 glasskerfu

2020-10-28

Find the gradient vector field $\stackrel{\to }{F}=\mathrm{\nabla }f$ for each of the function $f\left(x,y,z\right)=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$ Corben Pittman

$\stackrel{\to }{F}<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}$
$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}=\frac{1}{2}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}^{-\frac{1}{2}}\left(2x\right)=\frac{x}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}$
$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}=\frac{1}{2}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}^{-\frac{1}{2}}\left(2y\right)=\frac{y}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}}$
$\frac{\partial f}{\partial z}=\frac{\partial }{\partial z}\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}=\frac{1}{2}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}^{-\frac{1}{2}}\left(2z\right)=\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}$
Therefore,
$\stackrel{\to }{F}<\frac{x}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}},\frac{y}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}},\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}>$

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