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9.1 - Review / Preview
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<h2>[9.1] - Review / Preview
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<p>We started in on <a href="https://activecalculus.org/multi/S-9-1-Functions.html">section 9.1 in Schlicker's "Active Calculus"</a>.  With what we discussed about the distance formula, you should be able to answer questions 1 and 2 below.
<p>Now <b>Read section 9.1.4 and 9.1.5</b> on Traces and Contour Maps/Level curves. Then complete question 3.

<ol type="1">
<li><img src="../gco/0g/9.1.projections.jpg" style="width:300px" class="rightalign">Consider the point $P=(x,y,z)$  [I'm using an older diagram here.  Just translate these
parameters in the diagram to this notation: $a\to x; b\to y; c\to z$.]

<ol type='a'>
<li>What are the coordinates of the closest point to $P$ <i>on the $x$-axis</i>?  And what is the distance between these two points?
<p class="ustme">
Closest point, $Q$, has the same $x$ coordinate.  But to be on the $x$ axis $y=0$ and $z=0$, so <span class="answer">$Q=(x,0,0)$ </span><br><br>

Distance $PQ$ is $\sqrt{(x-x)^2 + (y-0)^2 +(z-0)^2}=$ <span class="answer"> $\sqrt{y^2+z^2}$ </span>.


<li>What are the coordinates of  the closest point to $P$ <i>on the $yz$ plane</i>? And what is the distance between these two points?

<div class="ustme">The $yz$ plane means $x=0$.  Closest point is <span class="answer"> $R=(0,y,z)$ </span>.  <br><br>

The distance $PR$ is $\sqrt{(x-0)^2+(y-y)^2+(z-z)^2}=$ <span class="answer"> $|x|$ </span> (absolute value of $x$).
</div>
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<li>The work above should  help you answer
Schlicker's  question 8 [in section 9.1].  Record your answers to question 8 below.

<p class="ustme">
With what we did above, we can answer part b) first:  The points whose distance from the $x$-axis equals the distance from the $yz$ plane means...
$$\begineq \sqrt{y^2+z^2} = |x|\\
y^2+z^2 = x^2\endeq$$
This is the <span class="answer">last equation in b.)</span><br><br>

a.) The term on the left represents a circle in the $yz$ plane centered on $(y,z)=(0,0)$ (which is where the $x$ axis is...) and the term on the right means that the radius is $x$.  So as the $x$ coordinate grows, the radius of the circle grows too.  So this is <span class="answer">A cone opening along the $x$-axis</span>.<br><br>




<li>Answer 
Schlicker's question 13 and record your answers below. [Hint: You can use Geogebra to graph an equation and a plane together.]
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