Problem 33 - Calculus for Biology and Medicine

Question

Suppose the size of a population at time \(t\) is \(N(t)\) and its growth rate is given by the logistic growth model $$ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right), \quad t \geq 0 $$ where \(r\) and \(K\) are positive constants. (a) Graph the growth rate of the population \(\frac{d N}{d t}\) as a function of population size, \(N\), assuming that \(r=2\) and \(K=100\), and find the population size for which the growth rate is maximal. (b) Show that whatever the value of the parameters \(N\) and \(K\), \(f(N)=r N(1-N / K), N \geq 0\), is differentiable for \(N>0\), and compute \(f^{\prime}(N)\).

Step-by-Step Solution

Answer

The growth rate is maximal at a population size of 50. For all \(N > 0\), \(f(N)\) is differentiable, and \(f'(N) = r(1 - \frac{2N}{K})\).

Step 1: Understanding Logistic Growth Model

The logistic growth model is an equation that describes how a population changes over time. It is given by \( \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right) \), where \( r \) is the growth rate and \( K \) is the carrying capacity. As \( N \) increases, the term \( 1-\frac{N}{K} \) decreases, limiting growth as \( N \) approaches \( K \).

Step 2: Graphing Growth Rate as a Function of Population Size

For part (a), set \( r=2 \) and \( K=100 \). The equation becomes \( \frac{d N}{d t}=2N\left(1-\frac{N}{100}\right) \). The function is a parabola opening downwards with a vertex (maximum point). To find the point of maximum growth rate, complete the square or use the vertex form formula for quadratics.

Step 3: Finding Maximum Growth Rate

The vertex of the quadratic \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Here, \( a = -\frac{2}{100} \), \( b = 2 \), and \( x = N \). So, the population size that maximizes growth is \( N = -\frac{2}{2 \times -\frac{2}{100}} = 50 \). The maximum growth rate occurs when the population size is 50.

Step 4: Proving Differentiability of Logistic Growth Function

For part (b), the function \( f(N) = rN\left(1 - \frac{N}{K}\right) \) must be differentiable for \( N > 0 \). Since both \( rN \) and \( \frac{N}{K} \) are polynomials, their product is differentiable.

Step 5: Computing the Derivative \( f'(N) \)

Differentiate \( f(N) = rN\left(1 - \frac{N}{K}\right) \) to find \( f'(N) \). Use the product rule: \( f'(N) = r\left(1 - \frac{N}{K}\right) + rN\left(-\frac{1}{K}\right) = r - \frac{2rN}{K} \). Thus, the derivative is \( f'(N) = r\left(1 - \frac{2N}{K}\right) \).

Key Concepts

Differentiation

Differentiation is a key concept in calculus used to find the rate at which a function is changing at any given point. In the context of the logistic growth model, differentiation allows us to understand how the population growth rate changes with respect to time.
When we differentiate the function that describes the growth rate of a population, we can determine critical information about when the population grows the fastest and when it starts to slow down.
The logistic growth rate function is given by:

• \( \frac{d N}{d t} = r N\left(1 - \frac{N}{K}\right) \)

Differentiating this function with respect to \(N\), using the product rule, helps to examine changes in growth rate:

• By finding the derivative \(f'(N) = r\left(1 - \frac{2N}{K}\right)\), we can identify at what population level the rate of change is maximized or minimized.

This mathematical process reveals the dynamics within the growth model by showing how changes in population size affect growth rate. Keeping the principles of differentiation in mind aids in predicting and managing real-life populations effectively.

Population Dynamics

Population dynamics involve studying how populations change over time, particularly in regard to size and composition. The logistic growth model is a classic example of examining these dynamics, as it incorporates essential factors like growth rate and carrying capacity.
As modeled by the logistic equation:

• \( \frac{d N}{d t} = r N\left(1 - \frac{N}{K}\right) \)

This equation captures typical behavior seen in populations: an initial rapid growth that slows and stabilizes as resources become limited.
The model accounts for:

• Exponential Growth: When \(N\) is much smaller than \(K\), the population grows at a rate close to \(rN\).
• Resource Limitation: As \(N\) approaches \(K\), growth slows due to increased competition for resources.
• Self-Regulation: Population sizes stabilize around the carrying capacity, preventing over-exploitation of resources.

Understanding these dynamics provides insights into real-world population issues, such as ecological balance, conservation efforts, and sustainable resource management.

Carrying Capacity

Carrying capacity, denoted as \(K\) in the logistic growth model, is an important concept that refers to the maximum population size that an environment can sustain indefinitely.
This limit is determined by available resources such as food, habitat, water, and other ecological factors that a species needs to survive.
In the logistic model:

• The term \(1 - \frac{N}{K}\) acts as a limiter: when \(N\) nears \(K\), this term approaches zero, slowing the growth rate \(\frac{d N}{d t}\).

Carrying capacity has crucial implications:

• Stability: Populations close to their carrying capacity exhibit stable dynamics over time.
• Ecosystem Health: Ensures that populations do not exceed what their environment can provide, maintaining ecological balance.
• Sustainable Practices: Helps in managing species for conservation, allowing for adjustments in land and resource use to maintain balance.

By understanding carrying capacity, policymakers and ecologists can create better strategies to preserve biodiversity and ensure sustainable population management.