If u = sin^{-1} sqrt{frac{x^2 + y^2}{x + y}}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $f(u) = \sin u = \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $z = f(u) = \sqrt{\frac{x^2 + y^2}{x

Vedclass · Accepted Answer

$\frac{1}{2} 	an u$

Vedclass · Answer

(C) Given $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $f(u) = \sin u = \sqrt{\frac{x^2 + y^2}{x + y}}$.Let $z = f(u) = \sqrt{\frac{x^2 + y^2}{x + y}}$.We check the degree of homogeneity of $z$. Replacing $x$ with $tx$ and $y$ with $ty$:$z(tx, ty) = \sqrt{\frac{(tx)^2 + (ty)^2}{tx + ty}} = \sqrt{\frac{t^2(x^2 + y^2)}{t(x + y)}} = t^{1/2} \sqrt{\frac{x^2 + y^2}{x + y}} = t^{1/2} z$.Thus,$z$ is a homogeneous function of degree $n = 1/2$.By Euler's theorem for homogeneous functions,$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z$.Since $z = \sin u$,we have $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = \frac{1}{2} \sin u$.$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = \frac{1}{2} \sin u$.Dividing by $\cos u$,we get $x u_x + y u_y = \frac{1}{2} \frac{\sin u}{\cos u} = \frac{1}{2} 	an u$.

If $u = \sin^{-1} \sqrt{\frac{x^2 + y^2}{x + y}}$,then $x u_x + y u_y$ is

Similar Questions

If $u = (x^2 + y^2 + z^2)^{3/2}$,then $\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial u}{\partial z} \right)^2 = $

If $u(x, y)=y \log x+x \log y$,then $u_x u_y-u_x \log x-u_y \log y+\log x \log y$ is equal to :

If $u=\log \left(x^3+y^3+z^3-3 x y z\right)$,then $(x+y+z)(u_x+u_y+u_z)$ is equal to

If ${z^2} = \frac{{x^{1/2} + y^{1/2}}}{{x^{1/3} + y^{1/3}}}$,then $x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module