No time to check it thoroughly now, but 2 cooents:I uploaded 2 parts of the answer
No. The gradient of a function is a vector. Therefore you need the three unit vectors. Do't use i j k unless you're in an xyz cartesian coordinate system. Use the ones I gave you previously.Sorry that was one long run on sentence. I meant to say will it be ok if I take off I,j, and k?
If wikipedia listed it I'm sure it was right. But they must have had the unit vectors too?Ok is my formula ok? I'm gonna go look again but I believe Wikipedia had that listed I can add unit vectors that's no problem.
Never mind, I screwed you up & forgot you're now doing divergences, not gradients.Ugggg that is all Wikipedia lists I'm gonna go to my instructors web page and check there
Oh, OK. I think I mistook this for another of your threads whichI think did have div's in them.actually I screwed up the problem never said anything about divergence! That being said I corrected what I did and have uploaded it hopefully its ok!!
Stand by, I'll give it a shot, haven't done that myself yet ...ok how do I use wolfram for gradients?
Unfortunately, B and C are incorrect. They are tricky so work carefully. And check your results with wolfram.(A) y+2y-x
These are my answers
look at what I came up with for the gradient and improvise! Play around with it for a while. I pointed out that it gave the gradient in comma-separated chunks, omitting the unit vectors. Wolfram will have examples of how to enter a vector to get the divergence.div B=pz^2+psin^2(phi)+2pzsin^2(phi) I entered that in wolfram it didnt work what syntax should I use.