\documentclass{proc-l}
\usepackage[margin=0.5in]{geometry}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{graphicx,changepage,xcolor, float, array}
\usepackage{fancyhdr}
\setlength{\parindent}{0in}
\pagestyle{fancy}
\fancyhf{}
\lhead{Sohail Farhangi, Pan Yan, Yilong Zhang}
\rhead{Solutions to the Recitation Handout for 3/11/2021}
\rfoot{Page \thepage}
%\lfoot{Our work for this problem}
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
  \hskip -\arraycolsep
  \let\@ifnextchar\new@ifnextchar
  \array{#1}}
\makeatother
%\makeatletter
%\newcommand*{\rom}[1]{\expandafter\@slowromancap\romannumeral #1@}
%\makeatother
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{xca}[theorem]{Exercise}
\setlength{\headheight}{12pt}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\definecolor{auburn}{rgb}{0.43, 0.21, 0.1}
\definecolor{darkgreen}{rgb}{0.0, 0.4, 0.13}
\definecolor{darkpastelgreen}{rgb}{0.01, 0.75, 0.24}
\newcommand{\dotrule}[1]{%
   \parbox[t]{#1}{\dotfill}}
\makeatletter
% new style {footmark}
\fancypagestyle{footmark}{
  \ps@@fancy % use {fancy} style as a base of {footmark}
  \fancyfoot[C]{\footmark}
}
% sets value of \footmark and sets the correct style for this page only
\newcommand\markfoot[1]{\gdef\footmark{#1}\thispagestyle{footmark}}
\makeatother
%    Absolute value notation
\newcommand{\abs}[1]{\lvert#1\rvert}

%    Blank box placeholder for figures (to avoid requiring any
%    particular graphics capabilities for printing this document).
\newcommand{\blankbox}[2]{%
  \parbox{\columnwidth}{\centering
%    Set fboxsep to 0 so that the actual size of the box will match the
%    given measurements more closely.
    \setlength{\fboxsep}{0pt}%
    \fbox{\raisebox{0pt}[#2]{\hspace{#1}}}%
  }%
}

\begin{document}
\Huge
\begin{center}Math 2173 Spring 2021 Recitation Handout 8 Solutions\end{center}
\huge
\vskip 15mm

Group Member 1: Sohail Farhangi

\vskip 15mm

Group Member 2: \underline{\hspace{140mm}}

\vskip 15mm

Group Member 3: \underline{\hspace{140mm}}

\vskip 15mm

Group Member 4: \underline{\hspace{140mm}}

\vskip 15mm

Group Member 5: \underline{\hspace{140mm}}

\vskip 15mm

Group Member 6: \underline{\hspace{140mm}}

\vskip 10mm
\normalsize
Below is a checklist of instructions to follow when completing this assignment. Failure to follow these directions will result in penalty on your final score and/or in some problems not being graded. If multiple directions are not followed, then it is also possible that the assignment will not be accepted for any credit at all. Please contact your TA or make a post on the discussion boards for this course if you have any questions about this assignment or these directions.
\vskip 2mm
Sohail Farhangi: farhangi.3@osu.edu, \quad Pan Yan: yan.669@osu.edu, \quad Yilong Zhang: zhang.6100@osu.edu
\vskip 2mm
\begin{tabular}{|m{5mm}|m{185mm}|}
\hline
\multicolumn{2}{|c|}{Checklist of Instructions}\\
\hline
& Please clearly write the names of all group members working on this assignment in the spaces allotted above. \\
\hline
& This assignment must be completed by a group of 3, 4, 5, or 6 members. \\
\hline
& This assignment is to be uploaded to gradescope as a pdf file no later than 11:59 PM EST on Sunday, March 14.\\
\hline
& The assignment will be uploaded by 1 group member, and that group member will be responsible for manually entering the names of all other collaborators into gradescope.\\
\hline
& This assignment must be completed using this template. You may either print this template to write on it and then scan it (pages ordered correctly) into a pdf file, or you may write directly on the template using programs such as notability.\\
\hline
& If you need more space than what is given to solve a given problem, then you will find blank pages provided at the end of this template. At the end of each problem section of this assignment you will find a space in which to indicate on what page your work is continued in case you used additional pages to complete your solution. You must provide the page number on which your work is continued in the alloted space, or write 'N/A' incase you did not use any additional pages.\\
\hline
& On the additional pages, you will also find space in which to indicate which problem the page is being used for, and if the page is used then that space must also be filled.\\
\hline
& To complete this handout, you may use your textbook, class notes, discussions with your TA and group members, and any resources that are available on Carmen. You should not receive any help from the MSLC or people outside of your group when solving these problems. You may discuss these problems on the Carmen discussion boards, but you should not provide your entire solution when answering a such question, you should only give a hint or a helpful idea.\\
\hline
\end{tabular}


\newpage 
\huge
{\bf \textcolor{red}{Ungraded Optional} Problem 14.7.43:} Let $D$ be the region that is bounded by the paraboloid $z = 16-x^2-4y^2$ and the $xy$-plane. Use the change of variables $x = 4u\cos(v), y = 2u\sin(v), z = w$ in order to evaluate

\begin{equation}
\iiint_DzdV.
\end{equation}
\vskip 5mm
\hrule
\vfill
\hrule
\markfoot{Our personal notes for the \textcolor{red}{Ungraded Optional} Problem 14.7.43 is continued on page}
\newpage
{\bf \textcolor{red}{Ungraded Optional} Problem:} Change variables for the following integral with the given transformation. 

\begin{center}
$\displaystyle \iint\limits_{R}$  
$xy$ $dA$, where $R$ is the square with vertices $(0,0), (3,3), (6,0), (3,-3)$.

Use $x=u+v$, $y=u-v$.
\end{center}

(a) Sketch the original region of integration $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane using the given change of variables.


(b) Find the limits of integration for the new integral with respect to $u$ and $v$.
\begin{center}
    $\rule{1cm}{0.15mm} \leq u \leq \rule{1cm}{0.15mm}$, \hspace{0.5cm}
    $\rule{1cm}{0.15mm} \leq v \leq \rule{1cm}{0.15mm}$. 
\end{center}

(c) Compute the Jacobian $J(u,v)$. 

(d) Set up and evaluate the new integral over the region $S$.
\vskip 5mm
\hrule
\vfill
\hrule
\markfoot{Our personal notes for this \textcolor{red}{Ungraded Optional} Problem are continued on page}
\clearpage
{\bf Problem 14.7.32 (8 points):} Let $R$ be the region bounded by the lines $y-x = 0, y-x = 2, y+x = 0, y+x = 2$. Use a change of variables to evaluate

\begin{equation}
\iint_R\sqrt{y^2-x^2}dA.
\end{equation}
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution:} We use the substitution $u = y-x$ and $v = y+x$ as suggested by the defining equations of the boundary curves. We also see in the picture below that this substitution results in a simple tesselation of our region $R$, which shows us that the new region of integration in the $uv$-plane is just $R' = \{(u,v)\ |\ 0 \le u \le 2, 0 \le v \le 2\}$. 

\begin{figure}[H]
\includegraphics[width=\textwidth]{Problem14-7-32xyPlane.PNG}
\caption{A picture of the region $R$ and the tesselation that results from our given change of variables.}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{Problem14-7-32uvPlane.PNG}
\caption{A picture of the region of integration in the $uv$-plane $R'$.}
\end{figure}

In order to calculate the Jacobian $J(u,v)$, we need to solve for $x$ and $y$ in terms of $u$ and $v$. To this end, we see that

\begin{equation}
\begin{array}{rcl}
u & = & y-x\\
v & = & y+x
\end{equation}
\rightarrow 
\begin{array}{rcccl}
x & = & \frac{1}{2}\Big((y+x)-(y-x)\Big) & = & \frac{1}{2}(v-u)\\
y & = & \frac{1}{2}\Big((y+x)+(y-x)\Big) & = & \frac{1}{2}(v+u)
\end{array}.
\end{equation}

We now see that

\begin{equation}
\renewcommand*{\arraystretch}{1.5}
J(u,v) = \begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix}-\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{vmatrix} = (-\frac{1}{2})\cdot\frac{1}{2}-\frac{1}{2}\cdot\frac{1}{2} = -\frac{1}{2}.
\end{equation}

It follows that $|J(u,v)| = |-\frac{1}{2}| = \frac{1}{2}$. We now see that

\begin{equation}
\text{Area}(R) = \iint_R\sqrt{y^2-x^2}dA = \iint_{R'}\sqrt{(y-x)(y+x)}\cdot|J(u,v)|dA 
\end{equation}

\begin{equation}
= \int_0^2\int_0^2\sqrt{uv}\frac{1}{2}dudv = \frac{1}{2}\int_0^2\int_0^2u^{\frac{1}{2}}v^{\frac{1}{2}}dudv = \frac{1}{2}\int_0^2\frac{2}{3}u^{\frac{3}{2}}v^{\frac{1}{2}}\Big|_{u = 0}^2dv
\end{equation}

\begin{equation}
= \frac{1}{2}\int_0^2\frac{2}{3}2^{\frac{3}{2}}v^{\frac{1}{2}}dv = \frac{2\sqrt{2}}{3}\int_0^2v^{\frac{1}{2}}dv = \frac{4\sqrt{2}}{9}v^{\frac{3}{2}}\Big|_0^2 = \boxed{\frac{16}{9}}.
\end{equation}
\newpage
{\bf Problem 14.7.42 (12 points):} Find the volume of the solid $D$ that is bounded by the planes $y-2x = 0, y-2x = 1, z-3y = 0, z-3y = 1, z-4x = 0,$ and $z-4x = 3$.
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution:} We use the substitution $u = y-2x, v = z-3y,$ and $w = z-4x$ as suggested by the defining equations of the boundary curves. We also see in the pictures below that this substitution results in a simple tesselation of our region $R$, which shows us that the new region of integration in the $uvw$-space is just $R' = \{(u,v,w)\ |\ 0 \le u \le 1, 0 \le v \le 1, 0 \le w \le 3\}$. 

\begin{figure}[H]
\includegraphics[width=\textwidth]{Problem14-7-42P1.PNG}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{Problem14-7-42P2.PNG}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{Problem14-7-42P3.PNG}
\end{figure}

\begin{figure}[H]
\includegraphics[width=.7\textwidth]{Problem14-7-42R'.PNG}
\caption{The region of integration $R'$ in the $uvw$-space.}
\end{figure}

In order to calculate the Jacobian $J(u,v,w)$, we need to solve for $x, y,$ and $z$ in terms of $u, v,$ and $w$. To this end, we see that

\begin{equation}
\begin{array}{rcl}
u & = & y-2x\\
v & = & z-3y\\
w & = & z-4x
\end{array}
\rightarrow
v-w = 4x-3y
\end{equation}

\begin{equation}
\rightarrow (v-w)+3u = -2x \rightarrow \textcolor{blue}{x} = \textcolor{blue}{\frac{1}{2}(-3u-v+w)}
\end{equation}

\begin{equation}
\rightarrow 
\begin{array}{rcccccl}
\textcolor{orange}{y} & = & u+2x & = & u-3u-v+w & = & \textcolor{orange}{-2u-v+w}\\
\textcolor{darkpastelgreen}{z} & = & w+4x & = & w-6u-2v+2w & = & \textcolor{darkpastelgreen}{-6u-2v+3w}
\end{array}.
\end{equation}

We now see that

\begin{equation}
\renewcommand*{\arraystretch}{1.5}
J(u,v,w) = \begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}\end{vmatrix} = \begin{vmatrix}-\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} \\ -2 & -1 & 1 \\ -6 & -2 & 3\end{vmatrix}
\end{equation}

\begin{equation}
= -\frac{3}{2}\begin{vmatrix}-1 & 1\\ -2 & 3\end{vmatrix}-(-\frac{1}{2})\begin{vmatrix}-2 & 1\\ -6 &3\end{vmatrix}+\frac{1}{2}\begin{vmatrix}-2 & -1\\-6 & -2\end{vmatrix}
\end{equation}

\begin{equation}
= -\frac{3}{2}(-3+2)+\frac{1}{2}(-6+6)+\frac{1}{2}(4-6) = \frac{1}{2}.
\end{equation}

It follows that $|J(u,v,w)| = \frac{1}{2}$. We now see that

\begin{equation}
\text{Volume}(D) = \iiint_D1dV = \iiint_{D'}1\cdot|J(u,v,w)|dV
\end{equation}

\begin{equation}
= \int_0^1\int_0^1\int_0^3\frac{1}{2}dudvdw = (1-0)(1-0)(3-0)\frac{1}{2} = \boxed{\frac{3}{2}}.
\end{equation} 
\clearpage
{\bf Problem 14.7.57 ({\bf Parabolic coordinates}, 30 points):} This problem has parts {\bf a.}-{\bf \textcolor{red}{g}.} spread out across the following pages. Your solutions to parts {\bf a, b,} and {\bf f} should include (hand drawn or computer generated) pictures. Part {\bf \textcolor{red}{g}.} is \textcolor{red}{optional and ungraded}.  Please clearly separate your work for the different parts of this problem and label the parts of your work accordingly.
\vskip 5mm
Consider the Transformation $T$ from the $uv$-plane to the $xy$-plane given by $T(u,v) = (u^2-v^2, 2uv)$.
\begin{itemize}
\item[{\bf a.}] Show that the lines $u = a$ in the $uv$-plane map to parabolas in the $xy$-plane that open in the negative $x$-direction with vertices\footnote{The vertex of the parabola $y = x^2$ is the point $(0,0)$ and the vertex of the parabola $x = y^2$ is also $(0,0)$.} on the positive $x$-axis.\footnote{You have to show that the curve $\vec{r_1}(v) = (a^2-v^2,2av)$ represents the same curve as $x-c = -by^2$ for some positive numbers $b$ and $c$.} Compare the images of the lines $u = a$ and $u = -a$ under $T$.

\item[{\bf b.}] Show that the lines $v = b$ in the $uv$-plane map to parabolas in the $xy$-plane that open in the positive $x$-direction with vertices on the negative $x$-axis.\footnote{You have to show that the curve $\vec{r_2}(u) = (u^2-b^2,2ub)$ represents the same curve as $x-c = by^2$ for some positive number $b$ and some negative number $c$.} Compare the images of the lines $v = b$ and $v = -b$ under $T$.

\item[{\bf c.}] Evaluate $J(u,v)$.
\end{itemize}
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution to part a:} We see that $T(a,v) = (a^2-v^2,2av)$. Setting $x = a^2-v^2$ and $y = 2av$, we see that $v = \frac{1}{2a}y$, so $x = a^2-(\frac{1}{2a}y)^2 = a^2-\frac{1}{4a^2}y^2$, or equivalently, $x-a^2 = -\frac{1}{4a^2}y^2$. Since $a^2 > 0$ and $-\frac{1}{4a^2} < 0$ when $a \neq 0$, we see (as mentioned in the footnote) that $T(a,v)$ is the parameterization of a parabola that opens in the negative $x$-direction and has its vertex on the positive $x$-axis. We see that $T(-a,v) = (a^2-v^2,-2av) = (a^2-(-v)^2,2a(-v) = T(a,-v)$, so $T(a,v)$ and $T(-a,v)$ parameterize the same parabola in the $xy$-plane, but the parameterizations are in opposite directions (if $a \neq 0$). We also see that $T(0,v) = (-v^2,0)$, which is a parameterization (with repitition) of the negative $x$-axis, which can be viewed as a degenerate parabola that opens in the negative $x$-direction and has its vertex at $(0,0)$. 

\begin{figure}[H]
\includegraphics[width=\textwidth]{uLines.PNG}
\caption{Vertical lines in the $uv$-plane.}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{ImageOfu=a.PNG}
\caption{The parabolas in the $xy$-plane corresponding to vertical lines in the $uv$-plane under the transformation $T$.}
\end{figure}

\vskip 5mm

{\bf Solution to part b:} We see that $T(u,b) = (u^2-b^2, 2ub)$. Setting $x = u^2-b^2$ and $y = 2ub$, we see that $u = \frac{1}{2b}y$, so $x = (\frac{1}{2b}y)^2-b^2 = \frac{1}{4b^2}y^2-b^2$, or equivalently, $x-(-b^2) = \frac{1}{4b^2}y^2$. Since $-b^2 < 0$ and $\frac{1}{4b^2} > 0$ when $b \neq 0$, we see (as mentioned in the footnote) that $T(u,b)$ is the parameterization of a parabola that opens in the postive $x$-direction and has its vertex on the negative $x$-axis. We see that $T(u,-b) = (u^2-b^2,-2ub) = ((-u)^2-b^2,2(-u)b) = T(-u,b)$, so $T(u,b)$ and $T(u,-b)$ parameterize the same parabola in the $xy$-plane, but the parameterizations are in opposite directions (if $b \neq 0$). We also see that $T(u,0) = (u^2,0)$, which is a parameterization (with repitition) of the positive $x$-axis, which can be viewed as a degenerate parabola that opens in the positive $x$-direction and has its vertex at $(0,0)$. 


\begin{figure}[H]
\includegraphics[width=\textwidth]{vLines.PNG}
\caption{Horizontal lines in the $uv$-plane.}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{ImageOfv=b.PNG}
\caption{The parabolas in the $xy$-plane corresponding to horizontal lines in the $uv$-plane under the transformation $T$.}
\end{figure}
\vskip 5mm

{\bf Solution to part c:} Since $x = u^2-v^2$ and $y = 2uv$, we see that

\begin{equation}
\renewcommand*{\arraystretch}{1.5}
J(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix} 2u & 2v \\ -2v & 2u\end{vmatrix} = 2u\cdot2u-(-2v)\cdot2v = \boxed{4u^2+4v^2}.
\end{equation}

We also observe that $|J(u,v)| = J(u,v) = 4u^2+4v^2$ since squares are always bigger than or equal to $0$.
\newpage
\begin{itemize}
\item[{\bf d.}] Use a change of variables into parabolic coordinates to find the area of the region $R$ in the $xy$-plane bounded by the curves $x = 4-\frac{1}{16}y^2$ and $x = \frac{1}{4}y^2-1$. Sketch a picture of the new region of integration as well.

\begin{figure}[H]
\includegraphics[width=\textwidth]{ParabolicRegion.PNG}
\end{figure}
\end{itemize}
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution to part d:} We begin by using parts {\bf a} and {\bf b} to see that the parabola $x = 4-\frac{1}{16}y^2 = 2^2-\frac{1}{4\cdot2^2}y^2$ in the $xy$-plane is the image under $T$ of the line $u = 2$ (or $u = -2$) in the $uv$-plane and the parabola $x = \frac{1}{4}y^2-1 = \frac{1}{4\cdot1^2}y^2-1^2$ in the $xy$-plane is the image under $T$ of the line $v = 1$ (or $v = 1$) in the $uv$-plane. Note that for a postive number $p$, $T(2,p)$ is on the upper half of the parabola $x = 4-\frac{1}{16}y^2$ and $T(2,-p)$ is on the lower half. Similarly, for $T(p,1)$ is on the upper half of the parabola $x = \frac{1}{4}y^2-1$ and $T(-p,1) = T(p,-1)$ is on the lower half. We also recall that $T$ bijects the closed right (or left) half of the $uv$-plane to the $xy$-plane.\footnote{The map $T$ from the $uv$-plane to the $xy$-plane is a one-to-one map if you restrict yourself to the closed right (or left) half of the $uv$-plane, but $T$ is not one-to-one on the entire $uv$-plane since $T(a,b) = T(-a,-b)$.} The picture below puts together all of the previous discussion to show that the region $R$ in the $xy$-plane is the image under $T$ of the region rectangle $R' = \{(u,v)\ |\ 0 \le u \le 2, -1 \le v \le 1\}$ in the $uv$-plane. 

\begin{figure}[H]
\includegraphics[width=\textwidth]{ParabolicRegionTesselation.PNG}
\end{figure}

\begin{equation}
\text{Area}(R) = \iint_R1dA = \iint_{R'}1\cdot|J(u,v)|dA = \int_0^2\int_{-1}^1|J(u,v)|dvdu
\end{equation}

\begin{equation}
= \int_0^2\int_{-1}^1(4u^2+4v^2)dvdu = \int_0^2(4u^2v+\frac{4}{3}v^3\Big|_{v = -1}^1)du = \int_0^2(8u^2+\frac{8}{3})du 
\end{equation}

\begin{equation}
= \frac{8}{3}u^3+\frac{8}{3}u\Big|_{u = 0}^2 = \boxed{\frac{80}{3}}.
\end{equation}
\newpage
\begin{itemize}
\item[{\bf e.}] Use a change of variables into parabolic coordinates to find the area of the curved rectangle $R$ above the $x$-axis bounded by $x = 4-\frac{1}{16}y^2, x = 9-\frac{1}{36}y^2, x = \frac{1}{4}y^2-1,$ and $x = \frac{1}{64}y^2-16$. Sketch a picture of the new region of integration as well.

\begin{figure}[H]
\includegraphics[width=\textwidth]{ParabolicRectangle.PNG}
\end{figure}
\end{itemize}
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution to part e:} We proceed as we did in part {\bf d}. We note that $\textcolor{red}{x = 4-\frac{1}{16}y^2}$ corresponds to $\textcolor{red}{u = 2,-2}$, $\textcolor{blue}{x = 9-\frac{1}{36}y^2}$ corresponds to $\textcolor{blue}{u = 3,-3}$, $\textcolor{darkpastelgreen}{x = \frac{1}{4}y^2-1}$ corresponds to $\textcolor{darkpastelgreen}{v = 1,-1}$, and $\textcolor{purple}{x = \frac{1}{64}y^2-16}$ corresponds to $\textcolor{purple}{v = 4,-4}$. Since $y = 2uv$ is positive when $u$ and $v$ are both positive (or both negative), we obtain the parabolic rectangle above the $x$-axis as the image of the region $R' = \{(u,v)\ |\ 2 \le u \le 3, 1 \le v \le 4\}$ under $T$. 

\begin{figure}[H]
\includegraphics[width=\textwidth]{Problem14-7-57-NewRegionPe.PNG}
\caption{The new region of integration $R'$ in the $uv$-plane.}
\end{figure}

We now see that

\begin{equation}
\text{Area}(R) = \iint_R1dA = \iint_{R'}1\cdot|J(u,v)|dA = \int_2^3\int_1^4(4u^2+4v^2)dvdu
\end{equation}

\begin{equation}
= \int_2^3(4u^2v+\frac{4}{3}v^3\Big|_{v = 1}^4)du = \int_2^3(12u^2+84)du = 4u^3+84u\Big|_2^3 = \boxed{160}.
\end{equation}

\newpage
\begin{itemize}
\item[{\bf f.}] Describe the effect of the transformation $(u,v)\mapsto(2uv, u^2-v^2)$ on horizontal and vertical lines in the $uv$-plane.\footnote{Remember that the transformation $(x,y)\mapsto(y,x)$ reflects points in the $xy$-plane across the line $y = x$. It will also help to use the results of parts {\bf a.} and {\bf b.} of this problem.}
\end{itemize}
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution to part f:} Let $S(u,v) = (2uv,u^2-v^2)$. If $P$ is a parabola that opens in the negative $x$-direction and has its vertex on the positive $x$-axis, then upon reflection over the line $x = y$, we obtain a parabola $P'$ that opens in the negative $y$-direction and has its vertex on the positive $y$-axis. It follows that the image of the vertical line $u = a$ in the $uv$-plane under the transformation $S$ gives a parabola in the $xy$-plane that opens in the negative $y$-direction and has its vertex on the positive $y$-axis. Similarly, if $P$ is a parabola that opens in the positive $x$-direction and has its vertex on the negative $x$-axis, then upon reflection over the line $x = y$, we obtain a parabola $P'$ that opens in the positive $y$-direction and has its vertex on the negative $y$-axis. It follows that the image of the horizontal line $v = b$ in the $uv$-plane under the transformation $S$ gives a parabola in the $xy$-plane that opens in the positive $y$-direction and has its vertex on the negative $y$-axis. 

\newpage

\begin{figure}[H]
\includegraphics[width=\textwidth]{uLines.PNG}
\caption{Vertical lines in the $uv$-plane.}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{AnotherImageOfu=a.PNG}
\caption{The parabolas in the $xy$-plane corresponding to vertical lines in the $uv$-plane under the transformation $S$.}
\end{figure}

\newpage

\begin{figure}[H]
\includegraphics[width=\textwidth]{vLines.PNG}
\caption{Horizontal lines in the $uv$-plane.}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{AnotherImageOfv=b.PNG}
\caption{The parabolas in the $xy$-plane corresponding to horizontal lines in the $uv$-plane under the transformation $S$.}
\end{figure}
\newpage
\begin{itemize}
\item[{\bf \textcolor{red}{g.}}] Show that the parabolas that are the images of the lines $u = a$ and $v = b$ under $T(u,v) = (u^2-v^2, 2uv)$ are orthogonal to eachother.
\end{itemize}
\vskip 5mm
\hrule
\vskip 5mm
{\bf Solution to part \textcolor{red}{g}:} We have already seen in parts {\bf a} and {\bf b} that $T(a,v)$ is the parabola $x = a^2-\frac{1}{4a^2}y^2$ and $T(u,b)$ is the parabola $x = \frac{1}{4b^2}y^2-b^2$. We will first find the intersection points of these 2 parabolas, then we will calculate the slope of the tangent lines at the intersection points in order to see that the tangent lines (and hence the curves) are orthogonal. 

\begin{figure}[H]
\includegraphics[width=\textwidth]{PerpendicularParabolas.PNG}
\end{figure}

\begin{figure}[H]
\includegraphics[width=\textwidth]{PerpendicularParabolasZoomIn.PNG}
\end{figure}

To this end, we see that

\begin{equation}
\renewcommand*{\arraystretch}{1.3}
\begin{array}{rcl}
x & = & a^2-\frac{1}{4a^2}y^2\\
x & = & \frac{1}{4b^2}y^2-b^2
\end{array}
\rightarrow
a^2-\frac{1}{4a^2}y^2 = \frac{1}{4b^2}y^2-b^2 \rightarrow a^2+b^2 = (\frac{1}{4a^2}+\frac{1}{4b^2})y^2
\end{equation}

\begin{equation}
\rightarrow y^2 = \frac{a^2+b^2}{\frac{1}{4a^2}+\frac{1}{4b^2}} = 4a^2b^2 \rightarrow y = \pm2ab\rightarrow x = a^2-b^2.
\end{equation}

It follows that $T(a,b) = T(-a,-b) = (a^2-b^2, 2ab)$ and $T(a,-b) = T(-a,b) = (a^2-b^2, -2ab)$ are the intersection points of the 2 parabolas. Noting that

\begin{equation}
x = a^2-\frac{1}{4a^2}y^2 \rightarrow dx = -\frac{1}{2a^2}ydy \rightarrow \frac{dy}{dx} = -2\frac{a^2}{y},\text{ and}
\end{equation}

\begin{equation}
x = \frac{1}{4b^2}y^2-b^2 \rightarrow dx = \frac{1}{2b^2}ydy \rightarrow \frac{dy}{dx} = 2\frac{b^2}{y},
\end{equation} 

We see that at the point $(a^2-b^2,2ab)$, the tangent line to the curve $x = a^2-\frac{1}{4a^2}y^2$ has a slope of $-\frac{a}{b}$ and the tangent line to the curve $x = \frac{1}{4b^2}y^2-b^2$ has a slope of $\frac{b}{a}$. Since $-\frac{a}{b}\cdot\frac{b}{a} = -1$, we see that the tangent lines at the point $(a^2-b^2,2ab)$ are indeed orthogonal to each other. Similarly, we see that at the point $(a^2-b^2,-2ab)$, the tangent line to the curve $x = a^2-\frac{1}{4a^2}y^2$ has a slope of $\frac{a}{b}$ and the tangent line to the curve $x = \frac{1}{4b^2}y^2-b^2$ has a slope of $-\frac{b}{a}$. Since $\frac{a}{b}\cdot(-\frac{b}{a}) = -1$, we see that the tangent lines at the point $(a^2-b^2,2ab)$ are indeed orthogonal to each other.

{\bf Remark:} We see that the parabolas produced by $S$ in part {\bf f} also share this orthogonality property since orthogonality is preserved under reflections.

\end{document}