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) \[\frac{t(2-{{t}^{3}})}{1+2{{t}^{3}}}\]
Correct Answer: B
Solution :
\[x=\frac{3at}{1+{{t}^{3}}},\ \ \ y=\frac{3a{{t}^{2}}}{1+{{t}^{3}}}\]. Clearly, \[y=tx\]. Differentiate w.r.t. x, we get \[\frac{dy}{dx}=t.\ \ 1+x.\frac{dt}{dx}\] ?..(i) Now, \[\frac{dx}{dt}=3a.\frac{1+{{t}^{3}}-t.\,3{{t}^{2}}}{{{(1+{{t}^{3}})}^{2}}}=\frac{3a.(1-2{{t}^{3}})}{{{(1+{{t}^{3}})}^{2}}}\] .....(ii) \[\therefore \frac{dy}{dx}=t+\frac{3at}{1+{{t}^{3}}}.\,\,\frac{{{(1+{{t}^{3}})}^{2}}}{3a(1-2{{t}^{3}})}=\frac{t(2-{{t}^{3}})}{1-2{{t}^{3}}}\], {by (ii)}.
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