Understanding Partial Differentiation with Polar Coordinates

In summary, when given z = f(x,y) and x = rcos(\theta) and y = rsin(\theta), we can find the partial derivatives \frac{\partial z}{\partial r} and \frac{\partial z}{\partial\theta} by using the chain rule and substituting the values of x and y. To find the second order partial derivative \frac{\partial^{2}z}{\partial r \partial \theta}, we take the derivative of \frac{\partial z}{\partial \theta} with respect to r or vice versa. It is important to note that the function z must be "well behaved" for the mixed partial derivatives to be equal.
  • #1
tandoorichicken
245
0
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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:
 
  • #6
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.
 

Related to Understanding Partial Differentiation with Polar Coordinates

1. What is partial differentiation?

Partial differentiation is a mathematical process used to find the rate of change of a multi-variable function with respect to one of its variables, while holding the other variables constant.

2. Why is partial differentiation important?

Partial differentiation is important because it allows us to analyze how a function changes with respect to specific variables, which is useful in many fields such as physics, economics, and engineering.

3. How is partial differentiation different from ordinary differentiation?

Ordinary differentiation is used to find the rate of change of a single-variable function, while partial differentiation is used to find the rate of change of a multi-variable function with respect to one of its variables.

4. What is the symbol used for partial differentiation?

The symbol used for partial differentiation is ∂ (the partial derivative symbol).

5. Can partial differentiation be applied to any function?

Yes, partial differentiation can be applied to any function that contains multiple variables, as long as the function is continuous and differentiable.

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