Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x=a\sin t-b\cos t,y=a\cos t+b\sin t \\ \mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \frac{dx}{dt}=\frac{d}{dt}\left(a\sin t-b\cos t\right)=a\cos t+b\sin t \\ \mathrm{and} \\ \frac{dy}{dt}=\frac{d}{dt}\left(a\cos t+b\sin t\right)=-a\sin t+b\cos t \\ \mathrm{Now},\left(\frac{dy}{dx}\right)=\frac{{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}}}{{\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}}}=\frac{-a\sin t+b\cos t}{{\displaystyle a\cos t+b\sin t}} \\ \mathrm{Therefore}, \\ \frac{d^{2}y}{d{x}^2}=\frac{\mathrm{d}}{dx}\left(\frac{dy}{dx}\right)=\frac{\mathrm{d}}{dx}\left(\frac{-a\sin t+b\cos t}{{\displaystyle a\cos t+b\sin t}}\right) \\ =\frac{\mathrm{d}}{dt}\left(\frac{-a\sin t+b\cos t}{{\displaystyle a\cos t+b\sin t}}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x} \\ =\frac{\left(a\cos t+b\sin t\right){\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\left(-a\sin t+b\cos t\right)-\left(-a\sin t+b\cos t\right){\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\left(a\cos t+b\sin t\right)}{{\left(a\cos t+b\sin t\right)}^2}\times \frac{1}{a\cos t+b\sin t} \\ =\frac{\left(a\cos t+b\sin t\right)\left(-a\cos t-b\sin t\right)-\left(-a\sin t+b\cos t\right)\left(-a\sin t+b\cos t\right)}{{\left(a\cos t+b\sin t\right)}^3} \\ =\frac{-{\left(a\cos t+b\sin t\right)}^2-{\left(-a\sin t+b\cos t\right)}^2}{{\left(a\cos t+b\sin t\right)}^3} \\ =\frac{-{\left(a\cos t+b\sin t\right)}^2-{\left(a\sin t-b\cos t\right)}^2}{{\left(a\cos t+b\sin t\right)}^3} \\ =\frac{-y^2-x^2}{y^3} \\ \mathrm{Hence},\frac{d^{2}y}{d{x}^2}=-\frac{x^2+y^2}{y^3}. \end{gathered}$$