Question

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

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

Answer 1

Answer: (A)

Given, $y=\frac{3a{t}^2}{1+t^3},x=\frac{3at}{1+t^3}$
On differentiating w.r.t. t respectively, we get
$\frac{dy}{dt}=\frac{(1+t^3)(6at)-3a{t}^2(3t^2)}{{(1+t^3)}^2}$
$=\frac{6at-3a{t}^4}{{(1+t^3)}^2}$
and $\frac{dx}{dt}=\frac{(1+t^3)(3a)-3at(3t^2)}{{(1+t^3)}^2}$
$=\frac{3a-6a{t}^3}{{(1+t^3)}^2}$
$\therefore$ $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3at(2-t^3)}{3a(1-2t^3)}$
$=\frac{t(2-t^3)}{(1-2t^3)}$