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} $