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Homework Help: Change in Temperature of a 2-D Temperature Field

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A two-dimensional temperature field is given by the expression [tex]T=4x^2y^2+3y^3[/tex]
    (a) What is the magnitude and direction of the temperature gradient at the point (2,3)?
    (b) What is the change in temperature [tex]dT/dS[/tex] at the point (2,3) in a direction along the curve [tex]3x-4y^2=-30[/tex] passing through the point (2,3)?

    2. Relevant equations
    [tex]T=4x^2y^2+3y^3[/tex]
    [tex]3x-4y^2=-30[/tex]

    3. The attempt at a solution

    So part (a) is pretty easy. I just found the gradient and evaluated it at the given point. However, I'm stuck on part (b). I think I have to take the dot product of the gradient of T with some unit vector of that curve. However, I don't know how to formulate this dot product. I figure that I first evaluate the gradient of T at (2,3), and then get a unit vector for that curve, evaluate that result at (2,3), then take the dot product to get dT/dS. Is this correct? If so, how do I handle the RHS of the curve? Would I move it over to the LHS and then find the unit vector? Would the 30 have to be squared when finding the magnitude of the curve too? I guess I'm a little confused with how to handle the nonzero RHS.
     
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  3. Jan 24, 2010 #2

    Dick

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    Since you can find x=(4y^2-30)/3, you can write the points (x,y) on the curve as a function only of the parameter y, like ((4y^2-30)/3,y). Now take d/dy to find a tangent. The RHS of the curve disappears when you differentiate.
     
  4. Jan 24, 2010 #3
    Sorry, I'm still a little confused. I see how you got x=(), but then are you saying I differentiate that equation, x(y), or do I plug that into T for x, and take grad(T)?
     
  5. Jan 24, 2010 #4

    Dick

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    ((4y^2-30)/3,y) parametrizes the curve in terms of the variable y. If I take d/dy of that vector I get a another vector that is tangent to the curve. You want to find the tangent vector at (2,3) to dot with your gradient at (2,3).
     
  6. Jan 24, 2010 #5
    But if I parametrize the equation in terms of only y, there will be no x to plug in the 2 of the (2,3) point. I would have curve=(4y^2-30)/3 i + 1 j. Differentiating that would give me (8/3)y i evaluated at (2,3).
     
  7. Jan 24, 2010 #6

    Dick

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    No, the curve is i*(4y^2-30)/3+j*y. You get (8/3)yi+j for the derivative. You don't need to plug x=2 into anything. If you know y, then you know x. Just plug in y=3. That gives you 2i+3j or (2,3) on the original curve.
     
  8. Jan 24, 2010 #7
    x=(4y^2-30)/3*i
    y=y*j

    So, S=(4y^2-30)/3*i + y*j

    dS=(8/3)y*i+1*j

    eval @ (2,3),

    dS=8*i+1*j

    So, I do (8xy^2*i + (8x^2y+9y^2)*j) DOT (8*i + 1*j) to get dT/dS

    Is this correct?
     
  9. Jan 24, 2010 #8

    Dick

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    I thought you were planning on normalizing the tangent vector so it was a unit vector.
     
  10. Jan 24, 2010 #9
    Oh,right. So it would be 8/sqrt(65)*i + 1/sqrt(65)*j

    Then dot grad(T) with this, and eval @ (2,3) to get dT/dS = 1152/sqrt(65)*i + 177/sqrt(65)*j?

    Is this correct?
     
  11. Jan 24, 2010 #10

    Dick

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    dT/dS isn't a vector, is it? It's a real number. It's the dot product of two vectors.
     
  12. Jan 24, 2010 #11
    Oh yeah, that's what I meant to do. Thanks for your help.
     
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