eclayj
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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)
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)