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B What would the "correct" way of doing this integral be?

  1. May 29, 2016 #1
    v(x(t)), where v represents velocity and is a function of position which is a function of time.

    I have the equation: v dv/dx = 20x + 5 and want to solve for velocity. The way our professor solved it was by multiplying both sides by dx and integrating => ∫v dv = ∫20x+5 dx. I know doing this is non-rigorous since dv/dx can't be treated as a fraction. So how would I "correctly" do this?

    Generally ∫ ƒ(y) dy/dx dx = ∫ f(u) du by substitution or the "reverse" chain rule since dy/dx is the derivative of the composite function y inside f In my case however, ∫ v dv/dx , dv/dx is not the derivative of the composite inside v. How would I solve for v without doing something like multiplying both sides by dx?
     
    Last edited: May 29, 2016
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  3. May 29, 2016 #2

    Orodruin

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    You are just intrgrating over t. Note that v dt = (dx/dt) dt = dx so you can just apply substitution of variables to x instead of t.
     
  4. May 29, 2016 #3
    I made a few mistakes initially sorry. Corrected them. Please look at the post again.
     
  5. May 29, 2016 #4

    Orodruin

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    This does not really change the principle, it is still just integrating both sides with respect to a variable, in this case x, and changing the integral parametrisation.
     
  6. May 30, 2016 #5
    v dv/dx = 20x + 5 If I want to integrate both sides with respect to x, how do I do it?
    The RHS would be pretty straightforward.

    But how do I integrate this:
    ∫ v(x) dv/dx dx without "cancelling" the dx's
     
  7. May 30, 2016 #6

    Orodruin

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    You are not "cancelling" the dx. You are using the normal form for a change of variables.
     
  8. May 30, 2016 #7
    But for change of variables, the "inner" or composite function's derivative has to be present outside like: ∫f(x(t)) dx/dt dx = ∫f(y) dy.
    In this case ∫ v(x) dv/dx dx, dv/dx is not the derivative of the composite function but it is the derivative of v(x)
     
  9. May 30, 2016 #8

    Orodruin

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    You are changing variables to v ... Also, your expression for change of variables is wrong. Fix it with replacing dx by dt and it is exactly what you have.
     
  10. May 30, 2016 #9

    Orodruin

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    Of course, you could also just note that v dv/dx = d(v^2)/dx / 2 and integrate a total derivative.
     
  11. May 30, 2016 #10
    In the change of variable, ∫ g(f(x)) df/dx dx = ∫ g(u) du. But my case is more analogous to ∫ g(f(x)) dg/df df since I have ∫ v(x(t)) dv/dx dx
     
  12. May 30, 2016 #11

    Orodruin

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    No it is not. You just have g(f) = f.
    The fact that v and x generally depend on t here is irrelevant.
     
  13. May 30, 2016 #12
    Alright got it thanks
     
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