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}})}{(1-2{{t}^{3}})} $
• B. $ \frac{t(2+{{t}^{3}})}{(1-2{{t}^{3}})} $
• C. $ \frac{t(2-{{t}^{3}})}{(1+2{{t}^{3}})} $
• D. $ \frac{t(2+{{t}^{3}})}{(1+2{{t}^{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}}(3{{t}^{2}})}{{{(1+{{t}^{3}})}^{2}}} $
$ =\frac{6at-3a{{t}^{4}}}{{{(1+{{t}^{3}})}^{2}}} $
and $ \frac{dx}{dt}=\frac{(1+{{t}^{3}})\,(3a)-3at\,(3{{t}^{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(1-2{{t}^{3}})} $
$ =\frac{t(2-{{t}^{3}})}{(1-2{{t}^{3}})} $