If x = frac{3at}{1 + t^3} and y = frac{3at^2} | vedclass

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(B) Given $x = \frac{3at}{1 + t^3}$ and $y = \frac{3at^2}{1 + t^3}$.Observe that $y = tx$.Differentiating both sides with respect to $x$,we get:$\frac

Vedclass · Accepted Answer

$\frac{t(2 - t^3)}{1 - 2t^3}$

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(B) Given $x = \frac{3at}{1 + t^3}$ and $y = \frac{3at^2}{1 + t^3}$.Observe that $y = tx$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = t + x \frac{dt}{dx}$ ... $(i)$Now,differentiate $x$ with respect to $t$ using the quotient rule:$\frac{dx}{dt} = 3a \left[ \frac{(1 + t^3)(1) - t(3t^2)}{(1 + t^3)^2} \right] = \frac{3a(1 - 2t^3)}{(1 + t^3)^2}$.Thus,$\frac{dt}{dx} = \frac{(1 + t^3)^2}{3a(1 - 2t^3)}$.Substituting this into $(i)$:$\frac{dy}{dx} = t + \left( \frac{3at}{1 + t^3} \right) \left( \frac{(1 + t^3)^2}{3a(1 - 2t^3)} \right)$$\frac{dy}{dx} = t + \frac{t(1 + t^3)}{1 - 2t^3} = \frac{t(1 - 2t^3) + t + t^4}{1 - 2t^3} = \frac{t - 2t^4 + t + t^4}{1 - 2t^3} = \frac{2t - t^4}{1 - 2t^3} = \frac{t(2 - t^3)}{1 - 2t^3}$.

If $x = \frac{3at}{1 + t^3}$ and $y = \frac{3at^2}{1 + t^3}$,then $\frac{dy}{dx} = $

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