I am trying to find the time for different paths (functions) for a traversing ball between two points under the influence of gravity, but I am often getting strange answers where the integrand tends towards infinity, hence the time obtained is infinite, implying the ball never reaches the second point:
Time is equal to:
$T=\sqrt{\frac{1}{2g}}\int_0^1\sqrt{\frac{1+(y')^2}{y}}dx$
Where $y$ is the functions I am trying to test.
My two points are $(0,1)$ between $(1,0)$
For one of the functions, I am trying to use a circle:
$(x - h)^2 + (y - k)^2 = r^2$
To express $y$ as a function of $x$:
$y=k\pm \sqrt{r^2-(x-h)^2}$
And finally:
$y=1-\sqrt{1^2-(x-1)^2}$
So I input the above and its derivative into my time equation:
$T=\sqrt{\frac{1}{2g}}\int_0^1\sqrt{\frac{1+(\frac{x-1}{\sqrt{1-\left(x-1\right)^{2}}})^2}{1-\sqrt{1^2-(x-1)^2}}}dx$
I can do this for various functions giving the same result.
Trigonometric function:
$\sqrt{\frac{1}{2g}}\int_{0}^{1}\sqrt{\frac{1+\left(-\frac{\pi\sin\left(\pi x\right)}{2}\right)^{2}}{\frac{\cos\left(x\pi\right)}{2}+\frac{1}{2}}}dx$
Quadratic:
$\sqrt{\frac{1}{2g}}\int_{0}^{1}\sqrt{\frac{1+\left(2x-2\right)^{2}}{x^{2}-2x+1}}dx$
What could be the explanation for this? Why do they all give infinite as an answer, I've seen other papers that dive into the same idea (testing different functions between points), yet they don't mention similar issues?
I'd be really grateful for anyone who could help! Thank you in advance.
If the ball is released from rest at $y=1$, $x=0$ then your velocity $\sqrt{2gy}$ is not right. By energy conservation
$$mgy+\frac{1}{2}mv^2=mg$$ $$\implies v=\sqrt{2g(1-y)}$$
So the time is
$$T=\sqrt{\frac{1}{2g}}\int_0^1\sqrt{\frac{1+(y')^2}{1-y}}\,dx$$
The velocity $\sqrt{2gy}$ would correspond to a velocity of zero at $y=0$ i.e. the ball is at rest at the point $(1,0)$ but then the integral diverges because it is impossible for the ball to roll down and finish at rest at the lowest point.