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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).

here is the answer given:

-6.66ap - 2.77aphi + 9az

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

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    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](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)'s a basic tutorial and quick reference. There's an "edit" link under the question.2012-10-01
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    @joriki oh, thanks. I was trying to figure out how to use tex. thanks.2012-10-01

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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 $\phi$.

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    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?).2012-10-01
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    wait, by ∂r⃗ /ρ, ∂r⃗ /ϕ and ∂r⃗ /z. you meant gradient, right?2012-10-02
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    @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.2012-10-02
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    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.2012-10-02
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    @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 $\partial\vec r/\partial\rho$.2012-10-02
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    shoot. sorry. I just got back from summer break. I need a while to think this stuff through. As for gradients, I realize my mistake. I think I understand this better now. $\frac{\partial r}{\partial\rho} $ is the scalar projection of the vector r in the $\rho$ direction.2012-10-03
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    Now that I think about it, I don't really understand how something could be in the direction $\frac{\partial \vec r}{\partial \phi}$, as phi is an angle. Or rather, I can understand that, but how would the answer be found from a distance/magnitude? anyways, so far we're on to chapter 2 and it's back to electric fields, which doesn't seem to use cylindrical component vectors (I took a similar class last quarter which dealt with fields in cylindrical coordinates and I never came across something like this...). Thanks for your help, I really appreciate it.2012-10-03
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    @Jacob: I think you need to get a better book and go back to some of the basics to get a better foundation. $\partial\vec r/\partial\rho$ is not the scalar projection of the vector $\vec r$ in the $\rho$ direction. It is by definition the rate at which $\vec r$ changes as $\rho$ changes. Similarly, $\partial\vec r/\partial\phi$ is the rate at which $\vec r$ changes as $\phi$ changes. Since $\vec r$ is a vector, this rate is also a vector. If you figure out all three of these vectors you'll find that they're mutually orthogonal. Thus you can use them as a basis to expand any vector.2012-10-03
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    P.S.: To see that $\partial\vec r/\partial\rho$ can't be a projection of $\vec r$, note that if you double $\vec r$, then $\partial\vec r/\partial\rho$ stays the same, whereas a projection of $\vec r$ would have to double, too. This is because $\partial\vec r/\partial\rho$ corresponds to a local *change* in $\vec r$, not to $\vec r$.2012-10-03
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    I was confused because we learned that the formula you used: $\left(\pmatrix{-1\\-4}-\pmatrix{3\\2}\right)\cdot\pmatrix{3\\2}\Big/\left|\pmatrix{3\\2}\right|=\frac{-24}{\sqrt{13}}\approx-6.66\4\;.$ is the formula for finding the scalar projection of a matrix. Is there anywhere that I can read about this?2012-10-03
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    ^I was talking about the only formula involving the dot product, which results in $\approx -6.66$2012-10-03
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    @Jacob: I don't know how a matrix got into it. The formula you refer to does indeed yield a projection, namely the projection of the vector from $C$ to $D$ onto $\partial\vec r/\partial\rho$ at $C$. So $\partial\vec r/\partial\rho$ is not a projection, it's a vector that we project *onto* to find the corresponding component. I'm not sure where best to read about this; you could look at http://en.wikipedia.org/wiki/Curvilinear_coordinates.2012-10-03