- #1

- 20

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^{2}(t), -Cos

^{2}(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)