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\noindent
{\Large {\bf Mathematics 1210  \hspace{5ex} PRACTICE EXAM II
\hspace{5ex}Fall 2004}}
%\vspace{2ex}\\
\begin{center}
{\Large {\bf ANSWER KEY}}
\end{center}
\vspace{2ex}



\begin{enumerate}

\item[1.] 

\begin{enumerate}

\item[a.] Using the chain rule,
$\ds x''(t) = -\omega^2 A\sin({\omega t-\phi})=-\omega^2x(t)$

\item[b.] 

Using the chain rule and the quotient rule, 
$\ds f'(x)=3\left( \frac{x-2}{x-\pi} \right)^{2}
\frac{(2-\pi)}{(x-\pi)^2}$


\item[c.] Implicit differentiation treats $y$ as a
function of $x$ even though it is not known explicitly.
First differentiate the equation defining the relationship
between $x$ and $y$ with respect to $x$ and apply the
usual differentiation rules (chain rule and product rule
in the first term) to obtain
$-\sin(xy)(y+xy')=2yy'+2$. Then collect terms which include
a factor of $y'$ from the chain rule and factor it out.
$-y\sin(xy)-2= (2y+x\sin(xy))y'$
Finally, divide to solve for $y'$:
$\ds y' = -(2+y\sin{xy})/(2y+x\sin{xy})$

\item[d.] Multiply the top and bottom by $1+\cos x$ to get
$$\lim_{x \to 0} \frac{ \sin x \tan x}{1-\cos x} =
\lim_{x \to 0} \frac{ (1+\cos x)}{\cos x} = 2. $$
(or use $\sin{x}\sim x, \;\tan{x} \sim x,\;
\cos{x} \sim 1 - x^2/2$ as $x \rightarrow 0$)

\item[e.] This is of the form $f(g(\frac{u(x)}{v(x)})$,
where $f(x)=\sin x$, $g(x)=x^{1/2}$, $u(x)=\tan x$, and
$v(x)=1+x^2$,
so using the chain rule (twice) and quotient rule,
$\ds f'(x)= \cos{\sqrt{ \frac{\tan{x}}{1+x^2} }}\;\; \frac{1}{2}
\left( \frac{\tan{x}}{1+x^2}  \right)^{-1/2}\;\;
\frac{(1+x^2)\sec^2{x}-2x\tan{x}}{(1+x^2)^2}$



\item[f.] This is of the form $p(x) q(r(s(x)))$ where
$p(x)=x^2$, $q(x)=x^2$, $r(x)=\sin x$, $s(x)=x^3$,
so using the product rule
and the chain rule (twice)
$\ds f'(x)= 2x\sin^2{(x^3)} + 6x^4\sin{(x^3)}\cos{(x^3)} .$
I like to rewrite $\sin^2 (x^3)$ as $(\sin (x^3))^2$ either explicitly
or at least in my mind. So the second term in the derivative
comes from $x^2 \cdot 2 \sin (x^3) \cos (x^3) \cdot 3x^2$.



\end{enumerate}

\item[2.] 
Using the sine difference formula, 
$$A \sin (\omega t - \phi) = 
(A \cos \phi)\sin (\omega t) - (A \sin \phi) \cos (\omega t).$$
This is of the form 
$P \cos (\omega t) + Q \sin (\omega t)$ where $P$ and $Q$
may be set arbitrarily: Set $\phi$ using $\phi=\arctan(-P/Q)$
(or $\phi=\pi /2$ if $Q=0$) 
and set $A=-P/\sin \phi$. 
\medskip

We know that $\cos (\omega t)$ and $\sin (\omega t)$ are
both solutions of $x''=-\omega^2 x$. Since differentiation is linear,
$(cf+g)'=cf'+g'$, $u(t)=A\cos (\omega t) + B\sin (\omega t)$ is also
a solution for any $A$ and $B$, with $u(0)=A$ and $u'(0)=\omega B$. 

\medskip
These are the only solutions, because if 
$v(t)$ is a solution
with $v(0)=A$ and $v'(0)=\omega B$, then $w=u-v$ is a solution
of $w''=-\omega^2 w$ with $w(0)=0$ and $w'(0)=0$. By the chain rule,
$${d \over dt} 1/2 (\omega^2 w(t)^2+w'(t)^2) =
\omega^2 w(t)w'(t)+ w'(t)w''(t) = w'(t)(w(t)-\omega^2 w(t)) = 0.$$
So $1/2 (\omega^2 w(t)^2+w'(t)^2)$ is a constant, and since
it is equal to zero at $t=0$, it must equal zero for all $t$.
And since this quantity is a sum of squares, the individual
terms must then be zero too. So $w(t)^2=0$ so $w(t)=0$.

\medskip
By the same arguments, we can also write the general solution in the form
$x(t)=B\cos({\omega t-\psi)}$.

For the form $x(t)=A \sin (\omega t - \phi) $,
$A$ is called the amplitude, and gives the maximum and minimum heights
of the graph, and $\phi$ is a phase shift: the zero of $\sin t$ at
the origin is shifted to where $\omega t-\phi=0$, i.e., where
$t=\phi/ \omega$.

We know this is where the slope of the sine function is at its maximum
(or minimum, depending upon the sign of $A$) so the velocity is
maximized at 
$$t=\phi/ \omega + n \pi/ \omega, \quad n=\ldots -1,0,1, \ldots$$

The displacement is maximized or minimized where the argument of the
sine function is $\pi/2 + n \pi, \quad n=\ldots -1,0,1, \ldots$,
so setting this to $\omega t - \phi$ we get
$$t=\frac{\phi + \pi/2 + n \pi}{\omega}, \quad \quad n=\ldots -1,0,1, \ldots$$

For $x(t)=2\sin{(t-\pi/4)}$, velocity is extremized when $x(t)=0$,
or $t=n\pi + \pi/4$, position is extremized when
$v(t)=x'(t)=0$, or $t=(n+\frac{1}{2})\pi + \pi/4$.
A figure is at the end of these solutions.
%\vspace{2.3in}



%\item[3.] $\ds \frac{dr}{dt}=0.02 \;in/sec, \;A=\pi r^2, \;\frac{dA}{dt}
%          =2\pi r\frac{dr}{dt}=1.018\; in^2/sec.$

%\item[4.] The positions of the tips of the hands (so that 3:00
%         corresponds to $t=0$ hours), are given by
%         $ (x_1(t),y_1(t))=(4\cos{\omega_1t},-4\sin{\omega_1t})$
%         for the hour hand,
%        and $ (x_2(t),y_2(t))= (5\sin{\omega_2t},5\cos{\omega_2t})$
%         for the minute hand, where
%         $\omega_1=2\pi f = 2\pi$ and $\omega_2=2\pi f = 2\pi 60$.
%        The rate of change of the distance separating the tips
%       of the hands at 3:00 is changing at the rate $r'(t)$ at $t=0$,
%        where $r(t)=\sqrt{(x_1(t)-x_2(t))^2+(y_1(t)-y_2(t))^2}$.

\item[3.] The area and the radius are related statically by
$A=\pi r^2$ and in this situation they are related dynamically
by $A(t)=\pi r(t)^2.$ Taking derivatives, we get $A'(t)=2\pi r(t) r'(t)$.
Given $r'(t)=1.5$ feet/second and $t=2$ hours, we get
Converting to seconds, $A'(7200)=2 \pi (1.5) r(7200)$.
The problem implies that the initial radius of the slick is zero feet,
so that $r(t)=1.5 t$, and $r(7200)=10800$ and the slick is growing
at $32400 \pi$ square feet per second.
\medskip

Alternatively, we can directly substitute $r(t)=1.5t$ into the
area formula, $A(t)=2.25 \pi t^2$ and $A'(t)=4.5 \pi t$ so at
$t=2$ hours or $7200$ seconds, $A' = 32400$.


\item[4.] Using $t$ in hours, 
with the center of the face of the clock
at the origin in the $x-y$ plane
and units in inches,
the position of the minute hand is
$(m_x(t),m_y(t))=(5 \sin (2 \pi t), 5 \cos (2 \pi t))$
and the hour hand is $(h_x(t),h_y(t))=(4 \sin (2 \pi t/12), 4 \cos (2 \pi  t/12))$
(clock parametrization behaves like the inverse function of standard circular parametrization,
starting from $(0,1)$ and going clockwise instead of from $(1,0)$ and going counterclockwise,
hence the reversal of $x$ and $y$.)
\medskip
(Another option is to make $3$ o'clock $t=0$, and then
standard parameterizations of the form $(r \cos (-kt), r \sin(-kt))$
work nicely, and the first minus sign can be ignored by the evenness
of cosine. It would also be nicer if the hands had length $3$ and $4$
instead of $4$ and $5$ so things would work out in rational numbers!) 
\medskip

So 
$d(t)^2=(m_x(t)-h_x(t))^2 + (m_y(t)-h_y(t))^2 $.

Then
$2d(t)d'(t)= 2(m_x(t)-h_x(t) (m_x(t)-h_x(t))' + (m_y(t)-h_y(t)) (m_y(t)-h_y(t))')$.

Plug in $t=3$, and divide the left hand side by $2d(3)$ to get $d'(3)$.
The necessary values will be $h_x(3)=4, h_y(3)=0$ and $m_x(3)=0, m_y(3)=5$, and
$h'_x(3)=0, h'_y(3)=-8\pi/12$ and $m'_x(3)=10\pi, m'_y(3)=0$.

\medskip
So $d'(3)=\frac{-40\pi+40\pi/12}{\sqrt 41}$ inches per hour.

\medskip
To check this, we use the approximation that the hour hand
is stuck at 3 o'clock, at which time the tip of the minute hand 
moves at $10 \pi$ inches per hour to the right.
The diagonal velocity is the horizontal velocity over the
cosine of the angle between the hands, or $(10 \pi)\frac{4}{\sqrt 41}$.
If we freeze the minute hand, the same analysis of the hour hand
moving down at $8\pi/12$ inches per hour gives an opposite
contribution of vertical velocity times the sine of the
angle at the minute hand, or $(8 \pi/12)\frac{5}{\sqrt 41}$.
Combining these contributions, we have found another probably
easier route to the same answer.


\item[5.] 

Use 
$y(x+ dx) \approx y(x)+  dy  y(x)+y'(x) dx$.
$dy \frac{dy}{dx} dx$
In the first one,
$y=x^{1/2}$, $x=64$, $y=8$, $dx = 2$, $y'=1/2x^{-1/2}$, $y'(64)=1/16$
so $(66)^{1/2} \approx 8 \frac{2}{16}$.

\medskip
In the second one, 
$y=\sin x$, $x=0$, $dx = \frac{\pi}{100}$, $y'(x)=\cos x$
$y(0)=0$, $y'(0)=1$, and $\sin( \frac{\pi}{100} ) \approx \frac{\pi}{100}$.

\medskip
For $m(v)=m_o(1-(v/c)^2)^{-1/2}$, $m'(v)= m_o(1-(v/c)^2)^{-3/2}\frac{v}{c^2}$,
and we are asked to find $100\frac {dm}{m}= 100\frac{m'(v_o)dv}{m(v_o)}$.
Plugging in $v_o/c=.9$ and $dv /c =.02$ we get
$100(.02c)\frac{m'(v_o)}{m(v_o)}=2c \frac{v_o}{c^2}(1-(v_o/c)^2)^{-1}$.
Simplifying, the percent increase is $\frac{2(.9)}{(1-.81)} \approx 9.5\%$.

\item[6.] 

Similar to the previous problems, with $y(r)= 4/3 \pi r^3$, $y'(r)=4\pi r^2$
and $y(3)=36 \pi$, and coincidentally, $y'(3)=36 \pi$ and $dy = 36 \pi (.025)$ cubic inches.




\item[7.]
Consider $f (x) = \left \{
\begin{array}{ll}
|x|, x < 0, \\
\sin{x},  x \geq 0.
\end{array}
\right.$ Find all local maxima and minima of $f$, where $f$ is increasing
and decreasing, where $f$ is concave up and concave down, and 
all inflection points.
Does $f$
have a global maximum or a global minimum? Sketch the graph of $f(x)$. 

\medskip
For $x<0$, $f(x)=|x|=-x$ so $f'(x)=-1$ and $f$ is decreasing.
For $x>0$, $f(x)=\sin x$, so $f'(x)=\cos x$.
So $f'>0$ and $f$ is increasing when $0<x<\pi/2$ and when
$3\pi/2 + 2\pi n < x< 5 \pi /2 + 2 \pi n , \quad n=0,1,\ldots$.
and $f'<0$ and $f$ is decreasing when
$\pi/2 + 2\pi n < x< 3 \pi /2 + 2 \pi n , \quad n=0,1,\ldots$.
The local minima are where $f$ changes from decreasing to increasing,
at $x=0$ and at $x=3 \pi /2 + 2 \pi n , \quad n=0,1,\ldots$,
and the
local maxima are where $f$ changes from increasing to decreasing,
at $x=0$ and at $x=\pi /2 + 2 \pi n , \quad n=0,1,\ldots$.
For $x<0$, $f''(x)=0$, and for $x>0$, $f''(x)=-\sin x$, so
$f$ is concave up when $\sin x$ is negative, i.e., when
$ \pi + 2\pi n < x< 2 \pi + 2 \pi n , \quad n=0,1,\ldots$
and 
$f$ is concave down when $\sin x$ is positive, i.e., when
$ 2\pi n < x< \pi + 2 \pi n , \quad n=0,1,\ldots$
There is no global maximum since $f(x) \to \infty$ as
$x \to -\infty$. The global minimum value of $f(x)=-1$ is taken
at all of the local minima with $x>0$, at 
$x=3 \pi /2 + 2 \pi n , \quad n=0,1,\ldots$
There are inflection points where $f''=0$ and
$f$ changes concavity, at $x=n \pi, \quad n=1,2,\ldots$

\medskip

For
$f(x)=x^3-12x+1$, $f'(x)=3x^2-12$ and $f''(x)=6x$.
Then $f'=0$ when $x^2=4$ or $x=\pm 2$. For $x<-2$,
$f$ is increasing, $f$ is decreasing for $-2<x<2$
and $f$ is increasing for $x>2$. So $f$ has a local
maximum at $x=-2$ and a local minimum at $x=2$.
There is no global maximum or minimum. $f$ is concave
up where $f''>0$, i.e., for $x>0$, and concave down
for $x<0$. There is one inflection point where $f$
changes concavity, at $x=0$ where $f''(x)=0$. See Sec. 4.2, $\#19$.

\medskip
For $\ds f(x)=\frac{1}{1+x^2}\;$, $f'(x)=\frac{-2x}{(1+x^2)^2}$.
So $f$ is increasing where $f'>0$, i.e., for $x<0$ (the
denominator of $f'$ is always positive) and $f$ is decreasing
where $f'<0$, i.e., for $x>0$. There is one local maximum
at $x=0$ where $f$ changes from increasing to decreasing.
This is also a global maximum. There are no local or global
minima as $f$ approaches zero but remains positive
as $x$ tends to infinity in both directions.
By the quotient rule, $f''(x)=\frac{6x^2-2}{(1+x^2)^3}$,
so $f''=0$ and $f$ has inflection points when $x=\pm \frac{\sqrt 3}{3}$,
$f''<0$ and $f$ is concave down between these points, and
$f''>0$ and $f$ is concave up outside these points.
See example 4, p. 170, section 4.2 for a similar problem.

\item[8.]
$f'(x)= 2 \cos (x - \frac{\pi}{4})$ which is positive for
$\frac{\pi}{2} < x < \frac{3\pi}{4}$ (where $f$ is increasing)
and negative 
for $\frac{3\pi}{4} < x < \frac{5\pi}{4}$ (where $f$ is decreasing.)
There is a local maximum at $x=\frac{3\pi}{4}$ where $f=2$, $f'=0$,
and $f''<0$, and no other critical points in the interval. Comparing
with the endpoints $f( \frac{\pi}{2} ) = \sqrt 2$ and
$f( \frac{5\pi}{4} ) = 0$, the global maximum is $2$ and the
global minimum is $0$ on this interval.



\item[9.] Since the corners are $(0,x), (0,-x), (x,12-x^2), (-x,12-x^2)$
the area will be $A=2xy=2x(12-x^2)=-2x^3+24x$. 
$\ds \frac{dA}{dx}=2(12-3x^2)=0$ which is zero when $x=\pm 2$.
We take $x$ to be the vertex on the right, so $x=2, y=8$,
the dimensions are $4$ by $8$ (and the maximum area is $32$.)

\item[10.]

The total travel time $T(x)$ is the sum of the times in the two media.
Using velocity = distance/time, we get $t_1=d_1/c_1$ and
$t_2=d_2/c_2$. We minimize $T(x)=t_1+t_2=d_1(x)/c_1+ d_2(x)/c_2$,
by setting $T'=0=\frac{d_1'}{c_1}+\frac{d_2'}{c_2}$ or
$\frac{d_1'}{c_1}=\frac{-d_2'}{c_2}$.
By Pythagoras and the diagram,
$d_1^2=x^2+a^2$ and $d_2^2=(d-x)^2+b^2$, so implicit differentiation
gives
$2d_1 d_1'=2x$ and $2d_2 d_2'=-2(d-x)$. So $d_1'= \frac{x}{d_1}$
and $d_2'= -\frac{d-x}{d_2}$.
From the diagram, $\frac{x}{d_1}=\sin \theta_1$ and $\frac{d-x}{d_2}=\sin \theta_2$.
Putting this all together, we get Snell's Law:
$\frac{\sin \theta_1}{c_1}= \frac{\sin \theta_2}{c_2}$.

\eject

\begin{figure}  %[htp]
\centering
\includegraphics{prac_2.eps}
\caption{Graph of $f(x)$ for \#2.}        
\end{figure}

\end{enumerate}

\end{document}




