Can someone help me with cylindrical coordinate systems?

[beer]

I'm having a hard time plotting them by hand.

If I have a fixed rho =3, a fixed z = 2, and a phi of 3(cos(phi)-sin(phi)), what does that plot look like?

I can't get the gnuplot command working right.

 

[dighn]

are you talking about spherical system? because cylindirical uses r, theta and z

and a fixed rho and a fixed z implies a fixed phi too. your equation doens't make sense.

 

[beer]

Well, the texbook we're using uses [rho,phi,z] for cylindrical, and then [r,theta,phi] for spherical. Is that odd?

Anyways, this is in a review chapter, so it isn't covered in any detail for me to do the HW with....

The exact problem is:

A field is given in cylindrical coordinates as F:

F= [40/(rho^2+1)+3(cos[phi]+sin[phi])]*unit_rho + 3(cos[phi]-sin[phi])*unit_phi-2*unit_z.

Prepare simple sketches of abs(F):

a) vs phi with rho = 3;

b) vs rho with phi = 0;

c) vs rho with ph = 45 deg

 

[dighn]

i guess (rho, phi, z) = (r, theta, z)

but anywa since z is always constant, and one of the other is always given, then it's just a straight x vs y plot

y being the magnitude of that vector and x the parameter being plotted against

 

[beer]

Originally posted by: dighn
i guess (rho, phi, z) = (r, theta, z)

but anywa since z is always constant, and one of the other is always given, then it's just a straight x vs y plot

y being the magnitude of that vector and x the parameter being plotted against

Alright.

So per the definition:

x = rho*cos[phi];
y = rho*sin[phi];

by definition
X = 3cos[3cos(phi)-3sin(phi)] from the problem statement;
Y = 3sin3cos(phi)-3sin(phi)] from the problem statement;

unfortunately I still have phi in the equations and I can't get them out. This is where I have been getting stuck at, and I have two variables, which doesn't make sense....

 

[dighn]

err

 

[dighn]

you don't need x or y
you are given F - a vector field, and you are supposed to find the magnitude of that vector field

you have F= [40/(rho^2+1)+3(cos[phi]+sin[phi])]*unit_rho + 3(cos[phi]-sin[phi])*unit_phi-2*unit_z.

for example for a) vs phi with rho = 3;

you sub 3 for rho given you
F=(4+3(cos(phi)+sin(phi)), 3(cos(phi)+sin(phi), - 2)
so F deends only on phi, yo uget the magnitude equation and plot aganst phil.. i thnk you can just use pythag for that since those 3 directions are always perpentiduclar

 

[beer]

so |F|=sqrt([4+3(cos(phi)+sin(phi))]^2,+3[(cos(phi)+sin(phi)]^2+(-2)^2))

How am I supposed to hand-plot that via 'simple sketch??'

 

[dighn]

you can probably simply it alot for the phi = 0 and phi = 45 deg cases

though for row = 3... i dunno maybe you can work it out and see if something cancesl to make it simpler