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Position Function and its antiderivative

  1. Oct 27, 2007 #1
    What is the physical siginificance of a position function x(t) and its antiderivative [tex]\int x(t)[/tex].


    Ex. a particles motion is defined by the function x(t) = [tex]x^{3} + \2x^{2} + 3x - 1[/tex]

    if its antiderivative is [tex]\int x(t)[/tex] = [tex]\frac{x^{4}}{4} + \frac{2x^{3}}{3} + \frac{3x^{2}}{2} - x[/tex]

    The values in which these two functions intersect is :
    [tex] x = .241732 [/tex] and [tex]x = 3.298644 [/tex]

    The area between these two curves is given by:

    [tex]\int^{3.298644}_{.241732} (x^{3} + \2x^{2} + 3x - 1) dx - \int^{3.298644}_{.241732} (\frac{x^{4}}{4} + \frac{2x^{3}}{3} + \frac{3x^{2}}{2} - x) dxdx [/tex] which is equal to 14.906747

    So i was just wondering what this meant if anything at all.
     
    Last edited: Oct 27, 2007
  2. jcsd
  3. Oct 28, 2007 #2
    Well... it seems that something is a bit off here. When you say you have a position function x(t), you say that the position is given by x, and it depends on some variable t. But, first of all, when you list the antiderivative, you actually are using incorrect and incomplete notation; shouldn't there be a "dt" at the end of that integral?

    Your notation is just totally incorrect. In your example, you define x(t) in terms of x; this does not make sense. When you write x(t), you imply that x(t) will be written in terms of the variable t. For example, x(t) = t, or x(t) = t^2 + t, and so on. Instead, you have written x(t) as a function of x; in other words, you're not actually writing x(t) at all, you're writing some other function, a function which depends on x, say, f(x). You then give the antiderivative of f(x) with respect to x.

    I think these are issues which you should think about and fix before moving on.
     
  4. Oct 28, 2007 #3

    arildno

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    In addition to mordechai's good post, I'd like to ask you:

    Do YOU know of any interesting physical quantity having units length*time?
     
  5. Oct 28, 2007 #4
    I think the point made by mordechai is irrelavent, any intelligent human being should be able to substitute t for x, seriously its not worth devoting an entire paragraph about. I made up the example off the top of my head, its not like i was going off of a problem from a book. so substitute t for x and we're good to go and for arildno, i can think of ft/sec, m/sec, and all of the other units of those kinds ft/sec^2 and so on....
     
    Last edited: Oct 28, 2007
  6. Oct 29, 2007 #5

    arildno

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    Where none of those is of the form (length unit)*(time unit).
     
  7. Oct 29, 2007 #6

    Integral

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    I think that the simple answer to your final question is: No. It does not mean anything.
     
  8. Oct 29, 2007 #7

    rcgldr

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    Perhaps you were thinking of a line integral? If the component of force in the direction of the line can be expressed as a function of position, then the line intergral would calculate the work done bewteen two points on the line.
     
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