Question

If $x=a\sin t-b\cos t,y=a\cos t+b\sin t$, show that $\frac{d^{2}y}{d{x}^2}=-\frac{x^2+y^2}{y^3}$.

Answer 1

$x=a\sin t-b\cos t;y=a\cos t+b\sin t$

$\frac{dx}{dt}=a\cos t+b\sin t;\frac{dy}{dt}=b\cos t-a\sin t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{b\cos t-a\sin t}{a\cos t+b\sin t}$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right).\frac{dt}{dx}$

$=\frac{d}{dt}\left(\frac{b\cos t-a\sin t}{a\cos t+b\sin t}\right).\frac{1}{a\cos t+b\sin t}$

$=\frac{(a\cos t+b\sin t)(-a\cos t-b\sin t)-(b\cos t-a\sin t)(b\cos t-a\sin t)}{(a\cos t+b\sin t{)}^3}$

$=\frac{-y^2-x^2}{y^3}$

$\therefore \frac{d^{2}y}{d{x}^2}=-\left(\frac{x^2+y^2}{y^3}\right)$