\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 116 Lecture 3/6/00} \vspace{1pt} \subsection{Chain Rule for Partial Derivatives} Recall that if $y=f\left( u\right) $ and and $u=g\left( x\right) ,$ then \vspace{1pt} \begin{equation*} \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=f^{\prime }\left( u\right) g^{\prime }\left( x\right) =f^{\prime }\left( g\left( x\right) \right) g^{\prime }\left( x\right) \end{equation*} \vspace{1pt} This is called the \emph{chain rule }for a function of one variable. \vspace{1pt} The chain rule may be extended to functions of two or more variables. Let $% z=f\left( x,y\right) $ and suppose that $x=g\left( r,s\right) $ and $% y=h\left( r,s\right) ,$ that is, $x$ and $y$ are functions of the variables $% r$ and $s.$ Then $z$ may be thought of as a function of $r$ and $s,$ since $% z=f\left( x,y\right) =f\left( g\left( r,s\right) ,h\left( r,s\right) \right) =F\left( r,s\right) .$ Then the chain rule for partial derivatives gives \vspace{1pt} \begin{eqnarray*} \dfrac{\partial z}{\partial r} &=&\dfrac{\partial z}{\partial x}\dfrac{% \partial x}{\partial r}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{% \partial r}=\dfrac{\partial F}{\partial r} \\ \dfrac{\partial z}{\partial s} &=&\dfrac{\partial z}{\partial x}\dfrac{% \partial x}{\partial s}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{% \partial s}=\dfrac{\partial F}{\partial s} \end{eqnarray*} \vspace{1pt} \paragraph{Example:} Let $z=f\left( x,y\right) =e^{xy}$ and $x=r\cos s$ and $y=r\sin s.$ We shall find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial s}.$ Now $\frac{\partial x}{\partial r}=\cos s,$ $\frac{\partial x}{\partial s}% =-r\sin s,$ $\frac{\partial y}{\partial r}=\sin s,$ and $\frac{\partial y}{% \partial s}=r\cos s.$ Thus \vspace{1pt} \begin{eqnarray*} \frac{\partial z}{\partial r} &=&ye^{xy}\cos s+xe^{xy}\sin s=2r\sin s\cos se^{r^{2}\sin s\cos s} \\ \frac{\partial z}{\partial s} &=&ye^{xy}\left( -r\sin s\right) +xe^{xy}\left( r\cos s\right) =r^{2}e^{r^{2}\sin s\cos s}\left( \cos ^{2}s-\sin ^{2}s\right) \end{eqnarray*} \begin{example} \vspace{1pt}Suppose that \begin{equation*} z=f\left( u,v\right) \text{ \ \ \ \ \ }u=2x+y,\text{ \ \ \ }v=3x-2y \end{equation*} \end{example} \vspace{1pt} Given the values $\frac{\partial z}{\partial x}=3$ and $\frac{\partial z}{% \partial v}=-2$ at the point $\left( u,v\right) =\left( 3,1\right) ,$ find $% \frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at the corresponding point $\left( x,y\right) =\left( 1,1\right) .$ \vspace{1pt} \emph{Solution: }% \begin{equation*} \dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}% =3\cdot 2+\left( -2\right) \cdot 3=0 \end{equation*} \vspace{1pt} \begin{equation*} \dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}% =3\cdot 1+\left( -2\right) \cdot \left( -2\right) =7 \end{equation*} \subsection{The Directional Derivative and the Gradient} \vspace{1pt}Let $\Phi \left( x,y,z\right) $ be a scalar function with first partial derivatives $\Phi _{x},\Phi _{y},$ and $\Phi _{z}$ in some region of $x,y,z-$space. Let $\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$ be the vector drawn from the origin to the point $P=\left( x,y,z\right) .$ Suppose that we move from $P$ to a nearby point $Q=\left( x+\Delta x,y+\Delta y,z+\Delta z\right) .$ \begin{center} \FRAME{dtbpF}{3.2993in}{2.0764in}{0pt}{}{}{r_delta_r.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.2993in;height 2.0764in;depth 0pt;original-width 13.3337in;original-height 9.1878in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/r_delta_r.bmp';file-properties "XNPEU";}} \vspace{1pt} \end{center} Then $\Phi $ will change by an amount $\Delta \Phi $ where \vspace{1pt} \begin{equation*} \Delta \Phi =\Phi _{x}\Delta x+\Phi _{y}\Delta y+\Phi _{z}\Delta z+\epsilon _{1}\Delta x+\epsilon _{2}\Delta y+\epsilon _{3}\Delta z \end{equation*} \vspace{1pt} where $\epsilon _{1},\epsilon _{2},$ and $\epsilon _{3}\rightarrow 0$ as the point $Q\rightarrow P.$ If we divide the change $\Delta \Phi $ by the distance $\Delta s=\left| \Delta \vec{r}\right| $ between $P$ and $Q$, we obtain a measure of the rate at which $\Phi $ changes when we move from $P$ to $Q:$ \vspace{1pt} \begin{equation*} \dfrac{\Delta \Phi }{\Delta s}=\Phi _{x}\frac{\Delta x}{\Delta s}+\Phi _{y}% \frac{\Delta y}{\Delta s}+\Phi _{z}\frac{\Delta z}{\Delta s}+\epsilon _{1}% \frac{\Delta x}{\Delta s}+\epsilon _{2}\frac{\Delta y}{\Delta s}+\epsilon _{3}\frac{\Delta z}{\Delta s} \end{equation*} \vspace{1pt} \paragraph{Example:} If $\Phi \left( x,y,z\right) $ represents the temperature at any point $% P\left( x,y,z\right) $ then $\frac{\Delta \Phi }{\Delta s}$ is the average rate of change in temperature per unit length at the point $P$ in the direction in which $\Delta s$ is measured. \vspace{1pt} The limiting value of $\frac{\Delta \Phi }{\Delta s}$ as $\Delta s\rightarrow 0,$ that is, as $Q\rightarrow P$ along the segment $PQ$ is called the derivative of $\Phi $ in the direction $PQ$ or simply the \emph{% directional derivative} of $\Phi .$ Since $\epsilon _{1},\epsilon _{2},\epsilon _{3}\rightarrow 0$ as $Q\rightarrow P,$ we have that \vspace{1pt} \begin{equation*} \dfrac{d\Phi }{ds}=\dfrac{d\Phi }{dx}\dfrac{dx}{ds}+\dfrac{d\Phi }{dy}\dfrac{% dy}{ds}+\dfrac{d\Phi }{dz}\dfrac{dz}{ds} \end{equation*} \vspace{1pt} \vspace{1pt}The first factor in each term of the products in the expression above for the directional derivative depend only on $\Phi $ and the point $P.$ The second factors in the products are independent of $\Phi $ and depend on the direction in which the derivative is being computed. We may rewrite the expression above in the form \vspace{1pt} \begin{eqnarray*} \dfrac{d\Phi }{ds} &=&\left( \Phi _{x}\vec{i}+\Phi _{y}\vec{j}+\Phi _{z}\vec{% k}\right) \cdot \left( \dfrac{dx}{ds}\vec{i}+\dfrac{dy}{ds}\vec{j}+\dfrac{dz% }{ds}\vec{k}\right) \\ &=&\left( \Phi _{x}\vec{i}+\Phi _{y}\vec{j}+\Phi _{z}\vec{k}\right) \cdot \dfrac{d\vec{r}}{ds} \end{eqnarray*} \vspace{1pt} The vector $\Phi _{x}\vec{i}+\Phi _{y}\vec{j}+\Phi _{z}\vec{k}$ is known as the \emph{gradient} of $\Phi $ or $grad\Phi .$ Thus \vspace{1pt} \begin{equation*} grad\Phi =\Phi _{x}\vec{i}+\Phi _{y}\vec{j}+\Phi _{z}\vec{k} \end{equation*} \vspace{1pt} The notation $\nabla \Phi $ is often used for $grad\Phi .$ In this notation the operator $\nabla $ is defined as \begin{equation*} \nabla =\vec{i}\dfrac{\partial }{\partial x}+\vec{j}\dfrac{\partial }{% \partial y}+\vec{k}\dfrac{\partial }{\partial z} \end{equation*} \vspace{1pt} \paragraph{Example:} Let $\Phi \left( x,y,z\right) =xyz+3x^{4}y^{2}z^{3}.$ The $\nabla \Phi =\left( yz+12x^{3}y^{2}z^{3}\right) \vec{i}+\left( xz+6x^{4}yz^{3}\right) \vec{j}+\left( xy+9x^{4}y^{2}z^{2}\right) $ \vspace{1pt} \paragraph{Example:} We may use SNB to find the gradient of a function. However, SNB writes vectors as ordered triples instead of the form given in the previous example. Thus $\nabla \left( xyz+3x^{4}y^{2}z^{3}\right) =\allowbreak \left( yz+12x^{3}y^{2}z^{3},xz+6x^{4}yz^{3},xy+9x^{4}y^{2}z^{2}\right) \allowbreak $ \vspace{1pt} With this notation we may write the directional derivative of $\Phi $ in the form \vspace{1pt} \begin{equation*} \frac{d\Phi }{ds}=grad\Phi \cdot \dfrac{d\vec{r}}{ds}=\nabla \Phi \cdot \frac{d\vec{r}}{ds} \end{equation*} Remark: Since $\Delta s$ is the length of $\Delta \vec{r}$ then $\frac{% \Delta \vec{r}}{\Delta s}$ and hence $\frac{d\vec{r}}{ds}$ are unit vectors. Therefore, $\nabla \Phi \cdot \frac{d\vec{r}}{ds}$ is the projection of \ $% grad\Phi $ in the direction of $\frac{d\vec{r}}{ds}.$ Thus $\nabla \Phi $ has the property that its projection in any direction is equal to the derivative of $\Phi $ in that direction. Since the maximum projection of a vector is the vector itself, it is clear that $grad\Phi $ extends in the direction of the greatest rate of change of $\Phi $ and has that rate of change for its length. \begin{center} \FRAME{dtbpF}{3.039in}{2.2632in}{0pt}{}{}{project_grad.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.039in;height 2.2632in;depth 0pt;original-width 3.8752in;original-height 2.8746in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/project_grad.bmp';file-properties "XNPEU";}} \end{center} \paragraph{Example:} What is the directional derivative of the function $\Phi =xy^{2}+yz^{3}$ at $% \left( 2,-1,1\right) $ in the direction of the vector $\vec{i}+2\vec{j}+2% \vec{k}$ ? \ $\nabla \left( xy^{2}+yz^{3}\right) =\allowbreak \left( y^{2},2xy+z^{3},3yz^{2}\right) $ and a unit vector in the given direction is $\dfrac{1}{3}\left( 1,2,2\right) .$ Thus \begin{equation*} \dfrac{d\Phi }{ds}=\left( y^{2},2xy+z^{3},3yz^{2}\right) \cdot \dfrac{1}{3}% \left( 1,2,2\right) =\allowbreak \frac{1}{3}y^{2}+\frac{4}{3}xy+\frac{2}{3}% z^{3}+2yz^{2} \end{equation*} Hence \vspace{1pt}% \begin{equation*} \left. \dfrac{d\Phi }{ds}\right| _{\left( 2,-1,1\right) }=\frac{1}{3}\left( -1\right) ^{2}+\frac{4}{3}\left( 2\right) \left( -1\right) +\frac{2}{3}% \left( 1\right) ^{3}+2\left( -1\right) \left( 1\right) ^{2}=\allowbreak -% \frac{11}{3}. \end{equation*} \vspace{1pt} \end{document} %%%%%%%%%%%%%%%%%%%%% End /document/lec_3_6_00.tex %%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% Start /document/graphics/r_delta_r.bmp %%%%%%%%%%%%%%% BudwMH@@@@@@@xC@@@@J@@@@OG@@@HR@@@P@@D@@@@@@@`P@@@@@@@@@@@@@H@@@@`@@@@@@@@ @@|Cp o@@p o@@p o@@p o@@p o@@p o@@p o@@p o@@p o@@p o@@p o@@p o@@p o@@p o @@p o@@p o@@p o@@p o@@p{G o@@p{G o@@p{G o@@p{G o@@p{ Go@@p {Go@@p {Go@@p {Go@@p {Go@@p {Go@@p {Go @@p{G o@@p{G o@@p{G o@@p{A o@@p{A o@@p{O@} o@@p{O@} o@@p o@@p o@@p yo@@p yo@@p _~o@@p _~o@@p cO@_|o@@p 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