# Conversion vectors in cylindrical to cartesian coordinates

1. Feb 12, 2017

### ForTheGreater

1. The problem statement, all variables and given/known data
It's just an example in the textbook. A vector in cylindrical coordinates.
A=arAr+aΦAΦ+azAz
to be expressed in cartesian coordinates.
Ax=A⋅ax=Arar⋅ax+AΦaΦ⋅ax

ar⋅ax=cosΦ
aΦ⋅ax=-sinΦ

Ax=ArcosΦ - AΦsinΦ

Looking at a figure of the unit vectors I get it. At the same time I just don't understand why ArcosΦ isn't enough to get the magnitude of the Ax component.

2. Feb 12, 2017

### kuruman

Because unit vector $\hat{x}$ has two components, one in the $\hat{r}$ direction and one in the $\hat{\phi}$ direction both of which depend on the angle $\phi$. So when you express $\hat{x}$ as in $A_x\hat{x}$, in terms of cylindrical unit vectors, you get two components which means you need the Pythagorean theorem to find the magnitude of $A_x$.

3. Feb 12, 2017

### ForTheGreater

If I have a point r=4, phi=pi/8 and z=z; in cylindrical coordinates.
sqrt( (r*cos(phi) )2 + (r*sin(phi) )2 ) = 4
So taking r*cos(phi) as the x component and r*sin(phi) as the y component seems to be enough to get the same point represented in both coordinate systems? What am I missing?

4. Feb 12, 2017

### kuruman

Suppose I gave you vector
$$\vec{A}=3\hat{a}_r-2 \hat{a}_{\phi}+4 \hat{z}$$
How would you proceed to find the Cartesian x-component of this vector? You will need the equations that transform the cylindrical unit vectors into the Cartesian unit vectors.

5. Feb 13, 2017

### Staff: Mentor

You did this correctly. In terms of interpretation, $A_r \cos {\phi}$ is only the component of the r component of A in the x direction. You also need the component of the $\phi$ component of A in the x direction. This is your $-A_{\phi}s\sin{\phi}$

6. Feb 14, 2017

### ForTheGreater

Thank you, I think I just didn't think of that you'll need both the Ax component and the Ay component of the vector to get the x coordinate.

7. Feb 14, 2017

### Staff: Mentor

I think you meant Ar and Atheta