Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 44 / 4 / b / Solution

Write $A_i={\varphi }_{i1}$, $b_i={\varphi }_{i2}>0$ and $k_i=-{\varphi }_{i3}={\theta }_{i3}$. ThenFor $k_i>0$, $A_i$ is the limiting circumference and $1/(1+b_i)$ is the fraction of that asymptote attained at the centered day $x=\overline{x}$. Under the code's transformations,The latter expression is the logistic function. The curve also satisfies the logistic growth equation $d{m}_i/dx=(k_i/s_x)m_i(1-m_i/A_i)$. Its inflection point is at half the asymptote. The original prior permits $k_i\le 0$, in which case it does not describe increasing growth.