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  • Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 14

Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 14

Answers (1)

Answer:

        \frac{-cosec\: x\frac{4\theta }{2}}{4a}

Hint:

You must know about derivative of

cos\: \theta \: and\: sin\: \theta

Given:

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

Solution:

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

        \begin{aligned} &\frac{d x}{d \theta}=a[1-\cos \theta] \; \; \; \; \; ......(1)\\ &\frac{d y}{d \theta}=a(-\sin \theta)\; \; \; \; \; ......(2) \\ &\text { Now } \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{\sin \theta}{(1-\cos \theta)} \end{aligned}

Using identity

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

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

Now diff on both sides

        \begin{aligned} &\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^{2} y}{d x^{2}}=-\operatorname{cosec}^{2} \theta \cdot \frac{1}{2} \bullet \frac{1}{a(1-\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\operatorname{cosec}^{2} \frac{\theta}{2} \bullet \frac{1}{2} \bullet \frac{1}{a 2 \sin ^{2} \frac{\theta}{2}} \end{aligned}

        \begin{aligned} &=\frac{-\cos e c^{2} \frac{\theta}{2} \times \cos e c^{2} \frac{\theta}{2}}{4 a} \\&\frac{d^{2} y}{d x^{2}}=\frac{\cos e c^{4} \frac{\theta}{2}}{4 a} \end{aligned}

Hence proved

Posted by

Gurleen Kaur

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