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Chain rule / acceleration

  1. Mar 7, 2009 #1
    Hi, I'm new to these forums so not exactly sure where to place this question, although calculus seems a good bet, so here goes:

    I'm currently taking a mechanics course at my university (current subject is work/energy), and I'll just post this snippit from our textbook (Physics for Scientists and Engineers with Modern Physics 7th edition, Jewett/Serway):

    [snippit]

    [tex]
    \begin{equation*}
    W_{net}=\int^{x_f}_{x_i}\sum F\,dx
    \end{equation*}
    [/tex]

    Using Newton's second law, we substitute for the magnitude of the net force [itex]\sum F=ma[/itex] and then perform the following chain-rule manipulations on the integrand:

    [tex]
    \begin{align*}
    W_{net}&=\int^{x_f}_{x_i}ma\,dx=\int^{x_f}_{x_i}m\frac{dv}{dt}\,dx=\int^{x_f}_{x_i}m\frac{dv}{dx}\frac{dx}{dt}\,dx=\int^{v_f}_{v_i}mv\,dv\\
    W_{net}&=\frac{1}{2}{mv_f}^2-\frac{1}{2}{mv_i}^2
    \end{align*}
    [/tex]

    [/snippit]

    How is:

    [tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}[/tex]

    legal in this context?

    I know the Chain Rule states the following:

    [tex]
    \begin{equation*}
    \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
    \end{equation*}
    [/tex]

    But this is valid only (to my understanding) for the derivative of the composite of two functions.

    If anyone could help me sort this out, I'd be much obliged.
    Thanks.
     
  2. jcsd
  3. Mar 7, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi milesyoung! Welcome to PF! :smile:

    v can be defined as a function of t or of x

    for example, you're probably familiar with the constant acceleration equations

    v = u + at

    v2 = u2 + 2ax

    in the first, v is expressed as a function of t (only),

    in the second, v is expressed as a function of x (only) :smile:
     
  4. Mar 7, 2009 #3
    Thanks for the quick response.

    Forgive me if I'm being a bit dense about this, been trying to wrap my head around it for hours and I'm probably just missing something trivial.

    What I don't get is how:

    [tex]
    \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}
    [/tex]

    is possible.

    For me it would seem like you would need to be able to interpret v as a composite of two functions for the Chain Rule to have a valid use.
     
  5. Mar 7, 2009 #4

    tiny-tim

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    Homework Helper

    Nooo … the chain rule is for v a function of one variable (which itself is a function of another variable) …

    here, v is a function of the variable x, and x is a function of the variable t :wink:
     
  6. Mar 7, 2009 #5

    Nabeshin

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    Science Advisor

    You're probably having a hard time looking at it this way because it's not usually introduced in this form in most 1st semester calculus classes, so perhaps if you were to rewrite it like this:

    [tex]v(t)=v(x(t))[/tex]

    [tex]v'(t)=v'(x(t))*x'(t)[/tex]

    [tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}[/tex]

    Helps?
     
  7. Mar 7, 2009 #6
    @Tim: Didn't mean to imply that v needed to be interpreted as a function of two variables, just meant v as a composite function, as in [tex]v(x(t))[/tex] :smile:

    Thanks for the help. I think I just needed to mull over how v could be a function of x, which in turn is a function of t.
     
  8. Mar 7, 2009 #7
    You know that we have x(t) and v(t), so invert x(t) to get t(x), then plug this into v so you have v(t) = v(t(x)) where in my last expression v is a function of x alone.
     
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