Question

How can I derive the first equation to get the second equation?

Answer 1

Keeping in mind that "R" and "r" are constants whilst "s" is a variable and θ is a function in terms of "s", thus

$\bf s^2=R^2+r^2-2Rrcos(\theta )\implies 2s^1=0+0-2Rr\left[ \stackrel{chain~rule}{-sin(\theta )\cfrac{d\theta }{ds}} \right] \\\\\\ 2s=2Rrsin(\theta )\cfrac{d\theta }{ds}\implies 2s=\cfrac{2Rrsin(\theta )d\theta }{ds}\implies \cfrac{2s\cdot ds}{2Rr}=sin(\theta )d\theta \\\\\\ \cfrac{s\cdot ds}{Rr}=sin(\theta )d\theta$