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Chain rule say what?

  1. Sep 5, 2012 #1
    ok stupid question probably-
    take v(velocity) to be a function of x and x to be a function of t(time).
    then dv/dt=vdv/dx thats cool
    but in the hint in problem 2.12 classical mechanics by john r taylor he equates vdv/dx and 1/2(dv^2)/dx
    that is- vdv/dx=1/2(dv^2)/dx
    Could someone please explain this?
     
  2. jcsd
  3. Sep 5, 2012 #2
    Just assume, v2 = y , then differentiate both sides w.r.t x, you have,
    d(v2)/dx = dy/dx ...........(1)
    2v dv/dx = dy/dx....................(2)

    and then eliminate dy/dx to solve 1 and 2, and you have your result.....and I think this should be moved to homework or something.
     
  4. Sep 5, 2012 #3
    thanks universal i had just got that part,also in the next step of hint he separates the differentials like so-
    md(v^2)=F(x)dx
    now how is it possible to integrate the lhs wrt v^2??
     
  5. Sep 6, 2012 #4

    HallsofIvy

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    For any function, f, the integral of df is just f (plus the "constant of integration) - that's the "Fundamental Theorem of Calculus".

    [tex]\int m d(v^2)= m\int d(v^2)= mv^2+ C[/tex]
    assuming that m is a does not depend on v.
     
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