Understanding Partial Differentiation with Polar Coordinates

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Homework Help Overview

The discussion revolves around the application of partial differentiation in the context of polar coordinates, specifically focusing on the function z = f(x,y) where x and y are expressed in terms of polar coordinates r and θ. Participants are tasked with finding the first and second partial derivatives of z with respect to r and θ.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to compute the first partial derivatives of z with respect to r and θ using the chain rule. They question whether additional considerations are needed for dz/dx and dz/dy. For the second derivative, they express uncertainty about the correct approach.

Discussion Status

Some participants affirm the original poster's calculations for the first derivatives, while others provide guidance on how to approach the second derivative. There is an acknowledgment of the behavior of mixed partial derivatives, suggesting a deeper exploration of the topic.

Contextual Notes

Participants note that the function z depends explicitly on two variables, x and y, which may influence the differentiation process. There is mention of the potential for mixed partial derivatives to differ in certain functions, indicating a consideration of function behavior in advanced contexts.

tandoorichicken
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If z = f(x,y), where x = rcos([itex]\theta[/itex]) and y = rsin([itex]\theta[/itex]), find [itex]\frac{\partial z}{\partial r}[/itex], [itex]\frac{\partial z}{\partial\theta}[/itex], and [itex]\frac{\partial^2 z}{\partial r\partial\theta}[/itex]

Here's what I've done:
(a)
[tex]\frac{\partial z}{\partial r} = \frac{dz}{dx} \frac{\partial x}{\partial r} + \frac{dz}{dy} \frac{\partial y}{\partial r} = \frac{dz}{dx} \cos{\theta} + \frac{dz}{dy} \sin{\theta}[/tex]
(b)
[tex]\frac{\partial z}{\partial\theta} = \frac{dz}{dx} \frac{\partial x}{\partial\theta} + \frac{dz}{dy} \frac{\partial y}{\partial\theta} = -\frac{dz}{dx} r\sin{\theta} + \frac{dz}{dy} r\cos{\theta}[/tex]

My question is, for parts a and b, is this correct or must something also be done with the dz/dx and dz/dy, and for part c, I don't know how to do it. Can someone help please?
 
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you don't have to put the dz/dx and dz/dy. its just cos + sin.
for c, you just take the derivative with respect to r and theta. i forget which you are supposed to do first. but you just do one, than take the derivative of the new form with respect to the other variable.
 
Points a) and b) are solved wonderfully.
Point c) is a bit tricky,meaning that u'll have to differentiate one of the 2 expressions found at a) & b) wrt the other variable.
[tex]\frac{\partial^{2}z}{\partial r \partial \theta}=\frac{\partial}{\partial r}(\frac{\partial z}{\partial \theta})[/tex]

Try to do it this way and tell where you get stuck.

Daniel.
 
One more thing.It's still a partial derivative for "z" (or "f") too,becasue it depends explicitely on 2 variables,namely "x" and "y"...

Daniel.
 
Thank you dexter. I checked both
[tex]\frac{\partial}{\partial r}(\frac{\partial z}{\partial \theta})[/tex]
and
[tex]\frac{\partial}{\partial \theta}(\frac{\partial z}{\partial r})[/tex]

and they both come to the same answer. So it must be right. :biggrin:
 
It meant that the function "z" is "well behaved".There are functions for which the mixed partial derivatives are different one from another.In a more advanced way,the 2-nd rank hessian is not symmetric...

Daniel.
 

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