Toronto Math Forum
APM3462016F => APM346Tests => Qbonus => Topic started by: Shentao YANG on December 01, 2016, 08:46:16 PM

Problem: We need to construct the fastest slide from point $(0,0)$ to $(a,h)$, If $u(x)$ describes its shape then time is:
$$\begin{equation}
T= \int_0^a \frac{1}{\sqrt{2gu}} \sqrt{1+u^{\prime\,2}}\,dx.
\label{eq10.P.3}
\end{equation}$$
Find solution satisfying $u(0)=0$ and $u(a)=h$
Solution: Let $L = {1 \over {\sqrt u }}\sqrt {1 + {u^{\prime \,2}}} $, since the factor $2g$ does not matter. Note that here $L$ does not depend on $x$, so we have$$u'{{\partial L} \over {\partial u'}}  L = c$$, where $c$ is some constant. Then,
$$\eqalign{
& {1 \over {\sqrt {1 + {{(u')}^2}} }} = c\sqrt u \cr
& \Rightarrow {1 \over {1 + {{(u')}^2}}} = {c^2}u \cr
& \Rightarrow u' = {{du} \over {dx}} = \sqrt {{1 \over {{c^2}u}}  1} \cr
& \Rightarrow dx = {{du} \over {\sqrt {{1 \over {{c^2}u}}  1} }} \cr
& \Rightarrow x = \int {{1 \over {\sqrt {{1 \over {{c^2}u}}  1} }}du} \cr} $$
where the solution $u$ should satisfy $u(0)=0$ and $u(a)=h$

While I did not ask too calculate an integral on the Quiz, can you do this?
You can either write
$$x= \int \sqrt{\color{brown}{\frac{c^2 u}{1c^2 u}}}\,du$$
and make a standard substitution (colored expression $=t^2$) or simply use Mathematica, Maple or just WolframAlpha:
http://www.wolframalpha.com/input/?i=integrate+sqrt(c%5E2+u%2F(1c%5E2+u))+du (http://www.wolframalpha.com/input/?i=integrate+sqrt(c%5E2+u%2F(1c%5E2+u))+du)
Not that the formula is of any good, but taking $c=1$ f.e. (you can achieve it by $u:=c^2u$) you can
http://www.wolframalpha.com/input/?i=plot+(integrate+sqrt(1+u%2F(1+u))+du) (http://www.wolframalpha.com/input/?i=plot+(integrate+sqrt(1+u%2F(1+u))+du))
which results in the plot, $x$ is vertical (up), $u$ horizontal (to the right) and you are looking for the red line

The solution suggested is a perfect one. If we do not use the fact that $L$ does not depend explicitly on $x$ we end up with the second order equation which after transformations becomes ... I am too lazy to do it directly, so I do it from the fact that $u(1+u'^2)=C$: Differentiating we $u'(1+u'^2)+2 u u'u''=0$ or $2uu''+u'^2+1=0$. This is done only to show the correct 2nd order equation.