A) ½
B) 2/5
C) 3/2
D) 1/3
Correct Answer: D
Solution :
\[y={{\cos }^{-1}}\sqrt{1-{{t}^{2}}}={{\sin }^{-1}}t\] and \[\frac{du}{ds}=2x\,.\,\frac{dx}{ds}+2y\,.\,\frac{dy}{ds}\] \[\frac{{{d}^{2}}y}{d{{s}^{2}}}=0\] Þ \[y=2s-t\].You need to login to perform this action.
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