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  • Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 36 maths

Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 36 maths

Answers (1)

Answer:

\frac{-3}{2}

Hint:

You must know the derivative of cos and sin function

\begin{aligned} &x=2 \cos t-\cos 2 t \\ \end{aligned}

Given:

\begin{aligned} &y=2 \sin t-\sin 2 t , \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{2} \end{aligned}

Solution:

        \begin{aligned} &x=2 \cos t-\cos 2 t \\ &y=2 \sin t-\sin 2 t \\ &\frac{d x}{d t}=-2 \sin t+2 \sin 2 t \\ &\frac{d y}{d t}=2 \cos t-2 \cos 2 t \end{aligned}

        \begin{aligned} &\text { Now, }\\ &\frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \Rightarrow \frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\\ &=\frac{2 \sin \frac{3 t}{2} \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \sin \frac{t}{2}} \Rightarrow \tan \frac{3 t}{2} \end{aligned}

        \begin{aligned} &\text { Therefore, }\\ &\frac{d^{2} y}{d x^{2}}=\sec ^{2} \frac{3 t}{2} \times \frac{3}{2} \times \frac{d t}{d x}\\ &=3 \sec ^{2} \frac{3 t}{2} \cdot \frac{1}{2 \sin 2 t-2 \sin t}\\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{2}}=\frac{-3}{2} \end{aligned}

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Gurleen Kaur

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