Problem 63 - Calculus for Biology and Medicine

Question

A population of organisms grows according to a logistic growth model: $$ N(t)=\frac{K}{1+(K-1) e^{-r t}} $$ where \(r\) and \(K\) are positive constants. (a) Find \(\frac{d N}{d t}\). (b) Show that \(N(t)\) satisfies the equation $$ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right) $$ [Hint: Use the function \(N(t)\) for the right-hand side, and simplify until you obtain the derivative of \(N(t)\) that you computed in (a).] (c) Plot the per capita rate of growth \(\frac{1}{N} \frac{d N}{d t}\) as a function of \(N\), and note that it decreases with increasing population size.

Step-by-Step Solution

Answer

(a) \(\frac{d N}{d t} = -\frac{Kr(K-1)e^{-rt}}{(1 + (K-1)e^{-rt})^2}\). (b) Verified. (c) The rate decreases linearly.

Step 1: Differentiate with respect to time

To find \(\frac{dN}{dt}\), differentiate the given logistic growth function with respect to \(t\): \(N(t) = \frac{K}{1 + (K-1)e^{-rt}}\). Use the quotient rule: \(\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\). Let \(u = K\) (constant, so \(u' = 0\)) and \(v = 1 + (K-1)e^{-rt}\) (then \(v' = -(K-1)(-r)e^{-rt} = r(K-1)e^{-rt}\)).Therefore, \[\frac{dN}{dt} = \frac{0 \cdot v - K \cdot v'}{v^2} = -\frac{K \cdot r(K-1)e^{-rt}}{(1 + (K-1)e^{-rt})^2}. \]

Step 2: Re-arrange the differential equation

Compute the derivative in a familiar form by factoring and simplifying:Factor out common terms:\[ N(t) = \frac{K}{1 + (K-1)e^{-rt}}, \]\[-\frac{Kr(K-1)e^{-rt}}{(1 + (K-1)e^{-rt})^2} = r \cdot \frac{K}{1 + (K-1)e^{-rt}} \left(1 - \frac{K}{1 + (K-1)e^{-rt}}\right). \]This matches the form \( r N (1 - \frac{N}{K}) \). Thus the equation is verified when rearranged as required in part (b).

Step 3: Calculate the per capita growth rate

The per capita rate of growth is given by \(\frac{1}{N}\cdot \frac{dN}{dt}\).Substitute: \[\frac{1}{N} \frac{dN}{dt} = \frac{-Kr(K-1)e^{-rt}}{K} \cdot \frac{1}{1 + (K-1)e^{-rt}}, \]meaning: \[\frac{1}{N} \frac{dN}{dt} = r\left(1 - \frac{N}{K}\right). \]Notice that as \(N\) increases, \(\frac{1}{N} \frac{dN}{dt}\) decreases as this is equivalent to \(r(1 - \frac{N}{K})\), a decreasing linear function of \(N\).

Step 4: Plot the per capita growth rate

Plot the function \(\frac{1}{N} \frac{dN}{dt} = r(1 - \frac{N}{K})\). This is a linear function that intercepts the y-axis at \(r\) and has a slope of \(-\frac{r}{K}\).This plot will show a line decreasing from value \(r\) with a negative slope. It demonstrates how the per capita growth rate decreases as \(N\) approaches \(K\), confirming the slowdown in population growth as it nears the carrying capacity.

Key Concepts

Differential Equations

Differential equations play a crucial role in modeling real-world systems, including population dynamics. Specifically, when studying the logistic growth model, we use a differential equation to describe how a population changes over time. This equation encapsulates two core concepts: the rate at which the population grows and how this rate is affected by the size of the population itself.
In the context of the logistic growth model, we start with the function:\[ N(t) = \frac{K}{1 + (K-1) e^{-r t}} \]
This equation defines the population size, \(N(t)\), at any given time \(t\), based on the constants \(K\) (carrying capacity) and \(r\) (intrinsic growth rate).

To determine how quickly this population changes over time, we compute the derivative, \(\frac{dN}{dt}\). This involves the application of calculus, specifically the Quotient Rule, to determine how \(N(t)\) depends on \(t\). By differentiating it with respect to time, you obtain an equation describing the growth of the population.

• This derivative tells us the rate of change, showcasing that as \(N\) gets larger, the rate of change adjusts accordingly, often slowing down towards the maximum capacity \(K\).

Per Capita Growth Rate

In population dynamics, the per capita growth rate measures the growth rate of a population per individual. This gives insight into the efficiency or productivity of the population in replicating under given conditions.
For the logistic model, the per capita growth rate is derived from the equation:\[ \frac{1}{N}\frac{dN}{dt} = r\left(1-\frac{N}{K}\right) \]
This expression reveals that the increase in population size relative to the current population adjusts as more individuals are present.

• The term \(r\) sets the initial growth potential, whereas \(1-\frac{N}{K}\) scales this potential relative to how close the population is to its maximum capacity \(K\).
• As \(N\) approaches \(K\), the term \(1-\frac{N}{K}\) becomes smaller, reducing the growth rate per individual, a hallmark of the logistic growth pattern.

This aspect is important in biological systems, where resources such as food or space are finite, influencing how populations grow or shrink over time.

Quotient Rule in Calculus

The quotient rule is a fundamental tool in calculus, essential for differentiating functions that are ratios of two other functions. It becomes particularly useful when working with the logistic growth model, where the function relies on a fraction.
To apply the quotient rule, recall its formula:\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
In the logistic model scenario, we designate the numerator as \(u = K\) and the denominator as \(v = 1 + (K-1)e^{-rt}\).

• Here, \(u'\) is zero since \(K\) is a constant, while \(v'\) requires using the chain rule to find the derivative of \(v\).

By applying the quotient rule, we calculate the derivative \(\frac{dN}{dt}\) successfully, allowing us to explore how the population changes over time. This step is key to verifying the behavior proposed in part (b) of the original exercise as well as understanding how the growth rate decreases as \(N\) nears \(K\). This understanding highlights the subtle interplay between calculus and biological growth patterns.