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Counting Bacteria with Calculus

Date: 2/15/96 at 10:36:37
From: Daniel Peterson
Subject: Math Problem

The number of bacteria in a culture increases from 3000 to 9000 
in 8 hours.  If the rate of increase is proportional to the number 
present, estimate the number of bacteria at the end of 24 hours.

Daniel Peterson

Date: 2/15/96 at 16:1:38
From: Doctor Byron
Subject: Re: Math Problem

Hi Daniel,

For answering this problem I'm going to assume you know a 
little basic calculus, since that is the best way to really understand 
where the answer to this type of problem comes from.

Let's call the bacteria population P(t) where t is the time.
P(0) is our initial value and is equal to 3000.  We also know that
P(8) = 9000.

Finally, we are told that the rate of increase of bacteria is 
proportional to the number of bacteria currently present.  This is 
the same as saying that the derivative of the population function 
is equal to some constant times the population.  We therefore 
P = k * ----  where k is an arbitrary constant.

Rearranging this expression slightly, we end up with:

 dt     dP
---- = ----
 k      P

This now makes finding P (relatively) easy, since we can integrate 
both sides.  Equations in which each variable can be isolated on a 
separate side of the equation are called "separable differential 
equations".  They are often much easier to solve than those in which 
the variables cannot be separated.

But to continue with the problem... integrating both sides results in:

t/k = ln(P) + C

[Normally the integral of 1/P would be natural log of the absolute 
value of P.  In this case, however, we know that a population must 
be positive, so this distinction isn't necessary.]

Applying the exponential function to both sides (i.e., raising e to 
the power of each side of the equation):

 t/k    [ln(P) + C]    ln(P)     C
e   =  e            = e      *  e  

(let A = e^C)

e   =  P * A

P = Be     (where B = 1/A)

Now we can use the two conditions from the problem to find the 
constants B and k.
P(0) = 3000 = B(e^0) = B

Therefore B = 3000.
P(8) = 9000 = 3000e

Solving this, we find that k = 7.282 (approximately)

Now it is possible to find P(24)
P(24) = 3000e         = approx. 81,000 bacteria

You may have noticed that the three values are following a 
progession proportional to the powers of 3:  3*1000, 3^2*1000, 
3^4*1000.  It ispossible to show that the value at 24 hrs is 
_exactly_ 81,000 bacteria.You can either do this analytically, 
as above, or realize that from the problem description the 
population will triple every 3 hours!  Inother words, now that 
you have seen how the answer can be derived there is a very 
efficient shortcut for figuring this type of problem out.  If you 
know the population increases by a certain factor in a set period 
of time, then it must continue to increase by that factor for each 
additional time period.

I know it was a little mean of me to go through all those equations 
when there is a quick shortcut, but learning the above method 
will aid you in solving many other related problems where cutting 
corners is not so easy.

-Doctor Byron,  The Math Forum

Associated Topics:
High School Calculus

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