Question

If $x=\frac{a}{1+t^3},y=\frac{at}{1+t^3}$ then show that $\frac{dy}{dx}$ at $t=1$ is $\frac{1}{p}$ then $p=$

Answer 1

$\frac{dx}{dt}=a\frac{d}{dt}\left(\frac{1}{1+t^3}\right)$

$\frac{dx}{dt}=\frac{-3a{t}^2}{{(1+t^3)}^2}$

$y=\frac{at}{1+t^3}$

$\frac{dy}{dt}=a\frac{d}{dt}\left(\frac{t}{1+t^3}\right)$

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

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

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

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a\left(\frac{1-2t^3}{{(1+t^3)}^2}\right)}{\frac{-3a{t}^2}{{(1+t^3)}^2}}$

$\frac{dy}{dx}=\frac{1-2t^3}{-3t^2}$

At $t=1$

Given that, $\frac{dy}{dx}=\frac{1}{p}$

$\frac{1}{p}=\frac{1-2\times 1^3}{-3\times 1^2}$

$\frac{1}{p}=\frac{-1}{-3}=\frac{1}{3}$

$\therefore p=3$