\def\wsn{13}
\input worksheet.tex

\item{1.}  When you see the function $\theta\mapsto\sin \theta$ it is
assummed that $\theta$ is given in radians.  Define a {\bf different}
function $f$ to by setting $f(x)$ as the value of the sine of an angle
which measures $x$ {\bf degrees}.  Let $g$ be defined similarly for
cosine.

\smallskip
\item{}  NOTE: These are indeed different functions!  For most values
of $x$, $f(x)\neq\sin x$.  Why not?  

\medskip
\itemitem{a)}  Express $f(x)$ and $g(x)$ in terms of sin and cos.
\itemitem{}  Remember: $x$ has units of degrees and sin and cos want
to be given radians!

\medskip
\itemitem{b)}  What is $df\over dx$?  What is $dg\over dx$?  (Hint:
Use part a) and the chain rule.)

\medskip
\itemitem{c)}  Express $df\over dx$ and $dg\over dx$ in terms of
$f(x)$ and $g(x)$.  (No mention of sin or cos allowed.)

\medskip
\itemitem{d)}  Is it still true that $(f(x))^2+(g(x))^2=1$?

\medskip
\itemitem{e)}  Why don't we use the unit of degrees in calculus?

\bigskip
\itemitem{2.\hskip12pt a)}  Solve each of the following equations for
$y$, then find $dy\over dx$.
$$
 x^{1/2}+y^{1/2}=1\hskip.8in
 |x|+|y|=1\hskip.8in
 x^2+y^2=1\hskip.8in
 x^3+y^3=1
$$

\itemitem{b)}  Use implicit differentiation to find $dy\over dx$.

\bigskip
\item{3.}   Consider the relation between $x$ and $y$ given by
$$
(\cos x)y^2+(3\sin x-1)y+(7x-2)=0.
$$  

\itemitem{a)}  Check that the pair of values $x=0$ and $y=2$ satisfy
this relation. 

\medskip
\itemitem{b)}  Find $y'$ at the point $(0,2)$ using implicit
differentiation.

\medskip
\itemitem{c)}  Explicitly find an equation for $y$ in terms of $x$
which will be valid near the point $(0,2)$.  
\itemitem{}  (Hint:  When solving quadratic equations, it is sometimes
the {\bf most} difficult part of the problem to decide which of $\pm$
to use!)

\medskip
\itemitem{d)}  Would you like to find $dy\over dx$ at $x=0$ by
differentiating the expression found in c)?  Notice that you are {\bf
not} actually asked to do the computation following from c).

\bigskip
\itemitem{4.\hskip12pt a)}  For each of the following equations, find
$dy\over dx$:  
$$
\pt i x^2+y^2=1 \hskip1in
\pt {ii} y=\cot^2x 
$$

\itemitem{b)}  For each equation above, find $dx\over dy$.  Explain the
difference between the {\bf meaning} of this derivative and the one
you found in part a).

\medskip
\itemitem{c)}  Now suppose that $x$ and $y$ are both dependent on the
variable $t$ (perhaps $t$ denotes time).  For each equation above,
suppose also that these functions $x(t)$ and $y(t)$ satisfy the
equation for all times $t$.  Differentiate each expression with
respect to $t$.  Solve for {\bf both} $dx\over dt$ and $dy\over dt$.
Explain the {\bf meanings} of these derivatives.

\bigskip
\item{5.}  Suppose a particle is moving along a circle described by
the equation $x^2+y^2=1$ as in part a) of the previous problem.  Here
the unit length is one meter, and time $t$ is measured in seconds.
\medskip
\itemitem{a)}  Which derivative expresses the rate (in m/s) at which
$x$ is changing?

\medskip
\itemitem{b)}  Which derivative expresses the rate (in m/s) at which
$y$ is changing?

\medskip
\itemitem{c)}  Suppose the particle is moving through the point
$({1\over2},{\sqrt3\over2})$ with horizontal velocity $4$ m/s.  What
is the vertical velocity of the particle?

\bigskip
\item{6.}  A rectangular room measures 30 feet in length and 12 feet
in height, and the ends are 12 feet in width.  A fly, with a broken
wing, rests at a point one foot down from the ceiling at the middle of
one end.  A smudge of food is located one foot up from the floor at
the middle of the other end.  The fly has just enough energy to $walk$
40 feet.  Show that there is a path along which the fly can walk that
will enable it to get to the food.




\bye