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\def\MAPLE{{\tt Maple}}
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\begin{document}
\begin{center} \large\sf Week 3 Examples\end{center}

{\bf Example 1}: Solve: (1) $y''=-9.8$, (2) $y'' = -0.04y' - 9.8$, both with $y(0)=0$, $y'(0)=49$. \newline
{\bf Answers}: (1) $y=-9.8t^2/2+ 49t$; (2) $y=7350-245t-7350 e^{-t/25}$.

{\bf Example 2}: Let $w=v\sqrt{\rho/g}$ and $p=\frac{1}{\sqrt{g\rho}}$ to replace Newton's quadratic drag model $v'=-g-\rho v|v|$ by $pw'=-1-w|w|$. Explain rise model $pw'=-1-w^2$ and fall model $pw'=-1+w^2$.

{\bf Example 3}: Solve $pw'=-w^2+1$ and $pw'=w^2+1$ as separable equations. See the previous example.
{\bf Answers}: $w(t)=\tanh(c_1+t/p)$ and $w(t)=\tan(c_2+t/p)$
% de:=p*diff(w(t),t)=-w(t)^2+1;dsolve(de,w(t));
% de:=p*diff(w(t),t)=w(t)^2+1;dsolve(de,w(t));

{\bf Example 4}: Verify rise time $4.6$ and fall time $4.8$ for
Newton's quadratic drag model $v'=-9.8-0.0011 v|v|$, $v(0)=49$.
Use textbook formulas or the previous two examples.

{\bf Example 5}: Find the point $r=r^*$ of zero acceleration in the Jules Verne equation $r'' = -\frac{Gm_1}{(R_1+r)^2} + \frac{Gm_2}{(R_3-r)^2}$. The answer has symbols. Then calculate $r^*\approx 339,620,820$ meters for the earth-moon problem. Reference:\newline
\href{http://www.math.utah.edu/~gustafso/s2017/2280/julesVerneDE2008.pdf}{http://www.math.utah.edu/\~{}gustafso/s2017/2280/julesVerneDE2008.pdf}
%G:=6.6726e-11:
%m1:=5.975e24:
%m2:=7.36e22:
%R1:=6.378e6:
%R2:=3.844e8:R3:=R2-R1;
%R4:=1.74e6: # radius of moon
%
%                                    8
%                          3.78022 10
%eq:=(R1+r)^2/(R3-r)^2=m1/m2;
%                                2
%                 /        6    \
%                 \6.378 10  + r/
%                ------------------ = 81.18206522
%                                 2
%                /          8    \
%                \3.78022 10  - r/
%solve(eq,r);
%                              8                8
%                3.396208200 10 , 4.260113591 10
\end{document} 
