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Mathematics 23 Online
OpenStudy (anonymous):

How can I find the period of this? <10cos⁡(4πe^(-t)),10sin⁡(4πe^(-t))> I should have said for one cycle.

OpenStudy (anonymous):

$e^{-t}$ is periodic in $t$ with a period of $2\pi i $

OpenStudy (dumbcow):

for this function, i dont think the period is constant

OpenStudy (dumbcow):

well imagine a particle accelerating along a cyclical path .... each cycle would take less time

OpenStudy (anonymous):

or not, i am a bit slow with my terminology

OpenStudy (dumbcow):

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

OpenStudy (anonymous):

you can always reduce the exponent of t modulo 2pi though

OpenStudy (dumbcow):

@Charollete , just read your updated post which cycle? for 0<t<2pi

OpenStudy (anonymous):

Sorry yeah the question says this "over what time interval does the object complete one lap?"

OpenStudy (dumbcow):

\[e^{-t} = \frac{1}{2}\] \[t = \ln 2\]

OpenStudy (anonymous):

so the 4pi doesn't affect it then?

OpenStudy (dumbcow):

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

OpenStudy (anonymous):

Thank you! that makes perfect sense!

OpenStudy (dumbcow):

yw

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