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\markright{Chapter 3: Answers 2}
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\Bf{Chapter 3: Answers 2 \hfill Jack K. Cohen \hfill Colorado School of Mines}

\begin{enumerate}
\item $A(r) = \pi r^2$, so $\frac{dA}{dr} = 2 \pi r$.  This says that $\frac{dA}{dr} = C$, the circumference of the circle---if you stop to think, this makes geometric sense.

\item  Evaluate the previous result for $\frac{dA}{dr} \approx 21.99$ m when $r = 3.5$ m.

\item (3.1.31) $A = \pi r^2, C = 2 \pi r$.  Eliminate $r =C / (2 \pi)$ to get $A(C) = C^2/ (4 \pi)$.  So, $\frac{dA}{dC} = C / (2 \pi)$.

\item  $\frac{dA}{dC} \approx  0.56$ m when $C = 3.5$ m.

\item (3.1.32) The implication is that $r$ increases at a \Em{constant} rate, so $r = 5 t$ ft.  Thus $A(t) = \pi r^2 = 25 \pi t^2$ ft$^2$ and $\frac{dA}{dt} = 50 \pi t$  ft$^2$/sec.  When $t = 10$, we have $\frac{dA}{dt} = 500 \pi$  ft$^2$/sec.

\item (Generalization of Example 6) 
	\begin{enumerate}
	\item At the maximum height, $v = 0$, so $t = t_{top} = v_0 / g$ and $s_{top} = v_0^2/ ( 2 g)$.
	\item At ground level, $s = 0$, so $0 = -gt^2/2 + v_0 t = t(v_0 - g t/2)$.  The $t = 0$ root corresponds to the initial time, so the other root gives the final time, $t_f = 2v_0/g$.  Thus, the final velocity is $v_f = - g \cdot 2v_0/g + v_0 = - v_0$.
	\item  The initial and final velocities are equal in magnitude and opposite in direction (no matter what the initial velocity is---there is nothing special about the text's choice of 96 ft/sec).
	\item The dimensions of the result, $s _{top} =  v_0^2/(2 g)$ are $(\frac{L}{T})^2 \div \frac{L}{T^2} = L$ which is correct.   The dimensions of $v_f = - v_0$ are obviously the dimensions of a velocity, so this checks as well.	\end{enumerate}

\item  Re the inverse square law:
	\begin{enumerate}
	\item Yes!  When $s = 0$, we see from the given formula for $s$ that $v_f^2 = v_0^2$, so again the final and initial velocities have the same magnitude.
	\item At the maximum height, $v = 0$, so $s_{top} = \frac{R v_0^2}{2 g R - v_0^2}$.
	
	\item With the given numbers, the constant force law gives $s_{top} = 144$ ft and the inverse square result gives 144.001 ft.  Thus for velocities of this order of magnitude, we can ignore the effect of the inverse square law unless unusually high precision is demanded.
	\end{enumerate}

\item Re the water bucket problem:
	\begin{enumerate}
	\item  The given function expands to the quadratic $10 - t/5 + t^2/1000$, so $V'(t) = -1/5 + t/500$.  Since $t$ is measured in seconds, we evaluate $V(60) = 8/5 = 1.6$.
	\item The average change is $(V(100) - V(0)) / (100 - 0) =(0 - 10)/100 = - 1/10$.  Equating this to $V'(t)$ gives the linear equation $-1/10 = -1/5 + t/500$ with solution $t = 50$ sec.
	\item No, it would imply that the bucket magically refills to 10 liters after all the water has leaked out.   It wasn't really asked, but a good scientist routinely checks that problems makes sense.  In this case, for example, is $V(0) = 10$ as asserted?  Is $V(100) = 0$ as asserted?
	\end{enumerate}

\item (3.1.36) Your figure should look something like Figure 1.  The graph is virtually linear and you can get sufficient accuracy by just computing the change over the 2 year period 1974-1976 which gives a growth rate of 17 thousand/year.  Using some very fancy \Mma\ curve fitting tools I got \Tt{17.0031} which is, indeed, 17 to the accuracy of the given data.

\begin{figure}[htb]
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\centerline{\epsffile{ans2p8.eps}}
\caption{Curve fit of population data.} 
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\end{enumerate}
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