Mathematics T Coursework 2013 Answer Sem 1

 

Maths T Assignment: Logistic Growth Model ( GUIDE )

Question 1

Just find the answer in the internet. It's only a definition, you may rephrase it if you want.

Question 2

You simply need to play with the logistic growth equation given. dP/dt = rP (1 - P/k) isthe growth rate, so for the growth rate to be increasing, the sign must be positive and for the growth rate to be decreasing, the sign must be negative.Since rP will never give you a negative value (

is the positive constant and

 P 

, the population, would never be smaller than zero), for dP/dt to have a negative sign, (1 - P/k)must be negative.Hence if the population exceeds the carrying capacity (P > k), then P over k (P/k in theequation) would be bigger than 1, and therefore (1 - P/k) < 0, suggesting the growth rateis decreasing. If the population does not exceed the carrying capacity (P < k), then P/k <0 and hence (1 - P/k) > 0

Question 3

3(a)

Basically you change your answer in question 2 into word form. Describe how the population growth will vary. If the population exceeds the carrying capacity, then the population growth rate will decrease and subsequently the population will drop. If the population equals the carrying capacity, the growth rate is zero and if the population issmaller than the carrying capacity, then the growth rate is positive and the population insincreasing.

3(b)

What would the value of P for constant growth? And what would the values be if thegrowth rate is increasing and decreasing? You should already know the answers if you'vedone the previous 2 questions.

Question 4

Differentiate the logistic growth equation once, and you'll get

r/k (k - 2P)

. For dP/dt to bemaximum, the derivative of the first population must equal to zero. Hence for 

r/k (k - 2P)=

0

 ,

(

k - 2P 

) must equal to zero. You'll establish a relation between the carrying capacityand the population here.However, you need to verify that the

 P 

value you get is indeed the maximum value of dP/dt. To verify it, you differentiate its first derivative once again to get d3P/dt3, thesecond derivative of the logistic growth equation.When you've verified it, find the maximum value of dP/dt by substituting the value youfound into the logistic growth equation.

 

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