Chain rule substitution help

  • Thread starter Tony11235
  • Start date
  • #1
254
0
Let f: [tex] \Re^3 \rightarrow \Re [/tex] be differentiable. Making the substitution

[tex] x = \rho \cos{\theta} \sin{\phi}, y = \rho \sin{\theta} \sin{\phi}, z = \rho \cos{\phi} [/tex]

(spherical coordinates) into f(x,y,z), compute (partially) df/d(rho), df/d(theta), and df/d(phi) in terms of df/dx, df/dy, and df/dz.

I'm just not sure I understand the question. Does it involve pulling out a very long chain rule?
 
Last edited:

Answers and Replies

  • #2
lurflurf
Homework Helper
2,432
132
Tony11235 said:
Let f: [tex] \Re^3 \rightarrow \Re [/tex] be differentiable. Making the substitution

[tex] x = \rho \cos{\theta} \sin{\phi}, y = \rho \sin{\theta} \sin{\phi}, z = \rho \cos{\phi} [/tex]

(spherical coordinates) into f(x,y,z), compute (partially) df/d(rho), df/d(theta), and df/d(phi) in terms of df/dx, df/dy, and df/dz.

I'm just not sure I understand the question. Does it involve pulling out a very long chain rule?
It involves the chain rule, not sure what you mean about the very long part.
[tex]\frac{\partial f}{\partial\rho}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\rho}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\rho}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\rho}[/tex]
[tex]\frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\theta}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\theta}[/tex]
[tex]\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\phi}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\phi}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\phi}[/tex]
The general form of the chain rule being
[tex]\frac{\partial f}{\partial x}=\sum_{k=1}^n \frac{\partial f}{\partial u_k} \ \frac{\partial u_k}{\partial x}[/tex]
where
[tex]f=f(u_1(x),u_2(x),...,u_{n-1}(x),u_n(x))[/tex]
 

Related Threads on Chain rule substitution help

  • Last Post
Replies
2
Views
1K
Replies
1
Views
648
Replies
1
Views
11K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
2
Replies
25
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
6K
Replies
5
Views
5K
Top