# Partial Derivatives

1. Oct 30, 2006

### Poop-Loops

Ok, so not really partials, I know how to do those. But now in my math physics class we were introduced to a new notation where it's the partial with respect to a variable, with another variable held constant. This is the problem I am trying to do in the book:

$$(\frac{\partial z}{\partial \theta})_x$$

Where $$z = x^2 + 2y^2 , x = r cos(\theta) , y = r sin(\theta)$$

The answer in the book gives me $$4 r^2 tan(\theta)$$

Now, am I missing something here? All the examples in the book (a whole quarter of a page... it's a crappy book...) specifically set up the equation to be in terms of the two variables (denominator and subscript) before taking the partial derivative. But this answer gives me an r? And a TANGENT!? I can't conceive of how to do this. I've been at it for a few hours.

I've tried things like transforming everything into polar first. Or only the x's. Or only the y's. Or going $$2x^2 + 2y^2 - x^2$$ and then transforming it all. Nothing. I can't get close to a tangent.

2. Oct 30, 2006

### quasar987

I get

$$2r^2\sin2\theta$$

or

$$4r^2\sin\theta\cos\theta$$

3. Oct 30, 2006

### Poop-Loops

Could the book be wrong, then? Because I got your 2nd answer.

4. Oct 30, 2006

### quasar987

It could. But let's wait for someone knowledgable's answer. I learned about this stuff just rencently (in a thermodynamics class) and it wasn't even formally introduced.

It will pay for you to understand this "partial with respect to a variable, with another variable held constant" thing well because in thermodynamics, if your course is anything like mine you will play with them to a point that it becomes stunning.

5. Oct 30, 2006

### Pseudo Statistic

Does that notation mean you're keeping x or y constant?

6. Oct 31, 2006

### arildno

Note that if x is a constant, say $$x=x_{0}$$, then we have:
$$r=\frac{x_{0}}{\cos(\theta)}$$
Thus, we have:
$$z=x^{2}+2y^{2}=r^{2}(1+\sin^{2}(\theta))=x_{0}^{2}(\frac{1}{\cos^{2}(\theta)}+\tan^{2}(\theta))=x_{0}^{2}(2\tan^{2}(\theta)+1)$$

Differentiating with respect to the angle, and reinserting r yields the result.

Last edited: Oct 31, 2006
7. Oct 31, 2006

### arildno

To do this properly, let us consider the bijection from polars to Cartesian coordinates:
$$\vec{\Phi}(r,\theta)=(X(r,\theta),Y(r,\theta))=(x,y)$$
That x equals a constant can be written that there exists a function G(x,y), so that
$$G(x,y)=0$$
Thus, we have:
$$H(r,\theta)\equiv{G}(\vec{\Phi(r,\theta)})=0$$
By the implicit function theorem, there exists a function $R(\theta)$ so that:
$$h(\theta)=H(R(\theta),\theta)=0$$
vanishes identically.
Therefore, the derivative of h vanishes also, which implies:
$$0=\frac{\partial{H}}{\partial{r}}\mid_{r=R}\frac{dR}{d\theta}+\frac{\partial{H}}{\partial\theta}=(\frac{\partial{G}}{\partial{x}}\frac{\partial{X}}{\partial{r}}+\frac{\partial{G}}{\partial{y}}\frac{\partial{Y}}{\partial{r}})\frac{dR}{d\theta}+\frac{\partial{H}}{\partial\theta}$$
where the last term is to be expanded likewise.

Thus, we may eventually solve for $\frac{dR}{d\theta}$, which we will need when differentiating z with respect to the angle, holding x constant.

Last edited: Oct 31, 2006
8. Oct 31, 2006

### Poop-Loops

Wow, I would have NEVER done something like that...

Thanks a lot, guys. Now I have something to work off of. I saw the book said "only in terms of the two variables" and then the answer had different variables than asked for, so that's why I was confused. It sucks how the rest of the problems that have answers in the back of the book are incredibly easy, but you KNOW he'll give something like this on the test.

9. Oct 31, 2006

### arildno

I made a clumsy substitution in the first post.
Simpler is, with $r=\frac{x_{0}}{\cos\theta}$:
$$z=x_{0}^{2}+2x_{0}^{2}\tan^{2}(\theta)$$
by inserting directly for x and y.

10. Oct 31, 2006

### quasar987

I think the rule of thumb for calculating

$$(\frac{\partial f}{\partial x})_y$$

would be to try to express everything in terms of x and y before differentiating.

11. Oct 31, 2006

### Poop-Loops

It's little tricks like x/cos(theta) that I have trouble remembering.

It doesn't help that all of math is just a bunch of those little tricks. =/

12. Oct 31, 2006

### quasar987

But the good news is that there's only a finite number of those little tricks. And when you learn about them the "hard" way (i.e. not just by seeing them, but by actually trying and failing and trying and failing and finally you either discover the trick or you get to know of it by another source [in this case arildno]), you tend to remember them and before you know it, you know all the little tricks!