# Vector Derivatives

• Apr 4th 2010, 06:35 PM
AesTheBroken
Vector Derivatives
Given a vector field V(x,y,z) = (xi + yj + zk)/(r^3) where r = sqrt(x^2 + y^2 + z^2)
a) what is the x,y,z components of V(x,y,z)
b) partial derivatives of each component

do i substitute the r^3 with the sqrt thing or i just leave it as r^3 when i find the x,y,z components and derivatives?
• Apr 5th 2010, 08:23 AM
tonio
Quote:

Originally Posted by AesTheBroken
Given a vector field V(x,y,z) = (xi + yj + zk)/(r^3) where r = sqrt(x^2 + y^2 + z^2)
a) what is the x,y,z components of V(x,y,z)
b) partial derivatives of each component

do i substitute the r^3 with the sqrt thing or i just leave it as r^3 when i find the x,y,z components and derivatives?

I think you can safely leave as $r^3$: as long as it is clear what it is who cares?

Now, $r^3=\left(x^2+y^2+z^2\right)^{3\slash 2}\Longrightarrow \frac{d(r^3)}{dx}=3x\sqrt{x^2+y^2+z^2}$ , and the same with the other two variables but instead $3x$ we have $3y,\,3z$ , resp. , so:

$V(x,y,z)=\left(\frac{x}{r^3},\frac{y}{r^3},\frac{z }{r^3}\right)$ $\Longrightarrow \frac{dV}{dx}=\left(\frac{r^3-x\frac{d(r^3)}{dx}}{r^6},\frac{-y\frac{d(r^3)}{dy}}{r^6},\frac{-z\frac{d(r^3)}{dz}}{r^6}\right)$ .

Tonio
• Apr 8th 2010, 09:14 AM
AesTheBroken
if i wanna find the curl and divergence of v(x y z)? leave it as r too?