# The population of a certain bacteria grows with time t as p(t) = \frac{1}{1 + Ae^{-kt}} ...

## Question:

The population of a certain bacteria grows with time {eq}t {/eq} as

{eq}p(t) = \frac{1}{1 + Ae^{-kt}} {/eq}

where {eq}A \enspace and \enspace k {/eq} are positive constants.

a. Find {eq}\lim_{t\rightarrow \infty} p(t) {/eq}

b. Find the rate of population growth and evaluate it at {eq}t = 0 {/eq}

## Population Growth

In population dynamics, there are several functions that model the number of individuals in a population. There is the ordinary exponential growth function or logistic function. The given function for this problem is a type of a logistic function and it differs from an ordinary exponential growth because a logistic function has an upper limit or maximum value.

The population of a certain bacteria that grows with time, t, is given by the function

{eq}\displaystyle p(t) = \frac{1}{1 + Ae^{-kt}}. {/eq}

a. Calculating the limit {eq}\displaystyle \lim_{t \to \infty} p(t) {/eq}:

{eq}\displaystyle \begin{align*} \lim_{t\to \infty} p(t) &= \lim_{t\to \infty} \frac{1}{1 + Ae^{-kt}} \\ &= \frac{1}{1 + Ae^{-k(\infty)}} \\ &= \frac{1}{1 + Ae^{-(\infty)}} \\ &= \frac{1}{1 + A(0)} \\ \lim_{t\to \infty} p(t) &= \boxed{ \frac{1}{1} \text{ or } 1} \end{align*} {/eq}

Therefore, the limit of the function as {eq}t \to \infty {/eq} is {eq}\boxed{ 1} {/eq}.

b. To solve the rate of population growth, we need the take the derivative with respect to time of the function, p(t).

Solving for the derivative.

{eq}\displaystyle \begin{align*} \frac{dp}{dt} &= \frac{d}{dt} \frac{1}{1 + Ae^{-kt}}\\ &= \frac{d}{dt} (1 + Ae^{-kt})^{-1} \\ \text{solving the derivative using chain rule:} \\ &= (-1) -Ake^{-kt}(1 + Ae^{-kt})^{-2}\\ \frac{dp}{dt} &= \boxed{\frac{Ake^{-kt}}{(1 + Ae^{-kt})^2}} \end{align*} {/eq}

Evaluating the rate of growth at {eq}t =0 {/eq}.

{eq}\displaystyle \begin{align*} t &= 0,\ \frac{dp}{dt} = \boxed{\frac{Ake^{-kt}}{(1 + Ae^{-kt})^2}}\\ \frac{dp}{dt}\bigg|_{t=0} &= \frac{Ake^{-k(0)}}{(1 + Ae^{-k(0)})^2}\\ &= \boxed{ \frac{Ak}{(1 + A)^2}} \end{align*} {/eq}

Therefore, the growth rate is given by the function {eq}\displaystyle \boxed{\frac{dp}{dt} = \frac{Ake^{-kt}}{(1 + Ae^{-kt})^2}} {/eq} and its growth rate at {eq}t = 0 {/eq} is {eq}\boxed{\frac{Ak}{(1 + A)^2}} {/eq}. 