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\begin{document}

\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

March 10, 2003}
\end{center}

\begin{enumerate}



\item  Solve the following equation for $P$:
	%
	\begin{equation} \label{logistic}
	\frac{M-P}{P} = Ae^{-kt}
	\end{equation}
	%
	\\\sol
	\begin{eqnarray*}
	M-P & = & PAe^{-kt} \\
	M & = & PAe^{-kt} + P \\
	M & = & (Ae^{-kt} + 1) P \\
	P & = & \frac{M}{Ae^{-kt} + 1}
	\end{eqnarray*}


\item  Suppose you know that in equation~(\ref{logistic}) that $M$ and $k$ are constants given by $M=100$ and $k=2$.
	Suppose you also know that $P=20$ when $t=0$.  Find $A$.
	Using the previous problem, write $P$ as a function of $t$.
	\\\sol
	To find $A$, we plug in all the numbers: $M=100$, $k=2$, $P=20$ and $t=0$:
	\[
	\frac{100-20}{20} = Ae^{0}
	\]
	Which gives us that $A=4$.
	This gives the equation:
	\[
	P = \frac{100}{4e^{-2t} + 1}
	\]

\newpage

\item  (Review problem)
	Suppose you use Simpson's Rule for the following integral:  $\int_1^5 e^{3x} \; dx$.
	\begin{enumerate}
	\item  What is the error formula for Simpson's rule (and what does everything mean)?
		\\\sol \[ |E_S| \leq \frac{K(b-a)^5}{180n^4} \]
	\item  Find a good value for $K$ for out integral.
		\\\sol $f(x)=e^{3x}$ so $f^{(4)}(x)=3^4 e^{3x}$.
		For all $x$ between $1$ and $5$, we have $|f^{(4)}(x)| \leq 3^4e^{15}$ (draw the graph!).
	\item  If $n=100$, then what is the worst the error can be?
		\\\sol Just plug everything into the formula:
		\[
		|E_S| \leq \frac{3^4e^{15}(5-1)^5}{180\cdot 100^4} = 15.06
		\]
		(so, not too accurate)
	\item  Find $n$ so that the error is less then $0.01$.
		\\\sol
		Solve the inequality:
		\[
		\frac{3^4e^{15}(5-1)^5}{180 n^4} < 0.01
		\]
		to get $n> 622.99$.
	\end{enumerate}

\newpage

\lecture

\item  Consider the curve $y=x^2+1$.
	\begin{enumerate}
	\item  Draw this curve and rotate it around the $y$-axis.
	
	\item  Find the area function $A(y)$ (the area of a $y$-cross section).
	\\\sol 
	\[
	A(y)=\pi x^2 = \pi(\sqrt{y-1})^2 = \pi(y-1)
	\]

	\item  Set up an integral representing the volume of the solid of revolution from $y=2$ to $y=4$.  (What does this part of the solid look like?)
	\\\sol 
	\[
	V = \int_2^4 A(y) \; dy = \int_2^4 \pi(y-1)\; dy = 4\pi
	\]
	\end{enumerate}


\item  Repeat the previous exercise, but rotate the curve around the $x$-axis and find the volume from $x=-1$ to $x=3$.
	\\\sol
	\[
	A(x) = \pi y^2= \pi(x^2+1)^2
	\]
	\[
	V = \int_{-1}^3 \pi(x^2+1)^2 = \int_{-1}^3 \pi(x^4+2x^2+1) = \frac{1072\pi}{15}
	\]




\end{enumerate}

\end{document}