1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the gradient

  1. Mar 4, 2006 #1
    Suppose distances are measured in lightyears and that the temperature T of a gaseous nebula is inversely proportional to the distance from a fixed point, which is the origin. The temperature 1 lightyear from the origin is 100 degrees celsius. Find the gradient of T at (x,y,z).

    here's what I have:

    [tex]d=\sqrt{x^2+y^2+z^2}=1[/tex]
    [tex]d=x^2+y^2+z^2=1[/tex]
    [tex]T=\frac{1}{x^2+y^2+z^2}[/tex]
    so the gradient is:
    [tex]T_x=-\frac{2x}{(x^2+y^2+z^2)^2}[/tex]
    [tex]T_y=-\frac{2y}{(x^2+y^2+z^2)^2}[/tex]
    [tex]T_z=-\frac{2z}{(x^2+y^2+z^2)^2}[/tex]

    but this is not right, i where did I go wrong?
     
  2. jcsd
  3. Mar 4, 2006 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    First of all, read your info properly!
    You are told that T(x,y,z) is inversely proportional to the distance from the origin. That means there exist a constant K, so that we have:
    [tex]T(x,y,z)=\frac{K}{\sqrt{x^{2}+y^{2}+z^{2}}}[/tex]
    Furthermore, you have the condition:
    [tex]T(x,y,z)=100, \sqrt{x^{2}+y^{2}+z^{2}}=1[/tex]
    Can you determine K from this?
     
  4. Mar 4, 2006 #3

    0rthodontist

    User Avatar
    Science Advisor

    The first equation you have for distance is correct but I don't see why you are setting it to 1. Then you squared it but didn't write d^2...

    The distance from the origin is sqrt(x^2 + y^2 + z^2). T is inversely proportional to that distance so T(x, y, z) = k/sqrt(x^2 + y^2 + z^2) where k is some constant. You know that T(x, y, z) = 100 when sqrt(x^2 + y^2 + z^2) = 1, so what is k? Then you can find the gradient.
     
  5. Mar 4, 2006 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Do you need a confirmation on that question?
     
  6. Mar 4, 2006 #5
    nope, i got it! thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding the gradient
Loading...