1. ## mathematical models

Show that the solution of the logistic equation dN/dt = N(a-bN)
has an inflection point at a population equal to 1/2 the saturation level. (population level a/b called the saturation level)

Explain why the sketches in fig 38.2 do not have inflection points
( it is a simple logistic growth model but has no inflection points, only a line with only growth and another with only decay)

Thanks for any help!

2. Originally Posted by dixie
Show that the solution of the logistic equation dN/dt = N(a-bN)
has an inflection point at a population equal to 1/2 the saturation level. (population level a/b called the saturation level)

Explain why the sketches in fig 38.2 do not have inflection points
( it is a simple logistic growth model but has no inflection points, only a line with only growth and another with only decay)

Thanks for any help!
Option 1:

To find points of inflection you solve $\frac{d^2 N}{dt^2} = 0$ and test the nature of the solution.

$\frac{dN}{dt} = N(a - bN) \Rightarrow \frac{d^2 N}{dt^2} = \frac{d}{dt} \left[N(a - bN)\right] = \frac{d}{dN} \left[N(a - bN)\right] \cdot \frac{dN}{dt} = (a - 2bN) \cdot \frac{dN}{dt}$.

Therefore $0 = a - 2bN \Rightarrow N = \frac{a}{2b}$.

To justify the existence of a point of inflection at $N = \frac{a}{2b}$ you need to show that $\frac{dN}{dt}$ has a turning at $N = \frac{a}{2b}$.

Note that $\frac{dN}{dt} \neq 0$ (why?)

Option 2:

If $\frac{dN}{dt}$ has a turning point then N = N(t) has a point of inflection. The point of inflection occurs at the value of N corresponding to the N-coordinate of the turning point of $\frac{dN}{dt}$.

Since $\frac{dN}{dt}$ is a quadratic function it's not difficult to:

1. establish that $\frac{dN}{dt}$ has a turning point.
2. find the N-coordinate of the turning point.