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If $u=\large\frac{1}{\sqrt{x^{2}+y^{2}}},$ then $x\large\frac{\partial u}{\partial x}$$+y\large\frac{\partial u}{\partial y}$ is equal to

\[\begin{array}{1 1}(1)\frac{1}{2} u&(2)u\\(3)\frac{3}{2} u&(4)-u\end{array}\]

1 Answer

$u= \large\frac{1}{\sqrt {x^2+y^2}}$
Replace x by tx and by ty
$u =\large\frac{1}{\sqrt {(tx)^2+(ty)^2}}$
$\quad=\large\frac{1}{\sqrt {t^2x^2+t^2y^2}}$
$\quad= \large\frac{1}{\sqrt {t^2(x^2+y^2)}}$
$\quad= \large\frac{1}{t} \large\frac {1}{\sqrt {x^2+y^2}}$
$u= \large\frac{t^{-1}}{\sqrt {x^2+y^2}}$
$u$ is a homogeneous function of degree $=-1$
$x \large\frac{\partial u}{\partial x}$$+y \large\frac{\partial u}{\partial y}=$$(-1) u=-u$
Hence 4 is the correct answer.
answered May 20, 2014 by meena.p
 

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