Question

If $x=a{\cos }^3\mathrm{\theta },y=a{\sin }^3\mathrm{\theta },\mathrm{then}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=$
(a) ${\tan }^2\mathrm{\theta }$
(b) ${\sec }^2\mathrm{\theta }$
(c) $\sec \mathrm{\theta }$
(d) $\left|\sec \mathrm{\theta }\right|$

Answer 1

(d) $\left|\sec \mathrm{\theta }\right|$

$$\begin{gathered} \mathrm{We}\mathrm{have},x=a{\cos }^3\theta \\ \Rightarrow \frac{dx}{d\theta }=a\frac{d}{d\theta }\left({\cos }^3\theta \right) \\ \Rightarrow \frac{dx}{d\theta }=3a{\cos }^2\theta \frac{d}{d\theta }\left(\cos \theta \right) \\ \Rightarrow \frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta ...\left(1\right) \\ \mathrm{and}, \\ y=a{\sin }^3\theta \\ \Rightarrow \frac{dy}{d\theta }=a\frac{d}{d\theta }\left({\sin }^3\theta \right) \\ \Rightarrow \frac{dy}{d\theta }=3a{\sin }^2\theta \frac{d}{d\theta }\left(\sin \theta \right) \\ \Rightarrow \frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta ...\left(2\right) \\ \mathrm{Dividing}\left(2\right)\mathrm{by}\left(1\right),\mathrm{we}\mathrm{get}, \\ \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta } \\ \Rightarrow \frac{dy}{dx}=\frac{\sin \theta }{-\cos \theta } \\ \Rightarrow \frac{dy}{dx}=-\tan \theta \\ \mathrm{Now},\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\sqrt{1+{\tan }^2\theta }=\sqrt{{\sec }^2\theta }=\left|\sec \theta \right| \end{gathered}$$