Why Does the Gradient Behave Differently in Cylindrical Coordinates?

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Discussion Overview

The discussion revolves around the behavior of the gradient of a function expressed in cylindrical coordinates, specifically examining the function z = r cos(2θ). Participants explore the implications of expressing the gradient in both cylindrical and Cartesian coordinates, particularly at the origin.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the behavior of the gradient at the origin, noting that the partial derivatives dz/dx and dz/dy are both equal to one, leading to confusion when translating this back to cylindrical coordinates.
  • Another participant points out that the function may have partial derivatives at the origin but is not differentiable, referencing examples of functions with directional derivatives that lack a Jacobian differential.
  • A third participant clarifies that the gradient is a derivative, not a second derivative, challenging the notion of d(dz/dr)/dθ as a valid gradient expression.
  • The initial participant reiterates their confusion, suggesting that the continuity of the gradient is a necessary condition for their argument to hold.

Areas of Agreement / Disagreement

Participants express differing views on the differentiability of the function at the origin and the validity of certain gradient expressions. There is no consensus on the implications of the gradient behavior in this context.

Contextual Notes

Participants acknowledge the complexity of differentiability in higher dimensions and the conditions under which gradients behave as expected, but do not resolve the specific mathematical nuances involved.

bmrick
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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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