Question

If x=a(cost+tsint) and y=a(sint-tcost),find \(\frac{d^2y}{dx^2}\)

Answer 1

It is given that x=a(cost+tsint) and y=a(sint-tcost)
∴\(\frac{dx}{dt}\)=a.\(\frac{d}{dt}\)(cost+sint)
=a[-sint+sint.\(\frac{d}{dt}\)(t)+t.\(\frac{d}{dt}\)(sint)]
=a[-sint+sint+tcost]=atcost
\(\frac{dy}{dt}\)=a.\(\frac{d}{dt}\)(sint-tcost)
=a[cost-{cost.\(\frac{d}{dt}\)(t)+t.\(\frac{d}{dt}\)(cost)}]
=a[cost-{cost-tsint}]=atsint

∴\(\frac{dy}{dx}\)=\(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)=\(\frac{atsint}{atcost}\)=tant

Then,\(\frac{d^2y}{dx^2}\)=\(\frac{d}{dx}\)(\(\frac{dy}{dx}\))=\(\frac{d}{dt}\)(tant)=sec²t.\(\frac{dt}{dx}\)
=sec²t.\(\frac{1}{atcos\,t}\)     [\(\frac{dx}{dt}\)=atcost\(\Rightarrow\)\(\frac{dt}{dx}\)=\(\frac{1}{atcos\,t}\)]
=\(\frac{sec^3t}{at}\), 0<t<\(\frac{\pi}{2}\)