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  • Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 49 maths textbook solution

Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 49 maths textbook solution

Answers (1)

Answer:

\frac{-(x^{2}+y^{2})}{y^{3}}

Hint:

You must know the derivative of cos t and sin t function

Given:

\begin{aligned} &x=a \sin t-b \cos t \\ &y=a \cos t-b \sin t \quad \text { Prove } \frac{d^{2} y}{d x^{2}}=\frac{-\left(x^{2}+y^{2}\right)}{y^{3}} \end{aligned}

Solution:

        \begin{aligned} &\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a \sin t+b \cos t}{a \cos t+b \sin t}\\ &\text { Use quotient rule }\\ &\frac{U}{V}=\frac{U V^{\prime}-U^{\prime} V}{V^{2}} \end{aligned}

        \begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d\left(\frac{d y}{d x}\right)}{d t}}{\frac{d x}{d t}}\\ &=\frac{(-a \cos t-b \sin t)(a \cos t+b \sin t)-(a \sin t+b \cos t)(-a \sin t+b \cos t)}{(a \cos t+b \sin t)^{2}(a \cos t+b \sin t)}\\ &=\frac{(-a \cos t+b \sin t)^{2}-(b \cos t-a \sin t)^{2}}{y^{3}}\\ &=\frac{-y^{2}-x^{2}}{y^{3}}=\frac{-\left(x^{2}+y^{2}\right)}{y^{3}} \end{aligned}

Posted by

Gurleen Kaur

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