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  • Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 47

Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 47

Answers (1)

Answer:

\frac{(\sin 2 t+2 a t \cos 2 t)}{(\cos 2 t-2 a t \sin 2 t)}

Hint:

You must know the derivative of cos and sin function

Given:

\begin{aligned} &x=a(\cos 2 t+2 t \sin 2 t) \\ &y=a(\sin 2 t-2 t \cos 2 t) \quad \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

        \begin{aligned} &x=a(\cos 2 t+2 t \sin 2 t) \\ &\frac{d x}{d t}=a(-2 \sin 2 t+2 \sin 2 t+4 t \cos 2 t) \\ &=4 a t \cos 2 t \\ &\frac{d^{2} x}{d t^{2}}=4 a \cos 2 t-8 a t \sin 2 t \end{aligned}

        \begin{aligned} &y=a(\sin 2 t-2 t \cos 2 t) \\ &\frac{d y}{d t}=a(2 \cos 2 t-2 \cos 2 t+4 t \sin 2 t) \\ &=4 a t \sin 2 t \\ &\frac{d^{2} y}{d t^{2}}=4 a \sin 2 t+8 a t \cos 2 t \end{aligned}

        \begin{aligned} &\text { So, } \frac{d^{2} y}{d x^{2}}=\frac{4 \sin 2 t+8 a t \cos 2 t}{4 \cos 2 t-8 a t \sin 2 t} \\ &=\frac{4(\sin 2 t+2 a t \cos 2 t)}{4(\cos 2 t-2 a t \sin 2 t)} \\ &=\frac{(\sin 2 t+2 a t \cos 2 t)}{(\cos 2 t-2 a t \sin 2 t)} \end{aligned}

Posted by

Gurleen Kaur

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