Question

# Find the gradient vector field vec F = grad f for each of the function f(x,y,z)=sqrt(x^2+y^2+z^2)

Vectors
Find the gradient vector field $$\displaystyle\vec{{F}}=\nabla{f}$$ for each of the function $$\displaystyle{f{{\left({x},{y},{z}\right)}}}=\sqrt{{{x}^{{2}}+{y}^{{2}}+{z}^{{2}}}}$$

2020-10-29

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