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Cylindrical coordinates

Screenshot 2018-11-20 at 22.11.50.png
can anyone show why e(theta) is that?
It was quick to see why e(r) is that though.
Okay. Here's the method that we were given for our course on LA&SVC, and the method supported in the RHB Mathematical Methods book;

Consider you have the point A in terms of the angle theta, as you've clearly shown on the diagram.

A = rcos(theta)i + rsin(theta)j + kz

What we want are unit vectors, unit vectors radial, unit vectors tangential to circular motion, and unit vectors vertical. I.e. as shown in the diagram. So, how do we find those? Well, consider the rate of change of A with respect to each direction.

If we are changing radius, we change along the radial direction. Consider the z component as a constant, and theta constant. Since I lack a partial derivative delta on my keyboard, I'll just have to use d/dz/standard differential notation.

dA/dr = cos(theta)i + sin(theta)j

That gives us a vector in the radial direction, i.e. parallel to the rate of change of r when we consider the vertical and angular position a constant. So that's a unit vector radial.

Now then, we want a unit vector where r is a constant, height from x-y is a constant. Angular position changes, though.

dA/d(theta) = -rsin(theta)i + rcos(theta)j

Normalise this for a unit vector which is tangential and points in the rate of change of the angular position, i.e. theta. I call it angular position, but that's just what I mean, the position relative to some angle, i.e. 0 at the positive x axis as shown.

That's one method to do it.

Another method would be to just consider a unit vector normal to the er vector as shown; you can easily calculate that if you draw the x-y plane and try to compare the gradients of the two vectors er and e(theta) from above.

Hope I helped!
Original post by Feynboy
Screenshot 2018-11-20 at 22.11.50.png
can anyone show why e(theta) is that?
It was quick to see why e(r) is that though.


You can also draw a unit radius circle and find the unit vector tangent to the circle in terms of e^x \hat{e}_x and e^y \hat{e}_y .
polar_coordinates_02.JPG
Reply 3
Thanks folks

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