Find the gradient vector field F→=∇f for each of the function f(x,y,z)=x2+y2+z2

F→&lt;∂f∂x,∂f∂y,∂f∂z ∂f∂x=∂∂xx2+y2+z2=12x2+y2+z2−12(2x)=xx2+y2+z2 ∂f∂y=∂∂yx2+y2+z2=12x2+y2+z2−12(2y)=yx2+y2+z2 ∂f∂z=∂∂zx2+y2+z2=12x2+y2+z2−12(2z)=zx2+y2+z2 Therefore, F→&lt;xx2+y2+z2,yx2+y2+z2,zx2+y2+z2&gt;

Get the answer to "Find the gradient vector field F→=∇f for each..."

glasskerfu

Answered question

2020-10-28

Find the gradient vector field

\vec{F} = \nabla f

for each of the function

f (x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}

Answer & Explanation

Corben Pittman

Skilled2020-10-29Added 83 answers

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

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