\magnification=\magstep1
\def\sol{\medskip\item{} {\it Solution:} }
\centerline{MATH 131, section 1 -- Calculus for Physical and Life Sciences}
\centerline{Solutions for Sample Exam Questions -- Exam 3}
\centerline{November 19, 2013}
\bigskip
\noindent
I.  Find $y'$ and simplify.
\bigskip
\item{(a)}  $y = \ln(x)\left(x^7 - {4\over \sqrt{x}}\right)$
\sol
By the product rule, 
$$\eqalign{ 
y' &= \ln(x)(7x^6 + 2x^{-3/2}) + {1\over x}\left(x^7 - {4 \over x^{1/2} }\right)\cr
&= \ln(x)(7x^6 + 2x^{-3/2}) + x^6 - { 4\over x^{3/2} } \cr }$$
\item{(b)}  $y = (e^{2x} + 2)^3$.
\sol
By the chain rule (twice), 
$$y' = 3(e^{2x} + 2)^2 \cdot e^{2x} \cdot 2 = 6 e^{2x}(e^{2x} + 2)^2.$$
\item{(c)}  $y = {x + 1\over 3x^4 - 1}$.
\sol
By the quotient rule, 
$$\eqalign{y' &= {(3x^4 - 1)(1) - (x+1)(12x^3) \over (3x^4 - 1)^2}\cr
&= {-9x^4 - 12x^3 - 1\over (3x^4 - 1)^2}\cr}$$
\item{(d)}  $y = {\sin(x) \over 1 + \cos(x)}$
\sol
By the quotient rule, 
$$\eqalign{y' &= {(1+\cos(x))(\cos(x)) - \sin(x)(-\sin(x)) \over (1 + \cos(x))^2}\cr
&= {\cos(x) + \cos^2(x) + \sin^2(x) \over (1 + \cos(x))^2}\cr
&= {1\over 1 + \cos(x)}\cr}$$
\item{(e)}  $y = \tan^{-1}(e^{5x})$.
\sol
By the derivative rule for inverse tangent and the chain rule,
$$y' = {1 \over 1 + (e^{5x})^2} \cdot e^{5x} \cdot 5 = {5e^{5x} \over 1 + e^{10x}}.$$
\item{(f)}  $xy^2 - 3y^3 + 2x^4 = 2$.
\sol
Since the equation involves both $x$ and $y$ we use {\it implicit differentiation}.
Differentiating thinking of $y$ as a function of $x$,
$$2xyy' + y^2 - 9y^2 y' + 8x^3 = 0$$
Then solving for $y'$:
$$y' = {-y^2 - 8x^3\over 2xy - 9y^2}.$$
\item{(g)}  $y = \cos(x)^{x^3}$.
\sol
For functions of the form $u(x)^{v(x)}$, we use {\it logarithmic differentiation}.
First take $\ln$ of both sides
$$\ln(y) = \ln(\cos(x)^{x^3}) = x^3 \ln(\cos(x)),$$
then differentiate using the product and chain rules:
$${1\over y} y' = -x^3 \tan(x) + 3x^2 \ln(\cos(x)).$$
So 
$$y' = y(-x^3 \tan(x) + 3x^2 \ln(\cos(x))) = \cos(x)^{x^3}(-x^3 \tan(x) + 3x^2 \ln(\cos(x))).$$

\bigskip
\noindent
II.  The quantity of a reagent present in a chemical 
reaction is given by $Q(t) = t^3 - 3t^2 + t + 30$ grams at time $t$ seconds for all $t \ge 0$.
\item{(a)}  Over which intervals with $t \ge 0$ is the amount increasing? decreasing?
\sol
We need to determine the intervals where $Q'(t) = 3t^2 - 6t + 1$ is positive and 
negative.  By the quadratic formula, $3t^2 - 6t + 1 = 0$ when 
$$t = {6 \pm \sqrt{24}\over 6} = {3 \pm \sqrt{6} \over 3},$$
which are both positive numbers.  
By making a sign chart for $Q'$ we see 
$Q'(t) > 0$ for $t$ in $\left[0, {3 - \sqrt{6}\over 3}\right) \cup \left({3 + \sqrt{6}\over 3}, +\infty\right)$.  (Note the problem just said look at $t \ge 0$.)
$Q'(t) < 0$ for $t$ in $\left({3 - \sqrt{6}\over 3}, {3 + \sqrt{6}\over 3}\right)$.
\medskip
\item{(b)}  Over which intervals is the rate of change of $Q$ increasing? decreasing?
The rate of change is increasing (decreasing) when $Q''(t) = 6t - 6$ is positive (negative).
This is increasing for $t > 1$ and decreasing for $0\le t < 1$ (again, the problem
said look only at $t \ge 0$ so we are ignoring $t < 0$).  
\bigskip
\noindent
III.  A spherical balloon is being inflated at 20 cubic inches per
minute.  When the radius is $6$ inches, at what rate is the 
radius of the balloon increasing?  At what rate is the surface
area increasing?  (The volume of a
sphere of radius $r$ is $V = {4\pi r^3\over 3}$ and the surface
area is $4\pi r^2$.)
\sol 
This is a {\it related rates} problem.  
From the volume formula, since $V$ and $r$ are changing with time $t$, 
$${dV\over dt} = 4\pi r^2 {dr\over dt}.$$
We are given that ${dV\over dt} = 20$, 
so when $r = 6$:
$$20 = 4\pi (6)^2 {dr\over dt},$$
so 
$${dr\over dt} = {20\over 144\pi} = {5\over 36\pi}$$
(inches per minute).  The surface area is changing at the rate
$${dA\over dt} = 8\pi r {dr\over dt} = 8\pi \cdot 6 \cdot {5\over 36\pi} = {20\over 3}$$
(square inches per minute).  
\bigskip
\noindent
IV. All parts of this question refer to $f(x) = 4x^3 - x^4$.
\item{(a)}  Find and classify all the critical numbers of $f$
using the First Derivative Test.
\sol
The first derivative is $f'(x) = 12x^2 - 4x^3 = 4x^2(3 - x)$, 
so $f$ has critical numbers at $x = 0$ and $x = 3$.  The first derivative
is positive for $x < 0$, positive for $0 < x < 3$, and negative for $x > 3$.
Hence $x = 3$ is a local maximum, and $x = 0$ is neither a local max nor a local min.
\medskip
\item{(b)}  Over which intervals is the graph $y = f(x)$ concave
up?  concave down?
\sol
For concavity, we need $f''(x) = 24x - 12 x^2 = 12x(2 - x)$.
This is negative for $x < 0$ and $x > 2$, and positive for $0 < x < 2$.  
Hence $y = f(x)$ is concave up on $(0,2)$ and concave down on $(-\infty, 0) \cup (2,+\infty)$. 
(Since the graph changes concavity at $x = 0, 2$, these are {\it points of inflection}.)
\medskip
\item{(c)}  Sketch the graph $y = f(x)$.  
\sol Omitted -- Ask Gopal to sketch in the review session.
\medskip
\item{(d)}  Find the absolute maximum and minimum of $f(x)$
on the interval $[1,4]$.  
\sol
On this closed interval, we have a critical number at $x = 3$.  The
critical value is $f(3) = 4\cdot 27 - 81 = 27$.  The values at the endpoints
are $f(1) = 4 - 1 = 3$ and $f(4) = 4 \cdot 64 - 256 = 0$.  Hence the maximum
value on the interval is $f(3) = 27$ and the minimum is $f(4) = 0$.  
\bigskip
\noindent
V.  All three parts of this question refer to the function
$f(x) = x^{2/3} - {1\over 5}x^{5/3}$.
\item{(a)}  Find all the critical numbers of $f(x)$.
\sol
The derivative is 
$$f'(x) = {2\over 3}x^{-1/3} - {1\over 3}x^{2/3} = {1\over 3}x^{-1/3}(2 - x)$$
The function has critical numbers at $x = 0$ ($f'(0)$ is undefined -- the graph
has a cusp point there), and $x = 2$
($f'(2) = 0$).  
\item{(b)}  Find all the inflection points of $f(x)$.
\sol
For this we need 
$$f''(x) = {-2\over 9} x^{-4/3} - {2\over 9} x^{-1/3} = -{2\over 9}x^{-4/3}(1 + x)$$
There is just one point of inflection at $x = -1$.  Since $x^{-4/3} = {1\over (x^{1/3})^4}$,
that term is positive whenever it is defined, so the sign of $f''$ changes only at $x = -1$.  
\item{(c)}  For which of the critical numbers here 
is the Second Derivative Test applicable?  Why?  Determine
the type of each such critical number using the Second
Derivative Test.
\sol 
The Second Derivative Test is applicable at critical numbers where
the first derivative is equal to zero; here only at $x = 2$.  We have
$f''(2) = -{2\over 9}\cdot 2^{-4/3}(1 + 2) = -{2\over 3}\cdot 2^{-4/3} < 0$.
Hence $f$ has a {\it local maximum} at $x = 2$ since the graph is concave down there.
%\bigskip
%\noindent
%VI. 
%\item{(a)} A rectangular pen is to be constructed with one
%side along an existing long stone wall (so no fence is
%needed along that side), and the other three sides
%made of new fencing.  The pen must enclose $1000$ square feet
%of space.  What is the smallest total length of fencing that
%can be used to enclose the pen?  What are the dimensions
%of the pen that will use the smallest total length of fencing.
%\sol
%Let $x$ be the length of the side of the pen parallel to the existing wall,
%and let $y$ be the length of the other side.  We have the area of the pen is
%$A = xy = 1000$, which implies $y = {1000\over x}$.
%We want to minimize the total fencing used, which is 
%$$L = x + 2y = x + {2000\over x}.$$
%To minimize, we take the derivative of $L$ and set it equal to zero 
%to find the critical numbers:
%$$L' = 1 - {2000\over x^2} = 0$$
%when $x = \sqrt{2000} \doteq 44.72$ feet.  This is a minimum by the 
%Second Derivative Test: 
%$$L'' = {4000\over x^3} > 0$$
%for all $x > 0$.  (Note that $L \to +\infty$ as $x \to 0^+$ and as $x \to \infty$.)
%The smallest total length 
%of fencing is 
%$$L(44.72) = 44.72 + {2000\over 44.72} \doteq 89.44$$
%feet.  The dimensions of the pen using the smallest
%amount of fencing are $x = 44.72$ and $y = {1000\over 44.72} \doteq 22.36$
%feet.    
%\item{(b)} Consider all rectangles with one side along the $x$-axis
%that are inscribed under the curve $y = \sqrt{x}$, extending
%to $x = 4$ on the right.  Find the dimensions of the rectangle
%of this type of largest area.
%\sol 
%If the top left corner of the rectangle is at $(x,\sqrt{x})$, then 
%the area is $A(x) = \sqrt{x}(4 - x) = 4x^{1/2} - x^{3/2}$.  To find
%the maximum area, we take the derivative and set it equal to zero:
%$A'(x) = 2x^{-1/2} - (3/2)x^{1/2} = 0$.  This has just one solution:
%$x = {4/3}$.  This gives a local maximum since $A''(x) = -x^{-3/2} - {3\over 4}x^{-1/2}$
%is negative for all $x > 0$.  The dimensions of the rectangle are base = 
%$4 - 4/3 = 8/3$ and height = $\sqrt{4/3} = 2/\sqrt{3}$.
%\bigskip
%\noindent
%VII.  Evaluate the following limits.
%\item{(a)} $$\lim_{x\to 0} {1 - \cos(x)\over x^2}$$
%\sol
%This is indeterminate of the form $0/0$, so applying L'Hopital's Rule twice,
%$$\lim_{x\to 0} {1 - \cos(x)\over x^2} = \lim_{x\to 0} {\sin(x)\over 2x} = 
%\lim_{x\to 0} {\cos(x)\over 2} = {1\over 2}.$$
%(We could also use $\lim_{x\to 0}{\sin(x)\over x} = 1$ to see this after
%applying L'Hopital's Rule the first time.)
%\item{(b)} $$\lim_{x\to \infty} {\ln(1 + e^x)\over x}$$ 
%\sol
%This is indeterminate of the form $\infty/\infty$, so applying L'Hopital's Rule,
%$$\lim_{x\to \infty} {\ln(1 + e^x)\over x} = \lim_{x\to \infty} {{e^x\over 1 + e^x}\over 1} = 
%\lim_{x\to \infty} {1\over e^{-x} + 1} = 1.$$

\bye
