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

  • Thread starter eclayj
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  • #1
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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)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #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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