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for 0<=t<=21 the rate of change of the number of blakc flies on a coastal island at time t days is modeled by R(t)=3sqrt(t)cos(...Asked by Anon
for 0<=t<=21 the rate of change of the number of black flies on a coastal island at time t days is modeled by R(t)=3sqrt(t)cos(t/3) flies per day. There are 500 flies on the island at the time t=0. To the nearest whole #, what is the max # of flies for 0<=t<=21? plz solve this completely b/c the choices r: 1)500 2)510 3)520 4)530 5)540
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Answered by
drwls
dN/dt = 3*t^1/2*cos(t/3)
Use separation of variables.
dN = 3*t^1/2*cos(t/3)*dt
Integrate both sides.
N - N(t=0) = Integral of
3*t^1/2*cos(t/3)*dt, from t=0 to t
I don't have time to finish this. I already explained in a previous answer how to get the time of maximum fly population.
Use separation of variables.
dN = 3*t^1/2*cos(t/3)*dt
Integrate both sides.
N - N(t=0) = Integral of
3*t^1/2*cos(t/3)*dt, from t=0 to t
I don't have time to finish this. I already explained in a previous answer how to get the time of maximum fly population.
Answered by
Anon
is it from 0 to 21? or 0 to 4.7??
Answered by
drwls
The maximum fly population is at t = 3 pi/2 = 4.712 days. Integrate R to that limit.
The R function varies as follows: 0 at t=0, 2.835 at t=1, 3.334 at t=2, 2.807 at t=3, 1.411 at t=4 and 0 at t=4.712 The integral of R from 0 to 4.7 will be about ten. That would make the answer 510. Not much of a change.
The R function varies as follows: 0 at t=0, 2.835 at t=1, 3.334 at t=2, 2.807 at t=3, 1.411 at t=4 and 0 at t=4.712 The integral of R from 0 to 4.7 will be about ten. That would make the answer 510. Not much of a change.
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