\documentclass[12pt]{article}
\pagestyle{empty}

\usepackage{amssymb}
\usepackage{amsmath}


\newcommand{\adj}{\mathrm{Adj}}
\newcommand{\mbf}{\mathbf}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ra}{\rightarrow}
\newcommand{\proj}{\mathrm{proj}}
\newcommand{\dotprod}{\bullet}

\begin{document}
\input{epsf.sty}


\begin{center}
{\Large KEY: TEST 3}
\end{center}



\begin{enumerate}

\item  We want to compute the hydrostatic force on the gate (see page 555 of text).
\begin{eqnarray*}
F & = & \int \rho g d \;dA
\end{eqnarray*}
where $\rho = 1000 kg/m^3$, $g =9.8 m/s^2$, $d$ is the depth (not a contant), and $dA$ is area.
I made my coordinate system so that $y$ is the vertical axis, $y=0$ at the bottom of the pool and $y=0.3$ at the top of the gate.
With this choice of axes, the depth is given by $d = 6-y$.
The width of a slice is $x$, so $dA=x\;dy$.
The equation of the side of the gate is $y=0.3-(3/2)x$, or $x=0.2-(2/3)y$.
This means that $dA = (.2-2y/3)dy$ giving hydrostatic force as:
\begin{eqnarray*}
F & = & \int_0^{0.3} (1000)(9.8)(6-y)(.2-2y/3) \; dy \\
	& = & 1734.6 \; N \approx 390 \; lbs
\end{eqnarray*}


\item 
\begin{enumerate}
\item  You should draw the arrow, but the curve starts in quadrant III, travels to quadrant IV, then continues around.

\item  You can either solve equations or plot points.  In any case, you should fine that the points are:
\begin{eqnarray*}
t=-2 & \implies & x = 0 \qquad y = -4 \\
t=0 & \implies & x = 0 \qquad y = 0 \\
t=2 & \implies & x = 0 \qquad y = -4 \\
\end{eqnarray*}
Determine the $t$ and $x$ and $y$ values of the intersection point on the $y$-axis.

\item  Here we just solve $dy/dx=0$ to find horizontal tangents and $dx/dt$ to find vertical tangent lines.
$dy/dt= -2t$, so there is a horizontal tangent when $t=0$ or at $(0,0)$.
$dx/dt = 3t^2-4$ and so there are vertical tangents when $t= \pm 2/\sqrt{3}$:
\begin{eqnarray*}
\mbox{Horizontal:} & & t=0 \quad (0,0) \\
\mbox{Vertical:} & & t=-\frac{2}{\sqrt{3}} \quad \left( \frac{16\sqrt{3}}{9}, \frac{4}{3} \right) \\
	 & & t=\frac{2}{\sqrt{3}} \quad \left( -\frac{16\sqrt{3}}{9}, \frac{4}{3} \right) 
\end{eqnarray*}

\item  Notice that area is given by $\int x \; dy$.  $dy = -2t\;dt$.
\begin{eqnarray*}
\mbox{Area} & = & 2\int_{-4}^0 x \; dy \\
	& = & 2\int_{-2}^0 (t^3-4t)(-2t)\; dt
\end{eqnarray*}

\item  Length is given by $\int ds$ where $ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}\; dt$.
\begin{eqnarray*}
L & = & \int ds = \int \sqrt{(dx/dt)^2 + (dy/dt)^2}\; dt \\
	& = & \int_{-2}^{2} \sqrt{(3t^2-4)^2 + (-2t)^2}\; dt
\end{eqnarray*}

\item  Surface area is given by $A = \int 2\pi r ds$ where $ds$ is as above and $r=x$.
\begin{eqnarray*}
A & = &  \int 2\pi r \;ds = \int_{-2}^0 2\pi x \sqrt{(dx/dt)^2 + (dy/dt)^2}\; dt \\
	& = & \int_{-2}^0 2\pi (t^3-4t)\sqrt{(3t^2-4)^2 + (-2t)^2}\; dt
\end{eqnarray*}
\end{enumerate}


\item  
\[
r = \sin \theta \cos^2 \theta
\]
\begin{enumerate}
\item  Here we can substitute $\sin \theta = \frac{y}{r}$ and $\cos \theta = \frac{x}{r}$.
	After that substitution, we can substitute $r^2=x^2+y^2$:
\begin{eqnarray*}
r & = & \left(\frac{y}{r}\right) \left(\frac{x}{r}\right)^2 = \frac{x^2y}{r^3} \\
r^4 & = & x^2y \\
(x^2+y^2)^2 & = & x^2y
\end{eqnarray*}

\item  Notice that the two loop are given by $0 \leq \theta \leq \pi$.
Therefore the area is:
\begin{eqnarray*}
A & = & \int_0^\pi \frac{1}{2} r^2 \; d\theta \\
	& = & \int_0^\pi \frac{1}{2} \left( \sin \theta \cos^2 \theta \right)^2 \; d\theta 
\end{eqnarray*}

\item  The easiest formula for this is on page 673 of the text.
Compute $dr/d\theta$ (using the product rule and chain rule):
\[
\frac{dr}{d\theta} = \cos^3\theta - 2 \sin^2\theta \cos \theta
\]
Formula for length:
\begin{eqnarray*}
L & = & \int ds = \int \sqrt{r^2 + \left( \frac{dr}{d\theta} \right)^2} \; d\theta \\
	& = & \int_0^\pi \sqrt{(\sin \theta \cos^2 \theta)^2 + (\cos^3\theta - 2 \sin^2\theta \cos \theta)^2} \; d\theta
\end{eqnarray*}
\end{enumerate}


\end{enumerate}

\end{document}