Recent content by Stantoine

I see the difference! I'm a little new here for thanks for the tip. I believe I was able to solve the problem with your help. It is very much appreciated! Thanks for the expedient responses and have a nice holiday season.

I did differentiate the LHS wrt θ and the RHS wrt phi. Come to think of it, that doesn't make sense at all! If I were to solve for \theta, I think I'd get: \theta=2arcsin(ksin(\phi)) then differentiating with respect to phi, \frac{d\theta}{d\phi}=2\frac{kcos(\phi)}{\sqrt{1-k^2sin(\phi)}} so...

Differenting both sides and then doing some algebra gives me d\theta=\frac{2sin(\alpha/2)cos(\phi)d\phi}{cos(\theta/2)} Also thank you for pointing out that my denominator can be rewritten as kcos(θ). That seems potentially helpful. If sin(\theta/2)=ksin(\phi), then...

Homework Statement given: dt=-\frac{1}{2}\sqrt{\frac{l}{g}}\frac{d\theta}{\sqrt{sin^2(\alpha/2)-sin^2(\theta/2)}} make the change of variables sin(\theta/2)=sin(\alpha/2)sin(\phi) to show that: dt=-\sqrt{\frac{l}{g}}\frac{d\phi}{\sqrt{1-k^2sin^2(\phi)}} where k=sin(\alpha/2) Homework...