#### Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 15

$\frac{-1}{a}$

Hint:

\begin{aligned} &{{\text { }}\text { You must know about derivative of } \cos \theta\: \&\: \sin \theta}\\ \end{aligned}

Given:

\begin{aligned} &\text { If } x=a(1-\cos \theta), y=a(\theta+\sin \theta)\\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{-1}{a} \: and\: \theta=\frac{\pi}{4}\\ \end{aligned}

Solution:

\begin{aligned} &{{\text { Let }} x=a(1-\cos \theta), y=a(\theta+\sin \theta)} \end{aligned}

\begin{aligned} &\frac{d x}{d \theta}=a \sin \theta \\ &\frac{d y}{d \theta}=a(1+\cos \theta) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{1+\cos \theta}{\sin \theta} \end{aligned}

Use Quotient rule

\begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &\text { Where } u=1+\cos \theta \: \&\: v=\sin \theta \\ &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{\sin \theta(-\sin \theta)-(1+\cos \theta) \cos \theta}{\sin ^{2} \theta} \end{aligned}

\begin{aligned} &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{-1+\cos \theta}{(1-\cos \theta)(1+\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\frac{-1}{1-\cos \theta}}{a \sin ^{2} \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-1}{a \sin \theta(1-\cos \theta)} \quad\left(\theta=\frac{\pi}{2}\right) \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-1}{a \sin \frac{\pi}{2}\left(1-\cos \frac{\pi}{2}\right)} \\ \frac{d^{2} y}{d x^{2}} &=\frac{-1}{a} \end{aligned}

Hence proved