Exponential Growth of Bacteria

Date: 7 Aug 1995 08:19:31 -0400
From: Greg Sharpe
Subject: Growth & Decay

A colony of N bacteria increases with time according to the formula

                    dN
                    -- = kN       where K is constant.
                    dT

If initially there are N  bacteria show that the number at any time, N,
                        0

                   kt
is given by N = N e
                 0

If the number of bacteria doubles every 2 hours, how long (to the nearest
minute) will the colony take to treble?

Date: 8 Aug 1995 22:35:26 -0400
From: Dr. Ken
Subject: Re: Growth & Decay

Hello there!

This is an example of a differential equation.  To solve it, we'll use
integration:

dN
-- = kN
dt

dN = kN * dt

1/N * dN = k * dT                 Then integrate:

log[N] = k*t + C
N = e^(k*t + C) = e^C * e^kt

To find C (or rather e^C), we plug in known simultaneous values of N and t:

N  = e^C * e^0 = e^C
 0

So if you plug back in, you get the desired equation.

>If the number of bacteria doubles every 2 hours, how long (to the nearest
>minute) will the colony take to treble?

The information "doubles every 2 hours" lets us actually find k.  We'll plug
in k=0 and k=1.

Since we know N = N e^kt, we know 2*N  = N  , i.e. 2*N e^0 = N e^k, so
                   0                 0    1           0       0

e^k = 2.  Therefore k = Log[2].  Right?  So then we have

N = N e^(Log[2] * t).  We can find out how long it will take to treble by 
     0
figuring out when the amount of N is 3*N .  So we have the equation
                                        0   
3N = N e^(Log[2] * t), which you can tackle with a calculator.
  0   0

Hope this helps!

-K