*A Tale of two cheeses - *Adrian Oldknow a_oldknow@compuserve.com

*Chapter 2 : The growth of the Listeria - (expurgated version)**
*Scientists have shown that Listeria bacteria in unpasteurised cheeses at about 25&d;C increase at a rate of about 20% per second. The average number of dormant Listeria in such a piece of Camembert would be 1000. Assuming that this is the count

*t* 0 1 2 3 4 5

*c* 1000 1200

What would the graph of *c* against *t* look like? How many Listeria bacteria would be on the Camembert at the time the Cheddar broke the sound barrier? Is this likely? Give some reasons ...

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This is exactly the same as the compound interest model! If we assume that the rate is constantly changing then we have the differential equation: _{ } with initial condition *c*= *c*_{0} at time *t*= 0. Otherwise we have the difference equation:_{} . With a time step of _{} we just have a geometric progression (multiplicative sequence): _{ } with the solution: _{} . Again we can illustrate this with either sequence mode plotting or by using lists.

The blue graph is the solution of the difference equation, the red graph is the solution of the differential equation _{} If we take smaller time steps then the finite approximation should approach the theoretical solution. Move the slider to change the value of the time step dt and to see the resulting change to graph..

So here t = n.dt , so _{,
}so we need to explore the limit:

Here TI InterActive! has used computer algebra to find the analytic solution, but you can use a spreadsheet or graphical calculator or lists to explore the limits of both (1+1/*n*)* ^{n}* and (1+

If we work from the differential equation we have a problem because the right hand-side is a function of the dependent variable *c*, not the independent variable *t*! We can get round this by thinking of time as a function of count, and instead of using _{} we can use_{ } .

So the gradient function of the graph of *t* against *c* is a reciprocal function, and its anti-derivative will be a logarithmic function:_{ } with initial condition:*c*= *c*_{0} at time *t*= 0, so _{}, and _{} . We must then invert this to recapture *c* as a function of *t*:

_{}. The analysis is exactly the same if we use definite integrals to calculate the area under the *t* against *c* graph: _{ }