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Question
Find \[\frac{dy}{dx}\] , when \[x = \frac{3 at}{1 + t^2}, \text{ and } y = \frac{3 a t^2}{1 + t^2}\] ?
Solution
\[\text{ We have, } x = \frac{3at}{1 + t^2}\]
Differentiating with respect to t,
\[\frac{dx}{dt} = \left[ \frac{\left( 1 + t^2 \right)\frac{d}{dt}\left( 3at \right) - 3at\frac{d}{dt}\left( 1 + t^2 \right)}{\left( 1 + t^2 \right)^2} \right] ............\left[\text{using quotient rule }\right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{\left( 1 + t^2 \right)\left( 3a \right) - 3at\left( 2t \right)}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{3a + 3a t^2 - 6a t^2}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{3a - 3a t^2}{\left( 1 - t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \frac{3a\left( 1 - t^2 \right)}{\left( 1 + t^2 \right)^2} . . . \left( i \right)\]
\[\text{ and }, y = \frac{3a t^2}{1 + t^2}\]
Differentiating it with respect to t,
\[\frac{dx}{dt} = \left[ \frac{\left( 1 + t^2 \right)\frac{d}{dt}\left( 3a t^2 \right) - 3a t^2 \frac{d}{dt}\left( 1 + t^2 \right)}{\left( 1 + t^2 \right)^2} \right] ..............\left[\text{using quotient rule } \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{\left( 1 + t^2 \right)\left( 6at \right) - 3a t^2 \left( 2t \right)}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \left[ \frac{6at + 6a t^3 - 6a t^3}{\left( 1 + t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \frac{6at}{\left( 1 + t^2 \right)^2} . . . \left( ii \right)\]
\[\text{ Dividing equation} \left( ii \right) by \left( i \right), \]
\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6at}{\left( 1 + t^2 \right)^2} \times \frac{\left( 1 + t^2 \right)^2}{3a\left( 1 - t^2 \right)} = \frac{2t}{1 - t^2}\]
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