# Thread: Implicit Function Theorem for a System of Equations

1. ## Implicit Function Theorem for a System of Equations

Consider the equations that relate cylindrical and Cartesian coordinates in $\displaystyle \mathbb{R}^3$;
$\displaystyle x=rcos\theta$
$\displaystyle y=rsin\theta$
$\displaystyle z=z$

a. Near which points of $\displaystyle \mathbb{R}^3$ can we solve for r, $\displaystyle \theta$, and z in terms of the Cartesian coordinates?

I started off by saying

$\displaystyle F: x-rcos\theta=0$
$\displaystyle G: y-rsin\theta=0$

but then what do you do for the third equation since $\displaystyle z-z=0$? If I use H:0 then that makes $\displaystyle \Delta J=0$ everywhere, meaning that it is not in general possible to solve for r, $\displaystyle \theta$, and z in terms of x, y, and z. That seems so odd that I assume it's wrong, and I have to think that my mistake is in H.

Any thoughts?

2. The equations F, G express the Cartesian coordinates x,y in terms of the cylindrical coordinates r, θ. But the question asks for how to do it the other way round: express the cylindrical coordinates in terms of the Cartesian coordinates.

The third coordinate z (cylindrical) is the same as z (Cartesian), so there is no problem finding z. There is also no problem finding r, because it is given by the formula $\displaystyle r = \sqrt{x^2+y^2}$. So the question is really about discussing whether θ can be reconstructed in a neighbourhood of the point (x,y,z).

Originally Posted by davesface
what do you do for the third equation since $\displaystyle z-z=0$? If I use H:0 then that makes $\displaystyle \Delta J=0$ everywhere, meaning that it is not in general possible to solve for r, $\displaystyle \theta$, and z in terms of x, y, and z. That seems so odd that I assume it's wrong, and I have to think that my mistake is in H.
Ah, I see, you're trying to use a more sophisticated approach, inverting a Jacobian matrix. (I should have realised that, from the title of the thread.) In that case, your third equation should be $\displaystyle z_{\text{Cart}} - z_{\text{cyl}} = 0$. The two z's are conceptually different, and the relation between them should result in a 1 somewhere in the Jacobian matrix.

In this problem, the sophisticated method can obscure the simple geometry of the transformation, and a naive approach may give more insight. Maybe that is why part b. of the question asks you to think about the geometric meaning of your answer.

3. That makes a lot of sense, because $\displaystyle \frac{\delta H}{\delta z_{Cart}}=1$. So then the determinate evaluated at $\displaystyle P_0=(\theta_0, r_0, z_0)$ ends up being $\displaystyle \theta_0 r_0 (cos^2(\theta_0)+sin^2(\theta_0))= \theta_0 r_0 \neq 0$.

So we can solve for $\displaystyle (x,y,z)$ in terms of $\displaystyle (r,\theta,z)$ near all points except $\displaystyle \theta =0$ or $\displaystyle r=0$.

For part b, this would seem to correspond to being unable to solve for $\displaystyle (r,\theta,z)$ only at the origin. This makes sense for $\displaystyle \theta$, since $\displaystyle \theta=arctan(\frac{y}{x})$, but $\displaystyle r= \pm \sqrt{x^2+y^2}$ has no such restrictions. Is there something else wrong in my calculations somewhere?