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How does these two equations relate?

  1. Oct 5, 2008 #1
    Hi, I'm in first year engineering and I have trouble understanding where this formula comes from in determining the relationship between the coefficient of volume expansion and the coeff of linear expansion.

    dV=(dV/dL)*dL = 3L^2*dL

    now i know that they changed the dV in the brackets to dL^3, but then I don't understand how dL^3/dL became 3L^2? Is there a part of calculus I am missing? does it have to do with a rate of change divided by a rate of change?

    Thanks!
     
  2. jcsd
  3. Oct 5, 2008 #2

    George Jones

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    Can you express V as a function of L?
     
  4. Oct 5, 2008 #3
    Volume = L^3, for a cube? i think that's the relationship there
     
  5. Oct 5, 2008 #4

    George Jones

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    What is dV/dL?
     
  6. Oct 5, 2008 #5
    delta V / delta L times delta L = delta V the delta L's cancel. they do it on wikipedia too like all of a sudden http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion it becomes 3Lo^2 and I don't understand
     
  7. Oct 5, 2008 #6

    George Jones

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    If y = x^3, what is dy/dx?
     
  8. Oct 5, 2008 #7
    then it is 3x^2, thats would be dy/dx. :S how about dX^3/dX? is it the same? OH or is it like dy/dx is the same as d/dX so doesn't matter what it is at the top?
     
  9. Oct 5, 2008 #8

    George Jones

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    Right, dy/dx = 3x^2.

    Now treat dy and dx as symbols small quantities that can be manipulated like other quantities.

    Multiplying both sides of dy/dx = 3x^2 by dx gives?
     
  10. Oct 5, 2008 #9
    umm dy = (3x^2)dx?
     
  11. Oct 5, 2008 #10

    George Jones

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    Right. Now change y to V and x to L throughout.

    Even though engineers and physicists routinely do these types of manipulations, they make pure mathematicians cringe.
     
  12. Oct 5, 2008 #11
    >.< ok that makes more sense now. I will try to hatch this in my brain. Thanks George Jones!
     
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