\begin{homework}\
\label{homework:10}

\medskip

\exercise %1.
Find parametric equations $x(t)$, $y(t)$ for the point $P$ if:

(a) $P$ travels at constant velocity from (2,3) to (11,$-3$) between $t=0$
and $t=3$;

(b) $P$ travels at constant speed $6.5$ units/sec from (1,2) in the
direction of (13,7);

(c) $P$ travels at constant speed 10 units/sec from the origin at an angle
of $30^\circ$ above the $x$-axis.

\smallskip


\exercise %2.
A bicycle has wheels of radius 0.4 meter.  Find:

(a) the angular velocity of the wheels in rad/sec if the bicycle is
traveling at 10 m/sec;

(b) the angular velocity of the wheels in rev/sec if the bicycle is
traveling at 10 m/sec;

(c) the speed of the bicycle in m/sec if the wheels are turning at 5
rev/sec;

(d) the speed of the bicycle in km/hr if the wheels are turning at 15
rad/sec.

\smallskip


\exercise %3.
A truck has wheels of radius 1.6 ft.  Find:

(a) the speed of the truck in mph if the wheels are turning at 5 rev/sec;

(b) the angular velocity of the wheels in rad/sec and in rev/sec if the
truck is traveling at 60 mph.

\smallskip

\exercise %4.
Find a formula for the distance in feet that a car has traveled during the
first $t$ sec if its 10-inch-radius wheels:

(a) are turning at the constant rate of 24 rad/sec;

(b) are turning at the constant rate of 9 rev/sec;

(c) start from rest, and accelerate at the rate of 18 rad/sec per second.

\smallskip

\exercise %5.
A wheel of radius 3 is centered at the origin.  Find parameteric equations
for the point $P$ on the rim if:

(a) the wheel is spinning counterclockwise at 0.5 rad/sec, and $P$ is at
(3,0) at time $t=0$;

(b) the wheel is spinning clockwise at 2 rev/sec, and $P$ is at (0,3) at
time $t=0$;

(c) the wheel is spinning clockwise at 1 revolution each 5 seconds, and $P$
is at $(0,-3)$ at time $t=0$;

(d) the wheel is spinning counterclockwise at 10 rev/sec, and $P$ is at
$(0,-3)$ at time $t=0$.
 
\smallskip

\exercise %6.
A girl is playing on a flatcar which is moving in the positive
$x$-direction at 10 m/sec.  At time $t = 0$ she throws the ball straight
upward (as viewed from the moving train)
with an initial velocity 24.5 m/sec from a height 1 meter above the
flatcar.  From her point of view, the ball goes straight up and down,
and lands on the flatcar directly below the point from which she threw it.
However, an observer on the ground sees the ball travel in an arc.

Set up a coordinate system which is fixed to the earth (i.e. one which is
not moving along with the train), with the $y$-axis pointing up and the
$x$-axis pointing in the direction of motion of the train.  Choose the
origin to be the point in space which is 1 meter below the point from which
the ball was released.  Neglect air resistance, and take $g=9.8$
m/sec${}^2$.

(a) Find parametric equations for the motion of the ball in this fixed
coordinate system.

(b) Find the instant when the ball reaches the peak of its trajectory.

(c) Find the instant when the ball hits the flatcar.

(d) Describe in words how you would find the angle at which the ball hits
the flatcar in these coordinates.

\smallskip

\exercise %7.
Suppose that an object is thrown at 70 ft/sec at an angle of 45${}^\circ$
{\it below} the horizontal from a window 163 ft above the ground.

(a)~Express both the horizontal distance $x$ from the window and the height
$y$ above the ground as functions of $t$.

(b)~Find $dy/dx$ in terms of $t$.

(c)~Find the {\bf angle} below the horizontal that the object is pointed
when $t=0.5$ sec and when $t=1$ sec.

(d)~Find the {\bf velocity vector} (its magnitude and angle below the
horizontal) of the object at the instant when it hits the ground.

\smallskip

\exercise %8.
Suppose that a ball is thrown from ground level at $v$ m/sec at an angle of
$\alpha$ above the horizontal.  Take the origin to be the point from which
the ball is thrown.

(a)~Find formulas for $x(t)$ and $y(t)$.  The constants $v$ and $\alpha$
should appear in your formulas.

(b)~Find the time $t$ when the ball hits the ground, expressing your answer
in terms of $v$ and $\alpha$.  There are two ways to do this: (i) Set
$y(t)=0$ and solve for $t$.  (ii) Find the time when the ball reaches its
peak (see Example~\ref{eg:10.8}), and double it (because it takes an equal time to fall
back to the ground).  Use both methods, and see that your answers agree.

(c)~Find the horizontal distance traveled by the time the ball hits the
ground, expressing your answer in terms of $v$ and $\alpha$.
\end{homework}

\newpage

\vspace{1cm}

\begin{center}
{\Large \bf Supplement: Differentiation Formulas}
\end{center}
\vspace{1cm}

\noindent
\large
In the following formulas, $f(x)$ and $g(x)$ are differentiable functions
and $c$ is a constant:

\begin{eqnarray*}
(f+g)'(x) & = & f'(x) + g'(x)\\
(f-g)'(x) & = & f'(x) - g'(x)\\
(fg)'(x) & = & f(x)g'(x) + f'(x)g(x)\\
(cf)'(x) & = & cf'(x)\\
(f/g)'(x) & = & \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}\\
(f(g(x))' & = & f'(g(x))g'(x)\\
g'(x) & = & 1/f'(g(x)) \mbox{\ if\ } g(x) \mbox{\ is the inverse of\ } f(x)\\
\end{eqnarray*}

\noindent
Formulas for common functions:

\vspace{.5cm}
\begin{tabular}{ll}
\vspace{.1cm}
$\frac{d}{dx} x^a = a x^{a-1}$ & $\frac{d}{dx} a^x = (\ln a) a^x$\\
$\frac{d}{dx} \ln x = 1/x$ & $\frac{d}{dx} \log_a x = \frac{1}{(\ln a)x} = \frac{\log_a e}{x}$\\
$\frac{d}{dx} \sin x = \cos x$ & $\frac{d}{dx} \cos x = -\sin x$\\
$\frac{d}{dx} \tan x = \sec^2 x$ & $\frac{d}{dx} \cot x = -\csc^2 x$\\
$\frac{d}{dx} \sec x = \sec(x)\tan(x)$\hspace{1cm} & $\frac{d}{dx} \csc x = -\csc(x)\cot(x)$\\
$\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}}$ & $\frac{d}{dx} \arccos x = \frac{-1}{\sqrt{1 - x^2}}$\\
$\frac{d}{dx} \arctan x = \frac{1}{1 + x^2}$ & $\frac{d}{dx} \mbox{arccot\ } x = \frac{-1}{1 + x^2}$\\
$\frac{d}{dx} \mbox{arcsec\ } x = \frac{1}{x\sqrt{x^2-1}}$ & $\frac{d}{dx} \mbox{arccsc\ } x = \frac{-1}{x\sqrt{x^2-1}}$\\
$\frac{d}{dx} \sinh x = \cosh x$ & $\frac{d}{dx} \cosh x = \sinh x$ \vspace{.1cm}\\
\end{tabular}
\newpage

\begin{center}
{\Large \bf Supplement: Derivative Skills Exercises}
\end{center}

\vspace{1cm}
\begin{enumerate}

\item $f(x) = x^{100} + 50x +1$

\item $f(x) = x^e$

\item $f(x) = (16x)^3$

\item $f(x) = (x^2+1)(2x-7)$

\item $f(x) = x^2 + \frac{1}{x^2}$

\item $f(x) = \frac{1-x^2}{1+x^2}$

\item $f(x) = \cos(x) - 2\: \tan(x)$

\item $f(x) = \frac{\sin(x)}{1+\cos(x)}$

\item $f(x) = x\; \csc(x)$

\item $f(x) = \csc(x)\; \cot(x)$

\item $f(x) = x\: \sin{(x)} \cos{(x)}$

\item $f(x) = e^{\sqrt{x}}$

\item $f(x) = x\; e^{2x}$

\item $f(x) = e^{x\: \cos(x)}$

\item $f(x) = \frac{e^{-x^2}}{x}$

\item $f(x) = \frac{e^{3x}}{1+e^x}$

\item $f(x) = \ln(2-x-x^2)$

\item $f(x) = \ln(\ln(x))$

\item $f(x) = \ln(\cos(x))$

\item $f(x) = \cos(\ln(x))$

\item $f(x) = \log_3(x^2-4)$

\item $f(x) = {(x+2)}^8{(x+3)}^6$

\item $f(x) = \frac{1}{{(x^2-2x-5)}^4}$

\item $f(x) = {\cos}^3(x)$

\item $f(x) = \cos(x^3)$

\item $f(x) = \cos(\tan(x))$

\item $f(x) = x\: \sin(\frac{1}{x})$

\item $f(x) = \frac{{\sin}^2(x)}{\cos(x)}$

\item $f(x) = -\frac{2}{\sqrt[4]{x^3}}$

\item $f(x) = \sqrt[5]{x\; \tan(x)}$

\item $f(x) = \tan(\sqrt{1-x})$

\item $f(x) = \frac{\sin(mx)}{x}$

\item $f(x) = {\cos}^2(\tan\; x)$

\item $f(x) = \sqrt{x\sqrt{x\sqrt{x}}}$

\item $f(x) = \arcsin(2x-1)$

\item $f(x) = \arctan(x^3)$

\item $f(x) = ({\arcsin\,x)}^2$

\item $f(x) = \frac{\arccos(x)}{x}$

\item $f(x) = (1+x^2)\; \arctan(x)$

\item $f(x) = \log_{10}\:(x^2-x)$

\item $f(x) = \arcsin(e^x)$

\item $f(x) = \ln(x^2\,e^x)$

\item $f(x) = x^{\sin(x)}$

\item $f(x) = x^{e^x}$

\item $f(x) = \sin(\tan\sqrt{1+x^3})$

\item $f(x) = \ln(\sqrt[4]{\frac{x^2+x+1}{x^2-x+1}})$

\hspace{-.9cm} Use implicit differentiation:

\item $x^2-xy+y^3 = 8$

\item $x^4+y^4 = 16$

\item $\sqrt{xy} - 2x = \sqrt{y}$

\item $\cos(x-y) = y\: \sin(x)$

\end{enumerate}




