If u = tan^{-1}(frac{y}{x}),then by Euler's T | vedclass

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

Question

(A) Given $u = 	an^{-1}(\frac{y}{x})$.We can rewrite this as $	an u = \frac{y}{x}$.Let $f(u) = 	an u = \frac{y}{x}$.Here,$f(u)$ is a homogeneous fu

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $u = 	an^{-1}(\frac{y}{x})$.We can rewrite this as $	an u = \frac{y}{x}$.Let $f(u) = 	an u = \frac{y}{x}$.Here,$f(u)$ is a homogeneous function of $x$ and $y$ with degree $n = 0$ because $\frac{ty}{tx} = (\frac{y}{x})^0 \cdot \frac{y}{x}$.According to Euler's Theorem for homogeneous functions,if $f(u)$ is a homogeneous function of degree $n$,then $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(u)$.Here,$f(u) = 	an u$ and $n = 0$.So,$x \frac{\partial (	an u)}{\partial x} + y \frac{\partial (	an u)}{\partial y} = 0 \cdot 	an u = 0$.Using the chain rule,$x \sec^2 u \frac{\partial u}{\partial x} + y \sec^2 u \frac{\partial u}{\partial y} = 0$.Dividing by $\sec^2 u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$.

If $u = \tan^{-1}(\frac{y}{x})$,then by Euler's Theorem,the value of $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is:

Similar Questions

If $z = \frac{(x^4 + y^4)^{1/3}}{(x^3 + y^3)^{1/4}}$,then $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = $

If $u = e^{-x^2 - y^2}$,then

If $z = \tan^{-1}\left(\frac{x}{y}\right)$,then $z_x : z_y = $

If $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$,then $x \frac{\partial z}{\partial x}$ is equal to

If $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$,then $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module