How Do You Convert Cartesian Vector Coordinates to Cylindrical Coordinates?

AI Thread Summary
To convert the Cartesian vector coordinates to cylindrical coordinates, the vector \vec A = (x + y)\hat x is evaluated at point P_1 (1, 2, 3). The calculations yield r = x, φ = 0, and z = 0, leading to \vec A being expressed as x\hat r at P_1. It's noted that \hat{x} represents the unit vector in the x direction, and the vector is identified as a vector field rather than a single vector. The discussion highlights the importance of understanding vector projections in cylindrical coordinates for accurate conversion.
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Homework Statement


Transform the following vector into cylindrical coordinates and then evaluate them at the indicated points:

\vec A = (x + y)\hat x

at

P_1 (1, 2, 3)


Homework Equations


r = \sqrt{x^2 + y^2}

\phi = \tan^{-1}(\frac{y}{x})

z = z

The Attempt at a Solution


r = \sqrt{x^2 + 0^2} = x

\phi = \tan^{-1}(\frac{0}{x}) = 0

z = z = 0

\vec A = x\hat r at point P_1 (1, 2, 3) \Longrightarrow \hat r

Could someone please check if this is correct? There are a few more of these, but if I can do this one, then the rest are no problem. Thanks.
 
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Can I assume that \hat{x} is the unit vector in the x direction? If so then (x+ y)\hat{x} is not a "vector", it is a "vector field"- a vector at each point in the xy-plane. At (1, 2, 3) (surprising how often that point shows up!), that is the vector 3 \hat{x}, of length 3 pointing in the x-direction. That vector has no z-component. The projection of the vector <2, 0> in the direction of the <1, 2> vector will be the \hat{r} component. <2, 0> minus that projection will be the component in the \theta direction.
 
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