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\begin{center}
MATH 120 -- EXAM II Hints and Answers

Version 1

Autumn 2005
\end{center}

\begin{enumerate}

\item HINT:  $$u(t-5) = \left\{ \begin{array}{cl}
0, & \mbox{if $t<5$} \\
1, & \mbox{if $5\leq t \leq 6$} \\
0, & \mbox{if $t>6$}
\end{array} \right. $$

$$t^2\cdot u(t+2) = \left\{ \begin{array}{cl}
0, & \mbox{if $t<-2$} \\
t^2, & \mbox{if $-2\leq t \leq -1$} \\
0, & \mbox{if $t>-1$}
\end{array} \right. $$

ANSWER:  $$w(t) = \left\{ \begin{array}{cl}
0, & \mbox{if $t<-2$} \\
t^2, & \mbox{if $-2\leq t \leq -1$} \\
0, & \mbox{if $-1<t<5$} \\
1, & \mbox{if $5\leq t \leq 6$} \\
0, & \mbox{if $t>6$}
\end{array} \right. $$

\item HINT:  Do the following to the graph of $f(x)$ in this order:
\begin{itemize}
\item Compress toward the $y$-axis by a factor of 2 to give the graph of $f(2x)$.  This graph contains the points $(-3,8)$, $(-2,0)$, $(-1.5, -1)$, $(-1,0)$, $(0,8)$, and $(4.5,0)$.
\item Compress toward the $x$-axis by a factor of 4 to give the graph of $\frac14f(2x)$.  This graph contains the points $(-3,2)$, $(-2,0)$, $(-1.5, -\frac14)$, $(-1,0)$, $(0,2)$, and $(4.5,0)$.
\item Reflect across the $x$-axis to give the graph of $-\frac14f(2x)$.  This graph contains the points $(-3,-2)$, $(-2,0)$, $(-1.5, \frac14)$, $(-1,0)$, $(0, -2)$, and $(4.5,0)$.
\item Shift the graph up 3 to give the graph of $g(x)=-\frac14f(2x)+3$.  This final graph contains the points $(-3,1)$, $(-2,3)$, $(-1.5,3.25)$, $(-1,3)$, $(0,1)$, and $(4.5,3)$.
\end{itemize}

\item HINT:  Let $x$ be Pam's distance from the base of the statue after she moved in 7~feet and let $y$ be the height of the statue.  Then $\tan 49^{\circ} = \frac{y}{7+x}$ and $\tan 55^{\circ} = \frac{y}x$.  This gives two equations in two unknowns.  Solve for $x$ and $y$.

ANSWER:  41.40 feet

\item  HINT:  Since $f(t)$ is a linear-to-linear rational function, $f(t)=\frac{at+b}{t+d}.$  We have that $a=5$ and thus $f(t)=\frac{5t+b}{t+d}$.  We also have that $f(50)=\frac{250+b}{50+d}=3.85$ and $f(200)=\frac{1000+b}{200+d}=4.7$.  This gives two equations in two unknowns.  Solve for $b$ and $d$.

ANSWER:  $f(t)=\frac{5t-46.18}{t+2.94}$

\item 

\begin{enumerate}

\item HINT:  Pete's linear velocity is 6 mph, which is equivalent to 528~ft/min.  This means that his angular velocty is $\omega = 2.64$~rad/min (use $v=r\omega$).  In two minutes, then, Pete goes 5.28~radians around the track to the point $H$.  This means that $\theta$ is $2\pi-5.28$~radians.

ANSWER:  $\theta = 2\pi - 5.28 = 1.003185307$ radians

\item HINT:  The coordinates of the point $P$ are $(200\cos(2\pi-5.28), 200\sin(2\pi-5.28))$ or $(107.5238452, 168.6375484)$.  After one hour, Pete has traveled 2.64~radians per minute for 60 minutes, or 158.4~radians.  So, after an hour, Pete's angular location is $2\pi+153.12$ and his coordinates are $(200\cos(2\pi+153.12), 200\sin(2\pi+153.12))$, or $(-136.7305846, 145.9614581)$.  Use the distance formula to find the distance between these two points.

ANSWER:  245.30 feet

\end{enumerate}







\end{enumerate}

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