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Geometric meaning of Mean Value Theorem

  1. May 12, 2008 #1
    I'd like the geometric meaning of the Mean Value Theorem. Say for instance I had a function of velocity that varied as [tex]t^{3}[/tex] + [tex]3t^{2}[/tex] + 3t +1. I consider the interval [0,4].
    So by MVT, I have a number c in [0,4] such that f'(c)(4) = f(4) - f(0). What does that mean? That there is an accelaration in that interval equal to the net change in accelaration? Meaning that the slope at 4 minus the slope at 0 is equal to the net change in slope?

    Could someone PLEASE explain?
  2. jcsd
  3. May 12, 2008 #2
    The physical meaning you are looking for is that, between 0 and 4 seconds, the instantaneous acceleration must be equal to the average acceleration at some time. This can happen multiple times, but the mean value theorem guarantees it happens at least once. Geometrically, the average acceleration is the slope of the secant line between (0, 1) and (4, 125) in your example. The instantaneous acceleration at a point c is the slope of the tangent line at c, or f'(c).

    Intuitively, this makes sense, because the average acceleration should be less than the maximum acceleration and more than the minimum acceleration in that time period (unless it is constant). It seems reasonable that at some time in that time period, the instantaneous acceleration would hit every value in between the minimum and maximum, thus it will hit the average at some time c.
  4. May 12, 2008 #3
    Another way of thinking about it is this. Let's say that there were some function f (that fulfills all necessary requirements for MVT to hold) such that MVT didn't hold. Then since f' is continuous (or assume that if that isn't in the hypotheses, since it's always true in "normal" phenomena), it must either always be less than the average acceleration or always greater than the average acceleration (since it can't jump from one side to the other). But in that case it would have to shoot too short or too far from where the average would take you...but that contradicts where the object ends (by choice of f).
  5. May 12, 2008 #4
    Yet another even more geometric way of looking at it would be just straight integrate your unknown function and the "average" function. If you assume (like my last post) that f' is always less than the average or always greater than the average, then the integrals of the f and the average function will (strictly) differ over the time period (because we're talking continuous functions) and thus would once again give us a contraction.
  6. May 13, 2008 #5
    Can I find the average of a function using this method?
    Last edited: May 13, 2008
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