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?
$ds = \sqrt{1+y'(x)^2}\, dx$ 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 $y(0)-y(x)$.
$\Delta h = y(0)-y(x)$