Help with Partial Derivatives and Implicit Differentiation

[end3r7]

The problem is as follows:

Cartesian and polar coordinates are related by the formulas
$$x = r\cos\theta$$
$$y = r\sin\theta$$
Determine $$\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial\theta}{\partial x}, and \frac{\partial\theta}{\partial x}$$. Differentiate the equations above implcitly adn then solve the resulting system of 4 equations in the four unknowns $$\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial\theta}{\partial x}, and \frac{\partial\theta}{\partial x}$$.

Now, this seems a whole lot harder than just differentiating $$r^2 = x^2 + y^2$$, but I did anyway and the 4 equations are:

$$\frac{\partial x}{\partial x} = \frac{\partial r}{\partial x}\cos\theta - \sin\theta\frac{\partial\theta}{\partial x}r$$

$$\frac{\partial x}{\partial y} = \frac{\partial r}{\partial y}\cos\theta - \sin\theta\frac{\partial\theta}{\partial y}r$$

$$\frac{\partial y}{\partial x} = \frac{\partial r}{\partial x}\sin\theta + \cos\theta\frac{\partial\theta}{\partial x}r$$

$$\frac{\partial y}{\partial y} = \frac{\partial r}{\partial y}\sin\theta + \cos\theta\frac{\partial\theta}{\partial y}r$$

Now, if I try to solve for $$\frac{\partial r}{\partial x}$$, I know what the answer should be by differentiating $$r^2 = x^2 + y^2$$, and it's $$\frac xr$$ or $$\cos\theta$$, however I can't seem to get that from combining the 4 equations.

What am I doing wrong?

Thanks in advance.

 

[0rthodontist]

Use the fact that dx/dx = 1 and dx/dy = 0, etc.

 

[end3r7]

I did... it still doesn't give me $$\cos\theta$$ as an answer.

For example, to calculate dr/dx
using eq 1
I get $$1 = \frac{\partial r}{\partial x}\cos\theta$$
using eq 3
$$0 = \frac{\partial r}{\partial x}\sin\theta$$

If I use both separatedly or try to combine them, I don't get the same answer as deriving r^2 = x^2 + y^2

 

[0rthodontist]

You seem to have cut off the end parts of those two equations. Try multiplying eq. 1 by cos theta / sin theta and adding it to eq. 3.

 

[end3r7]

Why mutliply times tangent will help?
I cut them off because I multiplied the other stuff by zero.
I can solve it using the Jacobian, but I just want to know the logic behind it.

 

[0rthodontist]

You multiply eq. 1 by cos theta/sin theta so that when you add it to eq. 3, you cancel the terms involving dtheta/dx.

When you say you cut them off because you "multiplied the other stuff by 0" you can't just multiply one term of an equation by 0 and assume it will still be correct. The stuff you cut off was not in general zero.

 

[end3r7]

Oh ****, you are right. Thanks a lot man, really appreciate it! =D