Question

If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), prove that $\frac{d^{2}x}{d{\mathrm{\theta }}^2}=a\left(\cos \mathrm{\theta }-\mathrm{\theta }\sin \mathrm{\theta }\right),\frac{d^{2}y}{d{\mathrm{\theta }}^2}=a\left(\sin \mathrm{\theta }+\mathrm{\theta }\cos \theta \right)\mathrm{and}\frac{d^{2}y}{d{x}^2}=\frac{{\sec }^3\mathrm{\theta }}{a\mathrm{\theta }}.$

Answer 1

We have, $x=a\left(\cos \theta +\theta \sin \theta \right)$
$$\begin{gathered} \frac{dx}{d\theta }=\frac{d\left[a\left(\cos \theta +\theta \sin \theta \right)\right]}{d\theta } \\ \Rightarrow \frac{dx}{d\theta }=-a\sin \theta +a\sin \theta +a\theta \cos \theta =a\theta \cos \theta .....\left(\mathrm{i}\right) \\ \frac{d^{2}x}{d{\theta }^2}=\frac{d}{d\theta }\left(\frac{dx}{d\theta }\right)=\frac{d\left(a\theta \cos \theta \right)}{d\theta } \\ \Rightarrow \frac{d^{2}x}{d{\theta }^2}=a\cos \theta -a\theta \sin \theta =a\left(\cos \theta -\theta \sin \theta \right) \end{gathered}$$

$$\begin{gathered} y=a\left(\sin \theta -\theta \cos \theta \right) \\ \frac{dy}{d\theta }=\frac{d\left[a\left(\sin \theta -\theta \cos \theta \right)\right]}{d\theta } \\ \Rightarrow \frac{dy}{d\theta }=a\cos \theta -a\cos \theta +a\theta \sin \theta =a\theta \sin \theta .....\left(\mathrm{ii}\right) \\ \frac{d^{2}y}{d{\theta }^2}=\frac{d}{d\theta }\left(\frac{dy}{d\theta }\right)=\frac{d\left(a\theta \sin \theta \right)}{d\theta } \\ \Rightarrow \frac{d^{2}y}{d{\theta }^2}=a\sin \theta +a\theta \cos \theta =a\left(\sin \theta +\theta \cos \theta \right) \end{gathered}$$
From (i) and (ii), we have
$$\begin{gathered} \frac{dy}{dx}=\frac{{\displaystyle \raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$d\theta $}\right.}}{{\displaystyle \raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$d\theta $}\right.}}=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta \\ \frac{{\displaystyle d^{2}y}}{{\displaystyle d{x}^2}}=\frac{{\displaystyle d}}{{\displaystyle dx}}\left(\frac{{\displaystyle dy}}{{\displaystyle dx}}\right)=\frac{{\displaystyle d\left(\tan \theta \right)}}{{\displaystyle dx}}={\sec }^2\theta \frac{d\theta }{dx} \\ \Rightarrow \frac{{\displaystyle d^{2}y}}{{\displaystyle d{x}^2}}=\left({\sec }^2\theta \right)\left(\frac{1}{a\theta \cos \theta }\right) \\ \Rightarrow \frac{{\displaystyle d^{2}y}}{{\displaystyle d{x}^2}}=\frac{{\sec }^3\theta }{a\theta } \end{gathered}$$
Hence proved.