A) \[\sqrt{1-{{x}^{2}}}\,\,dx\,\,+\sqrt{1-{{y}^{2}}}\,\,dy=0\]
B) \[\sqrt{1-{{x}^{2}}}\,\,dy\,\,+\sqrt{1-{{y}^{2}}}\,\,dx=0\]
C) \[\sqrt{1-{{x}^{2}}}\,\,dy\,\,-\sqrt{1-{{y}^{2}}}\,\,dx=0\]
D) \[\sqrt{1-{{x}^{2}}}\,\,dx\,-\sqrt{1-{{y}^{2}}}\,\,dy=0\]
Correct Answer: B
Solution :
Given equation is \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\] .....(i) On differentiating w.r.t. to x, we get \[\frac{1}{\sqrt{1-{{x}^{2}}}}+\frac{1}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}=0\]Þ\[\frac{dy}{dx}=-\frac{\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] Þ \[\sqrt{1-{{x}^{2}}}\,\,dy+\sqrt{1-{{y}^{2}}}\,\,dx=0\].
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