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Temperature Gradient Questions

  1. Oct 10, 2006 #1
    Is there a direction A in which the rate of change of the temperature function [tex]T(x,y,x)=2xy - yz[/tex] at P(1,-1,1) is -3ºC/ft? Give reasons for your answer.

    For this problem I found the gradient of at the point P. So
    [tex]\nabla f = 2y|_p \mathbf{i} + 2x-z|_p \mathbf{j} - y|_p \mathbf{k}[/tex]
    which I then found was
    [tex] \nabla f = -2 \mathbf{i} + 1 \mathbf{j} + 1 \mathbf{k} [/tex]
    Then we want to know if the gradient, in the direction of A will be 3ºC, so
    [tex] \nabla f \cdot \frac{A}{|A|} = \frac{3º}{ft}[/tex]
    which means that we want an x, y, z, such that
    [tex]\frac{-2x +y +z}{\sqrt{x^2 + y^2 + z^2}} = \frac{3º}{ft}[/tex]
    So I belive this is right so far, but I don't think this eqn can be solved for.

    Here is the other one:
    The Celsius temperature in a region in space is given by [tex]T(x,y,z) = 2x^2 -xyz[/tex]. A particle is moving in this region and its position at time t is given by [tex]x=2t^2[/tex], [tex]y=3t[/tex], [tex]z=-t^2[/tex], where time is measured in seconds and distanace in meters.
    a) How fast is the temp experienced by the particle changing in ºC/m when the particle is at the point P(8,6,-4)?
    b) How fast is the temp experienced by the particle changing in ºC/sec at P?

    So, using the chain rule
    [tex] \frac{dT}{dt} = (4x-yz)4t \mathbf{i} - (xz)3 \mathbf{j} + (xy)2t \mathbf{k} [/tex]
    For the change per meter I will just solve for one of the t eqns, and then substitute back in to the equation to find the gradient?
    Then for the change per second, it will be the opposite?
     
    Last edited: Oct 10, 2006
  2. jcsd
  3. Oct 11, 2006 #2
    Nevermind, I figured them out.
     
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