# Polar Partials

1. Nov 3, 2008

### dtl42

1. The problem statement, all variables and given/known data
Why are the following not contradictory?
$$r=\sqrt{x^2+y^2}$$
$$\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=cos{\theta}$$

and

$$r=\frac{x}{cos{\theta}}$$
$$\frac{\partial r}{\partial x}=\frac{1}{cos{\theta}}$$

2. Relevant equations

3. The attempt at a solution
I understand that we got these results from differentiating two separate relations, but I don't see why they're different, and seemingly contradictory.

2. Nov 3, 2008

### gabbagabbahey

Hint: what is the definition of $\theta$?...If $\theta$ has any explicit dependence on $x$, then is $$\frac{\partial}{\partial x} \left(\frac{x}{cos(\theta)} \right)$$ Really just $\frac{1}{cos(\theta)}$?Don't you have to use the product rule?

3. Nov 3, 2008

### HallsofIvy

Staff Emeritus
They are contradictory because the second is simply wrong. Yes,
$$r= x/cos(\theta)$$ but the derivative is wrong: $\theta$ is not a constant, it depends on x itself. A correct calculation would be
[tex]\frac{\partial r}{\partial x}= \frac{cos(\theta)+ xsin(\theta)\frac{\partial \theta}{\partial x}}{cos^2(\theta}[/itex]