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Homework Help: DuBois Formula Problem

  1. Jun 2, 2006 #1
    Ok, my teacher did this problem today, but there is 1 step I don't understand. If someone could explain to me how it's done, that'd be great.

    Here's the problem:

    Using the DuBois formula: S = 0.01W^(0.25) H^(.75)

    Solve for H(Height) as a function of S (Surface Area) for people of fixed weight(W) 70

    Answer:

    S = 0.01(70)^(.25) H^(.75)

    S = 100/70^(.25) = H^(.75)
    //This is the step I don't understand. Where does the 100 come from and what happened to 0.01? Am I an idiot?

    34.57 S = H^(.75)

    34.57^(4/3) S^(4/3) = H

    H = 112.6 S^(4/3)
     
    Last edited: Jun 2, 2006
  2. jcsd
  3. Jun 2, 2006 #2

    arildno

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    Dearly Missed

    Evidently, you don't understand it, since you are too sloppy about how you write stuff.
     
  4. Jun 2, 2006 #3

    AKG

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    It should not be "S =" at the beginning, it should be "S x", i.e. "S times". The whole thing should be:

    S = 0.01(70)^(.25) H^(.75)

    S x 100/70^(.25) = H^(.75)

    34.57 S = H^(.75)

    34.57^(4/3) S^(4/3) = H

    H = 112.6 S^(4/3)
     
  5. Jun 2, 2006 #4
    Thanks for the support
     
  6. Jun 2, 2006 #5
    AGK, how did you swap S and H^(.75)
     
  7. Jun 2, 2006 #6

    Hootenanny

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    [tex]S = 0.01(70^{0.25}) \cdot H^{0.75}[/tex]

    [tex]S = \frac{70^{0.25}}{100}\cdot H^{0.75}[/tex]

    [tex]\frac{S}{H^{0.75}} = \frac{70^{0.25}}{100}[/tex]

    [tex]\frac{1}{H^{0.75}} = \frac{70^{0.25}}{100S}[/tex]

    [tex]H^{0.75} = \frac{100S}{70^{0.25}}[/tex]

    [tex]H^{0.75} = S \times \frac{100}{70^{0.25}}[/tex]

    Does that make more sense now?

    ~H
     
  8. Jun 2, 2006 #7
    Yes, that makes more sense, thanks.

    I'm not sure how you got the 1 in step 4 though.
     
  9. Jun 2, 2006 #8

    Hootenanny

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    From here;

    [tex]\frac{{\color{red}S}}{H^{0.75}} = \frac{70^{0.25}}{100}[/tex]

    Just divide both sides 'S'

    [tex]\frac{1}{H^{0.75}} = \frac{70^{0.25}}{100{\color{red}S}}[/tex]

    Do you see?

    ~H
     
  10. Jun 2, 2006 #9
    Yes, Thanks.
     
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