Question

If x = a cos θ, y = b sin θ, show that $\frac{d^{2}y}{d{x}^2}=-\frac{b^4}{a^2{y}^3}$.

Answer 1

Here,

$$\begin{gathered} x=a\cos \theta \mathrm{and}y=b\sin \theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{\theta },\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=-a\sin \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=b\cos \theta \\ \therefore \frac{\mathit{d}y}{\mathit{d}x}=\frac{b\cos \theta }{-a\sin \theta }=\frac{-b}{a}\cot \theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=-\frac{b}{a}\times \left(-{\mathrm{cosec}}^2\theta \right)\frac{\mathit{d}\theta }{\mathit{d}x} \\ =\frac{b}{a}\times {\mathrm{cosec}}^2\theta \times \frac{1}{-a\sin \theta } \\ =-\frac{b}{a^2}\times \frac{1}{{\sin }^3\theta } \\ =-\frac{b}{a^2}\times \frac{b^3}{y^3}\left[\because y=b\sin \theta \right] \\ =\frac{-b^4}{a^2{y}^3} \end{gathered}$$

Hence proved.