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jlmac2001
Jan27-04, 06:53 PM
Here's the question:

Suppose we have a scalar field phi(r)=xy defined on the plane (x,y). Find th vector field gradient phi, and plot the result in a single graph atthe four points (1,1), (-1,1) (1,-1), (-1,-1) and at the four points (1,0),(0,1),(-1,0) and (0,-1).

Ans.

For the gradient of phi(r)=xy, I got yi + xj. Is this right?

jlmac2001
Jan27-04, 06:59 PM
how will I plot it. Do i have to solve the eequation at each of those points and plot the answer that I get?

Tom Mattson
Jan27-04, 07:01 PM
Originally posted by jlmac2001
For the gradient of phi(r)=xy, I got yi + xj. Is this right? [/B]

Yes.

Tom Mattson
Jan27-04, 07:02 PM
Originally posted by jlmac2001
how will I plot it. Do i have to solve the eequation at each of those points and plot the answer that I get?

No, it's not that involved. For each of the 4 points, you plug in the values of x and y (edit: plug them into the gradient). From those, you will get 4 vectors, which you need to sketch on the graph.

jlmac2001
Jan27-04, 07:18 PM
Okay. So at (1,1) it would be i + j
at(-1, 1)= -i
at (1,-1) = i-j
at (-1,-1) = -i-j

How would I plot it?

Tom Mattson
Jan27-04, 08:07 PM
Originally posted by jlmac2001
Okay. So at (1,1) it would be i + j


Right.


at(-1, 1)= -i
at (1,-1) = i-j


These are wrong. Remember that for your vector v, grad(v)=yi+xj.


at (-1,-1) = -i-j


This one is OK.


How would I plot it?


Go to the point in question and let that be the origin of a vector whose components are what you calculated them to be. For instance, in the first case go to the point (1,1) and draw the vector i+j.

Also, don't forget to do the same for the other 4 points listed. You will have 8 vectors in all.