<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Home on Nels Parenteau</title><link>http://www.nelsparenteau.com/</link><description>Recent content in Home on Nels Parenteau</description><generator>Hugo</generator><language>en-us</language><atom:link href="http://www.nelsparenteau.com/index.xml" rel="self" type="application/rss+xml"/><item><title>Chapter 12: Vectors and the Geometry of Space</title><link>http://www.nelsparenteau.com/math/multivariable/chapter_12/</link><pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate><guid>http://www.nelsparenteau.com/math/multivariable/chapter_12/</guid><description>&lt;h3 id="121-three-dimensional-coordinate-systems">
 12.1: Three-Dimensional Coordinate Systems
 
 &lt;a class="anchor" href="#121-three-dimensional-coordinate-systems">#&lt;/a>
 
&lt;/h3>

&lt;blockquote class='book-hint note'>
 &lt;p>&lt;strong>Problem 3&lt;/strong>: Which of the points $A(-4, 0, -1)$, $B(3, 1, -5)$, and
$C(2, 4, 6)$ is closest to the $yz$-plane? Which point lies in the $xz$-plane?&lt;/p>
&lt;/blockquote>&lt;p>To find the distance between a point and a plane, we find the distance between
that point and it&amp;rsquo;s projection to the plane in question. For example, $A$&amp;rsquo;s
projection to the $yz$-plane is $(0, 0, -1)$ as the projection removes the $x$
term. Note now that the distance between these two is very simple, it is&lt;/p></description></item><item><title>Chapter 14: Partial Derivatives</title><link>http://www.nelsparenteau.com/math/multivariable/chapter_14/</link><pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate><guid>http://www.nelsparenteau.com/math/multivariable/chapter_14/</guid><description>&lt;h3 id="141-functions-of-several-variables">
 14.1: Functions of Several Variables
 
 &lt;a class="anchor" href="#141-functions-of-several-variables">#&lt;/a>
 
&lt;/h3>

&lt;blockquote class='book-hint note'>
 &lt;p>&lt;strong>Problem 13&lt;/strong>: Find and sketch the domain of the function,&lt;/p>
&lt;p>$$
f(x,y) = \sqrt{x - 2} + \sqrt{y - 1}
$$&lt;/p>
&lt;/blockquote>&lt;p>Each sqrt root cannot go negative, so $x - 2 \geq 0$ and $y - 1 \geq 0$, so the
domain in set-builder notation is,&lt;/p>
&lt;p>$$
\{ (x, y) \in \R^2 : x \geq 2 \ \text{and,} \ y \geq 1 \}
$$&lt;/p></description></item><item><title>Chapter 15: Multiple Integrals</title><link>http://www.nelsparenteau.com/math/multivariable/chapter_15/</link><pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate><guid>http://www.nelsparenteau.com/math/multivariable/chapter_15/</guid><description>&lt;h3 id="151-double-integrals-over-rectangles">
 15.1: Double Integrals over Rectangles
 
 &lt;a class="anchor" href="#151-double-integrals-over-rectangles">#&lt;/a>
 
&lt;/h3>

&lt;blockquote class='book-hint note'>
 &lt;p>&lt;strong>Problem 1&lt;/strong>: Define the solid that lies below the surface $f(x, y) = xy$ and
above the rectangle&lt;/p>
&lt;p>$$
R
= {(x, y) : 0 \leq x \leq 6, 0 \leq y \leq 4 }
$$&lt;/p>
&lt;p>$(a)$ Estimate the volume by using a Riemann sum with $m = 3$, $n = 2$, where
the sample point is the upper right corner of each square.&lt;/p></description></item></channel></rss>