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<$\frac{\pi}{2}$