### Author Topic: Q-bonus  (Read 2626 times)

#### Shentao YANG

• Full Member
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• Karma: 0 ##### Q-bonus
« 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:
$$\begin{equation} T= \int_0^a \frac{1}{\sqrt{2gu}} \sqrt{1+u^{\prime\,2}}\,dx. \label{eq-10.P.3} \end{equation}$$
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 $u(a)=-h$

#### Victor Ivrii ##### Re: Q-bonus
« Reply #1 on: December 02, 2016, 03:17:16 AM »
While I did not ask too calculate an integral on the Quiz, can you do this?

You can either write
$$x= \int \sqrt{\color{brown}{\frac{c^2 u}{1-c^2 u}}}\,du$$
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(1-c%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)
which results in the plot, $x$ is vertical (up), $u$ horizontal (to the right)  and you are looking for the red line
« Last Edit: December 02, 2016, 03:27:56 AM by Victor Ivrii »

#### Victor Ivrii ##### Re: Q-bonus
« Reply #2 on: December 13, 2016, 08:11:21 AM »
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 2-nd order equation.