- #1

JasonHathaway

- 115

- 0

I've some points I want to make sure of.

1- When converting a "POINT" from a coordinate system to another, I'll just use the derived equation to convert (e.g. (1,2,3) from cartestian to cylindrical: [itex] \rho=\sqrt{x^{2}+y^{2}}, \phi=tan^{-1}\frac{y}{x}, z=z [/itex]

2- When converting an "EQUATION" (e.g. [itex] \sqrt{x^{2}+y^{2}}=\frac{x^{2}y}{y^{2}x} [/itex]), I'll use the same approach as 1.

3- When converting a "VECTOR" (e.g. [itex] 4\vec{i}+3\vec{j}+6\vec{k} [/itex]), I must convert the unit vector [itex]\vec{i}, \vec{j},\vec{k} [/itex] into their equivalent in my desired coordinate system, whether that vector is A unit vector or not, and I cannot put in the form of a point (e.g. (4,3,6) ) and use the same approach as 1.Okay, and for 3:

a) To convert from cartesian to cylindrical:

Since [itex]x=\rho cos(\phi), y=\rho sin(\phi), z=z [/itex], then:

[itex]\frac{dx}{d \rho}=cos(\phi),\frac{dx}{d \phi}=-\rho sin(\phi), \frac{dx}{dz}=0 [/itex]

[itex]\frac{dy}{d \rho}=sin(\phi),\frac{dy}{d \phi}=\rho cos(\phi), \frac{dy}{dz}=0 [/itex]

[itex]\frac{dz}{d \rho}=0,\frac{dz}{d \phi}=0, \frac{dz}{dz}=1 [/itex]

And then dot product my vector with each of rho, phi and z components.

But I've found in wikipedia that there's no rho in the derivative with respect to phi...?

http://en.wikipedia.org/wiki/Vector_fields_in_cylindrical_and_spherical_coordinates

The same thing for the cartesian to spherical.

b) If I want to convert a vector from cylindrical/spherical to cartesian, do I have to take the same approach as a)?

c) How can I convert unit vectors between cylindrical and spherical and vice versa?