Conversion vectors in cylindrical to cartesian coordinates

ForTheGreater
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Homework Statement


It's just an example in the textbook. A vector in cylindrical coordinates.
A=arAr+aΦAΦ+azAz
to be expressed in cartesian coordinates.
Start with the Ax component:
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.
 
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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##.
 
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?
 
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.
 
ForTheGreater said:

Homework Statement


It's just an example in the textbook. A vector in cylindrical coordinates.
A=arAr+aΦAΦ+azAz
to be expressed in cartesian coordinates.
Start with the Ax component:
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.
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}##
 
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.
 
ForTheGreater said:
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.
I think you meant Ar and Atheta
 

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