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  • Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 46

Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 46

Answers (1)

Answer:

\frac{8\sqrt{3}}{a}

Hint:

You must know the derivative of cos, sin, tan and logarithm function

Given:

\begin{aligned} &x=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &y=a \sin t \quad \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{3} \end{aligned}

Solution:

        \begin{aligned} &x=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right] \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{\cos \frac{t}{2}}{2 \sin \frac{t}{2}} \frac{1}{\cos ^{2} \frac{t}{2}}\right] \end{aligned}

        \begin{aligned} &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right] \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{\sin t}\right] \\ &\frac{d x}{d t}=\frac{a \cos ^{2} t}{\sin t} \\ &y=a \sin t \end{aligned}

        \begin{aligned} &\frac{d y}{d t}=a \cos t \\ &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a \cos t}{a \cos ^{2} t} \times \sin t=\tan t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=\sec ^{2} t \times \frac{\sin t}{a \cos ^{2} t} \end{aligned}

        \begin{aligned} &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{3}}=\frac{\sec ^{2} \frac{\pi}{3} \cdot \sin \frac{\pi}{3}}{a \cos ^{2} \frac{\pi}{4}}=\frac{(2)^{2} \cdot\left(\frac{\sqrt{3}}{2}\right)}{a\left(\frac{1}{2}\right)^{2}} \\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{3}}=\frac{8 \sqrt{3}}{a} \end{aligned}

Posted by

Gurleen Kaur

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