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\begin{document}
\centerline{MATH 131 -- Calculus for Physical and Life Sciences}
\centerline{Solutions for Exam 2}
\centerline{October 25, 2007}
\noindent
I.  A. Consider the parametric curve $x=4 \cos(t)$, $y=4\sin(t)$. 

\begin{itemize}
\item[(4) 1.]  Give a Cartesian equation for this curve (eliminate the parameter $t$ to obtain an equation in $x$ and $y$) and describe the shape of the curve. 
\vskip 10pt
\noindent
{\it Solution:}  We have $\frac{x}{4} = \cos(t)$ and $\frac{y}{4} = \sin(t)$, so
by the basic trigonometric identity, 
\[
\left(\frac{x}{4}\right)^2 + \left(\frac{y}{4}\right)^2 = \cos^2(t) + \sin^2(t) = 1.
\]
Hence $x^2 + y^2 = 16$.  The parametric curve lies on the circle with center $(0,0)$
and radius $4$.  
\vskip 10pt
\item[(4) 2.]  What portion of the curve is traced out for $0 \leq t \leq \frac{\pi}{4}$, 
and in which direction is the curve being traced?  (Either explain in words or draw a picture.)
\vskip 10pt
\noindent
{\it Solution:}  For $0 \le t \le \frac{\pi}{4}$, the parametric curve traces the
arc of the circle starting at $(x,y) = (4,0)$ and ending at $(x,y) = (2\sqrt{2},2\sqrt{2})$,
in the counterclockwise direction.
\end{itemize}
\vskip 10pt
\noindent
B.  Now consider the parametric curve $x= \cos(4t)$, $y=\sin(4t)$. 
\begin{itemize}
\item[ (3) 1.]  Give a Cartesian equation for this curve and describe the shape of the curve. 
\vskip 10pt
{\it Solution:}  By the basic trigonometric identity, 
\[
x^2 + y^2 = \cos^2(4t) + \sin^2(4t) = 1.
\]
Hence $x^2 + y^2 = 1$.  The parametric curve lies on the circle with center $(0,0)$
and radius $1$.  
\vskip 10pt
\item[ (4) 2.]
What portion of the curve is traced out of $0 \leq t \leq \frac{\pi}{4}$, and in which direction 
is the curve being traced?  (Either explain in words or draw a picture.)
\vskip 10pt 
\noindent
{\it Solution:}  For $0 \le t \le \frac{\pi}{4}$, this parametric curve traces the
arc of the circle starting at $(x,y) = (1,0)$ and ending at $(x,y) = (-1,0)$ (that is,
the top half of the circle), in the counterclockwise direction.
\end{itemize}

\noindent
II.  Apply the rules for differentiation to find the derivatives of the functions 
given below. Do NOT simplify your answers after completing the derivative (except 
to clean up your answers).
\begin{enumerate}
\item[(4) A.] $\displaystyle{f(x)=x^3+5x^\frac{2}{3}+\sqrt{2}}$
\vskip 10pt
\noindent
{\it Solution:}  By the sum and product rules, 
\[
f'(x) = 3x^2 + \frac{10}{3}x^{\frac{-1}{3}}
\]
(the $\sqrt{2}$ is constant, so the derivative of that term is zero).
\vskip 10pt
\item[(4) B.] $\displaystyle{g(x)= x^\frac{5}{2}-\frac{1}{x} + 7}$
\vskip 10pt
\noindent
{\it Solution:} Rewriting $g(x)$ first, $g(x) = x^\frac{5}{2} - x^{-1} + 7$, so
\[
g'(x) = \frac{5}{2}x^\frac{3}{2} + x^{-2}.
\]
\item[(4) C.] $R(z)=(5z^2-2z^4)(z^3-1)$
\vskip 10pt
\noindent
{\it Solution:}  Using the product rule, 
\begin{eqnarray*}
R'(z) &=& (5z^2 - 2z^4)\frac{d}{dz}{(z^3-1)} + (z^3 - 1)\frac{d}{dz}{(5z^2-2z^4)} \\
&=& (5z^2 - 2z^4)(3z^2) + (z^3 - 1)(10z-8z^3).
\end{eqnarray*}
The directions said not to simplify, so this is sufficient.  (This derivative could
also be computed by multiplying out $R(z)$ and applying the power rule repeatedly.
If this was done, the result would be $R(z) = 5z^5 - 2z^7 - 5z^2 + 2z^4$, so
$R'(z) = 25z^4 - 14z^6 - 10z + 8z^3$.)
\vskip 10pt
\item[(4) D.] $\displaystyle{s(t)=\frac{\sqrt{t}-2e^t}{e^t+t}}$
\vskip 10pt
By the quotient rule:
\begin{eqnarray*}
s'(t) &=& \frac{(e^t + t)\frac{d}{dt}(\sqrt{t} - 2e^t) - (\sqrt{t} - 2e^t)\frac{d}{dt}(e^t + t)}{(e^t+t)^2}\\
&=& 
\frac{(e^t + t)\left(\frac{1}{2}t^{-1/2} - 2e^t\right) - (\sqrt{t} - 2e^t)(e^t + 1)}{(e^t+t)^2}
\end{eqnarray*}
(again, no further simplification necessary as per the directions).
\vskip 10pt
\item[(4) E.] $P(x)=e^x-x^e+e^{\pi} $ 
\vskip 10pt 
\noindent
{\it Solution:}  The $e^x$ is an exponential function, the $x^e$ is a power function,
and the $e^\pi$ is constant:
$P'(x) = e^x - e x^{e-1}$.
\end{enumerate}
\vfil\eject

%%%%%%%%%% Combining functions and finding domains %%%%%%%%%%%%


\noindent
III. The following graph shows $y = f'(x)$ for some function $f(x)$.

\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{M2III.ps} 
\end{center}

\begin{itemize}
\item[(3) A.] Over what intervals is $f(x)$ increasing?  Over what intervals is $f(x)$ 
decreasing?
\vskip 10pt
\noindent
{\it Solution:}  $f(x)$ is increasing on intervals where $f'(x) > 0$, so on $(-2,0)$
and $(1,3)$.  Similarly $f(x)$ is decreasing on intervals where $f'(x) < 0$, so 
on $(0,1)$.  
\vskip 10pt 
\item[(3) B.] Over what intervals is $f(x)$ concave up?  Over what intervals is $f(x)$
concave down?
\vskip 10pt
\noindent
{\it Solution:}  $f(x)$ is concave up on intervals where $f''(x) > 0$ or equivalently
where $f'(x)$ is increasing.  This is true here on $(-2,-1)$.
Similarly $f(x)$ is concave down on intervals where $f''(x) < 0$, or 
equivalently where $f'(x)$ is decreasing.  This is true here on $(-1,1)$ and 
$(1,3)$. 
\vskip 10pt
\item[(3) C.] What happens at $x = 1$ on the graph $y = f(x)$?
\vskip 10pt
{\it Solution:}  At $x=1$, $f(x)$ is not differentiable because there
is a ``corner'' point where the slope suddenly changes from $-1$ to approximately
$2/3$.  
\item[ (6) D.]  Sketch the graph $y = f(x)$ if it is given 
that $f(0) = 0$.
\vskip 10pt
{\it Solution:}  The actual graph of the function $f(x)$ is at the top of the 
next page.
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{M2IIISol.ps} 
\end{center}
\noindent
{\it Note on how this will be graded}: The exact $y$-coordinates are
not important here (except for the given $f(0) = 0$).  We will be
looking for the correct signs of the slopes, the concavities, and the ``corner''
at $x = 1$.  
\end{itemize}
\vskip 10pt
\noindent
IV. 
\begin{itemize}
\item[(4) A.]  State the limit definition of the derivative $f'(x)$.
\vskip 10pt
\noindent
{\it Solution:}  The definition is:
\[
f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h},
\]
provided that the limit exists.
\vskip 10pt
\item[(6) B.] {\it Use the definition} to compute the derivative of $f(x)=x^2+2x+1$.
(Do not use the derivative rules.)
\vskip 10pt
\noindent
{\it Solution:}  Applying the definition to $f(x) = x^2 + 2x + 1$,
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{(x+h)^2 + 2(x+h) + 1 - (x^2 + 2x + 1)}{h}\\
&=& \lim_{h\to 0} \frac{x^2 + 2xh + h^2 + 2x + 2h + 1 - (x^2 + 2x + 1)}{h} \text{ (multiply out)}\\
&=& \lim_{h\to 0} \frac{2xh + h^2 + 2h}{h}\text{ (cancel terms on top)}\\
&=& \lim_{h\to 0} 2x + h + 2\text{ (cancel $h$'s)}\\
&=& 2x + 2.
\end{eqnarray*}

\item[(2) C.] Check your answer to part (b) by finding $f'(x)$ using the
rules for differentiation. State each rule that you use.
\vskip 10pt
\noindent
{\it Solution:} By the sum, constant multiple, and power rules:
\[
\frac{d}{dx}(x^2 + 2x + 1) = \frac{d}{dx}(x^2) + 2\frac{d}{dx}(x) + \frac{d}{dx}(1) = 2x + 2
\]
(the derivative of the constant $1$ is zero).  
\item[(4) D.] Find the equation of the tangent line to the curve  $y=x^2+2x+1$ at $x = -3$.
\vskip 10pt
\noindent
{\it Solution:}  The slope is $f'(3) = -6 + 2 = -4$ and the line goes through
the point $(-3,f(-3)) = (-3,4)$.  The equation is 
$y - 4 = (-4)(x + 3)$, so $y = -4x - 8$.  

\item[(4) E.] The position (in meters) of a moving object at time $t$ (in seconds) is given by 
$y = f(t) = t^2 + 2t + 1$.  What is the physical meaning of the statement $f'(2) = 6$
and what are the units of $f'(2)$?
\vskip 10pt
\noindent
{\it Solution:}  The physical meaning is that the {\it velocity} of the object
(the rate of change of position with respect to time) is 6 meters per second at time
$t = 2$ seconds.  
\end{itemize}
\vskip 10pt
\noindent
V.  The following graph shows $y = f(x)$ for some function $f(x)$.
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{M2VI.ps} 
\end{center}
\begin{itemize}
\item[(5) A.]  At which $x$ is $f(x)$ discontinuous?  Justify your answer
using the definition of continuity.  
\vskip 10pt
\noindent
{\it Solution:}  Recall that saying $f(x)$ is continuous at $x = a$ 
means that $\displaystyle{\lim_{x\to a} f(x) = f(a)}$.  In other words, in 
order for $f(x)$ to be continuous at $x = a$, $f(a)$ must exist, the limit
$\ds \lim_{x\to a} f(x)$ must exist, and the limit must equal $f(a)$.
  This fails 
for the given function at two points.  At $x = 2$, we have $f(2) = 1$, but
$\displaystyle{\lim_{x\to 2} f(x)}$ does not exist because 
$\displaystyle{\lim_{x\to 2^-} f(x) = 2,}$ while
$\displaystyle{\lim_{x\to 2^+} f(x) = +\infty}$.  At $x = 3$, we
have a similar but slightly different problem, 
$\displaystyle{\lim_{x\to 3^+} f(x) = f(3) = 0}$, but 
$\displaystyle{\lim_{x\to 3^-} f(x) \ne 0}$  (it's approximately $1/3$ from the graph).
Hence again $\displaystyle{\lim_{x\to 3} f(x)}$ does not exist, and 
$f(x)$ is not continuous at $x = 3$.
\vskip 10pt
\item[(5) B.] At which $x$ is $f(x)$ continuous but not differentiable?  Explain.
\vskip 10pt
\noindent
{\it Solution:}  $f(x)$ is continuous but not differentiable at the ``cusp'' point
at $x = 1$.  The derivative $f'(x)$ does not exist there because there is not a well-defined
tangent line there. (Another possible explanation: the slopes of the 
tangents are positive on one side of the cusp and negative on the other side.)   
\end{itemize}

\noindent
VI.  Compute the following limits. Show all work. 
\begin{itemize}
\item[(5) A.]  $\displaystyle \lim_{x \rightarrow \pi} \sin\left(\frac{x}{2}-\pi\right)$
\vskip 10pt 
\noindent
{\it Solution:}  $\sin(x)$ is a continuous function for all $x$, so
\[
\lim_{x \rightarrow \pi} \sin\left(\frac{x}{2}-\pi\right) = 
\sin\left(\lim_{x \rightarrow \pi} \frac{x}{2} - \pi\right) = 
\sin\left(\frac{-\pi}{2}\right) = -1.
\]
\item[(5) B.] $\displaystyle \lim_{x\rightarrow 10}\frac{\sqrt{x-6}-2}{x-10}$
\vskip 10pt 
\noindent
{\it Solution:}  For this one, we will do ``preliminary algebra'' that is similar to what
we use in computing the derivative of $\sqrt{x}$ by the limit definition -- we 
multiply the top and bottom by the ``conjugate radical'' expression
$\sqrt{x - 6} + 2$ to start:
\begin{eqnarray*}
\lim_{x\rightarrow 10}\frac{\sqrt{x-6}-2}{x-10} 
&=& \lim_{x\rightarrow 10}\frac{\sqrt{x-6}-2}{x-10}\cdot \frac{\sqrt{x-6}+2}{\sqrt{x-6}+2}\\
&=& \lim_{x\rightarrow 10}\frac{(x - 6) - 4}{(x-10)\sqrt{x-6}+2}\\
&=& \lim_{x\rightarrow 10}\frac{x - 10}{(x-10)\sqrt{x-6}+2}\\
&=& \lim_{x\rightarrow 10}\frac{1}{\sqrt{x-6}+2}\\
&=& \frac{1}{\sqrt{4} + 2}\\
&=& \frac{1}{4}.
\end{eqnarray*}
\item[(5) C.] $ \displaystyle \lim_{x \rightarrow 1^{-}}\frac{|x-1|}{x^2-3x+2}$ 
\vskip 10pt 
\noindent
{\it Solution:}  For the limit as $x \to 1^-$, we are interested only 
in $x$ with $x < 1$.  For $x < 1$, $|x - 1| = -(x - 1)$, and after factoring
the bottom and cancelling, we get:
\begin{eqnarray*}
\lim_{x \rightarrow 1^{-}}\frac{|x-1|}{x^2-3x+2}
&=& \lim_{x \rightarrow 1^{-}}\frac{-(x-1)}{(x-1)(x-2)}\\
&=& \lim_{x \rightarrow 1^{-1}}\frac{-1}{x-2}\\
&=& \frac{-1}{-1} = 1.
\end{eqnarray*}
\item[(5) D.] Find all vertical and horizontal asymptotes 
to the graph  $\displaystyle{y=\frac{x^2+x-2}{x^2+5x+6}}.$
\vskip 10pt 
\noindent
{\it Solution:}  The vertical asymptotes can be determined by factoring the 
top and bottom:
\[
\frac{x^2+x-2}{x^2+5x+6} = \frac{(x + 2)(x - 1)}{(x + 2)(x + 3)}.
\]
For $x \ne -2, -3$, we see 
\[
\frac{x^2+x-2}{x^2+5x+6} = \frac{x - 1}{x + 3}.
\]
This does have a finite limit as $x \to -2$ (the limit equals $-3$).  
That is, there is a {\it removable
discontinuity} of the original function at $x = -2$ (not a vertical asymptote).
On the other hand, $\ds \lim_{x\to -3^-} \frac{x - 1}{x + 3} = +\infty$, while
$\ds \lim_{x\to -3^+} \frac{x - 1}{x + 3} = -\infty$.
That leaves $x = -3$ as the only vertical asymptote.   The horizontal asymptote is $y = 1$, since
\[
\lim_{x\to \pm\infty} \frac{x^2+x-2}{x^2+5x+6} = 
\lim_{x\to \pm\infty} \frac{1+\frac{1}{x}-\frac{2}{x^2}}{1+\frac{5}{x}+\frac{6}{x^2}} = 1.
\] 
\end{itemize}
\end{document}
