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The time taken for a particle to descend a parametric curve under gravity $g$ is $$ \frac{1}{\sqrt{2g}}\int_0^{x_2}\sqrt{\frac{1+y'(x)^2}{y(x)}}\ \mathrm dx\tag{1} $$

I have the quadratic equation $y(x)=\frac{1}{9}\left(x-6\right)^{2}$, and I would like to find the time taken to move from $x=0$ to $x_2=6$, which is where the vertex of the parabola is. But when I plug the quadratic into equation 1, the integral diverges! This obviously can't be the case.

Am I missing something basic or going wrong anywhere?

Arctic Char
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SirCle
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  • At $x=6$, $y(x)=0$ – Prometheus Dec 01 '21 at 08:53
  • @Prometheus right. However the same thing happens for a linear function connecting (0,4) and (6,0), but the time integral doesn't diverge in that case. I'm wondering how this is supposed to work for a quadratic curve – SirCle Dec 01 '21 at 08:54
  • I found another source that states the denominator in the square root should be $x$, not $y(x)$: https://www.ucl.ac.uk/~ucahmto/latex_html/chapter2_latex2html/node7.html – David P Dec 01 '21 at 09:57

1 Answers1

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$ds = \sqrt{1+y'(x)^2}\, dx$ is the distance element.

$v =\sqrt{2\,g\,y(x)}$ is the velocity. It comes from the kinetic energy equaling the potential energy. $\frac{mv^2}{2} = mgh$

But $y(0)=\frac{36}{9} = 4$ so the initial velocity is not zero.

$y(6) = 0$ the velocity at the bottom is zero.This is incorrect. If the velocity is zero it has stopped moving so time will accumulate indefinitely.

Replace $y(x)$ in the denominator with $y(0)-y(x)$.

$\Delta h = y(0)-y(x)$