Question

If x = cos θ, y = sin³ θ, prove that $y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=3{\sin }^2\mathrm{\theta }\left(5{\cos }^2\mathrm{\theta }-1\right)$.

Answer 1

Here,
$$\begin{gathered} x=\cos \theta \mathrm{and}y={\sin }^3\theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\theta ,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=-\sin \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=3{\sin }^2\theta \cos \theta \\ \therefore \frac{\mathit{d}y}{\mathit{d}x}=\frac{3{\sin }^2\theta \cos \theta }{-\sin \theta }=-3\sin \theta \cos \theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\left(-3{\cos }^2\theta +3{\sin }^2\theta \right)\frac{\mathit{d}\theta }{\mathit{d}x}=\frac{\left(-3{\cos }^2\theta +3{\sin }^2\theta \right)}{-\sin \theta } \\ \mathrm{Now}, \\ \mathrm{LHS}=y\frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}+{\left(\frac{\mathit{d}y}{\mathit{d}x}\right)}^2 \\ ={\sin }^3\theta \times \frac{\left(-3{\cos }^2\theta +3{\sin }^2\theta \right)}{-\sin \theta }+{\left(-3\sin \theta \cos \theta \right)}^2 \\ =3{\sin }^2\theta {\cos }^2\theta -3{\sin }^4\theta +9{\sin }^2\theta {\cos }^2\theta \\ =12{\sin }^2\theta {\cos }^2\theta -3{\sin }^4\theta \\ =3{\sin }^2\theta \left(4{\cos }^2\theta -{\sin }^2\theta \right) \\ =3{\sin }^2\theta \left(5{\cos }^2\theta -1\right)[\because {\cos }^2+{\sin }^2\theta =1] \\ =\mathrm{RHS} \end{gathered}$$

Hence proved.