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
ANSWERED
asked 2020-10-28
Find the gradient vector field \(\displaystyle\vec{{F}}=\nabla{f}\) for each of the function \(\displaystyle{f{{\left({x},{y},{z}\right)}}}=\sqrt{{{x}^{{2}}+{y}^{{2}}+{z}^{{2}}}}\)

Answers (1)

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,
\(\displaystyle\vec{{F}}{<}\frac{{x}}{\sqrt{{{x}^{{2}}+{y}^{{2}}+{z}^{{2}}}}},\frac{{y}}{\sqrt{{{x}^{{2}}+{y}^{{2}}+{z}^{{2}}}}},\frac{{z}}{\sqrt{{{x}^{{2}}+{y}^{{2}}+{z}^{{2}}}}}{>}\)

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