A) ?1
B) \[\frac{1-t}{1+{{t}^{2}}}\]
C) \[\frac{1}{1+{{t}^{2}}}\]
D) 1
Correct Answer: D
Solution :
Obviously \[x={{\cos }^{-1}}\frac{1}{\sqrt{1+{{t}^{2}}}}\]and \[y={{\sin }^{-1}}\frac{t}{\sqrt{1+{{t}^{2}}}}\] Þ \[\frac{dx}{dt}=-\frac{1}{\sqrt{\frac{{{t}^{2}}}{1+{{t}^{2}}}}}.\frac{(-1)}{2{{(1+{{t}^{2}})}^{3/2}}}2t=\frac{1}{1+{{t}^{2}}}\] and \[\frac{dy}{dt}=\sqrt{1+{{t}^{2}}}.\frac{1}{{{(\sqrt{1+{{t}^{2}}})}^{3/2}}}=\frac{1}{1+{{t}^{2}}}\] Hence \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=1\].
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