Chain rule say what?

  • Thread starter vish22
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  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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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??
 
  • #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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