If u = tan^-1(x^4 + y^4/x^2 + y^2), prove that x ∂y/∂x + y ∂y/∂y = sin 2u.

Question

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.

Answer 1

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