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  • Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 13 maths textbook solution

Answers (1)

Answer:

        \frac{-a}{y^{2}}

HInt:

You must know about derivative of second order

Given:

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

Solution:

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

        \begin{aligned} &\left.\frac{d x}{d \theta}=a[1+\cos \theta] \quad \text { ( } \frac{d}{d \theta} \theta=1, \frac{d}{d \theta} \sin \theta=\cos \theta\right) \\ &\frac{d y}{d \theta}=a(-\sin \theta) \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-\sin \theta}{(1+\cos \theta)} \end{aligned}

        \begin{aligned} &\sin 2 \theta=2 \sin \theta \cos \theta \\ &\cos 2 \theta=2 \cos ^{2} \theta-1 \\ &\frac{d y}{d x}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \end{aligned}

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

        \begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{1}{a(1+\cos \theta)} \\ &=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)} \\ &\frac{-a}{y^{2}}=\frac{-a}{a(1+\cos \theta)^{2}}=\frac{-a}{a^{2}(1+\cos \theta)(1+\cos \theta)} \end{aligned}

        \begin{aligned} &\frac{-1}{a\left(1+2 \cos ^{2} \frac{\theta}{2}-1\right)(1-\cos \theta)} \\ &=\frac{-1}{2 a \cos ^{2} \frac{\theta}{2}(1+\cos \theta)}=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)} \end{aligned}

        \begin{aligned} &R H L:-\frac{-a}{y^{2}}=\frac{-a}{a(1+\cos \theta)^{2}}=\frac{-1}{a\left(1+2 \cos ^{2} \frac{\theta}{2}-1\right)(1-\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-\sec ^{2} \frac{\theta}{2}}{2 a(1+\cos \theta)}=\frac{-a}{y^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-a}{y^{2}} \end{aligned}

Hence proved

Posted by

Gurleen Kaur

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