Consider a particle sliding down $f(x)=x^2$ under the influence of gravity and without friction. The particle always remains on the curve - that is, it does not have any 'air-time'.
If we start the particle from position $(a, a^2), a>0$ with no initial velocity, how can we work out the time it takes to slide without friction to the origin?
I figured that I could just use its horizontal movement as means to work out total time:
Its horizontal acceleration can be worked out to be $$a_{x}=\frac{g}{2}\sin{2\theta}$$ where $\theta$ is the angle that the tangent to the curve makes with the $x$-axis at the point $(x,x^2)$.
We can simplify this to be:
$$a_{x}=g\left(\frac{2x}{1+4x^2}\right)$$
Have I done all of this right? If so, where do I go from here??
I would love any help with this.
We have that by energy conservation
$$\frac12mv^2=mgh \implies v=\sqrt{2gh}=\sqrt{2g(a^2-x^2)}$$
then
$$ds=vdt \implies dt = \frac{ds}v \implies T=\int_0^Tdt = \int_0^a \frac{\sqrt{1+4x^2}}{\sqrt{2g(a^2-x^2)}}dx$$
For example for $a^2= 1\,m$ and assuming $g=9.8 \,m/s^2$ we obtain
$$T= \int_0^1 \frac{\sqrt{1+4x^2}}{\sqrt{2\cdot9.8(1-x^2)}}dx \approx 0.60\, s$$
As suggested in the comments by Travis Willse, we can express the result in a more elegant way by $\xi =\frac x a$
$$T =\frac1{\sqrt{2g}} \int_0^1 \frac{\sqrt{1+4a^2\xi^2}}{\sqrt{1-\xi^2}}d\xi=\frac{E(-4a^2)}{\sqrt{2g}}$$
where E is the complete elliptic integral of the second kind.
Here is a graph to compare the time taken to fall down from a given height $h$ along the parabola and directly in a vertical path $(T_0=\sqrt{2h/g})$