Spherical coordinate derivatives

Click For Summary

Homework Help Overview

The discussion revolves around finding the derivatives of spherical coordinates in relation to Cartesian coordinates. The original poster presents a function defined in terms of spherical coordinates and seeks clarification on the correctness of their derivative calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive expressions for the partial derivatives of a function with respect to Cartesian coordinates, questioning their approach and the correctness of their expressions. They also inquire about the necessity of using Cartesian or spherical variables in their answers and express concern about the complexity of the algebra involved.

Discussion Status

Some participants have pointed out potential errors in the original poster's expressions for the spherical coordinates, suggesting corrections and clarifying the application of the chain rule for partial derivatives. There is an ongoing exploration of the relationships between the variables, with participants discussing the implications of these relationships on the derivatives.

Contextual Notes

Participants are addressing the definitions and relationships between spherical and Cartesian coordinates, with some confusion noted regarding the correct forms of the coordinate transformations. There is also a mention of the algebraic complexity involved in deriving the necessary expressions.

absci2010
Messages
10
Reaction score
0
1. Find the derivatives of the spherical coordinates in terms of df/dx, df/dy, and df/dz.

2. f(x,y,z)
x=rcos\thetasin\varphi
y=rsin\varthetacos\varphi
z=rcos\varphi

3. The Attempt at a Solution [/b]
I took the derivatives of the three equations and I got:
df/dx=rcos\thetacos\varphi(df/d\theta)-rsin\varphisin\theta(df/d\theta)+cos\thetasin\varphi(df/dr)
df/dy=rsin\thetacos\varphi(df/d\varphi)+rsin\varphicos\theta(df/d\theta)+sin\thetasin\varphi(df/dr)
df/dz=-rsin\varphi(df/d\varphi)+cos\varphi(df/dr)

I have three questions about this:
1) Am I taking the derivatives correctly?
2) Can my answer have x, y, and z in it, or does it have to be r, \theta, and \varphi?
3) I think the next step is just algebra. Is the algebra going to be really messy?

Thanks in advance!
 
Physics news on Phys.org
First of all, your expression for x is wrong, it should be

[PLAIN][PLAIN]http://latex.codecogs.com/gif.latex?[/URL][/URL] x = r \sin\theta \sin\varphi

to be consistent with the form of y and z.

That said, I'm confused by the statement of the question. Does it mean find the derivatives of a function f with respect to the spherical coordinates in terms of the derivatives with respect to the Cartesian coordinates? In any case, this question is dealing with applying the chain rule for partial derivatives, as you recognize. However, you seem to have incorrectly applied the chain rule.

[PLAIN][PLAIN]http://latex.codecogs.com/gif.latex?[/URL][/URL] \frac{\partial f}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial f}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial f}{\partial \theta} + \frac{\partial \varphi}{\partial x} \frac{\partial f}{\partial \varphi}

I believe you have the reciprocals of the correct factors in your expressions above. However, I believe that the problem is asking for you to express the inverse relationships

[PLAIN][PLAIN]http://latex.codecogs.com/gif.latex?[/URL][/URL] \frac{\partial f}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial f}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial f}{\partial z}

and so on. In this case we want to express everything on the right-hand side in terms of x,y,z. The algebra is not too bad. Just derive the relationships

[PLAIN][PLAIN]http://latex.codecogs.com/gif.latex?[/URL][/URL] r = \sqrt{x^2+y^2+z^2},

[PLAIN][PLAIN]http://latex.codecogs.com/gif.latex?[/URL][/URL] \cos\varphi = \frac{z}{\sqrt{x^2+y^2+z^2}}

[PLAIN][PLAIN]http://latex.codecogs.com/gif.latex?[/URL][/URL] \sin\theta = \frac{y}{z}

The other trig functions can be eliminated by using simple trig identities.
 
Last edited by a moderator:
absci2010 said:
1. Find the derivatives of the spherical coordinates in terms of df/dx, df/dy, and df/dz.

2. f(x,y,z)
x=rcos\thetasin\varphi
y=rsin\varthetacos\varphi
z=rcos\varphi

There's something wrong here.
Shperical coordinates have one radious and two angles, you got 1 radious and 3 angles.
I usually write:

x=r sin \varphi cos \theta

y=r sin \varphi sin \theta

z=r cos \varphi
 
Sorry, it should be:
x=rcos\thetasin\varphi
y=rsin\thetasin\varphi
z=rcos\varphi
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
26
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K