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 September 25th, 2011, 07:26 PM #1 Member   Joined: Mar 2011 Posts: 49 Thanks: 1 Vector Calculus Divergence of a Vector Field Given a vector field that describes which direction that gives you the highest temperature loss in a 3 dimensional space, with the origin of the function being the heat source. The function of the vector field is $-20 e^{-r^2}(xi + yj + zk)$ (i, k and z indicating the vector direction in the x, y, z direction's, r is distance from heat source.) This vector field is again derived from a scalar field for the temprature of the space. I've tried to find the divergence of this vector field, just to gain some intuitionon on the relation between different operation's in vector calculus, etc.. may not even be sensible to do in this case, so tell me if I'm screwing it all up already from the beginning. Also the function isn't necesairly any real accurate temprature function, the point is just the vector calculus mechanics, so nevermind that. I guess taking the divergence of this vector field would tell me something like the "de-compression-rate of the heat-density" ? Anyway, here's how I did the actual calculation. First just noting that $r^2=x^2+y^2+z^2$ And since all the variables have the same value, they are "symmetric" or what to call it, so I only need to do the actual calculation of one of them, and add them up. So the divergence of the vector field with respect to x is, $\frac{\partial}{\partial x} \ -20x e^{-r^2}$ $(-20)(e^{-r^2} )+(-20x)(e^{-r^2})(-2x)$ using the product rule and the chain rule, the -2x comes from the derivative of $r^2=x^2+y^2+z^2$ (Adding the y and z terms as well),this simplifies to $(40)(e^{-r^2})(x^2+y^2+z^2)-20e^{-r^2}$ I get another $r^2$ at the term in the end.. I guess that should make sense.. $(40)(e^{-r^2})(r^2)-20e^{-r^2}$ So, finally.. is what I have done here sensible and/ or correct?
September 26th, 2011, 05:48 AM   #2
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Re: Vector Calculus Divergence of a Vector Field

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I don't think that this is a high school problem!

$\vec{F}(x,y,z)=-20e^{-x^2-y^2-z^2}\left<x,y,z\right=>=$,

$div(\vec{F})=\vec{\nabla}\cdot \vec{F}=\frac{\partial}{\partial x }\left[x\cdot(-20e^{-x^2-y^2-z^2})\right]+\frac{\partial}{\partial y}\left[y\cdot(-20e^{-x^2-y^2-z^2})\right]+\frac{\partial}{\partial z}\left[ z\cdot(-20e^{-x^2-y^2-z^2})\right]=0$

something which can also be seen in the plot of the field, with the arrows heading to the origin (0,0,0).

[attachment=0:2w9tsn53]vecfield.gif[/attachment:2w9tsn53]

It would be a good exercise to try to prove that the field is conservative, $curl(\vec{F})=\vec{\nabla}\times \vec{F}=\left<0,0,0\right=>=$
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Attached Images
 vecfield.gif (31.0 KB, 165 views)

 September 26th, 2011, 10:17 AM #3 Member   Joined: Mar 2011 Posts: 49 Thanks: 1 Re: Vector Calculus Divergence of a Vector Field I wasn't sure where to post this, is there any way to move it? The Curl equals zero makes intuitive sense, $\frac{\partial Fz}{\partial y} - \frac{\partial Fy}{\partial z}$ in the x direction, and since all these partial derivatives are equal in all directions, it all cancels out to zero, right? Did the divergence I did make any sense at all?

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