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Why do we learn differentials?

  1. Nov 29, 2003 #1

    ShawnD

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    One application of derivatives from first year calculus is something called differentials. The intent is to find the change of something based on the derivative of a function and some sort of varialbe like time or a distance or something.
    Let's say you have this formula:
    y = x^2
    now here is the derivative:
    dy/dx = 2x
    now if you bring the dx over, it looks like this
    dy = 2x dx

    In math class, these are meant to find changes in things. Let's say you wanted to find the change in y when x changes from 5 to 10. you would just fill in the equation like this:
    dy = 2(5)(5)
    dy = 50

    the dy is your change in y. the first 5 is your original x value. the second 5 is your change in x.

    The differential said the change is 50. Now lets see what the original equation says the difference is:
    final - original
    = x^2 - x^2
    = 10^2 - 5^2
    = 100 - 25
    = 75

    The two different equations give VERY different answers. They're not even close. Knowing this, why do we still learn these?
     
  2. jcsd
  3. Nov 29, 2003 #2

    jamesrc

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    dy/dx is the instantaneous rate of change of y with respect to x. Since y(x) = x2 is nonlinear, you can't expect dy/dx to equal Δy/Δx
     
  4. Nov 29, 2003 #3

    ShawnD

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    Exactly. If they don't work then why the hell do we learn them?
     
  5. Nov 29, 2003 #4

    Hurkyl

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    Because differentials work well then the quantities involved are small.

    Let's stick to the [itex]f(x)=x^2[/itex] example, but with a smaller differential... how about [itex]x=10[/itex] and [itex]\delta x=1[/itex].

    In this case, we have [itex]f(11)=121[/itex] and the differential approximation gives [itex]f(11) \approx f(10) + 1 * f'(10) = 120[/itex] which is pretty darn close.

    The relevant theorem is:

    [tex]
    f(x + \delta x) = f(x) + f'(x) \delta x + \varepsilon (\delta x) \delta x \
    \mathrm{where} \lim_{\delta x \rightarrow 0} \varepsilon(\delta x) = 0
    [/tex]

    In other words, the error term in the approximation of [itex]f(x+\delta x)[/itex] shrinks "quickly" as [itex]\delta x[/itex] approaches 0. In fact, if [itex]f(x)[/itex] is twice differentiable, you can prove that there exists a constant [itex]c[/itex] such that [itex]|\varepsilon (\delta x)| < c |\delta x|[/itex], so the error term is quadratic in [itex]\delta x[/itex]. (This is the Taylor remainder theorem; a differential approximation is just a first degree Taylor polynomial!)


    edit: finally got the LaTeX right. :smile:
     
    Last edited: Nov 29, 2003
  6. Nov 29, 2003 #5

    ShawnD

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    oh ok, that makes sense.
     
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