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

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

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^{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

Posted by

Gurleen Kaur

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