A failing in 3d gradients?

[bmrick]

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.

 

[FactChecker]

You have an example of a function that has partial derivatives on the x and y axes at the origin, but is not differentiable. There are fairly simple functions that have directional derivatives on any streight line through the origin, but do not have a Jacobian differential. (see http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions)

 

[HallsofIvy]

The gradient, whether in one or several variables, is a derivative, not a second derivative, like your "d(dz/dr)/d@".

 

[WWGD]

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.