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Test 1
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<h3>
Math 213 - May 2023 - Test 1 <span style="margin-left: 4em">NAME ___________</span></h3>
<p class="justme answer">Results:<br>




<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>
Consider the vector function ${\bf r} (t)$
 describing the 2-d curve shown below, and the curvatures of the curve at positions at $A$, $B$, and $C$.  Put the positions in order from  smallest curvature 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>
Consider two vectors, ${\bf a}=\langle -5,1,1 \rangle$ and
${\bf b}=\langle 1,1,2
\rangle$.
<ol type="a">
<li>What's the angle between ${\bf a}$ and ${\bf b}$? (Give your answer in both
radians and degrees rounded to 3 digits.)
<p class="answer">$\myv a \cdot \myv b=-5*1+1*1+1*2=-2$.  The lengths of the vectors are $a=\sqrt(5^2+1^2+1^2=\sqrt{27}$ and $b=\sqrt(1^2+1^2+2^2)=\sqrt 6$. We can get the angle from:
$$\myv a\cdot\myv b=ab\cos\theta.$$
With our values:
$$\cos\theta=\frac{-2}{\sqrt{27}\sqrt{6}}=-\frac{\sqrt 2}{9}.$$
Taking the $\arccos$ of this angle gives us:
$$\theta=1.729\text{ Radians}=99.04\text{ degrees}.$$



<li>Find a vector,  ${\bf v}$, which is perpendicular to both $\bf a$ and $\bf b$.
<p class="answer">
$$\begineq \myv v &= \myv a \times \myv b =\det \begincv
 \uv i& \uv j &\uv k\\
-5 & 1 & 1\\
1 & 1 & 2
\endcv\\
&=[1*2-1*1]\uv i+[1*1-(-5)*2]\uv j+ [(-5)*1-1*1]\uv k\\
&=\langle 1, 11, -6\rangle.
\endeq
$$


</ol>






<!--------------------->

<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 ${\bf a} \perp {\bf b}$.
<div 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 $4u+5v=-4$.  E.g.:
<ul>
<li>$u=9;\ v= -8$.
<li>$u=-1;\ v=0$.
<li>$u=\frac14;\ v=-1$.
</div>


<!--------------------->
<li style="margin-top: 28ex">
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>

<div class="answer"><p>Some considerations:
<ul>
<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>#1 is the correct graph.
</div>


<!------------------------------->
<li style="margin-top:16ex">
Using the point $P=(0,-2,3)$ and a vector ${\bf b}=\myc{3,-2,1}$:
<ol type="a">
<li>Find an equation for the plane which is normal to ${\bf b}$ and includes the point $P$. The equation should be in the form $c_1x+c_2y+c_3z=c_4$.


<div class="answer">
<p><i>If you remembered<i> that the coefficients of $x$, $y$, and $z$ are 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:
$$3x-2y+z=7.$$

<p>Otherwise, you can start from the vector equation for a plane:
$$\begineq
0 &= (\myv r-\myv r_0)\cdot \myv n =(\myv r-\myvv{OP})\cdot \myv b\\

 &= (\myc{x,y,z}-\myc{0,-2,3})\cdot \myc {3,-2,1}\\
  &= \myc{x,y+2,z-3}\cdot \myc {3,-2,1}\\
     &= 3x-2y-4+z-3\\
     0&=3x-2y+z-7\\
  \endeq
  $$
  Which is the same as
  $$3x-27+z=7.$$


<li>Now consider a line, $L$, which is parallel to ${\bf b}$, and passes through $P$.  What are the coordinates of the point at which this line pierces the $xy$ plane?

</ol>









 

<!-------------------->
<li style="page-break-before:always">
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">$$\sqrt{2^2+2^2}=\sqrt 8=2.83$$

</ol>

<li style="margin-top: 8ex">
A crazed ostrich ("Oscar") runs along a mountain path
according to 
${\bf r} (t) =  \langle  3e^{−t}, 4(\cos t+t), t^2/18 \rangle$, with positions in meters and time in seconds.
<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 Oscar’s speed at $t$ = 0 seconds?
<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>



<!--------------------->
<li style="page-break-before:always">Beside each of the surface plots below, make a sketch of the trace $z(y)=z(x=0, y)$<br>

Then, 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>


<!--------------------->

<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>
What is the unit tangent vector, $\uv T$, at the origin (0,0,0)? (No complicated computations are needed.)  Give the components of $\uv T$, as well as sketching it on the diagram.
<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 
$$\uv N=\myc{0,1,0}.$$
<li>Give the vector form of a line which is tangent to the curve when $t=2$, and sketch the line on the diagram.
</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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