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If u = tan-1\((\frac{x^4+y^4}{x^2+y^2})\), prove that \(x\frac{∂y}{∂x}+y\frac{∂y}{∂y}\) = sin 2u. |
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Answer» u = tan-1\((\frac{x^4+y^4}{x^2+y^2})\) ⇒ tan u = \((\frac{x^4+y^4}{x^2+y^2})\).....(i) Partial differentiate (i) w. r. t x and y, we get sec2y = \(\frac{∂y}{∂x}=\frac{4x^3(x^2+y^2)-(x^4+y^4)2x}{(x^2+y^2)^2}\)\(=\frac{2x^5+4x^3y^2-2xy^4}{(x^2+y^2)^2}\) sec2y = \(\frac{∂y}{∂x}=\frac{4y^3(x^2+y^2)-(x^4+y^4)2y}{(x^2+y^2)^2}\)\(=\frac{2y^5+4y^3x^2-2x^4y}{(x^2+y^2)^2}\) sec2y \((x\frac{∂y}{∂x}+y\frac{∂y}{∂y})\)\(=\frac{2x^6+4x^4y^2-2x^2y^4+2y^6+4y^4x^2-2x^4y^2}{(x^2+y^2)^2}\) \(=\frac{2x^6+2x^4y^2+2x^2y^4+2y^6}{(x^2+y^2)^2}\) \(=\frac{2x^4(x^2+y^2)+2y^4(x^2+y^2)}{(x^2+y^2)^2}\)\(=2\frac{x^4+y^4}{x^2+y^2}\) = 2 tan u (by (i)) ⇒ \(x\frac{∂y}{∂x}+y\frac{∂y}{∂y}\) = \(\frac{2tan u}{sec^2y}\) \(=\frac{2sin u}{cos u}\times cos^2y\) = 2 sin u cos u = sin 2u Hence Proved |
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