Vector in cylindrical polar coordinates

In summary, the conversation is about writing a vector in cylindrical polar coordinates and finding the gradient of a function at a specific point. The vector components in cylindrical polar coordinates depend on position and can be expressed as (ρ,θ,z). The Cartesian components of a vector can be transformed into cylindrical polar coordinates using the equations provided. The conversation also includes a question about the gradient of a function at a specific point.
  • #1
ilikephysics
18
0
The problem is:

Write the vector V=i+j+k=(1,1,1) at the point (x,y,z)=(1,1,1) in cylindrical polar coordinates. What is the gradient of the function phi=x(x^2+y^2)z at this point?

Answer:

I don't know how to write the vector in cylindrical polar coordinates. I know that the coordinates are (r(perpendicular), theta, z). Can someone show me how to do this in cylindrical polar coordinates with an example?

Is the gradient of the funcition 4i+2j+2k at (1,1,1)?
 
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  • #2
The vector components in cylindrical polar coordinates depend on position. The position can be expressed in cylindrical polar coordinates as:

(ρ,θ,z)

where ρ is the perpendicular distance from the Cartesian z-axis, θ is the angle about the Cartesian z-axis that a line connecting the point to the Cartesian z-axis would make from the Cartesian x-axis, and z is the Cartesian z-coordinate. The Cartesian components of a vector transform into the cylindrical polar coordinates of a vector as:

vρ = (√(ux2+uy2))cos(θ-arctan(uy/ux))
vθ = -(√(ux2+uy2))sin(θ-arctan(uy/ux))
vz = uz

where

v = uxex+uyey+uzez = vρeρ+vθeθ+vzez




ilikephysics said:
Is the gradient of the funcition 4i+2j+2k at (1,1,1)?
That's what I got.
 
  • #3


Sure, I can help with that! To write the vector V=i+j+k=(1,1,1) in cylindrical polar coordinates, we first need to convert the given cartesian coordinates (x,y,z)=(1,1,1) to cylindrical polar coordinates. This can be done using the following equations:

r = √(x^2 + y^2)
theta = tan^-1(y/x)
z = z

Substituting the given coordinates, we get r = √(1^2 + 1^2) = √2, theta = tan^-1(1/1) = π/4, and z = 1. So the cylindrical polar coordinates for the point (1,1,1) are (r, theta, z) = (√2, π/4, 1).

Now, to write the vector V in cylindrical polar coordinates, we can use the following conversion formula:

V_cylindrical = V_cartesian · (cos(theta), sin(theta), 1)

Substituting the values, we get V_cylindrical = (1,1,1) · (cos(π/4), sin(π/4), 1) = (√2/2, √2/2, 1). So the vector V in cylindrical polar coordinates is (√2/2, √2/2, 1).

Moving on to the second part of the problem, the gradient of a function phi(x,y,z) is given by the following formula:

∇phi = (∂phi/∂x, ∂phi/∂y, ∂phi/∂z)

In this case, the function is phi = x(x^2 + y^2)z. So, we can find the gradient at the point (1,1,1) by substituting the coordinates in the formula:

∇phi = (∂phi/∂x, ∂phi/∂y, ∂phi/∂z) = (3x^2z + z, 3xy + x^2z, x(x^2 + y^2))

Substituting (x,y,z) = (1,1,1), we get ∇phi = (4, 4, 2). So the gradient of the function at the given point is 4i+4j
 

What is the formula for converting a vector from Cartesian to cylindrical polar coordinates?

The formula for converting a vector from Cartesian to cylindrical polar coordinates is:
Vr = Vx cosϕ + Vy sinϕ
Vϕ = -Vx sinϕ + Vy cosϕ
Vz = Vz

How do you find the magnitude and direction of a vector in cylindrical polar coordinates?

The magnitude of a vector in cylindrical polar coordinates is given by:
|V| = √(Vr² + Vϕ² + Vz²)
The direction of the vector can be found using the inverse tangent function:
θ = tan⁻¹(Vr/Vϕ)

What are the unit vectors in cylindrical polar coordinates?

The unit vectors in cylindrical polar coordinates are:
er = (cosϕ, sinϕ, 0)
eϕ = (-sinϕ, cosϕ, 0)
ez = (0, 0, 1)

What is the dot product of two vectors in cylindrical polar coordinates?

The dot product of two vectors in cylindrical polar coordinates is given by:
V•W = VrWr + VϕWϕ + VzWz

Can a vector in cylindrical polar coordinates be expressed as a linear combination of the unit vectors?

Yes, a vector in cylindrical polar coordinates can be expressed as a linear combination of the unit vectors:
V = Vr er + Vϕ eϕ + Vz ez

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