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\centerline{\textbf{\large Exam \#4 solutions $\cdot$ Thursday, April 26, 2007}}
\vspace*{.2in}
\centerline{MATH 124 $\cdot$ Calculus I $\cdot$ Section 8 $\cdot$ Spring 2007}
\vspace*{0.2in}

\emph{Note:  Some of my solutions are wordy, for the sake of explanation.  All
I expect of you is computation, unless a problem specifically requests a verbal
response.}

John Kerl (\texttt{kerl.john.r@gmail.com}).

\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 1.}
Let
\begin{eqnarray*}
	x(t) &=& 3t^2 - 6t \\
	y(t) &=& \frac{4}{3}\,t^3 - 4t.
\end{eqnarray*}

\textbf{Part (a).} Find the time(s) $t$, if any, when the particle comes to a stop.

\vspace*{0.1in}
\emph{Solution:}
We need to find when $dx/dt$ and $dy/dt$ are both zero.  We have
\begin{eqnarray*}
	dx/dt &=& 6t-6 \\
	dy/dt &=& 4t^2-4.
\end{eqnarray*}
The former is zero when $t=1$; the latter is zero when $t=\pm 1$.  They are
simultaneously zero only at $t=1$.
\vspace*{0.1in}


\textbf{Part (b).} Find an equation for the tangent line to this curve at
$t=3$.

\vspace*{0.1in}
\emph{Solution:}
Parametric equations for a line through a point $(x_0,y_0)$ in the direction
$a,b$ are
\begin{eqnarray*}
	x &=& x_0 + at \\
	y &=& y_0 + bt.
\end{eqnarray*}
The point $(x_0,y_0)$ is found by plugging $t=3$ into the original equations:
\begin{eqnarray*}
	x_0 &=& 3\cdot 3^2 - 6\cdot 3  = 9\\
	y_0 &=& \frac{4}{3}\,3^3 - 4\cdot 3 = 24.
\end{eqnarray*}
The $a$ and $b$ are $dx/dt|_{t=3}$ and $dy/dt|_{t=3}$:
\begin{eqnarray*}
	a &=& \frac{dx}{dt}\bigg|_{t=3} = 6\cdot 3 - 6 = 12 \\
	b &=& \frac{dy}{dt}\bigg|_{t=3} = 4\cdot 3^2 - 4 = 32 \\
\end{eqnarray*}
So, the tangent line has equations
\begin{eqnarray*}
	x &=& 9 + 12t\\
	y &=& 24 + 32t.
\end{eqnarray*}

\vspace*{0.1in}

%Parameteric equations:
%  * Write a parametric equation for a line between two specified points.
%  * Write a parametric equation for a line through a specified point,
%    in a specified direction.
%  * Given a curve specified by a parameteric equation and a point
%    on the curve, find an equation for the tangent line to the curve at
%    that point.
%  * Given a parameteric equation specifying the motion of a particle
%    along a curve, compute the speed of the particle at a given location or
%    time.
%  * Given a parameteric equation specifying the motion of a particle
%    along a curve, find the time(s) when the particle comes to a stop.
%  * Given parameteric equations specifying the motion of two particles,
%    find the time(s) when the particles collide.

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 2.}  The function $H(t)$ describes the growth rate in thousands
per month of flour beetles in a jar, where $t$ is measured in months since the
start of the year.

\vspace*{0.1in}

\textbf{Part (a).}  What are the units of $\int_{4}^{7} H(t) dt$?

\vspace*{0.1in}
\emph{Solution:}
The units of an integral are the units of the integrand times the units of the
dependent variable.  Since $H(t)$ is thousands of beetles per month and $t$ is
months, the units of the integral are thousands of beetles.
\vspace*{0.1in}

\textbf{Part (b).}  Give a practical interpretation of $\int_{4}^{7} H(t) dt$.

\vspace*{0.1in}
\emph{Solution:}
This is the population change between month 4 and month 7.
\vspace*{0.1in}
\vspace*{0.4in}

%% ----------------------------------------------------------------
\textbf{Problem 3.}  The air pressure within a chamber is given by
$$
	P(t) = 2.1 + 0.4 t^{0.5}
$$
where $P$ is in units called atmospheres and $t$ is measured in hours.  Find
the average pressure over the time between $t=2$ hours and $t=5$ hours.

\vspace*{0.1in}
\emph{Solution:}
Compute
\begin{eqnarray*}
	\frac{\int_2^5 P(t) }{5-2} &=& \frac{\int_2^5 (2.1+0.4t^{0.5}) }{5-2}.
\end{eqnarray*}
The integral is
$$
	2.1t + \frac{0.4t^{1.5}}{1.5}\bigg|_2^5
	= \left(
	10.5 + \frac{0.4\cdot 5^{1.5}}{1.5}
	\right) - \left(
	4.2 + \frac{0.4\cdot 2^{1.5}}{1.5}
	\right)
	\approx 8.53.
$$
Then the average value is
$$
	8.53/3 \approx 2.84.
$$
Note that if you graph $P(t)$ from 2 to 5, you'll see that it rises slowly from
$P(2)\approx 2.7$ to $P(5) \approx 3.0$.  Thus, an average of 2.84 is
reasonable.

\vspace*{0.1in}
\vspace*{0.4in}

%% ----------------------------------------------------------------
\textbf{Problem 4.}  Find the exact area between $f(x) = e^x-2$ and $g(x) = -1$
on the interval $[2, 4]$.

\vspace*{0.1in}
\emph{Solution:}
The area between $f(x)$ and $g(x)$ from 2 to 4 is $\int_2^4 (f(x)-g(x))\,dx$ so
we can compute
\begin{eqnarray*}
	\int_2^4 (e^x-2-(-1))\,dx &=& \int_2^4 (e^x-1)\,dx \\
	&=& e^x-x \bigg|_2^4 \\
	&=& (e^4-4)-(e^2-2) \\
	&=& e^4-e^2-2.
\end{eqnarray*}

%% ----------------------------------------------------------------
\textbf{Problem 5.}  Find the general antiderivative:
$$
	\int \left( \frac{y^{2.1}}{3} - \frac{7}{y} + 0.2 A e^y + B\right) \;dy.
$$

\emph{Solution:}
$$
	\frac{y^{3.1}}{9.3} - 7\ln(|y|) + 0.2 A e^y + By + C.
$$

% one big messy integral?

%Antiderivatives obtained symbolically:
%  * Know antiderivatives for constant functions, polynomial functions,
%    $1/x$, exponential function, sine and cosine.
%  * Remember that you always get a family of functions:
%    e.g. $\int \cos x = \sin x + C$.
%  * Use antiderivatives (indefinite integrals) to compute definite
%    integrals.  Example:  find $\int_2^8 e^x dx$.  Since $\int e^x dx =
%    e^x+C$, we have $\int_2^8 e^x dx = e^8-e^2$.  (What happened to the
%    integration constant?  It canceled out: $(e^8+C)-(e^2+C) = e^8-e^2$.)

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 6.} Let $g'(x)$ be given by the following graph,
and suppose $g(0) = 2$:
\vspace*{0.2in}
\begin{center}
\includegraphics{e4_g.eps}
\end{center}
\vspace*{0.2in}

\textbf{Part (a).}  What are the $x$-coordinates of the critical points of $g(x)$?

\vspace*{0.1in}
\emph{Solution:}
Critical points of $g(x)$ occur when $g'(x)$ is zero (or undefined).
From the graph we see $g'(x)=0$ at $x=0,2,4$.
\vspace*{0.1in}

\textbf{Part (b).}  What are the $x$-coordinates of the inflection points of $g(x)$?

\vspace*{0.1in}
\emph{Solution:}
Inflection points of $g(x)$ occur at extrema of $g'(x)$.
These are at $x=1,3$.
\vspace*{0.1in}

\textbf{Part (c).}  Find the values of $g(x)$ at the critical and inflection points.

\vspace*{0.1in}
\emph{Solution:}
We are given $g(0)=2$, and we are given $g'(x)$ in graphical format.  This suggests
the fundamental theorem of calculus, one landmark at a time.  By ``landmark''
I mean a critical or inflection point.
\begin{eqnarray*}
	g(1) &=& g(0) + \int_0^1 g'(x)\, dx =  2 + 10 = 12. \\
	g(2) &=& g(1) + \int_2^1 g'(x)\, dx = 12 + 10 = 22. \\
	g(3) &=& g(2) + \int_3^2 g'(x)\, dx = 22 -  5 = 17. \\
	g(4) &=& g(3) + \int_4^3 g'(x)\, dx = 17 -  5 = 12.
\end{eqnarray*}

\vspace*{0.1in}

\textbf{Part (d).}  Sketch a graph of $g(x)$.  Label
critical points and inflection points of $g(x)$.

\emph{Solution:}
Using the above information we can plot the critical points $(0,2)$, $(2, 22)$,
and $(4, 12)$, along with the inflection points $(1,12)$ and $(3, 17)$.
Then we can connect the dots as follows:
\begin{itemize}
\item Since $g'(x)$ is positive on $(0,2)$, $g(x)$ is increasing there.
\item Since $g'(x)$ is negative on $(2,4)$, $g(x)$ is decreasing there.
\item Since $g'(x)$ increases on $(0,1)$ and $(3,4)$, $g(x)$ is concave up there.
\item Since $g'(x)$ decreases on $(1,2)$ and $(2,3)$, $g(x)$ is concave down there.
\end{itemize}
(The sketch is on the next page.)
\newpage
Here is a sketch:
\begin{center}
\includegraphics{e4_gs.eps}
\end{center}

%% ----------------------------------------------------------------
\textbf{Problem 7.}  The quantity $A$ varies with time as specified by
$$
	\frac{dA}{dt} = 7.3\cos(t)-0.04.
$$
\vspace*{0.1in}

\textbf{Part (a).}  Write down a general solution for $A$.

\vspace*{0.1in}
\emph{Solution:}
The general antiderivative is
$$
	7.3\sin(t) - 0.04t + C.
$$
\vspace*{0.1in}

\textbf{Part (b).}  Given that $A(0) = 2.1$, write down a specific solution for
$A$.

\vspace*{0.1in}
\emph{Solution:}
Using the initial condition, we have
$$
	2.1 = 7.3\sin(0) - 0.04\cdot 0 + C = C.
$$
So, the solution is
$$
	7.3\sin(t) - 0.04t + 2.1.
$$
\vspace*{0.1in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 8.}

\textbf{Part (a).}
Let
$$
	G(x) = \int_1^x e^{t^2} dt.
$$
Determine (with justification) whether $G(x)$ is increasing,
decreasing, or constant.

\vspace*{0.1in}
\emph{Solution:}
A function is increasing, decreasing, or constant when its derivative is
positive, negative, or zero, respectively.  The derivative is
$$
	G'(x) = \frac{d}{dx} \int_1^x e^{t^2} dt = e^{x^2}
$$
using the second fundamental theorem of calculus.  The output of exponential
functions is always positive, so this function is increasing everywhere.

Here is another solution, which a student came up with:  $e^{t^2}$ is always
positive, so as the right endpoint of the integral slides rightward from 1,
only positive areas are added.  Thus, the integral function is increasing.
\vspace*{0.1in}

\textbf{Part (b).}
Now let
$$
	G(x) = \int_1^{ax+b} e^{t^2} dt.
$$
Find $G'(x)$.

\vspace*{0.1in}
\emph{Solution:}
Using the second fundamental theorem and the chain rule, we evaluate the
integrand at the right endpoint (upper limit), times the derivative of the
right endpoint.  This is
$$
	G'(x) = a e^{(ax+b)^2}.
$$
\vspace*{0.1in}

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