How can I find the period of this? <10cos(4πe^(-t)),10sin(4πe^(-t))> I should have said for one cycle.
$e^{-t}$ is periodic in $t$ with a period of $2\pi i $
for this function, i dont think the period is constant
well imagine a particle accelerating along a cyclical path .... each cycle would take less time
or not, i am a bit slow with my terminology
i think you could get it by saying... \[e^{-t} = \frac{n}{2}\] where n is the nth cycle notice this will give multiples of 2pi inside cos/sin functions
you can always reduce the exponent of t modulo 2pi though
@Charollete , just read your updated post which cycle? for 0<t<2pi
Sorry yeah the question says this "over what time interval does the object complete one lap?"
\[e^{-t} = \frac{1}{2}\] \[t = \ln 2\]
so the 4pi doesn't affect it then?
well thats where the 1/2 comes from t>0 so e^-t is getting smaller than 1 starting at cos(4pi) ...one completer cycle for cosine is 2pi how much time to get cos(2pi) thus e^-t must be 1/2
just verified with wolfram http://www.wolframalpha.com/input/?i=parametric+plot%2810*cos%284pi*e%5E%28-t%29%29%2C+10*sin%284pi*e%5E%28-t%29%29%29+for+t%3D0+to+log%282%29
Thank you! that makes perfect sense!
yw
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