A) \[\frac{1}{3}\]
B) \[\frac{2}{5}\]
C) \[\frac{3}{2}\]
D) \[\frac{2}{3}\]
E) \[0\]
Correct Answer: A
Solution :
\[x={{\sin }^{-1}}(3t-4{{t}^{3}})\] and \[y={{\cos }^{-1}}(\sqrt{1-{{t}^{2}}})\] Let \[t=\sin \theta \] \[\therefore \]\[x={{\sin }^{-1}}(3\sin \theta -4{{\sin }^{3}}\theta )\] \[={{\sin }^{-1}}(\sin 3\theta )=3{{\sin }^{-1}}t\] \[\therefore \] \[\frac{dx}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}}\] and \[y={{\cos }^{-1}}(\sqrt{1-{{\sin }^{2}}\theta })\] \[={{\cos }^{-1}}(\sqrt{{{\cos }^{2}}\theta })={{\sin }^{-1}}t\] \[\therefore \] \[\frac{dy}{dt}=\frac{1}{\sqrt{1-{{t}^{2}}}}\] \[\therefore \] \[\frac{dy}{dt}=\frac{dy/dt}{dx/dt}\] \[=\frac{1/\sqrt{1-{{t}^{2}}}}{3/\sqrt{1-{{t}^{2}}}}=\frac{1}{3}\]
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