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\leftline{Mathematics 133 -- Intensive Calculus for Science 1}
\leftline{Sample Exam II Solutions -- October 29, 2005}
\bigskip
\noindent
I. Refer to the graph on the handout showing {\it the derivative} $y = f'(x)$
for some function $f(x)$.
\item{A)} Using the information here,
construct a ``qualitative'' plot of $y = f''(x)$. 
\bigskip
\item{} {\it Solution:}  $f''(x)$ is the derivative of $f'(x)$, so
the $y$-coordinates of your graph should show the slopes of the
given graph.  $f''(x) = 0$ for all $x < 1$ since the given graph
is horizontal on that interval.  $f''$ starts close to zero for $x > 1$,
increases up to a positive maximum around $x = 2$, decreases to 
zero a bit after $x = 3$, then is negative the rest of the way.
It reaches its negative minimum around $x = 4$, then increases
at bit after that (but still stays negative).
\bigskip
\item{B)} Over which intervals is $f$ increasing?  
\bigskip
\item{} {\it Solution:}  Since $f$ is increasing where $f'$ is positive,
$f$ is increasing on the intervals where the $y$-coordinates of the given 
graph are positive:  $x < 1$ and $2 < x < 4$.
\bigskip
\item{C)} Is $f'$ continuous at $x = 1$?  Why or why not?
What happens on the graph $y = f(x)$ at $x = 1$?
\bigskip
\item{}{\it Solution:}  $f'$ is not continuous at $x = 1$ because
of the ``jump'' in the given graph.  We have $\lim_{x \to 1^-} f'(x) = 2$
but $\lim_{x \to 1^+} f'(x) = -1$.  This indicates that the
slope of the graph $y = f(x)$ changes suddenly from $+2$ to $-1$
at $x = 1$.  Assuming $f$ is continuous at
$x=1$, its graph would have a ``sharp corner'' at $x = 1$.
\bigskip
\noindent
II. 
\item{A)} The function $H(t)$ gives the number of hours of daylight
$t$ days after the start of the year in Worcester.  At $t = 304$ days
(October 31 in a non-leap year), $H'(304) = -0.083$.  Give the meaning of
this equation as a sentence, using appropriate units.
\bigskip
\item{} {\it Solution:}  At $t = 304$ (that is, on October 31), 
the number of hours of daylight per day in Worcester is 
decreasing at the rate of $0.083$ hours per day (about 
5 minutes per day).  (Note:  It's
decreasing because the derivative value is negative.)
\bigskip
\item{B)} The table below gives the position $s$ (in miles) of a 
freight train moving
along a straight line track as a function of time $t$ (in hours).  
$$\matrix{t & .5 & 1 & 1.5 & 2 & 2.5 \cr
          s & 10 & 25 & 42 & 50 & 55 \cr}$$
Estimate the train's instantaneous velocity at $t = 1.5$ hours
as closely as you can from this information.   
\bigskip
\item{} {\it Solution:}  From the table information, we can estimate
the instantaneous velocity by average velocities in many ways.
Thinking of the definition of the instantaneous velocity as
a limit, we expect the closest approximations would be for $h = \Delta t$ as
small as possible:
$$v \doteq {s(2) - s(1.5)\over 2 - 1.5} = {50 - 42\over 2 - 1.5} = 16$$
or
$$v \doteq {s(1) - s(1.5)\over 1 - 1.5} = {25 - 42\over 1 - 1.5} = 34$$
The actual instantaneous velocity at $t = 1.5$ is probably somewhere
in between, so the {\it average} is our best estimate:  
$v \doteq {16 + 34\over 2} = 25$
miles per hour.
\bigskip
\noindent
III.  Using the limit definition, find $f'(x)$ for $f(x) = 1/x$.
\bigskip
\item{} {\it Solution:}
$$\eqalign{
f'(x) &= \lim_{h\to 0} {f(x+h) - f(x)\over h}\cr
&= \lim_{h\to 0} {{1\over x+h} - {1\over x}\over h}\cr
&= \lim_{h\to 0} {x - (x+h)\over (x+h)x h}\cr
&= \lim_{h\to 0} {-h\over (x+h)x h}\cr
&= \lim_{h\to 0} {-1\over (x+h)x}\cr
&=  {-1\over x^2}\cr}$$
(Check with the power rule:  ${d\over dx} x^{-1} = -x^{-2} = {-1\over x^2}$.)

\noindent
IV.  Find derivatives of each of the following functions by applying
the appropriate ``shortcut'' derivative rules:
\bigskip
\item{A)} $f(x) = 5x^7 - {3\over \sqrt{x}} - 4^{2x}$
\medskip
\item{} {\it Solution:}  $f'(x) = 35x^6 + {3\over 2}x^{-3/2} - 4^{2x}\ln(4)\cdot 2$
\medskip
\item{B)} $g(x) = (x^2+1)^{12} 2^x$
\medskip
\item{} {\it Solution:} By the product and chain rules:
$$g'(x) = 12(x^2+1)^{11} 2^x + (x^2+1)^{12} 2^x\ln(2) = (x^2+1)^{11}2^x(12 + (x^2+1)\ln(2))$$
\item{C)} $h(x) = {x^2\over e^x - 1}$
\bigskip
\item{} {\it Solution:} By the quotient rule:
$$h'(x) = {(e^x-1)(2x) - x^2 e^x\over (e^x - 1)^2}$$
\noindent
V. (5)  Say whether the following statement is true or false,
and explain your reasoning:  If the time interval is short enough,
then we expect the average velocity of a car over the interval will be
close to its instantaneous velocity at any time in the interval.
\bigskip
\item{} {\it Solution:}  True:  Since 
$v_{\rm inst} = \lim_{\Delta t\to 0} {\Delta s\over \Delta t}$, 
the instantaneous velocity at each $t_0$
is the limit of average velocities
as the length of the time interval goes to zero.  If the 
time interval is short enough, we would expect the average
velocity over the whole interval to be very close to the limiting value.  
(Of course I am thinking of 
a ``usual'' physical situation where the position and the speed
are changing in a continuous fashion, so that even if the time
$t_0$ where the instantaneous velocity is computed is
not at one of the endpoints of the time interval, the 
average velocity over the whole interval
would still be close to the average velocities
over the smaller intervals that would be used to compute 
$v_{\rm inst}(t_0)$.)
\bye