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  • Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 21

Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 21

Answers (1)

Answer:               (c)

Hint:

We must know about the derivative of trigonometric function.

Given:

                x=f(t) \cos t-f^{\prime}(t) \sin t \\

                y=f(t) \sin t-f^{\prime}(t) \cos t \\

Explanation:

                x=f(t) \cos t-f^{\prime}(t) \sin t \\

                y=f(t) \sin t-f^{\prime}(t) \cos t \\

                \frac{d x}{d t}=f^{\prime}(t) \cos t-f(t) \sin t-f^{\prime \prime}(t) \sin t-f^{\prime}(t) \cos t \\

                \begin{aligned} & &\frac{d x}{d t}=-f(t) \sin t-f^{\prime \prime}(t) \sin t \end{aligned}

                \frac{d x}{d t}=-\sin t\left[f(t)+f^{\prime \prime}(t)\right] \\

                \frac{d y}{d t}=f^{\prime}(t) \sin t-f(t) \cos t-f^{\prime \prime}(t) \cos t-f^{\prime}(t) \sin t \\

                \begin{aligned} &\frac{d y}{d t}=-f(t) \cos t-f^{\prime \prime}(t) \cos t \end{aligned}

                \frac{d y}{d t}=-\cos t\left[f(t)+f^{\prime \prime}(t)\right]

                \\ \left(\frac{d x}{d t}\right)^{2}=\left[-\sin t\left[f(t)+f^{\prime \prime}(t)\right]\right]^{2} \\

                \begin{aligned} & &\left(\frac{d y}{d t}\right)^{2}=\left[-\cos t\left[f(t)+f^{\prime \prime}(t)\right]\right]^{2} \end{aligned}

                \left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=(\sin t)^{2}\left[f(t)+f^{\prime \prime}(t)\right]^{2}+(\cos t)^{2}\left[f(t)+f^{\prime \prime}(t)\right]^{2} \\

                \begin{aligned} & &\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=\left[f(t)+f^{\prime \prime}(t)\right]^{2}\left[(\sin t)^{2}+(\cos t)^{2}\right] \end{aligned}

                =\left[f(t)+f^{\prime \prime}(t)\right]^{2}                                                         \left[\because(\sin t)^{2}+(\cos t)^{2}=1\right]

Posted by

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