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Gradient of r

by aaaa202
Tags: gradient
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aaaa202
#1
Feb6-12, 04:43 PM
P: 1,005
Let r = √(x^2+y^2+z^2)
One can easily show that [itex]\nabla[/itex]r= [itex]\vec{r}[/itex]/r.
But I'm having a hard time understanding what this means geometrically - who can help? :)
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osnarf
#2
Feb6-12, 04:55 PM
P: 209
Not that this is a proof, but intuitively the direction of the gradient of a scalar function gives the direction of maximum change of the value of that function. Since your scalar function is the distance from the origin, and it makes sense that the direction of the gradient is in the direction of [itex]\vec{r}[itex] = [itex]r\vec{u_{r}}[itex].

Grad(r) = [itex]\frac{\vec{r}}{r} = \frac{r\vec{u_{r}}}{r} = \vec{u_{r}}[itex], the unit vector in the direction of the position vector. That makes sense, because the rate of change of r as you move in the direction of r is 1.


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