A bacteria population starts with 400 bacteria and grows at a rate of
r(t)= (450.268) e^1.12567t bacteria per hour.
How many bacteria will there be after 3 hours?
Oops... I just noticed this is the growth rate and not the population at time t.
r(t) = (450.268) e^(1.12567*t)
P(t) = INT r(t) dt = INT (450.268) e^(1.12567*t) dt
P(t) = (450.268/1.12567)e^(1.12567*t) + C
At t = 0, P(t) = 400
400 = (400)e^(1.12567*0) + C --> C = 0
P(t) = (400)e^(1.12567*t)
Solving for the population at t = 3, we get
P(3) = (400)e^(1.12567*3) = 11713.2334663