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Homework Help: Problem deriving an equation

  1. Sep 12, 2011 #1
    im doing my traffic analysis homework and i dont understand how they went from "1" to "2" in the picture below. more specifically i dont know if im taking the derivative wrong or if they are (those who wrote the solution). if you distribute the "k" thats at the beginning of the equation, you will get:

    q = uf (k - [(k2))/kj]3.5)=0

    then if i distribute the 3.5 power, i should get:

    q = uf (k-[(k5.5)/(kj3.5)] = 0

    then if i take dq/dk of this, shouldnt i get:

    dq/dk= uf(1-[(5.5k4.5)/(kj3.5)]


    the pic below is a screenshot of the solution, and theyre getting dq/dk to be:

    dq/dk= uf(1-[(4.5k3.5)/(kj3.5)]

    are they wrong?

    Last edited: Sep 12, 2011
  2. jcsd
  3. Sep 12, 2011 #2


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    Keep in mind that (ab)c = abc . So (k2)3.5 is k7.
  4. Sep 12, 2011 #3
    oh yeah thats right, how did they get 4.5k3.5 then? wouldnt it be 7k6?
  5. Sep 12, 2011 #4


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    There seems to be some inconsistency between what you've typed and what you show in the attachment. Is the first equation you wrote supposed to have that "k2" in it?

    Also, the first line in your attachment (what you call (1) ) is not an equation. Where does "q" come from? This isn't any way to get from (1) to (2) in what you displayed...
  6. Sep 13, 2011 #5
    sorry about the first post, that was completely wrong. i added the powers instead of multiplying them.

    "equation" (1) is what "q" is equal to. there should be a "q=" in front of it, i accidentally cut it off while cropping the screenshot. the second equation is a derivative of the first (1) with respect to "k."
  7. Sep 13, 2011 #6


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    All right, I think this is sorted out now. There should not be a "k2" in that first equation and the asterisk in (1) of your attachment is a multiplication sign.

    The first equation in your original post should read [itex]q = k \cdot u_{f} [ 1 - (\frac{k}{k_{j}})^{3.5} ] [/itex], so [tex]q = k \cdot u_{f} [ 1 - (\frac{k^{3.5}}{k_{j}^{3.5}}) ] = u_{f} [ k - (\frac{k^{4.5}}{k_{j}^{3.5}}) ]. [/tex] I think you'll find at this point that the differential equation for dq/dk is correct. (I'm assuming kj is a constant; it is not necessary for q to equal zero in the first line -- it could equal any real constant for the result in (2) to give dq/dk = 0 .)
    Last edited: Sep 13, 2011
  8. Sep 13, 2011 #7

    ahh now i get it. you cant distribute the "k" before distributing the 3.5 power to the fraction (k/kj).

    thank you for your help, i guess i need to refresh distributive laws:P
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