- #1
Shackleford
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I looked at my notes, but they're either incomplete or I simply forgot what the professor did to derive the gradient in spherical coordinates. Once I know that, deriving the divergence and curl given the supplementary equations listed is fairly straightforward. It was a little easier but certainly lengthy to do this for polar coordinates. We started by using the known formula for gradient in Cartesian coordinates and then using the chain rule with the Cartesian-to-polar coordinates equations to derive the polar coordinates gradient formula. However, consider this would be greatly more involved considering the Cartesian-to-spherical coordinates equations, I'm thinking there was a short-cut used that I didn't put in my notes. Thanks for any help.