This is a continuation of this question, where I asked how to find the time taken for a particle to slide down $y=x^2$ from some point $(a,a^2)$, $a>0$, to the origin.
The answer was found to be:
$$T=\int_0^a \frac{\sqrt{1+4x^2}}{\sqrt{2g(a^2-x^2)}}dx$$($g=9.8$)
My new question is as follows:
What is the value of $n$ such that the time taken for the ball to slide down $y=x^n$ from $(1,1)$ to the origin is minimised?
Our new time would look like:
$$T=\int_0^1 \frac{\sqrt{1+(nx^{n-1})^2}}{\sqrt{2g(1-x^n)}}dx$$
We would need to evaluate this integral to get an equation in terms of $n$ and then differentiate (and set $=0$) to find minimum time, however I was not able to evaluate this integral even after using online calculators.
Note: After messing around a bit in desmos, I found that $n\approx2.5$ will minimise time taken, but I would love an exact form.
Any help with this question would be greatly appreciated!
$$T_n=\frac 1{\sqrt{2g}} \int_0^1 \frac{\sqrt{1+(n\,x^{n-1})^2}}{\sqrt{1-x^n}}\,dx$$
If $n \gt 2$ is an integer, you can have explicit solutions but they are given in terms of bery nasty generalized hypergeometric functions.
For example ${\sqrt{2g}}\,T_3$ write $$-\frac{18}{5} \, _7F_6\left(\frac{1}{2},\frac{3}{4},\frac{5}{6},1,\frac{7}{6},\frac {5}{4},\frac{3}{2};\frac{7}{8},\frac{9}{8},\frac{4}{3},\frac{11}{8 },\frac{13}{8},\frac{5}{3};-729\right)+\frac{\Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{7}{6}\right) \, _6F_5\left(-\frac{1}{6},\frac{1}{12},\frac{1}{6},\frac{1}{2},\frac {7}{12},\frac{5}{6};\frac{5}{24},\frac{11}{24},\frac{2}{3},\frac{1 7}{24},\frac{23}{24};-729\right)}{\sqrt{\pi }}+\frac{\sqrt{\pi } \Gamma \left(\frac{2}{3}\right) \, _6F_5\left(\frac{1}{6},\frac{5}{12},\frac{1}{2},\frac{5}{6},\frac{ 11}{12},\frac{7}{6};\frac{13}{24},\frac{19}{24},\frac{25}{24},\frac {31}{24},\frac{4}{3};-729\right)}{\Gamma \left(\frac{13}{6}\right)}$$
Do you like this monster (this is nice compared to the next ones) ?
I think that the easiest is (as you implicitely did) to generate a table and, if you want to reuse the results, curve fit them.