%% Math 133 Midterm 3, Fall 2015
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\markright{MATH 133, section 1, Midterm 4}

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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2017} \\
\textbf{MATH 133, Solutions for Midterm 4} \\
\textbf{Thursday, December 7}
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\begin{enumerate}

\item{} Find $\frac{dy}{dx}$; do not simplify:
\begin{enumerate}
\item[A)] (7.5)  $y = \frac{\ln(x) + x^2}{\cos(x)}$
\item[] {\it Solution:}  By the Quotient Rule and the derivative rules for $\cos$ and $\ln$, 
$$\frac{dy}{dx} = \frac{\cos(x)(1/x + 2x) - (\ln(x) + x^2)(-\sin(x))}{\cos^2(x)}.$$
\item[B)] (7.5)  $y = \sin^{-1}(\sqrt{x})$
\item[] {\it Solution:} By the Chain Rule, 
$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}}\cdot \frac{1}{2\sqrt{x}}.$$
\item[C)] (10)  $x^3 y^2 - 4 y + x^4 = 0$  (use implicit differentiation)
\item[] {\it Solution:}  We have
$$x^3\cdot 2 y \frac{dy}{dx} + 3x^2y^2 - 4\frac{dy}{dx} + 4x^3 = 0$$
So 
$$\frac{dy}{dx} = \frac{-3x^2y^2 - 4x^3}{2x^3y - 4}.$$ 
\end{enumerate}

\vskip 10pt
\item{} (20) A stationary observer watches a weather balloon being launched from a point 600 feet 
away from her position.  The line of sight distance from the observer to the balloon
is changing at a rate of 20 feet per second.  How fast is the height of the balloon changing
when the balloon is 400 feet above the ground?
\vskip 10pt
\noindent
{\it Solution:}  Let $z$ be the distance from the observer to the balloon, and 
$y$ be the height of the balloon above the ground (both functions of time).  We have
$z^2 = 600^2 + y^2$ by the Pythagorean Theorem.  Hence differentiating with respect
to time:
$$2z \frac{dz}{dt} = 2y\frac{dy}{dt}.$$
At the given time $y = 400$, so $z = \sqrt{600^2 + 400^2} \doteq 721.1$.  We 
are given $\frac{dz}{dt} = 20$, so 
$$2 \cdot 721.1 \cdot 20 = 2 \cdot 400 \cdot \frac{dy}{dt}.$$
Then 
$$\frac{dy}{dt} = \frac{721.1 \cdot 20}{400} \doteq 36.1$$
(feet per second.) 

\vskip 10pt
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\item{} All parts of this question refer to the plot in Figure 1, \emph{which is} $y = f'(x)$
\emph{for some function} $f(x)$ .  Assume the whole domain
of the functions $f(x)$ and $f'(x)$ is the interval $[-3,3]$ shown. 
\begin{figure}
\begin{center}
  \includegraphics[width=70mm]{pieces.png}
\caption{$y = f'(x)$ for Problem 3}
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\begin{enumerate}
\item[(A)]  (10) Find the critical points of $f(x)$
in the interval shown:
\item[] Answer:  $x = -1, 1$ (where the graph of $f'(x)$ crosses the $x$-axis.   
\item[(B)]  (5) Briefly, in your own words, state how the First Derivative Test distinguishes
between local maxima, local minima, and critical points that are neither:
\item[] Answer:  If the sign of $f'$ changes from $+$ to $-$ at the critical
point, then $f$ has a local maximum there; if the sign of $f'$ changes from $-$ to 
$+$, then $f$ has a local minimum; if the sign of $f'$ does not change, then 
$f$ has neither a local maximum nor a local minimum.
\item[(C)]  (5) Identify each of the points you found in part (A) as a local maximum,
local minimum, or neither: 
\item[] Answer: $f$ has a local minimum at $x = -1$; $f$ has a local maximum at $x = 1$.
\item[(D)]  (5) Find all the inflection points of $f(x)$.
\item[] Answer:  Concavity of $f$ changes at $x = 0$, where $f'$ changes from increasing
to decreasing.
\item[(E)]  (5) Over which intervals is $y = f(x)$ concave up? concave down?
\item[] Concave up: $x < 0$ where $f'$ is increasing; Concave down:  $x > 0$ where
$f'$ is decreasing.  
\end{enumerate}

\item{}  A rectangular poster is to have total area 600 square inches, including
blank 1 inch wide margins on all four sides of a central printed area.  What overall
dimensions will maximize the printed area?  
The parts of this problem will lead you to the answer.
\begin{enumerate}
\item[(A)] (4) Draw a diagram clearly showing the whole poster, the central printed
area, and the 1 inch wide blank margins.
\item[] {\it Solution:} Appropriate diagram showing one rectangle inside
another one with 1-inch margins on all four sides of the inner regions is OK.
\item[(B)] (4) Call the horizontal side of the whole poster $x$ and the vertical side 
of the whole poster $y$. 
\item[] {\it Solution:}  Area of whole poster is $xy = 600$; area of 
printed region is $AP = (x - 2)(y - 2)$ (subtract one inch for the margins on top and 
bottom, one inch for the margins on left and right).
\item[(C)] (4) Solve for $y$ in terms of $x$ using the area of the whole poster and substitute
into the area of the printed region.
\item[] {\it Solution:}  $\displaystyle{y = \frac{600}{x}}$.  So
$$AP = (x - 2)\left(\frac{600}{x} - 2\right) = 604 - \frac{1200}{x} - 2x.$$
\item[(D)] (5) Determine a critical point of the area function.
\item[] {\it Solution:}  $\displaystyle{AP'(x) = \frac{1200}{x^2} - 2 = 0}$ when $x^2 = 600$, or
$x = 10\sqrt{6} \doteq 24.49$.  
\item[(E)] (4) How do you know your critical point is a maximum of the area?
\item[] {\it Solution:}  $\displaystyle{AP'' = \frac{-2400}{x^3}}$.  When you plug in 
$x = 10\sqrt{6}$,
the result is negative, so $AP(x)$ has a maximum there by the Second Derivative
Test.  This is the only critical point, so it must be an overall maximum too.
\item[(F)] (4) What are the dimensions $x$ and $y$ of the poster of maximum
printed area? 
\item[] {\it Solution:}
$$x = \sqrt{600} \doteq 24.49; \qquad y = \frac{600}{\sqrt{600}} = \sqrt{600}.$$
The poster of maximum area is a square.

\end{enumerate}


\end{enumerate}

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