A) \[1/2\]
B) \[2/5\]
C) \[3/2\]
D) \[1/3\]
Correct Answer: D
Solution :
Given that \[x={{\sin }^{-1}}(3t-4{{t}^{3}})\] and \[y={{\cos }^{-1}}\sqrt{1-{{t}^{2}}}\] Put \[t=\sin \theta \], we get \[x={{\sin }^{-1}}(3\sin \theta -4{{\sin }^{3}}\theta )\] and \[y={{\cos }^{-1}}\sqrt{1-{{\sin }^{2}}\theta }\] \[\Rightarrow \] \[x={{\sin }^{-1}}(\sin 3\theta )\] and \[y={{\cos }^{-1}}(\cos \theta )\] \[\Rightarrow \] \[x=3\theta \]and\[y=\theta \] \[\Rightarrow \] \[x=3{{\sin }^{-1}}t\]and\[y={{\sin }^{-1}}t\] On differentiating both sides \[\text{w}\text{.r}\text{.t}.t,\] respectively we get \[\frac{dx}{dt}=\frac{3}{\sqrt{1-{{t}^{2}}}}\]and\[\frac{dy}{dt}=\frac{1}{\sqrt{1-{{t}^{2}}}}\]
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