Cartesian vector to Cylindrical components

I recently started EE 1, and was confused by the jump from cylindrical coordinates to cylindrical vectors

In the past, I've just converted between cartesian and cylindrical via:

p = sqrt(X^2 + Y^2) theta = atan(y/x) z = z

and so on

I don't understand why you can't apply these equations to a cartesian vector in order to get a cylindrical vector. How exactly does a cylindrical vector work? My EE textbook is a piece of crap in terms of explaining math theory and we never learned cylindrical vectors in previous classes. I assumed they were pretty similar to cartesian vectors.

Here's the question I'm referring to:

"express in cylindrical components the vector from C(3, 2, -) to D(-1, -4, 2)"

I've tried finding the cartesian vector in between the two and converting to cylindrical (by formulas given above).

-6.66ap - 2.77aphi + 9az

can someone please either explain or suggest a better resource for figuring this out?

-
You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. – joriki Oct 1 '12 at 8:57
@joriki oh, thanks. I was trying to figure out how to use tex. thanks. – Jacob Sharf Oct 1 '12 at 22:24

The cylindrical coordinates don't form a vector. They're just a triple of numbers that you can use to describe a point; they don't have properties of vectors such as linearity, e.g. the cylindrical coordinates of the midpoint between two points aren't the average of the cylindrical coordinates of the two points.

What your text is likely referring to by "cylindrical vector" is a representation of a vector in terms of the orthogonal basis given by the unit vectors along $\partial\vec r/\rho$, $\partial\vec r/\phi$ and $\partial\vec r/z$. This is a local basis that depends on $\vec r$, so you're right in saying that your book does a bad job explaining these things; it's ambiguous to ask you to express a vector in this way without specifying the point whose basis is to be used. I'll assume that they mean "express the vector from $C(3,2,-7)$ to $D(-1,-4,2)$ in cylindrical components at $C$". (You omitted the $7$, but it can be reconstructed from the lengths of the vectors.)

Since $\partial\vec r/\partial z$ is just the canonical unit vector in the $z$ direction, the component in that direction is just $2-(-7)=9$. To find the other two components, note that $\partial\vec r/\partial\rho$ points in the direction from the $z$ axis to the point, so the corresponding component is given by

$$\left(\pmatrix{-1\\-4}-\pmatrix{3\\2}\right)\cdot\pmatrix{3\\2}\Big/\left|\pmatrix{3\\2}\right|=\frac{-24}{\sqrt{13}}\approx-6.66\;.$$

Then the remaining component along $\partial\vec r/\partial \phi$ is determined up to a sign from the length to be

$$\pm\sqrt{4^2+6^2-\frac{24^2}{13}}\approx\pm2.77\;,$$

and the sign is determined by your convention for $\phi$.

-
thank you, you're amazing. It puzzles me why my textbook couldn't explain it like that. instead it drew a diagram and referenced it as such: "by looking at diagram 1.7 and thinking mightily, can can infer...." (wtf?). – Jacob Sharf Oct 1 '12 at 22:27
wait, by ∂r⃗ /ρ, ∂r⃗ /ϕ and ∂r⃗ /z. you meant gradient, right? – Jacob Sharf Oct 2 '12 at 14:31
@Jacob: Your math got garbled there. I presume you mean $\partial\vec r/\partial\rho$ etc. No; I didn't mean gradient, I meant the partial derivatives just like I wrote them. – joriki Oct 2 '12 at 14:43
I mean, the partial derivatives are the components of the gradient. and you could claim that -6.66 is the multiple of the unit vector in the same direction as the gradient's rho component. It just seems more natural to call it that. – Jacob Sharf Oct 2 '12 at 15:17
@Jacob: I'm afraid that last comment indicates a more profound confusion than was apparent from the original question. There are several things wrong with it. a) A gradient is a derivative with respect to the position vector; what we're dealing with here are derivatives of the position vector. b) As I wrote, the partial derivatives with respect to non-Cartesian coordinates are not components of a vector; they're just a triple of numbers. c) The vector whose components we're dealing with here is $D-C$, and $-6.66$ is the coefficient of that vector along $\partial\vec r/\partial\rho$. – joriki Oct 2 '12 at 16:58