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Reparamterizing a curve

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    "Reparameterize the curve with respect to the arc length measured from the point where t=0 in the direction of increasing t. r(t)=e^(2t)*cos2t i +2 j + e^(t2)*sin2t k"


    2. Relevant equations



    3. The attempt at a solution

    the method we were taught is take r'(t) and then find |r'(t)| and then integrate that.

    but in every example we were given, the derivative was a real number, not a function (IE: 12, sqrt(23) etc)

    i took the derivative and got [-2e^(2t)sin(2t)+2te^(2t)cos(2t)]i +0j + [2e^(2t)cos(2t)+2te^(2t)sin(2t)] k

    then |r'(t)|=(after simplifying it i got) =2e^(2t)sqrt(1+t^2) which seems nice and simple now....but how to i integrate that...


    the form they used was s=s(t)=integral from 0-t of |r'(u)|du
    for every other problem of this nature
     
  2. jcsd
  3. Oct 14, 2007 #2

    Dick

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    Where did the t's in front of the exponentials in your derivative come from? They definitely shouldn't be there.
     
  4. Oct 14, 2007 #3
    this derivative? [-2e^(2t)sin(2t)+2te^(2t)cos(2t)]i +0j + [2e^(2t)cos(2t)+2te^(2t)sin(2t)] k


    r(t)=e^(2t)*cos2t i +2 j + e^(t2)*sin2t k

    ill just do the firts part as "k" is the same as "i"

    f'(t) "e^(2t)*cos2t"

    u'v*v'u corrrect?

    2t*e^t*cos2t + e(2t)*-sin2t

    or did i do that wrong?
    i thought the derivative of e^(2x)=2xe(2x) but maybe its juts 2e^(2x)? (this is calc 3 and i took calc 1 and 2 2 years ago so i kinda forget them! :(
     
  5. Oct 14, 2007 #4

    Dick

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    Yes, it's just 2e^(2x). It's a pretty easy chain rule.
     
  6. Oct 14, 2007 #5
    so doing it the RIGHT way i get the derivative is

    [-2e^(2t)sin(2t)+2e^(2t)cos(2t)]i +0j + [2e^(2t)cos(2t)+2e^(2t)sin(2t)] k

    which gives me a simplified |r'(t)| of 2e^(2t)sqrt2?
    that look right now
     
  7. Oct 14, 2007 #6
    and assuming thats right i would just integrate 2e^(2t)sqrt2? and get e^(2t)sqrt2? and then evaluate from o to t?
    which would give me e^(2t)sqrt2-sqrt(2)
     
    Last edited: Oct 14, 2007
  8. Oct 14, 2007 #7

    Dick

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    So what's your new parameter t? So far, you only have arc length.
     
  9. Oct 14, 2007 #8
    e^(2t)sqrt2-sqrt(2) would be the arc legnth

    so the new parameter is since s=e^(2t)sqrt2-sqrt(2)
    s+sqrt2=e^(2t)sqrt2
    (s+sqrt2)/sqrt2=e^(2t)
    ln((s+sqrt2)/sqrt2)=2tlne
    ln((s+sqrt2)/sqrt2) /2 =t


    original e^(2t)*cos2t i +2 j + e^(t2)*sin2t

    final e^(ln((s+sqrt2)/sqrt2))*cos(ln((s+sqrt2)/sqrt2)) i +2 j + e^((ln((s+sqrt2)/sqrt2)))*sin(ln((s+sqrt2)/sqrt2))
     
  10. Oct 14, 2007 #9

    Dick

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    Gotta confess, I'm not checking all the details for you. But it looks about right. You can check yourself. Take d/ds and see if you get a unit vector. BTW you could have skipped the integration constant. Why?
     
    Last edited: Oct 14, 2007
  11. Oct 14, 2007 #10
    what do you mean by the integration constant? the 2?
     
  12. Oct 14, 2007 #11

    Dick

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    No, the sqrt(2). The lower limit in your integral. It doesn't hurt you to include it though. Don't worry about it.
     
  13. Oct 14, 2007 #12
    you mean the sqrt2 is (s-sqrt2)/sqrt2 ??

    can you explain why i don't need it pleasE? :D
     
  14. Oct 15, 2007 #13

    Dick

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    I mean there is not much difference between writing s+sqrt(2) and just s. It's just a change in the origin of the parametrization.
     
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