# Lagrangian Problem

by forty
Tags: lagrangian
 HW Helper P: 1,495 The arc length is given by $s(t)=\int_a^b \sqrt{1+f'(x)^2}dx$, notice the square. Therefore $ds=\sqrt{1+f'(x)^2}dx$. Secondly y=f(x), dy/dx=f'(x). Can you take it from here?