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Why the curve r(t) approaches a circle as t approaches infinity

  1. Sep 25, 2013 #1
    Both statements 1 and 2 are given as an explanation of why the original statement is true, but I don't understand why you can use statement 2 (since in the original vector equation you do not have Sin2(t), -Cos2(t))

    Show why r(t) = <e-t, Sin(t), -Cos(t)> approaches a circle as t →∞.

    1. As t→∞, x(t)→0

    2. And as t→∞, (y(t))2+(z(t))2=1

    What I do not see, intuitively, is why the second statement is used and how it shows that r(t) = <e-t, Sin(t), -Cos(t)> approaches a circle as t →∞.

    I keep thinking that as t→∞, Sin(t) and -Cos(t) do not exist (the functions oscillate). And I don't understand how 2., (y(t))2+(z(t))2=1, relates beck to the original vector equation, Sin(t), -Cos(t)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2013 #2

    Office_Shredder

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    Try this simpler question. Explain why r(t) = <0, sin(t), -cos(t)> is a circle. If you can do that you're 90% of the way to answering this problem.
     
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