What is the Difference Between Partial Derivatives and Ordinary Derivatives?

[erok81]

Homework Statement

I am working on some PDE's where we are doing Laplacian's in various coordinate systems and got stuck on a partial derivative of all things. It's been a while and it seems I have forgotten how to do them.

Homework Equations

I have the equation

[tex]u(x,y)=\frac{1}{\sqrt{x^2 +y^2}}[/tex]

Which I converted to polar coordinates and ended up with

[tex]u(r,\theta )= \frac{1}{r}[/tex]

The Attempt at a Solution

I have to compute the following:

[tex]\frac{\partial ^{2} u}{\partial r^{2}}[/tex] (1)

[tex]\frac{\partial u}{\partial r}[/tex] (2)

[tex]\frac{\partial ^{2} u}{\partial \theta^{2}}[/tex] (3)

Here is what I have so far...

(1) 2r^-3
(2) -r^-2
(3) 0

I think I have (3) wrong. If they were normal derivatives, I wouldn't have an issue.

 

[Mark44]

Since u(r, [itex]\theta[/itex]) doesn't depend on [itex]\theta[/itex], the partials with respect to [itex]\theta[/itex] are all zero.

For 2, the sign in your exponent is wrong. You probably just neglected to put it in since your second partial is right.

 

[erok81]

Yeah, that was a typing error on my part.

Thanks for the help.

I was confused because if the term wasn't zero, I might have been able to satisfy another part of the problem.

Plus there was an example in the book that did this.

Starting with r²=x²+y² they differentiated with respect to x. Now if would have been me I would have gotten 0=2x+0 instead the book got:

[tex]2r \frac{\partial r}{\partial x}=2x[/tex]

So even though r² doesn't depend on x, they still got a non-zero answer.

Does that make sense?

 

[tiny-tim]

hi erok81!

So even though r² doesn't depend on x, they still got a non-zero answer.

Does that make sense?

nooo …

r does depend on x … r² = x² + y²

 

[erok81]

Why doesn't the y get any derivative then? I see what you mean with the r. I actually had the typed out originally, but decided against it. They are starting to come back to me at least. :)

So I guess why isn't the answer

[tex]
2r \frac{\partial r}{\partial x}=2x + 2y \frac{\partial y}{\partial x}
[/tex]

Actually I think I see why and my entire confusion. They didn't state what derivative they were taking, just the result. Since they were taking the derivative of r with respect to x, that is why nothing happened with the y...I think?

 

[Mark44]

Yes. In the equation r² = x² + y², r² is a function of both x and y, but x and y are independent of each other. IOW, dy/dx = 0 and dx/dy = 0. In the equation you asked about, they differentiated
with respect to x on both sides. On the left, because of the chain rule, they ended up with a factor of [tex]\frac{\partial r}{\partial x}[/tex]

 

[tiny-tim]

Since they were taking the derivative of r with respect to x, that is why nothing happened with the y...I think?

the partial derivative ∂/∂x means differentiating wrt the variable x while keeping the other variables constant …

that means that implicit (ie not stated) in the definition is the understanding of what the other variables are …

eg if you have a two-dimensional function F, you could use ordinary x and y coordinates, or you could use x and (x+y) coordinates …

in the first case, ∂F/dx would be keeping y fixed, ie going parallel to the x-axis

in the second case, ∂F/dx would be keeping x+y fixed, ie going at 45° to the x-axis ​