Question

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

• A. $\frac{dy}{dx}=\frac{t(2-t^3)}{1-2t^3}$
• B. $\frac{dy}{dx}=\frac{-t(2+t^3)}{1-2t^3}$
• C. $\frac{dy}{dx}=\frac{t(2+t^3)}{1+(2t{)}^3}$
• D. $\frac{dy}{dx}=\frac{t(2-t^3)}{1-(2t{)}^3}$

Answer 1

The correct option is A $\frac{dy}{dx}=\frac{t(2-t^3)}{1-2t^3}$

Given, $x=\frac{3at}{1+t^3}$ and $y=\frac{3a{t}^2}{1+t^3}.$

Clearly $y=tx.$ Differentiate w.r.t. $x$

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

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

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