A) \[\sqrt{1-{{x}^{2}}}dy+\sqrt{1-{{y}^{2}}}\,dx=0\]
B) \[\sqrt{1-{{x}^{2}}}dx+\sqrt{1-{{y}^{2}}}\,dy=0\]
C) \[\sqrt{1-{{x}^{2}}}dx-\sqrt{1-{{y}^{2}}}\,dy=0\]
D) \[\sqrt{1-{{x}^{2}}}dy-\sqrt{1-{{y}^{2}}}\,dx=0\]
Correct Answer: A
Solution :
Here we have \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\] Differentiating both side we get \[\frac{dx}{\sqrt{1-{{x}^{2}}}}+\frac{dy}{\sqrt{1-{{y}^{2}}}}=0\] \[=\frac{\sqrt{1-{{y}^{2}}}dx+\sqrt{1-{{x}^{2}}}dy}{\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}}=0\] \[\Rightarrow \] \[\sqrt{1-{{y}^{2}}}dx+\sqrt{1-{{x}^{2}}}\,dy=0\]
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