%% Math 135 Midterm 3, Fall 2016
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\markright{MATH 135, section 1, Midterm 3}

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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2016} \\
\textbf{MATH 135, section 1, Solutions for Midterm 3} \\
\textbf{Friday, December 2}
\end{center}

\begin{enumerate}

\item{} Find $\frac{dy}{dx}$ and simplify:
\begin{enumerate}
\item[A)] (10) $\displaystyle y = \frac{x^4 + 2x}{x^3 + x + 1}$
\item[] {\it Solution:}  By the quotient rule:
\begin{eqnarray*}
\frac{dy}{dx} &=& \frac{(x^3+x+1)(4x^3+2) - (x^4+2x)(3x^2+1)}{(x^3+x+1)^2}\\
&=& \frac{x^6 + 3x^4 + 2}{(x^3+x+1)^2}
\end{eqnarray*}
\item[B)] (10)  $y = \ln(3 \sin(x) - 4 \tan(x))$
\item[] {\it Solution:}  By the chain rule and the derivative formula for $\ln$:
$$\frac{dy}{dx} = \frac{3\cos(x) - 4\sec^2(x)}{3\sin(x) - 4\tan(x)}.$$
\item[C)] (10)  $y = \sin^{-1}(e^{-2x})$
\item[] {\it Solution:}  By the chain rule and the derivative formula for $\sin^{-1}$:
$$\frac{dy}{dx} = \frac{-2e^{-2x}}{\sqrt{1 - e^{-4x}}}.$$
(Note:  $(e^{-2x})^2 = e^{-4x}$.)
\item[D)] (10)  $x^2 y^4 - 4 \tan^{-1}(x^2) + y = 0$  (use implicit differentiation)
\item[] {\it Solution:}  Using implicit differentiation,
$$4x^2y^3\frac{dy}{dx} + 2x y^4 - \frac{8x}{1 + x^4} + \frac{dy}{dx} = 0.$$
So solving for $\frac{dy}{dx}$, we have
$$\frac{dy}{dx} = \frac{-2xy^4 + \frac{8x}{1 + x^4}}{4x^2y^3+1}
=\frac{-2xy^4 - 2x^5y^4 + 8x}{(1+x^4)(4x^2y^3 + 1)}$$
(either form is OK).  
\end{enumerate}
\vskip 10pt
\item{} (20) A stationary observer watches a weather balloon being launched from a point 
500 feet away from her position.  The balloon rises at a rate of 30 feet per second.  How
fast is the distance between the balloon and the observer changing when the balloon
is 375 feet above the ground?
\vskip 10pt 
\item[] {\it Solution:}  At all times, the observer, the balloon and the launch point
are at the corners of a right triangle.  Calling $y$ the height of the balloon, and
$z$ the distance from the observer to the balloon, the Pythagorean theorem gives
$$z^2 = y^2 + 500^2.$$
Taking $\frac{d}{dt}$ we get
$$2z\frac{dz}{dt} = 2y\frac{dy}{dt}.$$
At the time the question is asking about, $y = 375$ and $\frac{dy}{dt} = 30$.  
By the Pythagorean theorem, $z = \sqrt{375^2 + 500^2} = 625$.  Hence
$$\frac{dz}{dt} = \frac{375\cdot 30}{625} = 18 ({\rm ft/sec}).$$
\vskip 10pt
\item{}  All parts of this question refer to the plots in Figure 1.  
Assume the whole domain of the functions is the interval $[-2,8]$ shown (don't
try to extrapolate what might happen on a larger interval).

\begin{enumerate}
\item[(A)] (3)  Is $A'(4)$ positive or negative?  
Answer: $A(x)$ is increasing at $x = 4$, so $A'(4)$ is \emph{positive}.
\item[(B)] (3) At how many different points is $B'(x) = 0$?  Estimate the 
$x$-values from the graph.  Answer: There are two such points at about $x = 0, 2.7$
(anything between 2 and 3 closer to 3 is OK).  
\item[(C)] (3) On the $x$-interval $(0,2)$, is $A''(x)$ positive or negative?
Answer: $y = A(x)$ is \emph{concave up} on that interval so $A''(x) > 0$.
\item[(D)] (3) On the $x$-interval $(0,2)$ is $B'(x)$ positive or negative?
Answer: $y = B(x)$ is \emph{increasing on that interval}, so $B'(x) > 0$.
\item[(E)] (3)  One of the two functions $A(x)$ and $B(x)$ is the derivative of 
the other.  Which is which?  Answer: 
$B(x)$ is the derivative of $A(x)$.  (Note $B(x) = 0$ when $A$ has local maximum
and local minimum points -- for instance at $x \doteq -0.7, +0.8, 6$.)
\end{enumerate}
\vskip 10pt

\begin{figure}
\centering
  \subfloat[$y = A(x)$]{\includegraphics[width=70mm]{135P3IIIB.png}}
  \subfloat[$y = B(x)$]{\includegraphics[width=70mm]{135P3IIIA.png}}
\caption{Plots for Problem 3}
\end{figure}

\item{} All parts of this problem refer to $f(x) = x^4 - 2 x^2 - 1$. 

\begin{enumerate}
\item[(A)] (10) Find all critical points of $f(x)$. 
\item[] {\it Solution:}  $f'(x) = 4x^3 - 4x = 4x(x^2-1) = 4x(x+1)(x-1)$.
So $f(x)$ has critical points at $x = -1,0,1$.
\item[(B)] (10) Find the absolute maximum and minimum values of $f(x)$ 
on the interval $[0,3]$.  
\item[] {\it Solution:}  Of the $x$'s we found in part A, both $x = 0$ and
$x = 1$ are in this interval.  $f(0) = -1$, $f(1) = -2$ and $f(3) = 81 - 18 - 1 = 62$.  
The minimum is $f(1) = -2$ and the maximum is $f(3) = 62$. 
\item[(C)]  (5) Which of the critical points you found in part A are local maximum
points and which are local minimum points?  (Any correct method for this is OK.)
\item[] {\it Solution 1:}  By the First Derivative Test, $f'(x)$ changes from 
negative to positive at $x = \pm 1$ so those are local minima.  Then $f'(x)$
changes from positive to negative at $x = 0$, so that is a local maximum.
\item[] {\it Solution 2:}  By the Second Derivative Test, $f''(x) = 12x^2 - 4$.
We have $f''(-1) = 8 = f''(1)$.  These values are positive so $f$ has
local minima at $x = \pm 1$.  But $f''(0) = -4 < 0$.  So this is a local maximum.
\end{enumerate}
\end{enumerate}
\end{document}


