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Test 1
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<h3>Name-<br>
Math 213, May 2022, Test 1</h3>
<p class="justme answer">Raw Results 2022:<br>
10|&nbsp;8699 <br>
&nbsp;9|&nbsp; <br>
&nbsp;8|&nbsp; <br>
&nbsp;7|&nbsp;272 <br>
&nbsp;6|&nbsp; 

<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> (15 points)
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="result ustme">$$\myv b=\myc{2,2,0};\ \ \ \myv c=(0,0,\sqrt 2).$$

<li>
Find the distance of the point $P$ from the origin.
<p class="result ustme">$$\sqrt{2^2+2^2+\sqrt 2^2}=\sqrt{10}=3.16$$
<li>
Label a point $Q$ which is the nearest point to $P$ on the $z$-axis.
Find the distance of the point $P$ from $Q$.
<p class="result ustme">$$\sqrt{2^2+2^2}=\sqrt 8=2.83$$
<!--
<li>
What are the coordinates $(\rho, \theta, \phi)$ of $P$ in spherical coordinates?
<p class="result ustme">$\rho = \sqrt{10}=3.16$;
<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">
(10) 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="result ustme">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;">
(10) 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="result ustme">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;">
(10) 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="result ustme">$\myv a \cdot \myv b=-5*2+1*4+3*2=0$.  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="result ustme">
$$\myv v=\myv a \times \myv b =\det \begincv
 \uv i& \uv j &\uv k\\
-5 & 1 & 3\\
2 & 4 & 2
\endcv=\langle -10, 16, -22\rangle.$$
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="result ustme">
$$\myv r(t)=\myv r_0 +t\myv v=\langle -10t,3+16t,5-22t\rangle.$$
-->
</ol>
 
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<li style="page-break-before:always">
<b>Problem 5 was dropped from the test!</b> <strike>
<p><img src="0g/projection.jpg" class="rightalign" style="width:280px">
(10) Shown are two vectors, $\mathbf a = \langle 2,1 \rangle$ and
$\mathbf b = \langle\frac 35,\frac 45\rangle$ with their tails at the origin.
<ol type="a">
<li>Label $\mathbf a$ and $\mathbf b$ in the diagram, and sketch the vector projection, $\text{proj}_{\mathbf a}\mathbf b$, 
of $\mathbf b$ in the direction of $\mathbf a$. Indicate any right angles in your resulting diagram.
<li>Calculate the components of the vector $\text{proj}_{\mathbf a}\mathbf b$. 
<p class="result ustme">
$$\text{proj}_{\mathbf a}\mathbf b=b\cos\theta \uv a=\frac{a}{a}b\cos\theta\frac{a}{a}\uv a =\frac{ab\cos\theta}{a^2}\myv a=\frac{\mathbf a\cdot \mathbf b}{a^2}\mathbf a.$$
$$a=|\langle 2, 1\rangle|=\sqrt{2^2+1^2}=\sqrt 5.$$
$$\mathbf a\cdot\mathbf b=2\frac 35+1\frac 45=\frac{10}{5}=2.$$
<br>
So, the answer is
$$\frac{2}{(\sqrt 5)^2}\langle 2,1\rangle = \langle \frac 45, \frac 25 \rangle$$
</ol>
</strike>
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<li style="margin-top: 34ex">
(15) Oscar, a crazed ostrich, runs along a mountain path.  The vector equation for his trajectory is
${\bf r} (t) =  \langle  3e^{−t}, 4(\cos t+t), t^2/18 \rangle$.
<ol type="a">
<li>What is the change in Oscar’s altitude ($z$-component) from t = 0 to t = 6?
<p class="result ustme">$\Delta z= z(6)-z(0)=36/18=2.$
<li>Calculate Oscar's velocity and speed at $t=0$.  (Label them)
<p class="result ustme">$$\myv r'(t)
=-3e^{-t}\uv i+4(-\sin(t)+1)\uv j+(t/9)\uv k$, so $\myv r'(0)= \myc{-3,4,0}$, and 
speed = $|\myv r'(0)|=\sqrt{(-3)^2+4^2}=5$.
<li>Find a formula (but don't try to solve the integral) for the total distance that
Oscar travels from $t=0$ to $t=6$.
<p class="result ustme">$$\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;">
(10) Consider the vector function ${\bf r} (t)$
 describing the 2-d curve shown below. Sketch the "osculating circle" tangent to ${\bf r}$ at $A$.  Then, 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="result ustme">B</span><-- smallest<br><br>
<span class="result ustme">A</span><--<br><br>
<span class="result ustme">C</span><--largest
</p>

<br clear="both">

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

(20) Beside each of the surface plots below, make a sketch of $z(x)$ for the $y=0$ trace of that graph.<br>
Then match the equations with their graphs.  Give a reason for your choices.<br>
<span style="color:red">Yeah... (a) should be $8x+2y-3z=0$</span>
<img src="0g/1.matchmtext.jpg"><br>
<img src="0g/1.matchmgraphs.jpg">
<p class="result ustme">
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">
(10) Which curve below is the path traced out by ${\bf r}(t)=
\langle \sin \pi t, \cos \pi t, \frac14t^2\rangle$, $0\leq t \leq 2$?  <b>Justify</b> your answer.<br>
<img src="0g/T1curves.jpg"><br>

<p class="result ustme">Some considerations:
<ul class="result ustme">
<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="result ustme">#1 is the correct graph.

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<li style="margin-top: 25ex">
(10)
In the axes below, sketch and label the level curves for $z=0$, $z=1$, and $z=2$ of the function $z(x,y)=4x-2y+3$<br><br>
<img src="0g/graph.png">






</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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