From fe34de0fbc1d2554580bcc36920fb2f754a80215 Mon Sep 17 00:00:00 2001
From: Kyle Spaans
Date: Sat, 4 Jul 2009 20:51:37 -0400
Subject: [PATCH] lec 25 notes done
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+\documentclass{article}
+\usepackage{fullpage}
+\usepackage{amsmath}
+\author{Kyle Spaans}
+\date{July 3, 2009}
+\title{Calculus 3 Lecture Notes}
+\begin{document}
+\maketitle
+
+\section*{Lecture 25 -- Linear Approximation of a Mapping}
+We have a mapping $F: R^2 \rightarrow R^2$, or $(u,v) = \vec{F}(f(x,y),g(x,y))$.
+Therefore a point $(x,y)$, in a 2D plane gets mapped to another $(u,v)$ in a
+different 2D plane. What if we move a distance $(\Delta x, \Delta y)$ to
+$(x + \Delta x, y + \Delta y)$, where is the corresponding point
+$(u + \Delta u, v + \Delta v)$? If's possible that if we apply $F$ to the shift
+in coordinates that we will get something crazy like a curve, rather than a
+line.
+
+\subsection*{Linear Approximation}
+Recall the linear approximation
+\[ f(\vec{x}) \approx L_{\vec{a}}(\vec{x}) = f(\vec{a}) + f_x(\vec{a})(x - a)
++ f_y(y - b) \]
+We assume that $L_{\vec{a}}(\vec{x})$ is a good approximation to $f(\vec{x})$
+near $\vec{a}$ (and that the partials $f_x(\vec{a})$ and $f_y(\vec{a})$ are
+continuous). If this is true, we can say that
+\begin{eqnarray*}
+\Delta u &=& u(a + \Delta x, b + \Delta y) = u(a,b) \approx
+L_{\vec{u},\vec{a}}(a + \Delta x, b + \Delta y) - u(a,b) \\
+ &=& f(a,b) + f_x(a,b)(a + \Delta x - a) + f_y(a,b)(b + \Delta y - b) -
+f(a,b) \\
+ &=& f_x(a,b) \Delta x + f_y(a,b) \Delta y
+\end{eqnarray*}
+Similarly, $\Delta v = g_x \Delta x + g_y \Delta y$. We can represent this with
+a matrix!
+\[
+\left (
+\begin{array}{c}
+\Delta u \\
+\Delta v
+\end{array}
+\right )
+=
+\left (
+\begin{array}{cc}
+f_x & f_y \\
+g_x & g_y
+\end{array}
+\right )
+\left (
+\begin{array}{c}
+\Delta x \\
+\Delta y
+\end{array}
+\right )
+\]
+This matrix has a special name, we call it the \textbf{Jacobian Matrix}, or
+just the Jacobian (named after Jacoby, represented with $J$, the course notes
+use $DF$). It is analogous to $f'$ for $f(x)$ or $\nabla f$ for $f(x,y)$.
+
+\paragraph*{Example 1}
+A mapping $F$ is defined $F(x,y) = (2xy, x^2 - y^2)$. Find the approximate
+image of $(0.9, 2.1)$. \\
+A nearby point that's convenient is $(1,2)$, we start by finding the image of
+this point.
+\[ F(1,2) = (4,-3) \]
+Then we need to find $\Delta u$ and $\Delta v$, evaluated at our chosen point.
+The $\Delta x$ and $\Delta y$ are just the difference between our chosen
+point and the one whose image we are trying to find.
+\[
+\left (
+\begin{array}{c}
+\Delta u \\
+\Delta v
+\end{array}
+\right )
+=
+\left (
+\begin{array}{cc}
+2y & 2x \\
+2x & -2y
+\end{array}
+\right )
+\left (
+\begin{array}{c}
+\Delta x \\
+\Delta y
+\end{array}
+\right )
+=
+\left (
+\begin{array}{cc}
+4 & 2 \\
+2 & -4
+\end{array}
+\right )
+\left (
+\begin{array}{c}
+-0.1 \\
+0.1
+\end{array}
+\right )
+\]
+This gives $\Delta u = -0.4 + 0.2$, $\Delta v = -0.2 - 0.4$. We can now
+calculate the approximate image as $(u + \Delta u, v + \Delta v)$:
+\[ (4 - 0.2, -3 - 0.6) = (3.8, -3.6) \]
+
+\subsection*{Composite Mapping and the Chain Rule}
+This is the same as the linear approximation mapping, but we use the Chain Rule
+to compute the derivative. It's the same thing, it can just get more
+complicated with more variables. The composite mapping is given as two mappings,
+$F: p = p(u,v), q = q(u,v)$ and $G: u = u(x,y), v = v(x,y)$. So the mapping
+$F \circ G$ is defined by $p = p(u(x,y),v(x,y)), q = q(u(x,y),v(x,y))$.
+
+Question: Determine $D(F \circ G)$ in terms of $DF$ and $DG$. It's $DF \cdot
+DG$. Just like the chain rule. Using the Jacobian, we can guess the answer
+\[
+\left (
+\begin{array}{cc}
+\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\
+\frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}
+\end{array}
+\right )
+=
+\left (
+\begin{array}{cc}
+\frac{\partial p}{\partial u} & \frac{\partial p}{\partial v} \\
+\frac{\partial q}{\partial u} & \frac{\partial q}{\partial v}
+\end{array}
+\right )
+\cdot
+\left (
+\begin{array}{cc}
+\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
+\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
+\end{array}
+\right )
+\]
+Keep in mind this is just a guess. How do we know if it's true? We can use the
+Chain Rule on $p(u(x,y),v(x,y))$ to get
+\[ \frac{\partial p}{\partial x} = \frac{\partial p}{\partial u}
+\frac{\partial u}{\partial x} + \frac{\partial p}{\partial v}
+\frac{\partial v}{\partial x} \]
+(assuming that all partials are continous, we can do similarly for the other
+components)
+\end{document}
--
2.11.0