Question

If $x=3cot-2{\cos }^{3}t,y=3\sin t-2{\sin }^{3}t,$ find $\frac{d^{2}y}{d{x}^2}.$

Answer 1

We have,

$$\begin{gathered} x=3\cos t-2{\cos }^{3}t \\ \Rightarrow \frac{dx}{dt}=3\left(-\sin t\right)-6{\cos }^{2}t\left(-\sin t\right) \\ =-3\sin t+6\sin t{\cos }^{2}t \end{gathered}$$

Also,

$$\begin{gathered} y=3\sin t-2{\sin }^{3}t \\ \Rightarrow \frac{dy}{dt}=3\cos t-6{\sin }^{2}t\cos t \end{gathered}$$

Now,

$$\begin{gathered} \frac{dy}{dx}=\frac{\left({\displaystyle \frac{dy}{dt}}\right)}{\left({\displaystyle \frac{dx}{dt}}\right)} \\ =\frac{3\cos t-6{\sin }^{2}t\cos t}{-3\sin t+6\sin t{\cos }^{2}t} \\ =\frac{3\cos t\left(1-2{\sin }^{2}t\right)}{3\sin t\left(-1+2{\cos }^{2}t\right)} \\ =\frac{\cot t\left(\cos 2t\right)}{\left(\cos 2t\right)} \\ =\cot t \end{gathered}$$

So,

$$\begin{gathered} \mathrm{So},\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right) \\ =\frac{d}{dx}\left(\cot t\right) \\ =-{\mathrm{cosec}}^{2}t\frac{dt}{dx} \\ =\frac{-{\mathrm{cosec}}^{2}t}{\left({\displaystyle \frac{dx}{dt}}\right)} \\ =\frac{-{\mathrm{cosec}}^{2}t}{-3\sin t+6\sin t{\cos }^{2}t} \\ =\frac{-{\mathrm{cosec}}^{2}t}{-\sin t\left(1-2{\cos }^{2}t\right)} \\ =\frac{{\mathrm{cosec}}^{3}t}{\left(-\cos 2t\right)} \\ =\frac{-{\mathrm{cosec}}^{3}t}{\cos 2t} \end{gathered}$$