Partial derivatives, equation help

tweety1234
Messages
111
Reaction score
0

Homework Statement

Heat is being conducted radially through a cylindrical pipe. The temperature at a radius r is T(r). In Cartesian co-ordinates, [tex]r = \sqrt{(x^{2}+ y^{2}})[/tex]

show that [tex]\frac{\partial T}{\partial x} = \frac{x}{r} \frac{dT}{dr}[/tex]
 
on Phys.org
cant you just say [tex]\frac{\partial T}{\partial x} = \frac{\partial T}{\partial r} * \frac{\partial r}{\partial x}[/tex]
but since T is function of r you can write [tex]\frac{\partial T}{\partial r}[/tex] as dT/dr I am not sure this is correct though.
 
madah12 said:
cant you just say [tex]\frac{\partial T}{\partial x} = \frac{\partial T}{\partial r} * \frac{\partial r}{\partial x}[/tex]
but since T is function of r you can write [tex]\frac{\partial T}{\partial r}[/tex] as dT/dr I am not sure this is correct though.

oh I thought we might have to make use of the equation relating x and r given in the question ?
 
yes to get the partial of r with respect to x you need the equation right?
 
Last edited:
But the expression wants x/r?
 
I think that the [tex] \frac{\partial T}{\partial r} [/tex] will give you the x/r part
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
2K
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K