Question

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

• A. $ax$
• B. $a^2{x}^2$
• C. $\frac{x}{a}$
• D. $\frac{x}{2a}$
• E. x $2a$

Answer 1

Answer: (C)

$\because$ $x=\frac{2at}{1+t^3}$ and $y=\frac{2a{t}^2}{{(1+t^3)}^2}$
$\Rightarrow$ $y=2a{\left(\frac{t}{1+t^3}\right)}^2=2a.\frac{x^2}{{(2a)}^2}$
$\Rightarrow$ $y=\frac{x^2}{2a}$
On differentiating w.r.t. $x,$ we get
$\frac{dy}{dx}=\frac{2x}{2a}=\frac{x}{a}$