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 --- lec25-0703.tex | 145 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 145 insertions(+) create mode 100644 lec25-0703.tex diff --git a/lec25-0703.tex b/lec25-0703.tex new file mode 100644 index 0000000..4e01f6f --- /dev/null +++ b/lec25-0703.tex @@ -0,0 +1,145 @@ +\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