If $\mathrm{x}=\left(\mathrm{t}+\mathrm{sint}\right)$ and $\mathrm{y}=\mathrm{a}\left(1-\mathrm{cost}\right),$ then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=$

$$\begin{gathered} \mathrm{x}=\mathrm{a}(\mathrm{t}+\sin \mathrm{t}) \mathrm{y}=\mathrm{a}(1-\cos \mathrm{t}) \\ \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}(1+\cos \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \sin \mathrm{t} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin \mathrm{t}}{1+\cos \mathrm{t}} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\tan \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\sec }^2 \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.·\frac{1}{2} \frac{\mathrm{dt}}{\mathrm{dx}} \\ =\frac{1}{2{\cos }^2 {\displaystyle \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} ·\frac{1}{\mathrm{a}(1+\cos \mathrm{t})} \\ =\frac{1}{4\mathrm{a} {\cos }^4 {\displaystyle \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \end{gathered}$$