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

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

Answers (1)

Answer:

        \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

Posted by

Gurleen Kaur

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