Recent content by Divh

Sorry, I've already done that but i posted only part of my solution here. Thank you for your help.

I think i have solution. What I did is I derivated x equation to get dx as function of t: dx=1-\cos t dt . Next I substituted this to my integral and I got \int_{0}^{a} \frac{\sqrt{1+{y'}^2}*(1-\cos t)}{\sqrt{2g\,y}}\,dt Is that right?

So i don't have to integrate y' to get y? Should i use one of the parametric form equations, or all i have to do is to use integrated y' and then integrate by substitution whole expression? That kinda confuses me a little bit.

y'=\frac{dy}{dx}=\frac{r*(t-\sin t)}{-r*(1-\cos t}=-\frac{\sin t}{1-\cos t} \int -\frac{\sin t}{1-\cos t}\,dt=-\ln (1-\cos t) Is that what you meant?

So for y should I substitute y equation or integrate dy/dx and substitute solution of that?

Homework Statement I have to calculate minimum travel time between two points. I already have cycloid equations in parametric form: x=r*(t-\sin t) y=r*(1-\cos t) Homework Equations For calculating time i want to use following formula: \int_{0}^{a} \frac{\sqrt{1+{y'}^2}}{\sqrt{2g\,y}}dx My...