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Quiz 2 (Chapter 10)
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<h3>Name-<br>
Math 213, Fall 2016, Quiz - Review of Chapter 10</h3>
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<li>
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>Imagine that this space curve represents the trajectory of a particle moving as time, $t$, increases. In the diagram draw one arrow somewhere on the space curve
indicating the direction of motion of the particle at that point along the trajectory.


<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)? (It's possible to answer this question without complicated computations...)

<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 
$$\uv N=\myc{0,1,0}.$$
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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 not a possible <b>acceleration</b>
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> 



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