From 2fb7b45d11e589e889badee463fc390798f128a7 Mon Sep 17 00:00:00 2001 From: Kyle Spaans Date: Fri, 17 Jul 2009 10:10:18 -0400 Subject: [PATCH] lec 30 notes done, lec 26,27 notes half complete --- lec26-0706.tex | 35 +++++++++++++++++++++++++++++++++++ lec27-0708.tex | 13 +++++++++++++ lec30-0715.tex | 57 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 3 files changed, 105 insertions(+) create mode 100644 lec26-0706.tex create mode 100644 lec27-0708.tex create mode 100644 lec30-0715.tex diff --git a/lec26-0706.tex b/lec26-0706.tex new file mode 100644 index 0000000..4b7bb7e --- /dev/null +++ b/lec26-0706.tex @@ -0,0 +1,35 @@ +\documentclass{article} +\usepackage{fullpage} +\usepackage{amsmath} +\author{Kyle Spaans} +\date{July 6, 2009} +\title{Calculus 3 Lecture Notes} +\begin{document} +\maketitle + +\section*{Lecture 26 -- Mappings Continued} +Restate the Jacobian. + +\paragraph*{Example 1} +$(u,v) = G(x,y) = (\frac{x}{y}, x - y) \,\, (p,q) = F(u,v) = (uv, u + v)$ +Calculate $D(F \circ G)$ at the point $(x,y) = (3,1)$. + +\subsection*{Inverse Mappings} +We can find the mapping from $(u,v)$ back to $(x,y)$. + +\paragraph*{Example 2} +Image of domain $D$ under $F$. We have to areas in a Cartesian graph. One from +$y = 0$ to $y = \frac{\pi}{2}$ and the other from $y = 2\pi$ to $y = +\frac{5\pi}{2}$, both from $x = 0$ to $x = 1$. Both of these functions will map +to the same quarter of a unit circle in polar coordinates (defined by $F$). + +\subsubsection*{Definitions} +Mapping and one-to-one. + +\subsection*{The Jacobian of an Inverse Mapping} +$DF^{-1} = (DF)^{-1}$ valid if one-to-one and partials are continuous. + +\paragraph*{Example 3} +$(u,v) = F(x,y) = (x + y, \frac{x}{x + y}) \, y \neq -x$ Find $DF, DF^{-1}$ and +verify $DF \cdot DF^{-1} = I$. +\end{document} diff --git a/lec27-0708.tex b/lec27-0708.tex new file mode 100644 index 0000000..a5caac6 --- /dev/null +++ b/lec27-0708.tex @@ -0,0 +1,13 @@ +\documentclass{article} +\usepackage{fullpage} +\usepackage{amsmath} +\author{Kyle Spaans} +\date{July 8, 2009} +\title{Calculus 3 Lecture Notes} +\begin{document} +\maketitle + +\section*{Lecture 27 -- Inverse Mappings} +Last time we were calculating the inverse of composite mappings. We said +$D(F \circ G) = DF \cdot DG$. Jacobians $F : R^2 \rightarrow R^2$... +\end{document} diff --git a/lec30-0715.tex b/lec30-0715.tex new file mode 100644 index 0000000..57151d9 --- /dev/null +++ b/lec30-0715.tex @@ -0,0 +1,57 @@ +\documentclass{article} +\usepackage{fullpage} +\usepackage{amsmath} +\author{Kyle Spaans} +\date{July 15, 2009} +\title{Calculus 3 Lecture Notes} +\begin{document} +\maketitle + +\section*{Lecture 30 -- Change of Variables in Double Integrals} +$\iint \limits_{D_{xy}} \! f(x,y) \, dA$ +Why do we want to change the variables? Two reasons: $f(x(u,v),y(u,v))$ is +sometimes easier to integrate, and $D_{uv}$ can be easier to derive than +$D_{x,y}$. + +\paragraph*{Example 1} +Change $\displaystyle \iint \limits_D \! f(x,y) \, dx \, dy$ into polar +coordinates. Use the Jacobian: +$\frac{\partial(x,y)}{\partial(r,\theta)} = + \left | \begin{array}{cc} + x_r & x_{\theta} \\ + y_r & y_{\theta} + \end{array} \right | = + \left | \begin{array}{cc} + \cos \theta & -r \sin \theta \\ + \sin \theta & r \cos \theta + \end{array} \right |$ +giving $r \cos^2 \theta + r \sin^2 \theta -r$ to get +$\iint \limits_{D_{xy}} \! f(x,y) \, dx \, dy = + \iint \limits_{D_{r \theta}} \! f(x(r,\theta),y(r,\theta)) r \, dr \, + d\theta$ + +\paragraph*{Example 2} +Use the polar coordinates to calculate the volume of a sphere of radius $R$. We +know the centre of the sphere is at $(0,0,0)$ (for simplicity). Let's calculate +only the upper half of the volume (doable because the shape is symmetric) and +multiply the volume by 2. +$V = 2 \iint \limits_{z \ge 0} \! f(x,y) \, dx \, dy$ +We know that $x^2 + y^2 + z^2 = R^2$ means that $f(x,y) = z = \sqrt{R^2 - x^2 - +y^2}$. Substitute in polar coordinates to get $\sqrt{R^2 - r^2 \cos^2 \theta - +r^2 \sin^2 \theta} = \sqrt{R^2 - r^2}$. +$V = 2 \int_0^R \int_0^{2\pi} \! \sqrt{R^2 - r^2} r \, dr \, d\theta$ +We get the new change-of-variables $D$ from $0 \le r \le R$ and $0 \le \theta +\le 2\pi$. Since $r$ does not depend on $\theta$, and vice-versa, the order of +integration is not important. Let's integrate $r$ first: +$2 \left. \int_0^{2\pi} (R^2 - r^2)^{\frac{3}{2}} \frac{2}{3} \frac{-1}{2} + \right |_0^R \, d\theta$ +$2 \left. \int_0^{2\pi} (R^2)^{\frac{3}{2}} \frac{1}{3}\, d\theta = 2R^3 \theta + \right |_0^{2\pi} = \frac{4\pi}{3} R^3$ + +\subsection*{Tips for changing variables} +\begin{enumerate} +\item Substitute $r,\theta \rightarrow x,y$ +\item Dont't forget $r$, the Jacobian when changing variables +\item Change the limits too +\end{enumerate} +\end{document} -- 2.11.0