\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}{Definition} \newtheorem{example}{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.6in,top=0.7in,bottom=0.7in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Implicit Differentiation} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright \; Hidegkuti, Powell, 2009} \rfoot{\small Last revised: February 14, 2015} \textwidth 7.4in \textheight 9.3in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\Large Sample Problems\bigskip } \end{center} \begin{enumerate} \item Find the equation of the tangent line drawn to the graph of $\ -3x^{2}-16xy-2y^{2}+3y=178$ \ at the point $\left( -3,5\right) $. \item Consider the relation determined by the equation $xy^{2}-5x=2\left( y^{2}+x^{2}y-16\right) $. \ Find an equation for all tangent line(s) drawn to the graph of the relation at \ $x=3$. \item If $y=f\left( x\right) $ \ is a function, we define the curvature as \begin{equation*} C\left( x\right) =\dfrac{\left\vert y^{\prime \prime }\right\vert }{\left( 1+\left( y^{\prime }\right) ^{2}\right) ^{3/2}} \end{equation*}% Prove that if $f\left( x\right) =\sqrt{r^{2}-x^{2}}$ \ \ where $r>0,$ then the curvature is constant on the interval $\left( -r,r\right) $.\bigskip \end{enumerate} \begin{center} {\Large Practice Problems\bigskip }\bigskip \bigskip \end{center} \begin{enumerate} \item Find the slope of the tangent line drawn to the graph of $% x^{4}-y^{4}=2x^{2}y+23$ \ to the point $\left( 2,-1\right) $. \item Find an equation for the tangent line drawn to the graph of $% x^{3}+y^{3}-5y^{2}=6x^{2}+13x-42$ \ at the point \ $\left( -3,5\right) $. \item Find an equation for all tangent lines drawn to the graph of $% 2x^{2}+y^{2}=5y-x$ \ at \ $x=-2$. \item Find an equation of all tangent lines drawn to the curve $% x^{2}-xy+y^{2}=16$ at $x=0.$ \item Use implicit differentiation to compute $y^{\prime }$ in terms of $x$ and $y.$% %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion a) $\ 2x^{2}+4xy=10$ b) $\ x^{4}+y^{4}=20y\ $ c) \ $x^{3}+y^{3}=2xy$ d) \ $x^{3}+y^{3}=x^{2}+y^{2}$ e) $\ \ln x-2+y^{2}=y^{5}$ f) \ $x^{2}+y^{2}=\dfrac{1}{y}$ g) $\ \sin x+\cos y=-2y^{3}$ h) \ $x^{4}y-xy^{4}=y$ i) \ $x^{3}+y^{3}=\left( x-y\right) ^{5}$ j) \ $y^{3}+y=\sqrt{x^{2}-y^{2}}$ k) \ $2^{x+y}=xy^{3}$ l) \ $y+xy=\sqrt{xy-2}$ m) \ $\ln y=\sin \left( xy\right) -1$ n) \ $\left( \sin ^{3}x+\sin ^{3}y\right) ^{2}=x+y$\ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak \end{enumerate} \begin{center} {\Large Sample Problems - Answers}\bigskip \bigskip \bigskip \end{center} 1.) \ $y=2x+11$ \ \ \ \ \ \ \ \ \ \ 2.) \ $y=-5x+32$ \ \ \ and \ $y=-x+4$ \ \ \ \ \ \ 3.) \ see solutions\bigskip \begin{center} \bigskip {\Large Practice Problems - Answers} \end{center} \bigskip \begin{enumerate} \item $10$ \item $-2\left( x+3\right) =y-5$ \item $y=-7x-12\ $and $y=7x+17$ \item $y=\dfrac{1}{2}x+4$ and $y=\dfrac{1}{2}x-4$ \item a) $\ y^{\prime }=-\dfrac{x+y}{x}$ \ \ \ \ \ \ b) $\ y^{\prime }=-% \dfrac{x^{3}}{y^{3}-5}$ \ \ \ \ \medskip \medskip\ \ c) \ $y^{\prime }=% \dfrac{3x^{2}-2y}{2x-3y^{2}}$ \ \ \ \ \ \ d) \ $y^{\prime }=\dfrac{-3x^{2}+2x% }{3y^{2}-2y}\allowbreak $ e) $\ $\ $y^{\prime }=-\dfrac{1}{x\left( 2y-5y^{4}\right) }$ \ \ \ \ \ \ f) \ $y^{\prime }=-\dfrac{2xy^{2}}{2y^{3}+1}$\medskip \medskip\ \ \ \ \ \ \ g) $% \ y^{\prime }=\dfrac{\cos x}{\sin y-6y^{2}}$ \ \ \ \ \ \ h) \ \ $y^{\prime }=% \dfrac{y^{4}-4x^{3}y}{x^{4}-4xy^{3}-1}$ i) \ \ $y^{\prime }=\dfrac{-3x^{2}+5\left( x-y\right) ^{4}}{3y^{2}+5\left( x-y\right) ^{4}}$ \ \ \ \ \ \ \ \medskip \medskip\ \ \ j) \ \ $y^{\prime }=% \dfrac{x}{y+\left( y+y^{3}\right) \left( 3y^{2}+1\right) }$ \ \ \ \ \ \ \ k) \ \ $y^{\prime }=\dfrac{y^{3}-\left( \ln 2\right) 2^{x+y}}{-3xy^{2}+\left( \ln 2\right) 2^{x+y}}$ l) \ $y^{\prime }=\dfrac{y-2y\sqrt{xy-2}}{2\sqrt{xy-2}-x+2x\sqrt{xy-2}}$ \medskip \medskip\ \ \ \ \ \ \ \ \ m) \ $y^{\prime }=\dfrac{y^{2}\cos xy}{% -xy\cos xy+1}$ \ \ \ \ \ \ \ \ \ \ \ n) \ \ $y^{\prime }=\dfrac{-6\left( \cos x\sin ^{2}x\right) \left( \sin ^{3}x+\sin ^{3}y\right) +1}{6\left( \cos y\sin ^{2}y\right) \left( \sin ^{3}x+\sin ^{3}y\right) -1}$\pagebreak \end{enumerate} \begin{center} {\Large Sample Problems - Solutions\bigskip } \end{center} \begin{enumerate} \item Find the equation of the tangent line drawn to the graph of $\ -3x^{2}-16xy-2y^{2}+3y=178$ \ at the point $\left( -3,5\right) $.\newline Solution: \ We start with implicit differentiation. \ We first differentiate both sides: Then we solve for $y^{\prime }$.% \begin{eqnarray*} -3x^{2}-16xy-2y^{2}+3y &=&178 \\ -6x-16y-16xy^{\prime }-\allowbreak 4yy^{\prime }+3y^{\prime } &=&0 \\ -16xy^{\prime }-\allowbreak 4yy^{\prime }+3y^{\prime } &=&6x+16y \\ y^{\prime }\left( -16x-\allowbreak 4y+3\right) &=&6x+16y \\ y^{\prime } &=&\dfrac{6x+16y}{-16x-\allowbreak 4y+3}~~~~~~~\text{compute }% y^{\prime }\text{ \ when \ }x=-3\text{ \ and \ }y=5 \\ y^{\prime } &=&\dfrac{6\left( -3\right) +16\left( 5\right) }{-16\left( -3\right) -\allowbreak 4\left( 5\right) +3}=2 \end{eqnarray*}% The line must pass through $\left( -3,5\right) $ and have slope $2$. \begin{eqnarray*} y-5 &=&2\left( x+3\right) \\ y &=&2x+6+5=2x+11 \end{eqnarray*}% Thus the answer is $\ y=2x+11.$ \item Consider the relation determined by the equation $xy^{2}-5x=2\left( y^{2}+x^{2}y-16\right) $. \ Find an equation for all tangent line(s) drawn to the graph of the relation at \ $x=3$.~\newline Solution: $\ $We substitute \ $x=3$ into the equation and solve for $y$.% \begin{eqnarray*} 3y^{2}-15 &=&2\left( y^{2}+9y-16\right) \\ 3y^{2}-15 &=&2y^{2}+18y-32 \\ y^{2}-18y+17 &=&0 \\ \left( y-17\right) \left( y-1\right) &=&0\text{ \ \ \ \ \ }\Longrightarrow y_{1}=17\text{ \ \ \ \ \ \ \ \ }y_{2}=1 \end{eqnarray*}% Thus there are two points with tangent lines: \ $\left( 3,17\right) $ \ and \ $\left( 3,1\right) $.\newline For the slope of each tangent lines, we differentiate both sides and solve for $y^{\prime }$.% \begin{eqnarray*} xy^{2}-5x &=&2\left( y^{2}+x^{2}y-16\right) \\ y^{2}+x\left( 2yy^{\prime }\right) -5 &=&2\left( 2yy^{\prime }+2xy+x^{2}y^{\prime }\right) \\ y^{2}+2xyy^{\prime }-5 &=&4yy^{\prime }+4xy+2x^{2}y^{\prime } \\ y^{2}-4xy-5 &=&4yy^{\prime }+2x^{2}y^{\prime }-2xyy^{\prime } \\ y^{2}-4xy-5 &=&y^{\prime }\left( 4y+2x^{2}-2xy\right) \\ \dfrac{y^{2}-4xy-5}{4y+2x^{2}-2xy} &=&y^{\prime } \end{eqnarray*}% The slope of the tangent line drawn to $\left( 3,17\right) $% \begin{equation*} m_{1}=\dfrac{y^{2}-4xy-5}{4y+2x^{2}-2xy}=\dfrac{17^{2}-4\left( 3\right) \left( 17\right) -5}{4\left( 17\right) +2\left( 3\right) ^{2}-2\left( 3\right) \left( 17\right) }=\dfrac{80}{-16}=-5 \end{equation*}% We can easily find the point-slope form of the line with slope $-5,$ passing through $\left( 3,17\right) $,\ it is \newline $-5\left( x-3\right) =y-17$. \ Simplifying that, we obtain the slope intercept form which is $y=-5x+32$. \ The other tangent line, passing through $\left( 3,1\right) $ and has slope% \begin{equation*} m_{2}=\dfrac{y^{2}-4xy-5}{4y+2x^{2}-2xy}=\dfrac{1^{2}-4\left( 3\right) \left( 1\right) -5}{4\left( 1\right) +2\left( 3\right) ^{2}-2\left( 3\right) \left( 1\right) }=\dfrac{-16}{16}=-1 \end{equation*}% Thus the slope is $-1$ and the equation of this line is $y-1=-\left( x-3\right) $. \ The slope intercept form is then $y=-x+4.\bigskip $ \item If $y=f\left( x\right) $ \ is a function, we define the curvature as \begin{equation*} C\left( x\right) =\dfrac{\left\vert y^{\prime \prime }\right\vert }{\left( 1+\left( y^{\prime }\right) ^{2}\right) ^{3/2}} \end{equation*}% Prove that if $f\left( x\right) =\sqrt{r^{2}-x^{2}}$ \ \ where $r>0,$ then the curvature is constant on the interval $\left( -r,r\right) $.\newline Proof: \ Let us write $y$ for $f\left( x\right) .$ \ We can see that on $% \left( -r,r\right) $ \ $y$ is always positive and that $x^{2}+y^{2}=r^{2}$\ \begin{eqnarray*} x^{2}+y^{2} &=&r^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ differentiate both sides} \\ 2x+2yy^{\prime } &=&0 \\ x+yy^{\prime } &=&0\text{ \ \ \ \ \ }\Longrightarrow ~~y^{\prime }=-\dfrac{x% }{y} \end{eqnarray*}% For the second derivative, $y^{\prime \prime }$ we differentiate both sides of the statement $x+yy^{\prime }=0$ \begin{eqnarray*} x+yy^{\prime } &=&0 \\ 1+y^{\prime }y^{\prime }+yy^{\prime \prime } &=&0\text{ \ \ \ \ } \\ 1+\left( y^{\prime }\right) ^{2}+yy^{\prime \prime } &=&0 \\ y^{\prime \prime } &=&\dfrac{-1-\left( y^{\prime }\right) ^{2}}{y}=\dfrac{% -1-\left( -\dfrac{x}{y}\right) ^{2}}{y}=\dfrac{-1-\dfrac{x^{2}}{y^{2}}}{y}=% \dfrac{\dfrac{-y^{2}-x^{2}}{y^{2}}}{y}=\dfrac{-x^{2}-y^{2}}{y^{3}}=\dfrac{% -r^{2}}{y^{3}} \end{eqnarray*}% $\allowbreak $Notice that since $y$ is always positive, $y^{\prime \prime }=% \dfrac{-r^{2}}{y^{3}}$ is always negative. \ Thus $\left\vert y^{\prime \prime }\right\vert =-y^{\prime \prime }.$% \begin{eqnarray*} C\left( x\right) &=&\dfrac{\left\vert y^{\prime \prime }\right\vert }{\left( 1+\left( y^{\prime }\right) ^{2}\right) ^{3/2}}=\dfrac{-y^{\prime \prime }}{% \left( 1+\left( y^{\prime }\right) ^{2}\right) ^{3/2}}=\dfrac{\dfrac{r^{2}}{% y^{3}}}{\left( 1+\left( -\dfrac{x}{y}\right) ^{2}\right) ^{3/2}}=\dfrac{% \dfrac{r^{2}}{y^{3}}}{\left( 1+\dfrac{x^{2}}{y^{2}}\right) ^{3/2}} \\ &=&\dfrac{\dfrac{r^{2}}{y^{3}}}{\left( \dfrac{y^{2}+x^{2}}{y^{2}}\right) ^{3/2}}=\dfrac{\dfrac{r^{2}}{y^{3}}}{\left( \dfrac{r^{2}}{y^{2}}\right) ^{3/2}}=\dfrac{\dfrac{r^{2}}{y^{3}}}{\dfrac{r^{3}}{y^{3}}}=\dfrac{1}{r} \end{eqnarray*} \end{enumerate} \vspace{1in} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}