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Reparamterizing, I think i got the answer but whats up with the t = 0?

  1. Oct 8, 2005 #1
    Hello everyone, Did i do this all right? I don't get what the point of this part is...
    THe directions say, Reparameterize the curve r(t) which is listed on my image. With respect to arc elnth measured from the point (1,0,1) in the direction of incresing t. Well if u plug in t = 0, u will get the point (1,0,1), so what>? to make the problem correct, do i just have to note that somewhere? Here is my work and the final answer, is that acceptable? i can't see if i'm right or not and its just for studying for an exam. Thanks.
    http://show.imagehosting.us/show/775217/0/nouser_775/T0_-1_775217.jpg
     
  2. jcsd
  3. Oct 8, 2005 #2
    I'm curious why you wrote:

    [tex]s(t)=\int_0^t\sqrt{2}e^{t}\,du[/tex]

    Did you mean dt? If so, then it should equal [itex]\sqrt{2}\left(e^t-1\right)[/itex].

    Alex
     
  4. Oct 8, 2005 #3
    Oh yeah my bad, sorry! hm..how did u get e^t-1? if [tex]sqrt{2}[/tex] is a constant, can't u just bring it out? and get... sqrt(2)[tex]\int[/tex]e^t? and isn't the integral of e^t just e^t?
     
  5. Oct 8, 2005 #4
    Yes, but remember that you are integrating from u=0 to u=t. What is the value of e0? It isn't 0 like you assumed in your work :smile:

    Alex
     
  6. Oct 8, 2005 #5
    ohh, t hanks again alex for the help...
    so i'd get t = ln|s/sqrt(2) + 1|; do u know where the t = 0 comes in? or do i just state, t = 0?
     
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