Are the Paths of Reparameterized Functions Identical?

[jessawells]

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.

 

[Hurkyl]

to prove that f and h are equal

That's not what the problem asks you to prove...

 

[OlderDan]

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.

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

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

\vec g = g(t)\widehat u

\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

or something analogous in another representation in \Re ^3.

 

[AKG]

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.

 

[jessawells]

an additional piece 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.

 

[OlderDan]

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?