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Estimating partial derivatives

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    a metal plate is situated in the xy plane and occupies the rectangle 0<x<10 and 0<y<8 where x a y are measure in meters. The temperature oat the pooint x,y on the plate it T(x,y), where T is measured in degrees celcius.
    note the attached table

    a- estimate the values of teh partial derivatives of Tx(6,4) and Ty(6,4). What are the units of measure
    b - Estimate the values of Tu(6,4) where u = (i + j)/(2^1/2). Interpret your result
    c - Estimate the value of Txy(6,4)


    2. Relevant equations



    3. The attempt at a solution

    welp im not sure if correct but hers what i did

    a - t(x) = 6, because from point 6,4, moving down the x value the temp rises 6 degrees
    and so t(y) = -5. the units are degrees celsius i belive

    b - for b i pluged my x and y partials from part a into the i & j component and got u = 7.8 degrees celcius

    c - t(x) = 6 just as in part a. and from there point 8,4, i moved the right for a change of -6 degrees
     

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  3. May 5, 2009 #2

    HallsofIvy

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    What is "t"? Is it the same as T? If so, what do you mean by T(x)? T is a function of both x and y. Do you mean Tx(x,y), the derivate with respect to x?

    The temperature at (6,4) is 80 while the termperatures at (4, 4) and (8,4) are 78 and 96 respectively. you could estimate the rate of temperature there as (80- 78)/(6-4)= 1 or (96- 80)/(8- 6)= 3 or (96- 78)/(8- 4)= 2. Of those I would expect the last to be most accurate. You seem to have forgotten to divide by the "[itex]\Delta x[/itex]".

    Finally, the units are not "degrees celcius". Because this is a rate of change, the units are "degrees celcius per meter"- again, you did not divide by the "[itex]\Delta x[/itex]".

    Did you divide by [itex]\sqrt{2}[/itex]?

    This question is asking for a second derivative. Calculate the partial derivatives with respect to x at (6,4), (6,2), and (6,6), then do the same calculation for the partial derivative with respect to y.
     
    Last edited: May 5, 2009
  4. May 6, 2009 #3
    sorry i was being lazy t(x) was just the partial with respect to x

    ok so heres what i did

    T(x) = (86 - 80)/(8-6) = 3
    T(y) = (75-80)/(6-4) = -2.5

    T(u), where u = (i + j)/21/2 was i corect in saying that i = T(x) & j = t(y), so T(u) = .354

    for T(xy) dont i have to take the partial T(x), and then partial that with respect to y.

    how do i do that.
     
  5. May 11, 2009 #4
    Ok I think i got Txy(6,4)

    Tx(6,4) = 3
    Tx(6,6) = 2.5

    =(Tx)(6,6) - Tx(6,4))/delta T = (3-2.5)/2 = .25

    Can i get a ahmen
     
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