Question
If √(x2+y2)=a.etan−1(y/x) a>0,then y"(0) is equal to
Answer: Option C
:
C
√(x2+y2)=a.etan−1(y/x)
⇒12√x2+y2(2x+2yy′)=a.etan−1(y/x)×11+y2x2×xy′−yx2.....(i)
⇒x+yy′√x2+y2=√(x2+y2)×xy′−yx2+y2
[fromEq.(i)]
∴x+yy′=xy′−y⇒y′=x+yx−y
∴y"=2(xy′−y)(x−y)2
y"(0)=2(0−y(0)){0−y(0)}2=−2aeπ/2=−2ae−π/2
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:
C
√(x2+y2)=a.etan−1(y/x)
⇒12√x2+y2(2x+2yy′)=a.etan−1(y/x)×11+y2x2×xy′−yx2.....(i)
⇒x+yy′√x2+y2=√(x2+y2)×xy′−yx2+y2
[fromEq.(i)]
∴x+yy′=xy′−y⇒y′=x+yx−y
∴y"=2(xy′−y)(x−y)2
y"(0)=2(0−y(0)){0−y(0)}2=−2aeπ/2=−2ae−π/2
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