Why Does the Gradient Behave Differently in Cylindrical Coordinates?

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The discussion centers on the behavior of the gradient of a function expressed in cylindrical coordinates, specifically the function z = r cos(2θ). The user notes that at the origin, the partial derivatives dz/dx and dz/dy are both equal to one, leading to confusion about the gradient's implication in cylindrical coordinates. They highlight that while directional derivatives may exist, the function is not differentiable at the origin, which complicates the interpretation of the gradient. The conversation emphasizes that the continuity of the gradient is crucial for the expected behavior of partial derivatives to hold true. Overall, the key issue is the discrepancy between the gradient's behavior in different coordinate systems and the implications of differentiability.
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So my question is regarding the gradient of a function. Suppose we have a function expressed In cylindrical coordinates. Its expressed as z=rcos2@

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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The gradient, whether in one or several variables, is a derivative, not a second derivative, like your "d(dz/dr)/d@".
 
bmrick said:
So my question is regarding the gradient of a function. Suppose we have a function expressed In cylindrical coordinates. Its expressed as z=rcos2@

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.

If I understood correctly, this is true only when the gradient itself is continuous.
 

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