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  • Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 8

Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 8

Answers (1)

Answer:

            \frac{d y}{d x}=\frac{2 t}{1-t^{2}}

Hint:

            Use quotient rule

Given:

                       x=\frac{3 a t}{1+t^{2}} \\      ,    y=\frac{3 a t^{2}}{1+t^{2}} \\

Solution:

x=\frac{3 a t}{1+t^{2}} \\

\frac{d x}{d y}=\frac{d}{d t}\left(\frac{3 a t}{1+t^{2}}\right) \\

\begin{aligned} & &=\frac{\left(1+t^{2}\right) \frac{d(3 a t)}{d t}-3 a t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}                                                        [Using quotient rule]

=\frac{\left(1+t^{2}\right)(3 a)-(3 a t)(2 t)}{\left(1+t^{2}\right)^{2}} \\

\begin{aligned} & &=\frac{3 a\left(1+t^{2}\right)-6 a t^{2}}{\left(1+t^{2}\right)^{2}} \end{aligned}

=\frac{3 a+3 a t^{2}-6 a t^{2}}{\left(1+t^{2}\right)^{2}} \\

=\frac{3 a-3 a t^{2}}{\left(1+t^{2}\right)^{2}} \\

\begin{aligned} & &\frac{d x}{d t}=\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \end{aligned}                                                                                                           (1)

y=\frac{3 a t^{2}}{1+t^{2}} \\

\frac{d y}{d t}=\frac{d}{d t}\left(\frac{3 a t^{2}}{1+t^{2}}\right)

\begin{aligned} &\\ &=\frac{(1+t) \frac{d\left(3 a t^{2}\right)}{d t}-3 a t^{2} \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}                                                                                                                     [Using quotient rule]

=\frac{\left(1+t^{2}\right)\left(3 a \times \frac{d t^{2}}{d t}\right)-3 a t^{2}\left(\frac{d(1)}{d t}+\frac{d\left(t^{2}\right)}{d t}\right)}{\left(1+t^{2}\right)^{2}}

 

=\frac{(1+t) \cdot 3 a \cdot(2 t)-3 a t^{2}(0+2 t)}{\left(1+t^{2}\right)}                                                                     \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

=\frac{6 a t\left(1+t^{2}\right)-6 a t^{3}}{\left(1+t^{2}\right)^{2}} \\

=\frac{6 a t+6 a t^{3}-6 a t^{3}}{\left(1+t^{2}\right)^{2}} \\

\begin{aligned} & &\frac{d y}{d x}=\frac{6 a t}{(1+t)^{2}} \end{aligned}                                                                                                              (2)

Now

\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d t}}

Put the value of   \frac{dy}{dt}\:and\: \frac{dx}{dt}  from the equation (2) and (1)

In  \frac{dy}{dx}

\frac{d y}{d x}=\frac{\frac{6 a t}{\left(1+t^{2}\right)^{2}}}{\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}}

=\frac{6 a t}{3 a\left(1-t^{2}\right)}

\begin{aligned} & \\ &\frac{d y}{d x}=\frac{2 t}{1-t^{2}} \end{aligned}

Posted by

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