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

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

Which I converted to polar coordinates and ended up with

u(r,\theta )= \frac{1}{r}

The Attempt at a Solution

I have to compute the following:

\frac{\partial ^{2} u}{\partial r^{2}} (1)

\frac{\partial u}{\partial r} (2)

\frac{\partial ^{2} u}{\partial \theta^{2}} (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, \theta) doesn't depend on \theta, the partials with respect to \theta 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:

2r \frac{\partial r}{\partial x}=2x

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

<br /> 2r \frac{\partial r}{\partial x}=2x + 2y \frac{\partial y}{\partial x}<br />

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 \frac{\partial r}{\partial x}

 

[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 ​