Vector in cylindrical polar coordinates

[ilikephysics]

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)?

 

[turin]

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:

v_ρ = (√(u_x²+u_y²))cos(θ-arctan(u_y/u_x))
v_θ = -(√(u_x²+u_y²))sin(θ-arctan(u_y/u_x))
v_z = u_z

where

v = u_xe_x+u_ye_y+u_ze_z = v_ρe_ρ+v_θe_θ+v_ze_z

Is the gradient of the funcition 4i+2j+2k at (1,1,1)?

That's what I got.

 

[M98Ranger]

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