#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 16 maths

$\frac{-1}{a}$

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

Hence proved