Homework Help: Quick Question on Unusual Derivatives

1. Nov 27, 2013

Hertz

The problem I'm curious about is this:
$\frac{\partial}{\partial r}(\frac{\partial r}{\partial θ})$

I found that the answer is zero using WolframAlpha, but obviously I won't have that on a future test xD. Can someone please explain to me how to think about the derivative above? How can I look at it and intuitively say "Oh, that derivative is equal to zero!"

-edit
I do see that $\frac{\partial r}{\partial \theta}$ is usually only a function of theta, so therefore the partial with respect to r would be zero. But when you differentiate implicitly you can get r' in terms of both r and θ right?

2. Nov 27, 2013

BruceW

think about what the partial differentiation actually means. When you write
$$\frac{\partial r}{\partial \theta}$$
you're actually leaving out some crucial information. What is being held constant? This will depend on the context in which you used this partial differential.

As an example, I could have a function of 3 variables $f(x,y,z)$ and maybe the $z$ variable can be re-written as some other variable $u$. So then if I say
$$\frac{\partial f}{\partial x}$$
you cannot tell what this means. I also must say what is being held constant. In this case, it will either be the two variables $(y,z)$ or the two variables $(y,u)$ and the value of
$$\frac{\partial f}{\partial x}$$
will depend on which other two variables I am holding constant. Anyway, I'm just saying that you need to make sure you remember which variables you are holding constant when you take the partial differential. In your case, you can probably guess which variable is meant to be held constant.

3. Nov 27, 2013

Hertz

Sorry, but I'm quite confused by that.

There could be any number of things being held constant, but that seems irrelevant. $\frac{\partial r}{\partial theta}$ simply states "how does r change with a small change in theta", this disregards anything else given in the problem. So why do you have to explicitly acknowledge what's being held constant? It's the general identity that I'm confused about:

$\frac{\partial}{\partial r}(\frac{\partial r}{\partial \theta})=0$

This should hold regardless of what the function r looks like; however, I don't see why it does because assuming you need to do implicit differentiation, you would still get r terms in your derivative. Also, I'm not talking specifically about polar coordinates, that's just the case that was on my mind when I first posted this question.

4. Nov 27, 2013

BruceW

no, it does not state "how does r change with a small change in theta". The quantity that states "how does r change with a small change in theta" is the total derivative. As an example, if we have a function of two variables $f(x,y)$ then the total derivative with respect to $x$ is:
$$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}$$
and in the partial derivatives, I am holding the other variable constant. So for the total derivative, nothing is held constant, in a certain sense (since each of the terms accounts for one of the variables changing). But for a partial derivative, there is (at least) one thing being held constant, and the value of the partial derivative will generally depend on what is being held constant.

5. Nov 29, 2013

Hertz

http://www.wolframalpha.com/input/?i=d/dx+(dx/dy)

I'm clearly not understanding what you're hinting at...

I understand that the full derivative represents the change in dadada with a small change in dadada. dadada. What I'm confused about it why this general partial derivative always equals zero. Wolfram is not the first place I've seen this. IT DOESN'T MATTER WHAT ELSE r IS A FUNCTION OF.

Anyways, in regard to my original point. If you take a derivative of a function y implicitly, you might end up with y's in the derivative itself. However, if you use the original function to replace y in terms of the independent variable, say x, then technically the partial of y' with respect to y is zero. In this case though, the real question is whether or not every function can be solved explicitly for the dependent variable. If not, then can we really say that the partial of y' with respect to y is identically equally to zero always?

Last edited: Nov 29, 2013
6. Nov 29, 2013

BruceW

It does depend what else r is a function of. And it depends on what is being held constant when the partial derivative is taken. The reason wolfram alpha gives zero is because it assumes that x is a function of y only. And it assumes that the partial derivative should hold y constant. Therefore, it takes a partial derivative of a function $f(y)$ while holding $y$ constant, so naturally, this is zero.

maybe it is better if I give some examples of cases where this thing is not zero. OK, let's say we have a function $z(x,y)=y^2 \exp(xy)$ Now, if we take partial derivative with respect to x while holding y constant, we get:
$$\frac{\partial z}{\partial x} = yz$$ and now, we can re-write this as a function of z and y. It's a pretty simple function, since it is simply yz. So now, if we take the partial derivative of this with respect to z while holding y constant, then we get y. so
$$\frac{\partial}{\partial z} (\frac{\partial z}{\partial x}) = y$$
where in both partial derivatives, we are holding y constant.

So, this thing is not always zero. Now, why is it zero in the particular case of spherical polars? the equation is:
$$\frac{\partial}{\partial r}(\frac{\partial r}{\partial \theta})$$
and for the inner derivative, r is being held constant. Well, of course this is zero, since we are taking a partial derivative with respect to r, while holding r constant. So, we can even say this equation:
$$\frac{\partial r}{\partial \theta}$$
is zero, since we are holding r constant. And the reason that we are holding r constant is because in spherical polar coordinates, we use three variables $r,\theta ,\phi$ and we vary just one, while holding the other two constant. This is how we get the so-called "coordinate curves".