Question

If x = a (1 − cos³ θ), y = a sin³ θ, prove that $\frac{d^{2}y}{d{x}^2}=\frac{32}{27a}\mathrm{at}\mathrm{\theta }=\frac{\mathrm{\pi }}{6}$.

Answer 1

Here,

$$\begin{gathered} x=a\left(1-{\cos }^3\theta \right),y=a{\sin }^3\theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\theta ,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=3a{\cos }^2\theta \sin \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=3a{\sin }^2\theta \cos \theta \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{3a{\sin }^2\theta \cos \theta }{3a{\cos }^2\theta \sin \theta }=\tan \theta \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}={\sec }^2\theta \frac{\mathit{d}\theta }{\mathit{d}x} \\ =\frac{{\sec }^2\theta }{3a{\cos }^2\theta \sin \theta } \\ =\frac{{\sec }^4\theta }{3a\sin \theta } \\ \therefore \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}\mathrm{at}\theta =\frac{\mathrm{\pi }}{6} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{{\left(\sec \frac{\mathrm{\pi }}{6}\right)}^4}{3a\sin \frac{\mathrm{\pi }}{6}}=\frac{32}{27a} \end{gathered}$$