Question

If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find $\frac{d^{2}y}{d{x}^2}\mathrm{at}t=\frac{\mathrm{\pi }}{2}$.

Answer 1

Here,
$$\begin{gathered} x=2\cos t-\cos 2t\mathrm{and}y=2\sin t-\sin 2t \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.t\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=-2\sin t+2\sin 2t\mathrm{and}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=2\cos t-2\cos 2t \\ \therefore \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{2\cos t-2\cos 2t}{-2\sin t+2\sin 2t}=\frac{\cos t-\cos 2t}{-\sin t+\sin 2t} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}\mathit{y}}{\mathit{d}{\mathit{x}}^{\mathit{2}}}=\frac{\left(-\sin t+2\sin 2t\right)\left(-\sin t+\sin 2t\right)-\left(\cos t-\cos 2t\right)\left(-\cos t+2\cos 2t\right)}{{\left(-\sin t+\sin 2t\right)}^2}\frac{dt}{dx} \\ =\frac{\left(-\sin t+2\sin 2t\right)\left(-\sin t+\sin 2t\right)-\left(\cos t-\cos 2t\right)\left(-\cos t+2\cos 2t\right)}{{\left(-\sin t+\sin 2t\right)}^2\left(-2\sin t+2\sin 2t\right)} \\ \mathrm{At}t=\frac{\mathrm{\pi }}{2}: \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{\left(-1+0\right)\left(-1+0\right)-\left(0+1\right)\left(-0-2\right)}{{\left(-1+0\right)}^2\left(-2+0\right)}=\frac{1+2}{-2}=\frac{-3}{2} \end{gathered}$$