Logistic+Equation


 * Logistic Equation **

So far we have looked at exponential growth and bounded growth (like Newton's Law of Cooling). The logistic equation combines both of these together. If we think about a population of rabbits on a small island the population will increase exponentially as they breed. At somepoint however there will be too many for the available food so pure exponential growth is not a good model. We can refine it by having exponential growth that then slows as it approaches a bounded value or a carrying capacity (in the case of the island). This gives us a S shaped graph like the one to the right.

Mathematically we model this as a growth proportional to the product of exponential growth (kN) and bounded growth ( k(P-N) )to get: math \dfrac{dN}{dt}=cN(P-N) \\ \text{or} \\ \dfrac{dN}{dt}=kN(1-\frac{N}{P}) \\ math where N is the population at time t, c and k are the constants of proportionality for the given situation and P is the carrying capacity (maximum value of N)

For a given initial condition N(0)=N o and k>0, P>N o >0 we can get the general case:

math N(t)=\dfrac{PN_0}{N_0+(P-N_0)e^{-kt}} \\ \text{or} \\ N(t)=\dfrac{P}{1+(\frac{P}{N_0}-1)e^{-kt}} \\ math If we don't know all of P, No and k we can use simultaneous equations with the above general formula to find these values (using a CAS). If we are given P and N o and another point then we generally will want to use calculus to solve as shown below:

=Example= Mr Thompson has a petri dish with a certain bacteria colony in it. Initially the area of the bacteria is 2cm 2. After 4 days it has grown to 10cm 2. Given that the petri dish has a total area of 50cm 2 and the area of the bacteria grows at a rate proportional to its area **and** the area left find: a) The Area at time t days after the initial growth. b) The area after **another** 6 days to 3 decimal places c) Sketch the graph of N vs t

math \dfrac{dA}{dt}=kA(50-A), A_0=2 \\ \dfrac{dt}{dA}=\dfrac{1}{kA(50-A)} \\ math Integrating both sides gives math kt=\displaystyle{\int} \dfrac{1}{A(50-A)}dA \\ math Using partial fractions on the right hand side: math \dfrac{1}{A(50-A)} = \dfrac{B}{N}+\dfrac{C}{50-A} \\ \dfrac{1}{A(50-A)} = \dfrac{50B-BA+CA}{A(50-A)} \\ \dfrac{1}{A(50-A)} = \dfrac{(C-B)A+50B}{A(50-A)} \\ \text{so } C-B=0 \\ C=B \\ \text{and 50B =1 so} \\ B=C=\frac{1}{50} \\ math Now our integral becomes: math kt=\frac{1}{50} \displaystyle{\int} \frac{1}{A}+\frac{1}{50-A} \\ kt=\frac{1}{50} (log_e(A)-log_e(50-A))+c \text{ note modulus signs are not needed as } 2<A<50 \\ math Therefore using our initial condition to find c we get math 0=\frac{1}{50}log_e(\dfrac{1}{24})+c \\ c=\frac{1}{50}log_e(24) \\ \text{subbing back in then using our other piece of information to get k gives} \\ kt=\frac{1}{50}log_e(\dfrac{24A}{50-A}), \text{ N(4)=10} \\ 4k=\frac{1}{50}log_e(\dfrac{240}{40} \\ k=\frac{1}{200}log_e(6) \\ k=0.008958797...... \\ math We will leave this as k in our equation for the moment but we know its value for calculations. Our last step is to rearrange our expression to get A in terms of t. math 50kt=log_e( \dfrac{24A}{50-A}) \\ \dfrac{24N}{50-A} = e^{50kt} \\ 24A=(50-A)e^{50kt} \\ 24A=50e^{50kt}-Ae^{50kt} \\ 24A+Ae^{50kt}=50e^{50kt} \\ A(t)=\dfrac{50e^{50kt} }{24+e^{50kt} } \\ \text{or } A(t)=\dfrac{50}{1+24e^{-50kt}} \\ math math A(10)=\dfrac{50}{1+24e^{-500k}} \\ A(10)=39.303 cm^2 \\ math
 * a)** We can set up the differential equation:
 * b)** Here we simply need to sub in t=10 (note it means 6 days after the first 4 days)
 * c)** Using a CAS or similar gives the graph below