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Homework Help: Equality of functions

  1. Jun 8, 2005 #1
    hi,

    i was wondering if anyone could tell me how i should approach this problem:

    Let f:R->R^3 be a differentiable, vector-valued function and g:R->R be a strictly increasing vector-valued function. Let h = fog:R->R^3. Show that the paths traced by f and h are equal and that h'(t) = f'(g(t))g'(t). h is called the reparameterization of f by g.


    i have no idea where to start. i'm not sure how to put the info given in the question to use, to prove that f and h are equal, especially since it doesn't state the equations for f and g. any help is appreciated.
     
  2. jcsd
  3. Jun 8, 2005 #2

    Hurkyl

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    That's not what the problem asks you to prove...
     
  4. Jun 9, 2005 #3

    OlderDan

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    I'm not real fresh on this stuff, but it seems odd to me that g is being defined as a "vector function" in a 1-dimensional space, rather than calling it a scaler. In any case, it appears to me that as a starting point in a spatial representation you can write

    [tex] \vec f = f_x (x,y,x)\widehat i + f_y (x,y,x)\widehat j + f_z (x,y,x)\widehat k [/tex]

    [tex] \vec g = g(t)\widehat u [/tex]

    [tex] \vec h (t) = f_x \left[ x \circ g(t),y \circ g(t),x \circ g(t) \right] \widehat i + f_y \left[ x \circ g(t),y \circ g(t),x \circ g(t) \right] \widehat j + f_z \left[ x \circ g(t),y \circ g(t),x \circ g(t) \right] \widehat k [/tex]

    or something analogous in another representation in [tex]\Re ^3 [/tex].
     
  5. Jun 9, 2005 #4

    AKG

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    This doesn't seem to make any sense. Choose g = exp. Then h(R) need not be equal to f(R) so whatever "the path traced" means, I don't think the two can be equal. As for the other part, it's just the Chain Rule. If you only know the Chain Rule for "1-dimensional" functions, you can still use that rule to prove this more general case.
     
  6. Jun 9, 2005 #5
    an additional peice of info in the question, which i forgot to type, is:
    g is onto - that is, for every y in R, there is an x such that g(x) = y.

    anyway, i tried reparameterizing according to arclength. let s1, s2, s3 = arclength of f, g, and h.
    functions are f = f(t), and g = g(b).

    s1(t) = [integral]|f'(u)| du
    ds1/dt = |f'(t)|

    s2(b) = [integral]|g'(u)|du
    ds2/db = |g'(b)|

    s3(b) = [integral]|f'(u)|du
    s3/db = |f'(g(b))|
    = |f'(g(b))*g'(b)|

    i don't know where to go from here. i don't even know if i started off correctly. how do i put all that together to show that the path traced by f = path traced by h? any help would be appreciated.
     
  7. Jun 10, 2005 #6

    OlderDan

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    Perhaps there is some basic property of the function f that is being overlooked. What does f:R->R^3 tell you about the funtion f?
     
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