So my question is regarding the gradient of a function. Suppose we have a function expressed In cylindrical coordinates. Its expressed as z=rcos2@(adsbygoogle = window.adsbygoogle || []).push({});

I expressed the equation in cylindrical, but for the sake of my logic I'll now talk about it in cartesian. It appears that dz/dx and dz/dy at the origin are both equal to one, and so the gradient would imply (back into cylindrical coordinates) dz\Dr at the 45 degree angle is not what the graph implies, What am I missing here?

It seems to me that d(dz/ dr) \d@ is still valid as a gradient?

Isn't the argument of the gradient in cartesian that dz/dy should not change much over small changes in x, and yet this is not true in this example.

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# A failing in 3d gradients?

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