- #1
teslajet
- 2
- 0
So I'm trying to find out what the procedure is to convert a cartesian unit vector to a cylindrical unit vector. Any thoughts?
How to find the unit vector in cylindrical coordinates
- Thread starter teslajet
- Start date
In summary, the conversation discusses the procedure for converting a cartesian unit vector to a cylindrical unit vector. The cylindrical unit vector is represented as er and can be expressed in terms of the cartesian unit vectors and trigonometric functions. The question arises about the correct value of \phi to use in the transformation, with the solutions manual using a different value than what was calculated in part a.
- #2
Jerbearrrrrr
- 127
- 0
The cylindrical unit vector is er.
x/|x|
Where x is where ever we are. (not cartesian x)
x=r(e_r)+z(e_z)
and |x|^2 = x.x
I think that works. Sorry about not underlining vectors.
e_r can be expressed in terms of e_x and e_y and some trig things.
x/|x|
Where x is where ever we are. (not cartesian x)
x=r(e_r)+z(e_z)
and |x|^2 = x.x
I think that works. Sorry about not underlining vectors.
e_r can be expressed in terms of e_x and e_y and some trig things.
- #3
teslajet
- 2
- 0
Here is my understanding,
Given a unit vector A = x1,y1,z1 in cartesian, to transform into cylindrical just use the transform
A . [tex]\rho[/tex]
A . [tex]\phi[/tex]
Z(cartesian)=Z(cylindrical)
my question is, since x . [tex]\rho[/tex] = cos[tex]\phi[/tex], is the [tex]\phi[/tex] that I am supposed to use the tan^-1(y1/X1)?
If this is the case then I don't understand what the solutions manual did with the following problem
I understand part a.) but in part b.) they use 70 degrees as [tex]\phi[/tex] when according to part a.), [tex]\phi[/tex] should be -89 degrees. Am I missing something?
Given a unit vector A = x1,y1,z1 in cartesian, to transform into cylindrical just use the transform
A . [tex]\rho[/tex]
A . [tex]\phi[/tex]
Z(cartesian)=Z(cylindrical)
my question is, since x . [tex]\rho[/tex] = cos[tex]\phi[/tex], is the [tex]\phi[/tex] that I am supposed to use the tan^-1(y1/X1)?
If this is the case then I don't understand what the solutions manual did with the following problem
I understand part a.) but in part b.) they use 70 degrees as [tex]\phi[/tex] when according to part a.), [tex]\phi[/tex] should be -89 degrees. Am I missing something?
1. What are cylindrical coordinates?
Cylindrical coordinates are a type of coordinate system commonly used in mathematics and physics to describe the position of a point in 3-dimensional space. They consist of a radial distance from the origin, an angle from a reference plane, and a height or vertical distance from the reference plane.
2. How do I convert Cartesian coordinates to cylindrical coordinates?
To convert from Cartesian coordinates (x, y, z) to cylindrical coordinates (r, θ, z), you can use the following equations:
r = √(x^2 + y^2)
θ = arctan(y/x)
z = z
3. What is a unit vector?
A unit vector is a vector with a magnitude of 1 that indicates the direction of a vector in a particular coordinate system. In cylindrical coordinates, the unit vectors are typically denoted as ρ̂, θ̂, and ẑ for the radial, azimuthal, and vertical directions, respectively.
4. How do I find the unit vector in cylindrical coordinates?
To find the unit vector in cylindrical coordinates, you can use the following equations:
ρ̂ = cos(θ) x̂ + sin(θ) ŷ
θ̂ = -sin(θ) x̂ + cos(θ) ŷ
ẑ = ẑ
Note that these unit vectors are dependent on the value of θ, which represents the angle from the reference plane.
5. Can I use the unit vector in cylindrical coordinates to find the magnitude and direction of a vector?
Yes, you can use the unit vector in cylindrical coordinates to find the magnitude and direction of a vector. The magnitude of a vector is equal to the length of the vector multiplied by the unit vector in that direction. The direction of a vector is given by the unit vector in that direction. Therefore, by multiplying a vector by its corresponding unit vector in cylindrical coordinates, you can determine its magnitude and direction.
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