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jlmac2001
Feb3-04, 02:00 PM
Here's the problem:

Find the Laplacian of sqrt(x^-y^2) and ln(r^2).

Will i just take the gradient of each one of these twice?

Tom Mattson
Feb3-04, 02:10 PM
Originally posted by jlmac2001
Will i just take the gradient of each one of these twice?


No, the Laplacian is (in Cartesian coordinates):

∂2/∂x2+∂2/∂y2+∂2/∂z2

jlmac2001
Feb3-04, 06:36 PM
For sqrt(x^2-y^2), I got:

((x^2-y^2)^-1/2 + x^2(x^2-y^2)^-3/2)i + ((x^2-y^2)^-1/2 -y^2(x^2-y^2)^3/2)j

I'm not sure how to do the ln(r^2). Can someone help?

HallsofIvy
Feb3-04, 07:27 PM
ln(r^2)= ln(x^2+ y^2). Does that help?

jlmac2001
Feb3-04, 07:49 PM
First off, did I do the Sqrt(x^2-y^2) right? For ln(r^2), i'm not sure if I'm doing the derivative right. i got, -2/r^3 i