Answer
${\frac{ds}{dt}}={\frac{x}{\sqrt{(x^2+y^2)}}({\frac{dx}{dt}})}$
${\frac{ds}{dt}}={\frac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$
${\frac{dx}{dt}}={\frac{-y}{x}\frac{dy}{dt}}$
Work Step by Step
Given $s={\sqrt{(x^2+y^2)}}$a)
Taking y as constant and differentiating s, we get:
${\frac{ds}{dt}}={\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}2x{\frac{dx}{dt}}}$
${\frac{ds}{dt}}={\frac{x}{\sqrt{(x^2+y^2)}}({\frac{dx}{dt}})}$
Differentiating s for both variables x and y:
${\frac{ds}{dt}}={\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}(2x{\frac{dx}{dt}}+2y{\frac{dy}{dt}})}$
${\frac{ds}{dt}}={\frac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$
When s is constant then the above differentiation of s is zero:
$0={\frac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$
${\frac{1}{\sqrt{(x^2+y^2)}}{x\frac{dx}{dt}}}$ =-${\frac{1}{\sqrt{(x^2+y^2)}}{y\frac{dy}{dt}}}$
${\frac{dx}{dt}}={\frac{-y}{x}\frac{dy}{dt}}$
Thus the final answer is:
${\frac{ds}{dt}}={\frac{x}{\sqrt{(x^2+y^2)}}({\frac{dx}{dt}})}$
${\frac{ds}{dt}}={\frac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$
${\frac{dx}{dt}}={\frac{-y}{x}\frac{dy}{dt}}$
