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Confusion on chain rule substitution

  1. Aug 16, 2011 #1
    I'm confident in my math ability, but how is it that by using the chain
    W_{x_1 \rightarrow x_2} = \int^{x_2}_{x_1} m \frac{dv}{dt} dx
    can be turned into
    W_{x_1 \rightarrow x_2} = \int^{x_2}_{x_1} m \frac{dv}{dx} \frac{dx}{dt} dx = \int^{v_2}_{v_1}mv dv
    I understand the concept of using chain rule to make velocity depend on position which is dependent on time

    however even with the above in mind integration by substitution is defined as:
    [tex]\int^{b}_{a} f(g(t))g'(t)dt = \int^{g(b)}_{g(a)} f(x)dx [/tex]
    which means the integral "dx" should be a "dt" instead since time is the base independent variable.
    [tex] \int^{x_2}_{x_1} m v'(x(t))x'(t) dt [/tex]
    and after substitution should be in the form
    [tex] \int^{x(x_2)}_{x(x_1)} mv'(x)dx [/tex]
    but obviously that makes no sense either since the limits of integration are in terms of "x" and x(t) needs to have inputs of time. where did i go wrong?
  2. jcsd
  3. Aug 16, 2011 #2

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    Hi skyfire101! :smile:

    You're mixing up your x and t a bit.

    I'll try to make it a little clearer by putting it this way:
    \int^{x_2}_{x_1} m v'(t(x)) dx
    &=& \int^{x_2}_{x_1} m v'(x) x'(t(x)) dx \\
    &=& \int^{x_2}_{x_1} m v'(x) v(t(x)) dx \\
    &=& \int^{x_2}_{x_1} m v(t(x)) v'(x) dx \\
    &=& \int^{v(x_2)}_{v(x_1)} m v dv
  4. Aug 16, 2011 #3


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    He's mixing up his v's.

    If you're going to convert to notating things as functions, you either have be careful about the functions. If all variables are representing functions of time, then you have to introduce an entirely new function that expresses v as a function of x. If you call that function f, you have
    v(t) = f(x(t))​
  5. Aug 17, 2011 #4
    thank you so much for the responses, however im a little confused....
    specifically here;
    \int^{x_2}_{x_1} m v'(x) x'(t(x)) dx
    how did you get the
    it looks like you have x and t depending on each other, is that correct? please explain :)
  6. Aug 18, 2011 #5


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    This looks like a piece of the proof of the Work-Energy Theorem as it is often presented in Introductory Physics textbooks.

    My objection to this proof is that velocity may or may not be a "well defined" function of x. (If the particle reverses direction on its way from x1 to x2, then the velocity, v, is a so called "multi-valued function" of x.)

    I would rather use a different intermediate step.

    [itex]\displaystyle W_{x_1 \rightarrow x_2} = \int^{x_2}_{x_1} m \frac{dv}{dt} dx[/itex]
    [itex]\displaystyle =\int^{t_2}_{t_1} m \frac{dv}{dt}\frac{dx}{dt} dt\ ,\text{ where }x_1=x(t_1) \text{ and }x_2=x(t_2)\,.[/itex]

    [itex]\displaystyle =\int^{t_2}_{t_1} mv \frac{dv}{dt} dt[/itex]

    [itex]\displaystyle =\int^{v_2}_{v_1} mv dv\ ,\text{ where }v_1=v(t_1) \text{ and }v_2=v(t_2)\,.[/itex]

    [itex]\displaystyle =(1/2)m{v_2}^2-(1/2)m{v_1}^2[/itex]
  7. Aug 18, 2011 #6

    I like Serena

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    It's how you applied the chain rule.

    The chain rule is:
    (f o g)'(t) = f'(g(t)) g'(t)​

    You applied this to v as a function of x and x as a function of t:
    (v o x)'(t) = v'(x(t)) x'(t)​

    But since you are integrating with respect to x, you are implicitly using t as a function of x (which you actually shouldn't assume in general as SammyS already pointed out).
    Note that t(x) = x-1(x) if we're using the function x(t).
    Sorry about the mix-up caused by using x and t as variables, as well as functions, which is confusing as Hurkyl already pointed out.
    I've bolded the functions to distinguish them. :)

    So we have:
    (v o x)'(t(x)) = v'(x(t(x))) x'(t(x))​
    x(t(x)) = x​
    this becomes:
    (v o x)'(t(x)) = v'(x) x'(t(x))​
    Last edited: Aug 18, 2011
  8. Aug 19, 2011 #7
    thanx so much it all makes sense now! :D
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