordinates are (pos,neg,neg). <br>

<li style="margin-top:20ex;">
Consider the following regular hexagon.  (Regular means that all sides are of equal length and that all interior angles are equal).<br>
<img src="0g/T1.3.jpg"><br>
<br>
<img src="0g/T1.hexanswer.jpg"><br><br>

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<li style="page-break-before:always">
Let ${\bf a}=(x+y){\bf i}+2{\bf j}+y{\bf k}$, and
${\bf b}=3{\bf i}+(4x+y+1){\bf j} + 4{\bf k}$.  Find values of $x$ and $y$ such 
that ${\bf a} \perp {\bf b}$.

<p class="answer">If $\myv a \perp \myv b$, then $\myv a \cdot \myv b=0$:
$$\begineq
0&=&a_xb_x+a_yb_y+a_zb_z=(x+y)3+2(4x+y+1)+(y)4\\
&=&3x+3y+8x+2y+2+4y\\
0&=&11x+9y+2
\endeq
$$
Many solutions are possible.  One is $x=-1;\ y=1$.


<li style="margin-top:28ex;">
Find an equation of the plane through $(2,-3,6)$ with normal vector ${\bf n} = \myc{2,1, -4}$
<p class="answer">Using the components of the normal vector, the plane could have the equation $2x+y-4z=c$.  Substituting in the coordinates of the one point we know...
$$c=2(2)+1(-3)-4(6)=4-3-24=-23$$
So the equation is:
$$2x+y-4z=-23.$$

 <li style="margin-top:24ex;">
Given the points (1,4,-5) and (3,-1,5)
<ol type="a">
<li>
Find a parametric equation for the line that passes through the points.
<p class="answer">Let the first point be the position vector $\myv a$, and the second point have the position vector $\myv b$. The difference vector between these two points is
$$\myv b - \myv a =2\uv i-5\uv j+10\uv k.$$
So, the parametric equation could be
$$\myv a + t (\myv b-\myv a) = (1,4,-5) +t\myc{ 2,-5, 10}.$$

 <li style="margin-top:24ex;">
Find the point where the line intersects the $xz$-plane.</li>
<p class="answer">The intersection happens when $y=0$. Find $t$:
$$0=4-5t; \ \ \Rightarrow t=\frac 45.$$
The point of intersection is...
$$(1,4,-5)+\frac 45(2,-5,10)=(2.8,0,3).$$


</ol>


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<li style="page-break-before:always">
Match the equations with their graphs.<br>
<img src="0g/T1.matchchoices.jpg"><br>
<img src="0g/T1.matches-answers.jpg" style="width:70%;"><br>
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<li style="page-break-before:always">
Let $y = x^2$ be a parabola in the $xy$ plane parameterized by ${\bf r}(t)=
\myc{t,t^2,0}$ (as shown below).
<br><img src="0g/T1.xyparabola.jpg"></br>
<ol type="a">
<li>In the diagram, draw an arrow that indicates how a particle moving along this curve would travel when $t$ is increasing.

<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>


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<li style="page-break-before:always">

A particle is moving counterclockwise around a circle. Which of the
vectors in the figure below is <b>not</b> a possible acceleration
vector? Explain. For the remaining two possible acceleration vectors,
state whether the particle is speeding up or slowing down.<br>

<img src="0g/T1.circularacceleration.jpg"><br> 

<li style="margin-top:26ex;">
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>
</ol>
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<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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