%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  Answers to problem in the Koblitz 125 Notes
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newcommand{\Arctan}{\arctan}
\chapter*{Answers}\
%\setcounter{section}{0}
\section{Answers to Selected Homework Problems}

\begin{small}
%\newlabel{homework:1} {{1}{5}}
%\section{Answers to Selected Homework 1 Problems}
\centerline{Homework 1}
\noindent
%\begin{answers}{homework:1}
1.~(a) $8\sqrt{2}/15=0.754$, (b) $1/12$, (c) $(n-1)/(n+1)$, (d) $4.5$ (the
curves intersect at $(-3,2)$ and $(0,-1)$), 
(e) $4/3$, (f) $(2/3)-(2/\pi)=0.030$,
(g) $(3/\pi)(1-0.5\sqrt{3})-(1/8)=0.00294$. 


\noindent
2.~(a) $\sqrt{2}-1=0.414$, (b) $46/3$.

\noindent
3.~(a) net=$1/\pi$, total=$5/\pi$, (b) net=0, total=245, (c) net=20,
total=28, (d) net=$(1.5/\pi)-(1/2)=-0.0225$,
total=$0.25+(4.5-3\sqrt{3})/\pi= 0.0284$.

\noindent
4.~(a) $a_x(t)=(F/m)\cos(\omega t)$, $a_y(t)=(F/m)\sin(\omega t)$, (b)
$v_x(t)=(F/m\omega)\sin(\omega t)$, $v_y(t)=(F/m\omega)(1-\cos(\omega t))$,
(c) $x(t)=\left(F/m\omega^2\right)(1-\cos(\omega t))$,
$y(t)=\left(F/m\omega^2\right)(\omega t-\sin(\omega t))$, (d) a cycloid
running up the $y$-axis (rather than the $x$-axis), (e) a circle if
$v_0=-F/m\omega$.

%\end{answers}
%\newlabel{homework:2} {{2}{5}}
\centerline{Homework 2}

%\begin{answers}{homework:2}

\medskip

\noindent
1.~hits ground at (a) 3 sec, (b) 4.5 sec, (c) 2 sec. 2.~44.3 m/sec, 0.232
sec.

\noindent
3.~$v_0=-38$ ft/sec, $s_0=590$ ft. 4.~1032 ft.

\noindent
5.~(a) $8\pi/3$, (b) $\pi/30$, (c) $0.5\pi^2-\pi=1.793$.  6.~(a) $22.8\pi$,
(b) $14.8\pi$, (c) $20\pi$, (d) $4\pi$.

\noindent
7.~(a) $\frac{4}{ 3}\pi ab^2$, (b) $\frac{4}{ 3}\pi a^2b$.  8.~$\pi h^2\left(
r-\frac{1}{ 3}h\right)$. 9.~$\int_0^H\frac{\sqrt{3}}{ 4}(\frac{x}{ H}s)^2dx=
\frac{\sqrt{3}}{12}s^2H$.

%\end{answers}
%\newlabel{homework:3} {{3}{5}}
\centerline{Homework 3}

%\begin{answers}{homework:3}

\medskip

\noindent
1.~(a) 6 min 17 sec, (b) 0.03183 units/min.  2.~(a) 20 min 57 sec, (b)
0.03183 units/min.  3.~0.6283 ft${}^3$/min.
4.~$dV/dt=0.04\pi\left(4-(-2+0.04t)^2\right)$.
5.~$dV/dt=0.09\pi\left(4-(-2+0.04t)^2\right)$.  6.~1.21 cm/sec.
7.~cross-sectional area = $\frac{1}{ 4}\sqrt{3}(\frac{h}{ H}s)^2$.

%\end{answers}
%\newlabel{homework:4} {{4}{5}}
\centerline{Homework 4}

%\begin{answers}{homework:4}

\medskip

\noindent
1.~(a) $x^2\!/4+4y^2=1$, (b) $x^2\!/100+y^2\!/64=1$, (c)
$x^2\!/625+y^2\!/576=1$, (d) $x^2+y^2\!/9=1$, (e) $x^2\!/16+y^2\!/25=1$.

\noindent
2.~(a) $(\pm 3,0)$, $e=0.6$; (b) $(0,\pm 24)$, $e=0.96$; (c) $(\pm 0.3727,0
)$, $e=0.7454$.  3.~10.

\noindent
4.~(a) $\int_0^b\sqrt{1+n^2x^{2(n-1)}}\,dx$; (b) $\frac{8}{ 27}\left(
(1+9b/4)^{3/2}-1\right)$.
5.~(a) $\int_0^4\sqrt{98+(-9.8t+7\sqrt{2})^2}\,dt$; (b)
$\int_0^{\pi/2}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}\,d\theta$; (c)
r$\int_0^{2\pi}\sqrt{1.25-\cos\theta}\,d\theta$.
6.~(a) $1/\sqrt{7}=0.3780$; (b) $\frac{d}{ dt}(x+2.5y)=0.3780(1+2.5
\sqrt{3}/2) =1.1964$; (c) $u=x+y\cot\alpha$, so $\frac{du}{ dt}=\frac{dx}{ dt}+
\frac{dy}{ dt}\cot\alpha+y(-\csc^2\alpha)\frac{d\alpha}{ dt}=1.1964-y
(\csc^2\alpha)\left(-15\cdot\frac{\pi}{ 180}\right)=1.1964+0.5\cdot
7.2509\cdot\pi/12=2.1455$ units/hour.

%\end{answers}
%\newlabel{homework:5} {{5}{5}}
\centerline{Homework 5}

%\begin{answers}{homework:5}

\medskip
\noindent
1.~55.4 m/sec 2.~(a) 0.790766, (b) 1.183465, (c) 0.987116, (d) 0.997943.
3.~8.878.

\noindent
4.~(a) 0.58726, (b) 0.58622 (limits of integration are $\pi/3$ and
$\pi/2$).

\noindent
5.~(b) 0.2402 m, (c) 0.4541 m.

\noindent
6.~(a) $x(8)-x(0)=\int_0^8\dot{x}(t)dt\approx 34.45$, $y(8)-y(0)=\int_0^8
\dot{y}(t)dt\approx 46.35$, and finally $\sqrt{34.45^2+46.35^2}=57.75$; (b)
$\int_0^8\sqrt{(dx/dt)^2+(dy/dt)^2}dt\approx 59.29$.

%\end{answers}
%\newlabel{homework:6} {{6}{5}}
\centerline{Homework 6}

%\begin{answers}{homework:6}

\medskip

\noindent
1.~(a) (2,1), (b) (1.5,1.2), (c) $(0.4244r,0)$, (d) $(0,-0.8377)$.

\noindent
2.~at $39.37^\circ$ northern latitude \& $116.6^\circ$ western longitude
(right between Summit Mountain and Hickison Summit on the border between
Lander and Eureka counties). 3.~0.5466.

%\end{answers}
%\newlabel{homework:7} {{7}{5}}
\centerline{Homework 7}

%\begin{answers}{homework:7}

\medskip
\noindent
1.~(a) $2x^2/(4+x^2)\>+\>2x\,\arctan(x/2)$.

\noindent
(b) $\Bigl(\arcsin(x^2)
       -2x(x+1)(1-x^4)^{-1/2}\Bigr)\Bigm/\Bigl(\arcsin(x^2)\Bigr)^2$.

\noindent
(c) $-1/2\sqrt{x-x^2}$, (d) $\arcsin\,x$.  2.~(a) $0.25\,\arcsin(0.4x)+C$,
(b) $0.5\sqrt{2}\,\arctan(\sqrt{2}\,x)+C$.  3.~$31.32^\circ$. 4.~$\pi^2\!/4$.
5.~(0.2559,0).

\noindent
6.~(a) max: $(1,0.2854)$, min: $(-1,-0.2854)$, (b) max:
$(-\sqrt{3}, 0.685)$, min: $(\sqrt{3},-0.685)$.

\noindent
7.~(a) $\pi-(2/3)$, (b) $2\sqrt{3}-2\pi/3$. 8.~4.584 deg/sec.

\noindent
9.~(a) $\arccos\left(\bigl(a^2+b^2-x^2\bigr)/2ab\right)$, (b)
$2c\bigm/\sqrt{(2ab)^2-(a^2+b^2-c^2)^2}$.

%\end{answers}
%\newlabel{homework:8} {{8}{5}}
\centerline{Homework 8}

%\begin{answers}{homework:8}

\medskip

\noindent
1.~(a) $1/x$, (b) $(4\ln^3x)/x$, (c) $(2\ln(\ln x))/(x\ln x)$, (d) $1/(\sin
x
\cos x)$, (e) $-\arctan(x/4)$.

\noindent
2.~(a) both derivatives are 0; (b) use the fact that only a constant
function has derivative zero, find the constant by setting $x=1$, and then
replace $x$ by $b$ in the first function and by $a$ in the second function.

\noindent
3.~(a) $\ln 3$, (b) $\frac{3}{ 2}\ln 2$, (c) $\frac{2}{ 3}\ln 3$, (d) $\frac{3}{
2}\ln 3 -\ln 2$, (e) $\ln\pi$, (f) $\frac{1}{ 2}\ln(r+x)$.

\noindent
4.~(a) $-\ln 2$, (b) $\ln\sin x+C$, (c) $\ln(\ln x)+C$, (d) $(1/\ln
2)-(1/\ln 3)=0.5325$, (e) $2\ln 3$.

\noindent
5.~(a) 1.4427, (b) 1.3863, (c) 0.4413.  6.~(a) 0.4449, (b) 0.09436.
7.~4.3552.

\noindent
8.~11.18, 14.14, 22.36.

\noindent
9.~(a) 0, (b) 0, (c) 0, (d) 0.  10.~(a) $(1+\ln x)x^x$, (b)
$\bigl(\sqrt{x}\cot x+(\ln\sin x)/2\sqrt{x}\bigr)(\sin x)^{\sqrt{x}}$,

\noindent
(c) $(\frac{1}{\ln x}+\ln(\ln x))(\ln x)^x$, (d) $\left(\frac{2}{ x}\ln
x\right)x^{\ln x}$, (e) $\left(-(\arctan\,x)
\tan x+\frac{\ln\cos x}{ 1+x^2}\right)(\cos x)^{\arctan\,x}$.

\noindent
11.~Let $a$ be the number of hours before noon when it started snowing,
and let $s(t)$ be the distance traveled after $t$ hours of plowing.  Show
that $ds/dt$ is inversely proportional to $t+a$, and then use the method
in section 2 of Chapter 1 to find a formula for $s(t)$; you should find that
$s(t)=C\ln((t+a)/a)$.  (The value of the constant $C$ makes no difference.)
You're given that $s(2)=\frac{3}{ 2}\>s(1)$.  Use the properties of $\ln$ and
some algebra to get a quadratic equation for $a$, from which you find that
$a=(-1+\sqrt{5})/2$ (this number is called the ``golden ratio'').  Snow
started falling at 11:23 a.m.

%\end{answers}
%\newlabel{homework:9} {{9}{5}}
\centerline{Homework 9}

%\begin{answers}{homework:9}

\medskip

\noindent
1.~(a) 2, (b) $\infty$, (c) $\infty$, (d) 2, (e) $\pi/2$, (f) $\pi$, (g)
$\pi /3$, (h) $\pi/2$, (i) $\pi$, (j) 1, (k) $\infty$, (l) $\pi/4$.

\noindent
2.~$\pi n/(2-n)$, infinite volume if $n\ge 2$. 3.~$(n-1)/(n-2)$.

%\end{answers}
%\newlabel{homework:10} {{10}{5}}
\centerline{Homework 10}

%\begin{answers}{homework:10}

\medskip

\noindent
1.~(a) 1, (b) 0.9917.  2.~(a) $3x$, (b) $-x^2$, (c) $x/y^2$, (d) $x$, (e)
$\frac{1}{ 2}x^2$.  3.~(a) $20\ln x$, (b) $x^2+x$,

\noindent
(c) $\arcsin\,e^x$, (d) $\pm\sqrt{2\ln x}$, (e) $1/(1+e^{0.01x})$.
4.~(a)~$e^{2x}\left(\frac{1}{ 1+x^2}+2\,\arctan\,x\right)$,

\noindent
(b)~$e^{-\sin(\pi x)}\bigl(1-\pi x\,\cos(\pi x)\bigr)$, (c)~$e^{x+e^x}$,
(d)~$e^{-\cos x}(2x+x^2\sin x)$, (e)~$e^{\arcsin\,x}/\sqrt{1-x^2}$.
5.~(a) $\frac{d}{ dx}\cosh x=\sinh x$, $\frac{d}{ dx}\sinh x=\cosh x$, (b) 1.

\noindent
6.~(a) 3.6, (b) $\frac{1}{ 2}-\frac{1}{ 2e}$, (c) $e-\frac{1}{ e}$, (d) $1/\ln
10$.  7.~$\frac{15\pi}{ 16\ln 4} \approx 2.1245$.

\noindent
8.~(a) max at $(\pi/6,1.9477)$, inflection points at
$(\pi/18,1.3956)$ and $(5\pi/18,1.3956)$.

\noindent
(b) max at $(0.6065,0.1839)$, inflection at $(0.2231,0.0747)$.

%\end{answers}
%\newlabel{homework:11} {{11}{5}}
\centerline{Homework 11}

%\begin{answers}{homework:11}

\medskip
\noindent
1.~\$4932. 2.~(a) 11.55 yrs, (b) 6.93 yrs, (c) 3.85 yrs; (a) 11.90 years,
(b) 7.27 yrs, (c) 4.19 yrs.  3.~20480.  4.~1020201, 1491825, 34.66 yrs.
5.~$250e^{0.007675(t-1650)}$, A.D.~2250.  6.~11.33\%, 6.12 yrs.  7.~6.05
times 8.~(a) \$4034300, (b) $100e^{0.13t}$, (c) $\int_0^{10} 52\cdot
100e^{0.13t}= 40000e^{0.13t}|_0^{10}=\$106772$, (d) \$44241000, (e) about
2/3 day.  9.~$8033\ln(y_0/y_1)$. 10.~6.64 hrs.  11.~$100e^{-0.0112t}$, 62.1
yrs.  12.~\$11477. 13. (a) $v(t)=v_0e^{-kt}$;
(b) $s(t)=\frac{v_0}{ k}(1-e^{-kt})$; (c)$\arctan(ky_0/v_0)$.
%\end{answers}
%\newlabel{homework:12} {{12}{5}}
\centerline{Homework 12}

%\begin{answers}{homework:12}

\medskip
\noindent
1.~2.5587. 2.~(a) $-\frac{1}{\pi}\sqrt{2+\cos(2\pi t)}+C$, (b) $-1/\left(
7a\sin^7(ax+b)\right)+C$. 3.~2.0191.

\noindent
4.~(a) $4/(3\pi)=0.4244$, (b) $3/8=0.375$, (c) $16/(15\pi)=0.3395$, (d)
$5/16=0.3125$, (e) $32/(35\pi)=0.2910$.


%\end{answers}
%\newlabel{homework:13} {{13}{5}}
\centerline{Homework 13}

%\begin{answers}{homework:13}
\medskip

\noindent
1.~(a) $-\frac{x}{ a}\cos(ax+b)+\frac{1}{ a^2}\sin(ax+b)+C$, 
(b) $x^2\sin x+2x\cos x-2\sin x+C$,

\noindent
(c) $x\,\arctan\,x-\frac{1}{ 2}\ln\left(1+ x^2\right)+C$, 
(d) $\left(x^3-3x^2+6x-6\right)e^x+C$, 

\noindent
(e) $\frac{1}{ 2}x^2 \arctan\,x-\frac{x}{ 2}+\frac{1}{ 2}\,\arctan\,x+C$.

\noindent
2.~(a) $\left(-\frac{1}{ k},\frac{1}{ 4}\right)$, (b) $(9/16,0)$, (c)
$\left(\left(\frac{n+1}{ n+2}\right)^2,0\right)$, (d) 0.18484.  3.~0.60672.

%\end{answers}
%\newlabel{homework:14} {{14}{5}}
\centerline{Homework 14}
%\begin{answers}{homework:14}

\medskip
\noindent
1.~(a) 0.2747, (b) 0.09861, (c) 0.05371, (d) 0.4621, (e) 0.06112,
(f) 0.04251, (g) 0.2118, (h) 0.2, (i) 0.1903,
(j) $-\frac{1}{5}\ln(6)=-0.35835$.

\noindent
2.~(a)
(0.7095,0), (b) (0.7912,0).  3.~$\frac{1}{ 2}\ln\frac{1+\sin x}{ 1-\sin x}+C $.

%\end{answers}
%\newlabel{homework:15} {{15}{5}}
\centerline{Homework 15}

%\begin{answers}{homework:15}

\medskip
\noindent
1.~$\arcsin(x/r)+C$. 2.~(a) $\ln|x+\sqrt{x^2+r^2}|+C$, (b) $\ln|x+\sqrt{x^2-
r^2}|+C$.  3.~$(0.37616,0)$.  4.~0.93313.  5.~0.48121.

%\end{answers}
%\newlabel{homework:16} {{16}{5}}
\centerline{Homework 16}
%\begin{answers}{homework:16}

\medskip

\noindent
1.~\$27.98, \$1007.30. 2.~(a) \$379.15, \$27749, (b) 7.7 years, (c)
\$2613.91 (1st year), \$1574.15 (5th year).
3.~\$48633.80, \$9726.76. 4.~(a) \$5613.14, (b) only \$17.01 more, (c)
22.57 years.

\noindent
5.~$T-(\ln 2)/k=$ 13 years and 8.22 months.

\noindent
6.~(a) \$44754, (b) \$328.12, (c) 64.9 years, (d) \$298.36.  8.~9:43 p.m.
9.~7.27 hours

\noindent
10.~4:58 a.m.  11.~8:57 a.m. 12.~(a) 0.2, (c) 40.3 m/sec upward, (d) 1.71
sec.

\noindent
13.~$s(t)=\begin{cases}
    98t+980\left(e^{-0.1t}-1\right),& \text{for $0\le t\le10$;}\\ 
7t-39.25e^{-1.4(t-10)}+329.77,& \text{for $t>10$,}
\end{cases}$ 24.3 sec.

\noindent
14.~$v(t)=
 \begin{cases}
   9.8t,&\text{for $0\le t\le 7$;}\\
   19.6+49.0e^{-0.5(t-7)},&\text{for $7\le t\le 15$.}
   \end{cases}$

\noindent
15.~(a) $y=10-5e^{-0.03t}$, (b) $y=10+10e^{-0.03t}$, (c) $y=10$.  
16.~7.42 lb.  

\noindent
17.~(a) $y(t)=48\left(1-e^{-t/12000}\right)$, (b) 36 min.

\noindent
18.~(c) $Vk+VP/r$, (e) Superior 431 yr, Michigan 71.4 yr, Erie 6.05 yr,
Ontario 17.6 yr.

\noindent
19.~(a) $dy/dt=-ky+I$, (b) 339 million, 400 million. 20.~$R=80565$ per
year; if $R=300000$, then $y(t)=3147600-147600e^{0.0953t}$, and the salmon
will die out after 32.1 years.
 
%\end{answers}
%\newlabel{homework:17} {{17}{5}}
\centerline{Homework 17}

%\begin{answers}{homework:17}

\medskip

\noindent
1.~(a) $y=\sqrt{3x/(6+x)}$, (b) $y=1/(10-\frac{1}{3} x^3)$,
(c) $y=\sqrt{\frac{2}{\pi}\sin(\pi x)~+~1}$.

\noindent
2.~(a) $y(t)=M\!\!\bigm/\!\!\left(1+Ce^{-kt}\right)$, (b)
$100\!\!\bigm/\!\!
\left(1+19e^{-0.9474t}\right)$, 6.216 years.

\noindent
3.~$y_0(1-0.02t)^{2.5}$, domain is $0\le t\le 50$. 4.~(a) $v_\infty=49$
m/sec, (b) $dv/(100+v)(49-v)=0.002dt$, then get
$(100+v)/(49-v)=e^Ce^{0.298t}$, where $e^C=100/49$ (since $v(0)=0$),
finally end up with
$v(t)=49-\Bigl(73.01\!\!\bigm/\!\!\left(e^{0.298t}+0.49\right)\Bigr)$, (c)
$\int_0^349-\left(73.01\!\!\bigm/\!\!\left(e^{0.298t}+0.49\right)\right)$.

\noindent
(d) $ds/dv=v\!\!\bigm/\!\!\left(9.8-0.102v-0.002v^2
\right)$, so $\int_0^{48} v\,dv\!\!\bigm/\!\!\left(9.8-0.102v-
0.002v^2\right)=508$ m.

\noindent
5.~205 m. 6.~(a) $\frac{dy}{ dt}=-\,\frac{r(t)}{ V}\,y$, (b)
$y_0e^{-0.02t-0.0025 t^2}$, (c)
$y_0e^{-0.01t+\frac{0.01}{\pi}\cos(\pi(t-0.5))}$.

\noindent
7.~$dy/dt=-2y/(100+0.25t^2)$, $y(t)=y_0e^{-0.4\,\arctan(0.05t)}$,
$y(\infty)=y_0e^{-0.2\pi}=0.5335y_0$.

\noindent
8.~ (a) $\frac{dy}{ dt}=-\frac{r(t)}{ V}(y-kV)$, where $V=100$.
 (b) $y(t)=100k+(y_0-100k)e^{-0.01\int_0^tr(u)du}$.

\noindent (c) $\int_0^{10}r(u)du=26.73$; $e^{-0.01\cdot 26.74}=0.765$,
i.e., 23.5\% decrease.

\noindent (d) $y(10)=40+(30-40)e^{-0.01\cdot 26.74}=32.35$ gm.

\noindent
9.~(a) $\frac{dv}{ dt}=-8-0.4v-0.01v^2$,  
(b) $v(t)=20\bigl(\tan(1.19-0.2t)-1\bigr)$,
(c) 2.024 sec.

\noindent
10.~(a) $(m-\alpha t)\frac{dv}{ dt}=F-Kv$, (b) $v(t)=\frac{F}{ K}+\left(v_0-
\frac{F}{ K}\right)\,\left(1-\frac{\alpha t}{ m}\right)^{K/\alpha}$.

\noindent
11.~$y(t)=y_0\sin(\omega t+\frac{\pi}{ 2})=y_0\cos(\omega t)$.
12.~$y(t)=\frac{v_0}{\omega}\sin(\omega t)+y_0\cos(\omega t)$.

%\end{answers}
%\newlabel{homework:18} {{18}{5}}
\centerline{Homework 18}
%\begin{answers}{homework:18}

\medskip

\noindent
2.~(a) $y_0=D-Ak\alpha/(k^2+\alpha^2)$, (b) for $k$ small the time delay is
nearly 6 hours and the amplitude is small, while for $k$ large the time
delay is small and the amplitude is only a little less than $A$, (c) for
$k$ near zero, time delay$=\frac{1}{\alpha}\Bigl(\pi/2-\arctan\left(
\frac{k}{\alpha}\right)\Bigr)\approx 6-k/\alpha^2=6-144k/\pi^2$ (since
$\arctan(x)\approx x$ if $x$ is near 0), while for $k$ large, time delay$
=\frac{12}{\pi}\,\arctan\left(\frac{\pi}{ 12k}\right)\approx 1/k$, (d)
$\theta=60^\circ$, so time delay is 4 hrs, hottest: 7 p.m.

\noindent
3.~$dy/dt=-0.01y+(2+0.25t)$.
4.~$\frac{dy}{ dt}=-k\bigl(y-(b+at)\bigr)$.

\noindent
5.~$dy/dt=-2y/(100+0.25t^2)+(4+t)$.
6.~$dy/dt=-\frac{r}{ V}y+r(k+\alpha t)$.

\noindent
7.~(a) $dy/dt=ky+\left(18.5\sin\left(\frac{2\pi t}{ 365}\right)-20\right)$,
(b) \$29200, (c) you can use the solution to Example 18.2 by
replacing $k$ by $-k$ throughout Example 18.2, where
$k=0.00025\>\text{day}^{-1}$, and choosing $\alpha=-2\pi/365$,
$\theta=\arctan(2\pi/365k)=1.5563\,$rad, $D=80000$, $A=7400$; then
$y(t)=80000+(y_0-78925.5)e^{0.00025t}-1074.58\sin\left(
\frac{2\pi}{ 365}t+1.5563\right)$; (d) \$24136.

\noindent
8.~(a) $r(t)=12-at$, $V(t)=100-12t+\frac{1}{ 2}at^2$; (b) $\frac{dy}{
dt}=-\frac{r(t)}{ V(t)}y+l$.

\noindent
9.~(a) $x_0e^{-k_1t}$, (b) $dy/dt=-k_2y+k_1x_0e^{-k_1t}$.

\noindent
10.~(a) $x(t)=kV+(x_0-kV)e^{-rt/V}$, (b) $dy/dt=-\frac{r}{ W}y+\left(rk+
\left(\frac{x_0}{ V}-k\right)re^{-rt/V}\right)$.

%\end{answers}

\end{small}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Answers to Review Problem Sets}
\begin{small}

\centerline{\bf Answers to Review Problem Set \#1: First Midterm }
\medskip
\noindent
1.~$v(t)=-1.5+\int_0^t\sin(u/3)du=1.5-3\cos(t/3)$; velocity is at first
negative, then becomes positive after $v(t)$ crosses the $t$-axis, i.e.,
when: $1.5-3\cos(t/3)=0$; $\cos(t/3)=0.5$; $t/3=\pi/3$; $t=\pi$. So total
distance $=-\int_0^\pi(1.5-3\cos(t/3))dt+
\int_\pi^5(1.5-3\cos(t/3))dt=$
$=-(1.5t-9\sin(t/3))|_0^\pi+(1.5t-9\sin(t/3))|_\pi^5
=-1.5\pi+9(\sqrt{3}/2)+7.5-9\sin(5/3)-(1.5\pi-9(\sqrt{3}/2))=
-3\pi+9\sqrt{3}+7.5-9\cdot 0.9954=4.705$.

\noindent
2.
$$
\text{(a)}~\pi\int_0^{1/2}\cos^2(\pi x)-(1-2x)^2dx;\qquad\text{(b)}~
\overline{x}=0,~\overline{y}=
\frac{\int_0^4 1.5x^2dx+\int_4^5 2x\sqrt{25-x^2}dx }{
\int_0^4 1.5x\,dx+\int_4^5 2\sqrt{25-x^2}dx}.
$$  

\noindent
3.~(a)~$\frac{1}{ 5}\cdot\pi\int_{-3}^0x^2dy=\frac{\pi}{ 5}\int_{-3}^0
25(1-y^2/9)dy =5\pi(y-y^3/27)|_{-3}^0=10\pi$ sec.

\noindent
(b) $5/$(cross-sectional area)= $5/(\pi 5^2)=0.2/\pi$ units/sec; the
average rate is total rise divided by $T$, i.e., $3/10\pi=0.3/\pi$, which
is 1 1/2 times as great.

\bigskip
\noindent
4.~(a)~$r\int_0^\pi\sqrt{(1-0.5\cos\theta)^2+(0.5\sin\theta)^2}\,d\theta=
r\int_0^\pi\sqrt{1.25-\cos\theta}\,d\theta$.

\noindent
(b) $r\frac{\pi}{ 6}\Bigl(\frac{1}{ 2}\cdot
0.5+\sqrt{1.25-0.5\sqrt{3}}+\sqrt{1.25-0.5}+\sqrt{1.25}
+\sqrt{1.25+0.5} + \sqrt{1.25+0.5\sqrt{3}}$

\noindent
$+\frac{1}{ 2}\cdot 1.5\Bigr)=r\frac{\pi}{ 6}(0.25+0.6197+0.8660+1.1180+
1.3229 + 1.4547+0.75)=3.3412r.$

\bigskip

\centerline{\bf Answers to Review Problem Set \#2: First Midterm }

\medskip
\noindent
1.~$x(t)=(10\cos\alpha)t$, $y(t)=-4.9t^2-(10\sin\alpha)t+s_0$.  Since
$y(5)=0$ and $x(5)=25$, from the second equation we get $\cos\alpha=0.5$,
i.e., $\alpha=60^\circ$, and from the first equation we then get
$0=-4.9\cdot 5^2-(10\cdot 0.5\sqrt{3})5+s_0$, so that $s_0=165.8$ m.

\noindent
2.
$$
\text{(a)}~\pi\int_0^2\frac{1}{ 3-y}dy\qquad\qquad\text{(b)}~\int_0^{0.1}
5\sqrt{(1-\cos(12.5t))^2+\sin^2(12.5t)}\,dt.
$$

\smallskip
\noindent
3.~(a)~Taking $d/dt$ of $\frac{x^2}{ 25}+\frac{y^2}{ 16}=1$ gives
$y\,\dot{y}=-0.64x\,\dot{x}=-0.384$.  When $x=3$, we have $y=3.2$, so that
$\dot{y}=-0.384/3.2=-0.12$.  (b)~The position of the shadow is $x+y$, so
that its rate is $\dot{x}+\dot{y}=0.20-0.12=0.08$.  (c)~$\sqrt{
\dot{x}^2+\dot{y}^2}=\sqrt{0.20^2+0.12^2}=0.2332$.

\medskip
\noindent
4.~(a) $\int_0^\pi\sin(x)dx\Big/\int_0^\pi \frac{\sin x}{ x}dx$. Top integral
= 2.

\noindent
(b) Bottom integral $=\frac{\pi}{ 6}
\left(\frac{1}{ 2}\cdot 1+\frac{0.5}{ \pi/6}+
\frac{0.5\sqrt{3} }{ \pi/3}+\frac{1}{\pi/2}+\frac{0.5\sqrt{3} }{ 2\pi/3}
+\frac{0.5 }{ 5\pi/6}+0\right)=1.84465$.

\noindent
(c) $\overline{x}=1.0842$.

\bigskip

\centerline{\bf Answers to Review Problem Set \#3: First Midterm }
\medskip
\noindent
1.~$y=2\sin\left(\frac{\pi}{ 6}(x-1)\right)$, 
$\int_2^6\Bigl(2\sin\left(\frac{\pi}{ 6}(x-1)\right)-1\Bigr)dx$
$=12\sqrt{3}/\pi\>-\,4$ $=2.616$.

\noindent
2.~(a) $\int_0^1\sqrt{13.0^2+(-9.8t-7.5)^2}dt$.  

\noindent
(b)~$c=ea=6$, so $b=\sqrt{a^2-c^2}$ 
$=8,\>4\int_0^{\pi/2}\sqrt{100\sin^2\theta+64\cos^2\theta} \>d\theta$.  

\noindent
(c)~$\int_0^2\left(10-\frac{10}{ 0.1t+1}\right)dt$.

\noindent
3.~$\int_0^3\sqrt{\dot{x}(t)^2+\dot{y}^2}dt\approx$ 65.98 meters (in
practical terms 66 meters, since your data are given to only two places
of accuracy). 

\noindent
4.~(a) $x=3t$, $y=2-2t$. (b) $\sin\theta$ in $x$-direction, $\cos\theta$ in
$y$-direction. 

\noindent
(c) $\theta=\pi t/2$.
(d) $x=3t+\sin(\pi t/2)$, $y=2-2t+\cos(\pi t/2)$.  

\bigskip


\centerline{\bf Answers to Review Problem Set \#4: First Midterm }
\medskip
\noindent
1.~$\int_0^1\cos(\pi x/2)dx-\int_0^1(1-x)dx=(2/\pi)-(1/2)=0.13662$.

\noindent
2.~(a) $\int_0^{0.7}\sqrt{6.86^2+(-9.8t+6.86)^2}dt$.  (b)
$\pi\int_0^29(1-y^2\!/4)dy$.

\noindent
3.~$\overline{x}=\int_0^{\pi/2}x\sin x\,dx\Bigm/\int_0^{\pi/2}\sin x\,dx$,
bottom$=1$, 

\noindent
top$\approx\frac{\pi}{ 12}\Bigl(\frac{\pi}{ 12}\sin\frac{\pi}{ 12}+
\frac{\pi}{ 6}\sin\frac{\pi}{ 6}+\frac{\pi}{ 4}\sin\frac{\pi}{ 4}+\frac{\pi}{ 3}
\sin\frac{\pi}{ 3}+\frac{5\pi}{ 12}\sin\frac{5\pi}{ 12}+\frac{1}{ 2}\cdot
\frac{\pi}{ 2}
\Bigr)=1.00573$, so $\overline{x}=1.00573$.  (Note: Later, when we come to
integration by parts, we will be able to compute the top integral exactly,
and find that it is actually equal to 1.)

\noindent
4.~(a) linear distance car traveled $=3t^2+20t$, angular distance $\theta=
15t^2+100t$ radians.  (b) About $20^\circ$ beyond the top point
(clockwise).  (c) $dx/dt=(6t+20)(1-\cos(15t^2+100t))=44.661$ when $t=0.5$,
$dy/dt=(6t+20)\sin(15t^2+100t)=-7.734$ when $t=0.5$, so
speed$=\sqrt{44.661^2+7.734^2}=45.33$ m/sec.

\bigskip

\centerline{\bf Answers to Review Problem Set \#5: First Midterm }
\medskip

\noindent
1.~Note that $\overline{y}=\overline{x}$ because of symmetry; the bottom
integral in $\overline{x}$ is
$\int_0^1(0.5x+1)dx+\int_1^2(0.5x+1-(2x-2))dx= 5/4\>+\>3/4=2$; the top
integral in $\overline{x}$ is
$\int_0^1x(0.5x+1)dx+\int_1^2x(0.5x+1-(2x-2))dx=
\int_0^1(0.5x^2+x)dx+\int_1^2(-1.5x^2+3x)dx=5/3$, so that $\overline{x}=
(5/3)/2=5/6$.

\noindent
2.~(a) $c=\sqrt{a^2-b^2}=24$, foci at $(-24,0)$ and $(24,0)$.  (b)
$e=c/a=24/25=0.96$.  (c) $x^2\!/25^2+y^2\!/7^2=1$, so
$x^2=625(1-y^2\!/49)$, which is $625\cdot 33/49$ when $y=-4$; $\frac{dh}{
dt}=\frac{dV/dt}{ dV/dh}=\frac{0.7}{
\pi\cdot 625\cdot 33/49}=0.00052936$ m/sec.

\noindent
3.~(a) $x=6t-0.3\sin(15t)$, $y=0.4-0.3\cos(15t)$. (b) When $t=0.4$,
$x=2.4838$ and $y=0.1119$, which is below and slightly to the right of the
center, whose coordinates at that time are $(2.4,0.4)$.  (c)
$dx/dt=1.6792$, $dy/dt=-1.2573$, speed$=2.098$ m/sec directed $36.8^\circ$
below the horizontal.

\noindent
4.~(a) $v_{\text{horiz}}(t)=\frac{1}{2}t^2$, 
$v_{\text{vert}}(t)=\frac{6}{\pi}\sin(\pi
t/6)$.  (b) $\int_0^3\sqrt{\bigl(0.5t^2\bigr)^2+\bigl(\frac{6}{\pi}\sin(\pi
t/6)\bigr)^2 }dt$. 

\noindent
(c) $\frac{1}{ 2}\bigl(\frac{1}{ 2}f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+\frac{1}{ 2}
f(3)\bigr)$, where 

\noindent
$f(t)=\sqrt{0.25t^4+(36/\pi^2)\sin^2(\pi t/6)}$ (final answer $=6.007$).

\bigskip
 
\centerline{\bf Answers to Review Problem Set \#1: Second Midterm }
\medskip

\noindent
1.~If you take 2 as the hypotenuse of the triangle and $x$ as the leg
opposite $\theta$, then $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$,
$\sqrt{4-x^2}=2
\cos\theta$, and the limits of integration become $0$ to $\pi/6$.  The top
integral in $\overline{x}$ becomes: $\int_0^{\pi/6}8\sin^3\theta\,
d\theta=8\int_0^{\pi/6}\left(1-\cos^2\theta\right)\sin\theta\,d\theta=
8\int_1^{\sqrt{3}/2}(1-u^2)(-du)$ (where $u=\cos\theta$).  This leads to
the value $\frac{16}{ 3}-3\sqrt{3}=0.13718$ for the top integral.  The bottom
integral is: $\int_0^{\pi/6}4\sin^2\theta\,d\theta
=4\int_0^{\pi/6}\left(\frac{1-\cos(2\theta)}{ 2}\right)d\theta=
\bigl(2\theta-\sin(2\theta)\bigr)\Bigm|_0^{\pi/6}=
\frac{\pi}{ 3}-\frac{\sqrt{3}}{2}=0.18117.$ So
$\overline{x}=0.13718/0.18117=0.75718.$

\smallskip
\noindent
2.~Since $x^4/(4-x^2)=-x^2-4+16/(4-x^2)=-x^2-4-\frac{16}{ 4}\left(\frac{1}{
x-2}-\frac{1}{ x+2}\right)$, we have:

\noindent
$\text{Vol}$ $=\pi\int_0^1\left(-x^2-4-\frac{16}{ 4} \left(\frac{1}{
x-2}-\frac{1}{ x+2}\right)\right)dx$ $=\pi\left( -\frac{x^3}{
3}-4x-4\ln\left|\frac{x-2}{ x+2}\right|\right)\bigg|_0^1$
$=\pi\left(-\frac{13}{ 3}+4\ln 3\right)$ $=0.1920.$

\smallskip
\noindent
3.~Vol$=f(x)=\pi(e^{-x})^2x=\pi xe^{-2x}$.  Set
$0=f'(x)=\pi\left(-2xe^{-2x}+ e^{-2x}\right)=\pi e^{-2x}(-2x+1)$, so
$-2x+1=0$, i.e., $x=1/2$.  So the maximum volume is $\pi\cdot\frac{1}{
2}\cdot e^{-1}=\pi/2e=0.57786$.

\smallskip
\noindent
4.~(a) $(x_A,y_A)=(8t,40-4.9t^2)$, (b) $10-2t$, (c)
$\arctan\bigl((40-4.9t^2)/8t\bigr)-\arcsin\bigl((10-2t)/20\bigr)$, (d)
$$\frac{-8t(9.8t)-8(40-4.9t^2)}{(8t)^2+(40-4.9t^2)^2}+\frac{0.1}{
\sqrt{1-(0.5-0.1t)^2}}.$$

\bigskip
 
\centerline{\bf Answers to Review Problem Set \#2: Second Midterm }
\medskip
\noindent
1.~(a) $f(1)=0$ and $\lim_{x\rightarrow 0}f(x)=0$.  (b) $f'(x)=
\frac{\sqrt{x}}{ x}+\frac{\ln x}{ 2\sqrt{x}}=0$ gives $\ln x=-2$, so $x=1/e^2$
and the point is $(1/e^2,\>-2/e)$.  (c) The top integral is
$$\int_0^1x^{3/2}\ln x\,dx=\frac{2}{ 5}x^{5/2}\ln x\Bigm|_0^1-\int_0^1
\frac{2}{ 5}x^{5/2}\cdot\frac{1}{ x}\,dx=-\frac{2}{ 5}\int_0^1x^{3/2}dx=-\frac{4}{
25} .$$ Similarly, the bottom integral is $\int_0^1x^{1/2}\ln
x\,dx=-\frac{4}{ 9}$.  So $\overline{x}=9/25=0.36$.

\smallskip
\noindent
2.~(a) Write $1/(x^2+5x+4)=\frac{1}{ -1-(-4)}\left(\frac{1}{ x+1}-\frac{1}{
x+4}\right)$, so we get $\frac{1}{ 3}\ln\left(\frac{x+1}{ x+4}\right)
\Bigm|_0^\infty=\frac{1}{ 3}\ln 4=0.4621.$ (b) Write
$x^2+2x=(x+1)^2-1^2$; if you take $x+1$ for the hypotenuse of the triangle
and 1 for the leg adjacent to $\theta$, then you have $x+1=\sec\theta$,
$dx=\sec\theta\tan\theta\,d\theta$, $\sqrt{x^2+2x}=\tan\theta$, and the
limits of integration become 0 to $\pi/2$ (because the triangle has
$\theta=0$ when $x=0$ and has angle $\theta$ approaching $\pi/2$ as
$x\longrightarrow\infty$).  In the integral everything cancels, and you get
simply $\int_0^{\pi/2}d\theta=\pi/2=1.5708.$

\smallskip
\noindent
3. $\pi\int_0^{\pi/2}x\cos^2x\,dx=\pi\int_0^{\pi/2} x\frac{1+\cos(2x)}{
2}dx=\pi\left(\frac{\pi^2}{ 16}+\frac{1}{ 2}\int_0^{\pi/2} x\cos(2x)dx\right)=$

\noindent
$=\frac{\pi^3}{ 16}+\frac{\pi}{ 2}\left(\frac{1}{ 2}x\sin(2x)\Bigm|_0^{\pi/2}-
\frac{1}{ 2}\int_0^{\pi/2}\sin(2x)dx\right)=\frac{\pi^3}{ 16}-\frac{\pi}{ 4}=
1.1525.$

\smallskip
\noindent
4.~$y_A(t)=5e^{0.03t}$, $y_B(t)=15e^{-0.05t}$, so minimize
$f(t)=5e^{0.03t}+15e^{-0.05t}$.  Set
$0=f'(t)=0.15e^{0.03t}-0.75e^{-0.05t}$, so $0.15e^{0.03t}=0.75e^{-0.05t}$,
and $e^{0.03t}/e^{-0.05t}=0.75/0.15$, i.e., $e^{0.08t}=5$.  Take $\ln$ of
both sides and divide by $0.08$ to get $t=(\ln 5)/0.08=20.118$ hours.

\bigskip
 
\centerline{\bf Answers to Review Problem Set \#3: Second Midterm }
\medskip
\noindent
1.~(a) $\int_0^1(1+2y)\arctan(y)dy$ $=(y+y^2)\arctan(y)\Bigm|_0^1-
\int_0^1\frac{y+y^2}{ 1+y^2}dy$ 
$=\frac{\pi}{ 2}-\int_0^1\left(1+\frac{y}{ 1+y^2}-
\frac{1}{ 1+y^2}\right)dy$ $=\frac{\pi}{ 2}-\left(y+\frac{1}{ 2}\ln(1+y^2)
-\arctan\,y\right)\Bigm|_0^1$ $=\frac{3}{ 4}\pi-1-\frac{1}{ 2}\ln 2$.

\noindent
(b) Write $x-x^2=0.5^2-(x-0.5)^2$, take $0.5$ as the hypotenuse and $x-0.5$
as the leg opposite $\theta$; then $\int\frac{x\,dx}{\sqrt{x-x^2}}=
\int\frac{0.5(\sin\theta+1)0.5\cos\theta\,d\theta}{ 0.5\cos\theta}=
-0.5\cos\theta+0.5\theta+C=-\sqrt{x-x^2}+\frac{1}{ 2}\arcsin(2x-1)+C$.

\noindent
(c) $\int_\pi^{4\pi}\cos^3(t/4)dt=\int_\pi^{4\pi}\left(1-\sin^2(t/4)\right)
\cos(t/4)dt=\int_{\sqrt{2}/2}^0(1-u^2)(4\,du)$ (where $u=\sin(t/4)$); this
is computed to be $-5\sqrt{2}/3$.

\noindent
2.~$\ln y=\ln\left((\ln x)^{e^x}\right)=e^x\ln(\ln x)$, which leads to
$\frac{1 }{ y}\,\frac{dy}{ dx}=e^x\Bigl(\frac{1}{ x\ln x}+\ln(\ln x)\Bigr)$,
and then $\frac{dy}{ dx}=(\ln x)^{e^x}\left(\frac{1}{ x\ln x}+\ln(\ln
x)\right)$.

\noindent
3.~(a) $dv/dt=-kv$; (b) $v(t)=5e^{-kt}$; (c) $s(t)=\int_0^t5e^{-ku}du=
\frac{5}{ k}\left(1-e^{-kt}\right)$; (d) (1) $5e^{-0.8}=2.247$ m/sec, (2)
$50(1-e^{-0.8})=27.53$ m, (3) $s/r=137.67$ rad, (4) $v/r=11.23$ rad/sec;
(e) $$\frac{dL}{ dt}=\frac{dL}{ d\theta}\,\frac{d\theta}{ dt}=\left(\sqrt{
\left(\frac{dx}{ d\theta}\right)^2+\left(\frac{dy}{ d\theta}\right)^2}\right)
\cdot 11.23=11.23\sqrt{r^2(1-\cos\theta)^2+r^2\sin^2\theta}=1.247\>
\frac{\text{m}}{\text{sec}}$$ when $r=0.2$, $\theta=137.67$ rad.

\bigskip
 
\centerline{\bf Answers to Review Problem Set \#4: Second Midterm }
\medskip
\noindent
1.~(a) $\int_{10\sqrt{3}}^\infty\frac{1}{ 100}\frac{dx}{ 1+(x/10)^2}=\frac{1}{
10}\int_{\sqrt{3}}^\infty\frac{du}{ 1+u^2}=\frac{1}{ 10}\,\arctan
\,u\Bigm|_{\sqrt 3}^\infty=\frac{1}{ 10}\left(\frac{\pi}{ 2}-\frac{\pi}{ 3}\right)=
\pi/60$.

\noindent
(b) $\int(x+2)\left(\frac{1}{ x-2}-\frac{1}{ x-1}\right)dx=\int\frac{x+2}{
x-2}dx-
\int\frac{x+2}{ x-1}dx=\int\left(1+\frac{4}{ x-2}\right)dx-\int\left(1+\frac{3}{
x-1}\right)dx=4\ln|x-2|-3\ln|x-1|+C$.  (c) Set $u=2(x+1)$, so that:

\noindent
$\int_{-1/2}^{1/2}x\,e^{2(x+1)}dx=\frac{1}{ 2}\int_1^3 (\frac{1}{
2}u-1)e^udu=\frac{1}{ 4}\int_1^3ue^udu-\frac{1}{ 2}\int_1^3e^udu=\frac{1}{
4}\left(ue^u|_1^3-e^u|_1^3\right)-\frac{1}{ 2}e^u|_1^3=\frac{1}{ 2}e$.

\noindent
(d) $\int_0^\pi\sin^5(x/3)dx=\int_0^\pi\left(1-\cos^2(\frac{x}{ 3})\right)^2
\sin(\frac{x}{ 3})dx=\int_1^{1/2}(1-u^2)^2(-3\,du)$ (where $u=\cos(\frac{x}{
3})$).  This gives $3(-\frac{1}{ 5}u^5+\frac{2}{ 3}u^3-u)|_1^{1/2}=0.331$.

\noindent
2.~(a) $\arctan\Bigl((7-t-2\sin(2\pi t))\bigm/(3+1.4t-2\cos(2\pi t))\Bigr)$.
(b) Let $y(t)$ be the top function and $x(t)$ the bottom function in the
Arctan.  Then $\frac{d}{ dt}\arctan\left(\frac{y}{ x}\right)=\frac{1}{
1+(y/x)^2}\cdot \frac{x\dot{y}-y\dot{x}}{ x^2}=
\frac{x\dot{y}-y\dot{x} }{ x^2+y^2}$; multiply by $180/\pi$ to 
convert to degrees; the result is:
$$\frac{180}{\pi}\,\frac{(3+1.4t-2\cos(2\pi t))(-1-4\pi\cos(2\pi t))-
(7-t-2\sin(2\pi t))(1.4+4\pi\sin(2\pi t))}{ (3+1.4t-2\cos(2\pi
t))^2+(7-t-2\sin(2\pi t))^2}.$$

\noindent
3.~(a) $0=f'(x)=(1/2\sqrt{x})-(1/x)$ gives $x=4$, i.e., the point
$(4,0.6137)$.  (b) $f''(x)=-\frac{1}{ 4}x^{-3/2}+x^{-2}$, which is positive
if $x=4$, so that the point in part (a) is a minimum.  (c) $f''(x)=0$ when
$x=16$, and the point $(16,1.2274)$ is an inflection; the curve is concave
up for $x<16$ and concave down for $x>16$.  (d) $f(x)$ approaches $+\infty$
as $x\longrightarrow 0$ and as $x\longrightarrow\infty$.

\noindent
(e)

%-figure RD -----------------------------
\begin{psfigure}{4.0in}{1.3in}{figRD.ps}
%\putcrosshair{0,0}
\end{psfigure}
%------------------------------

 


 


\centerline{\bf Answers to Review Problem Set \#1: Final Exam } 
\medskip
\noindent
1.~(a) $0=f'(x)=\frac{1}{ x}-\frac{2.5\cdot 0.5}{ 1+(0.5x)^2}$, so that
$1/x=1.25/(1+0.25x^2)$, i.e., $1+0.25x^2-1.25x=0$, or $x^2-5x+4=0$, which
has roots $x=1$ and $x=4$; hence, the critical points are $(1,-1.159)$ and
$(4,-1.382)$; by the 2nd derivative test, the first is a max and the second
is a min; (b) $\ln x$; (c) $\ln x-(5\pi/4)=
\ln x-3.927$.  

\noindent
2.~(a) $\pi\int_0^{12}g(y)^2dy\approx 2\pi\bigl(\frac{1}{ 2}\cdot 5.4756+
6.9169+8.4100+9.9856+11.5600+13.1769+\frac{1}{ 2}\cdot 14.8225\bigr)=
378\>\text{cm}^3$; (b) $\frac{dV}{ dt}=\frac{dV}{
dh}\cdot\frac{dh}{ dt}=\pi(3.85)^2\cdot 0.05=2.33\>\text{cm}^3$/sec.

\noindent
3.~(a) $K\Delta x/(r^2+x^2)$, (i) $Kr\Delta x/(r^2+x^2)^{3/2}$, (ii) $K
x\Delta x/(r^2+x^2)^{3/2}$; (b) in the direction perpendicular to wire:
$\int_0^\infty\frac{Kr\,dx}{\left(r^2+x^2\right)^{3/2}}$, use inverse trig
substitution with $x=r\tan\theta$, $dx=r\sec^2\theta\,d\theta$,
$\sqrt{r^2+x^2}=r\sec\theta$, to transform the integral to
$\int_0^{\pi/2}\frac{Kr^2\sec^2\theta\,d\theta}{ r^3\sec^3\theta}=\frac{K}{ r}
\int_0^{\pi/2}\bigl(1/\sec\theta\bigr)d\theta=\frac{K}{ r}\int_0^{\pi/2}
\cos\theta\,d\theta=\frac{K}{ r}$;
in the direction parallel to wire: $\int_0^\infty
\frac{Kx\,dx}{\left(r^2+x^2\right)^{3/2}}$, use $u$-substitution
$u(x)=r^2+x^2$ to get:
$\int_{r^2}^\infty\frac{1}{
2}Ku^{-3/2}du=-Ku^{-1/2}\bigm|_{r^2}^\infty=\frac{K}{ r}$; (c) the horizontal
and vertical components of the force are the same ($K/r$ in both cases), so
the force points at a 45${}^\circ$ angle.

\noindent
4.~(a) $\frac{dv}{ dt}=g-0.07v$; (b) $v(t)=-140\left(1-e^{-0.07t}\right)$;
(c) 41.34 m/sec downward;

\noindent
(d) $\frac{dv}{ dt}=g-2.0v$, $v(t)=-4.9-36.44e^{-2t}$; (e) 1.75 sec.

\noindent
5.~(a) $2-0.2t$; (b) $10-0.1t^2$, empty when $t=10$ min; (c) $\frac{dy}{ dt}=
-\frac{2}{ 10-0.1t^2}y+0.56(1-0.1t)$.

\bigskip

\centerline{\bf Answers to Review Problem Set \#2: Final Exam }
\medskip
\noindent
1.~(a) $0=f'(x)=(-3x^2+1)e^{-1.5x^2}$, so $x=\pm 1/\sqrt{3}$, and the
critical points are $(0.57735,0.35018)$ and $(-0.57735,-0.35018)$; since
$f''(x)=(-3x(-3x^2+1)-6x)e^{-1.5x^2}=9x(x^2-1)e^{-1.5x^2}$, the second
derivative test shows that the first of these points is a max, the second
is a min; (b) inflections at $(0,0)$, $(1,0.22313)$ and $(-1,-0.22313)$;
(c) $f(x)$ rapidly approaches the $x$-axis; (d) (see the figure below).

\begin{centering}
%-figure RB -----------------------------
\begin{psfigure}{3in}{1.0in}{figRB.ps}
%\putcrosshair{0,0}
%\puttext{0,-1}{}
\end{psfigure}\\
%------------------------------
\end{centering}

\noindent
2.~(a) $\theta=\arctan(a/b)$; (b) $\frac{d\theta}{ dt}=\frac{180}{\pi}
(b\dot{a}-a\dot{b})/(a^2+b^2)$.

\noindent
3.~$\int_0^8\sqrt{\dot{x}^2+\dot{y}^2}dt\approx 2\Bigl(\frac{1}{ 2}
\sqrt{10.0^2+0^2}+\sqrt{9.8^2+8.0^2}+\sqrt{9.4^2+15.2^2}+
\sqrt{9.0^2+22.0^2}+\frac{1}{ 2}\sqrt{8.4^2+28.2^2}\Bigr)=148$ meters.

\noindent
4.~(a) $\frac{dy}{ dt}=-\frac{r}{ V}y$; (b) $0.01y_0=y_0e^{-(r/V)T}$, so
$T=-(V/r)\ln(0.01)=V\ln 100/r=4.605V/r$.

\smallskip
\noindent
5.~(a) $dy/dt=0.12y-1500$; (b) $y=12500-4500e^{0.12t}$; (c) 8.51 years; (d)
12770 rhinos; (e) if you buy an \$8000 car, no money down, 12\% interest on
the loan, making monthly payments of \$125, how long does it take you to
pay off the car loan, and how much do you end up having paid for the car in
all?

\smallskip
\noindent
6.~(a) $\frac{dv}{ dt}=-8-0.05(v+5t)$; (b) $\frac{dv}{ dt}=-0.05v-(8+0.25t)$.

\bigskip

\centerline{\bf Answers to Review Problem Set \#3: Final Exam }
\medskip
\noindent
1.~(a) $x=17.32t$, $y=-4.9t^2-10t+50$; (b) $\theta=\arctan\Bigl(\left(-4.9
t^2-10t+50\right)/17.32t\Bigr)$;

\noindent
(c) $\frac{d\theta}{ dt}=\Bigl(17.32t(-9.8t-10)-17.32(-4.9t^2-10t+50)\Bigr)/
\Bigl(300t^2+(-4.9t^2-10t+50)^2\Bigr)$; when $t=2$ this gives
$-0.9215$ rad/sec.

\noindent
2.~(a) $v(t)=55+\int_0^t\left(24/\sqrt{1-u^2}\right)du=24\arcsin(t)+55$;
(b) $v(\frac{1}{ 2})=67.57$ miles/hr;

\noindent
(c) $s(\frac{1}{ 2})-s(0)=
\int_0^{1/2}v(t)dt=\int_0^{1/2}24\arcsin(t)+55\>dt=$

\noindent
$=24\left(t\arcsin(t)\bigm|_0^{1/2}-\int_0^{1/2}t(1-t^2)^{-1/2}dt\right)
+55t\bigm|_0^{1/2}=$

\noindent
$=24\cdot\frac{1}{ 2}\cdot\frac{\pi}{
6}-24\left(-\sqrt{1-t^2}\bigm|_0^{1/2}\right) +\frac{1}{ 2}\cdot
55=2\pi+12\sqrt{3}+3.5=30.57$ miles.

\noindent
3.~(a) $y=(1/x)+\ln x$; $y'=(-1/x^2)+(1/x)=0$ gives the point $x=1$, $y=1$
(a min pt);

\noindent
$y''=(2/x^3)-(1/x^2)=0$ gives the point $x=2$, $y=0.5+\ln 2=1.193$;

\noindent
(b) approaches $+\infty$; (c) approaches $+\infty$; (e)
$\int_{1/2}^1\sqrt{1+\left(-(1/x^2)+(1/x)\right)^2}\,dx\approx$

\noindent
$\approx\frac{1}{ 8}\left(\frac{1}{ 2}f(\frac{1}{ 2})+f(\frac{5}{ 8})+f(\frac{3}{
4})+f(\frac{7}{ 8})+\frac{1}{ 2}f(1)\right)=\frac{1}{ 8}(1.1180+1.3862+1.0943+
1.0132+0.5)=0.6390$.

\noindent
4.~(a) $\frac{dy}{ dt}=-\frac{r}{ V}(y-kV)=-(0.6+0.4\sin(\frac{\pi t}{
10}))(y-5k)$; (b) separation of variables leads to
$y-5k=Ce^{-0.6t+(4/\pi)\cos(\pi t/10)}$, where $C=(y_0-5k)e^{-4/\pi}$,
i.e., $y(t)=5k+(y_0-5k)e^{-0.6t-(4/\pi)(1-
\cos(\pi t/10))}$;
(c) increases if $y_0<5k$, decreases if $y_0>5k$.

\noindent
5.~$dy/dt=0.025y+(1.0+0.1t)$.

\bigskip
 
\centerline{\bf Answers to Review Problem Set \#4: Final Exam }
\medskip
\noindent
1.~(a) $\theta=\arctan(\frac{5}{ y})$; (b) $\frac{d\theta}{ dt}=\frac{d\theta}{
dy}
\cdot\frac{dy}{ dt}=1/\left(1+\frac{25}{ y^2}\right)\cdot\frac{-5}{ y^2}\cdot 0.5
=-\frac{2.5}{ y^2+25}=-\frac{1}{ 260}$ rad/min $=-0.22$ deg/min.

\smallskip
\noindent
2.~$f'(x)=x^x(\ln x+1)$; critical point at $(1/e,0.692)$;
$f''(x)=x^x\left(\frac{1}{ x}+(\ln x+1)^2\right)$, which equals $0.692e$ at
the critical point --- so the critical point is a minimum.

\smallskip
\noindent
3.~$\Delta x\Bigl(\frac{1}{ 2}\sqrt{1+f'(0)^2}+\sqrt{1+f'(0.5)^2}+
\sqrt{1+f'(1)^2}+\sqrt{1+f'(1.5)^2}+\sqrt{1+f'(2)^2}+$

\noindent
$+\sqrt{1+f'(2.5)^2}+ \frac{1}{ 2}\sqrt{1+f'(3)^2}\Bigr)=\frac{1}{ 2}
(0.7071+1.3521+1.2440+1.1461+1.0808+$

\noindent
$+1.0218+0.5056)=3.529$.

\smallskip
\noindent
4.~position of shadow$=u=x+y=2\theta+2-\sin\theta-\cos\theta$;
$\frac{d\theta}{ dt}=\frac{1}{ r}\cdot\text{speed}=2.5$ rad/sec; $\frac{du}{
dt}=\frac{du}{ d\theta}\cdot\frac{d\theta}{ dt}=(2-\cos\theta+\sin
\theta)\cdot 2.5$; for $\theta=2\pi/3$ this gives 8.415 m/sec.

\smallskip
\noindent
5.~(a) $\frac{dy}{ dt}=ky-a$; (b) $y(t)=\frac{a}{ k}-(\frac{a}{ k}-y_0)e^{kt}$;
(c) $a=ky_0/(1-e^{-kT})$.  (d) Note that $ky_0=a(1-e^{-kT})$.  To find when
$ky(t)=\frac{1}{ 2}a$, we set $\frac{1}{ 2}a=ky=a-(a-ky_0)e^{kt}$.  This
gives $(a-ky_0)e^{kt}=a-\frac{1}{ 2}a=\frac{1}{ 2}a$.  Replacing $ky_0$
by $a(1-e^{-kT})$ and dividing through by $a$, we get
$(1-(1-e^{-kT}))e^{kt}=1/2$, i.e., $e^{k(t-T)}=1/2$.  Taking $\ln$ of
both sides and dividing through by $k$, we finally get $t=T-\frac{\ln 2}{ k}$.
In this problem, $T=20$ years and $k=10\%=0.1$, so that $t=20-10\ln 2
=13.07$ years.

\smallskip
\noindent
6.~$\frac{dy}{ dt}=-k_1y+I_0e^{k_2t}$.
7.~time units are months, so convert 12\% annual interest to 1\%
monthly interest; $\frac{dy}{ dt}=0.01y-\alpha(1-0.01t)$.

\end{small}


