# Integral

• Apr 29th 2007, 02:26 PM
camherokid
Integral
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?
• Apr 29th 2007, 03:11 PM
ecMathGeek
Quote:

Originally Posted by camherokid
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?

For this, you need only to plug in 3 for t:

r(3) = (450.268) e^(1.12567*3) = 13185.2355160

Round this number to as many digits as you need for you solution.
• Apr 29th 2007, 03:13 PM
ecMathGeek
Quote:

Originally Posted by camherokid
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?

How can the rate of growth be
r(t) = (450.268) e^(1.12567t)

The coefficient (450.268) of the exponential function (e^(1.12567t)) should be equal to the initial population (400).
I don't see why you would have 450.268 in front of the e.
• Apr 29th 2007, 03:19 PM
ecMathGeek
Quote:

Originally Posted by ecMathGeek
For this, you need only to plug in 3 for t:

r(3) = (450.268) e^(1.12567*3) = 13185.2355160

Round this number to as many digits as you need for you solution.

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