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?

Printable View

- April 29th 2007, 01:26 PMcamherokidIntegral
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? - April 29th 2007, 02:11 PMecMathGeek
- April 29th 2007, 02:13 PMecMathGeek
- April 29th 2007, 02:19 PMecMathGeek
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