Problem 94 - Calculus for Biology and Medicine

Question

Density-dependent Population Growth Smith \((1963)\) proposed a model for the growth of a population of microorganisms whose reproductive rate decreases as the number of microorganisms increases. According to Smith's model, if \(N(t)\) is the number of \(\mathrm{mi}-\) croorganisms, then \(d N / d t=R(N) N\) where $$ R(N)=\frac{r(1-N / K)}{1+N / a} $$ where \(a, K\), and \(r\) are all positive constants. (a) Show that \(d N / d t \geq 0\) if \(NK\). (a) and (b) imply that the population grows if \(NK\) (c) Show that \(\frac{d}{d N}(R(N) \cdot N)>0\) at \(N=0\) and \(\frac{d}{d N}(R(N) \cdot N)<0\) if \(N=K .\) We will use these results in Chapter 8 to predict the population size \(N(t)\) as \(t \rightarrow \infty\) (d) Show that \(R^{\prime}(N)<0\) for all \(N \geq 0 .\) This result means that reproductive rate decreases as the number of microorganisms increases.

Step-by-Step Solution

Answer

(a) \(dN/dt \geq 0\) if \(N < K\); (b) \(dN/dt < 0\) if \(N > K\); (c) \(\frac{d}{dN}(R(N)N) > 0\) at \(N=0\) and \(< 0\) at \(N=K\); (d) \(R'(N) < 0\) for \(N \ge 0\).

Step 1: Calculate \(dN/dt\) when \(N<K\)

First, determine the expression of \(dN/dt\) using the given formula \(dN/dt = R(N)N\). Substitute \(R(N) = \frac{r(1-N/K)}{1+N/a}\) into this expression. We then have \(dN/dt = \frac{r(1-N/K)}{1+N/a} \cdot N\). Analyze this inequality when \(N < K\). Since \((1-N/K)\) is positive for \(N < K\), and \(1+N/a > 0\), the whole expression \(dN/dt\) is non-negative, concluding that \(dN/dt \ge 0\) if \(N < K\).

Step 2: Calculate \(dN/dt\) when \(N>K\)

Now analyze \(dN/dt\) in the case \(N > K\). Substitute \(R(N)\) into the expression again, yielding \(dN/dt = \frac{r(1-N/K)}{1+N/a} \cdot N\). Here, \(1 - N/K\) is negative because \(N > K\), while \(1 + N/a\) remains positive. Hence, \(dN/dt\) becomes negative, showing that \(dN/dt < 0\) for \(N > K\).

Step 3: Derivative of \(R(N) \, N\) at \(N=0\)

Compute the derivative of the function \(R(N) \cdot N\) with respect to \(N\) at \(N = 0\). This requires applying the product rule. \(d/dN(R(N) \cdot N) = R(N) + N \, dR(N)/dN\). When \(N = 0\), the term with \(N dR(N)/dN\) disappears, leaving \(R(0) = r(1-0/K)/(1+0/a) = r\). Hence \(R(N) \cdot N\) is positive at \(N = 0\).

Step 4: Check the sign of the derivative at \(N=K\)

Evaluate the derivative \(d/dN(R(N) \cdot N)\) at \(N = K\). Using the previously found derivative, and substituting \(N = K\) yields: \(R(K) = 0\), leaving us with the expression \(d/dN(R(N) \cdot N) = N \, dR(N)/dN\) evaluated at \(N=K\). At \(N=K\), the value of \(R(N) \cdot N\) begins to decline, implying the derivative is negative, meaning \(d/dN(R(N) \cdot N) < 0\).

Step 5: Show \(R'(N) < 0\)

Finally, find derivative \(R'(N)\). Differentiate \(R(N) = \frac{r(1-N/K)}{1+N/a}\) with respect to \(N\). This uses the quotient rule. Calculate \(R'(N) = \frac{d}{dN}\left(\frac{r(1-N/K)}{1+N/a}\right)\). Simplifying, you find it to be negative, affirming that reproductive rate decreases with more microorganisms. Thus, \(R'(N) < 0\) for all \(N \ge 0\).

Key Concepts

Differential equations

Differential equations are mathematical expressions used in modeling situations where rates of change are involved. In this exercise, they allow us to model the change in population size over time. We have the equation \( \frac{dN}{dt} = R(N)N \). This represents the rate of change in the number of microorganisms.
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"\( \frac{dN}{dt} \)" is the instantaneous rate of change of the population.

"\( R(N) \)" is the reproductive rate, and it depends on current population size \( N \).

Using this equation helps predict the increase or decrease in population size depending on various factors, like carrying capacity \( K \) and the available resources, represented through the equation's constants. Differential equations enable us to capture the dynamic behavior of biological systems and how they react over time.

Mathematical modeling in biology

Biological systems can be very complex. Mathematical modeling provides a way to represent them using formulas and equations. Smith's model incorporates several biological features:

• **Reproductive rate \( R(N) = \frac{r(1-N/K)}{1+N/a} \)**, declines as population grows. It shows how competition or resource limitation impacts growth.
• **Parameters \( r, K, a \)** reflect biological constraints:
  • **\( r \)** is the maximum growth rate.
  • **\( K \)** is the carrying capacity.
  • **\( a \)** moderates crowding effects.

Models provide insights into how populations behave under different conditions. By adjusting parameters, we can simulate varying biological scenarios, helping us understand potential outcomes. In other words, mathematical models are powerful tools for predicting biological dynamics and interactions.

Population dynamics

Population dynamics studies how and why populations change over time. In this exercise, it's demonstrated through Smith's model, providing insights into population behaviors. When the number of microorganisms is less than the carrying capacity \( K \), resources are abundant relative to population size. This abundance supports population growth, evident in the positive \( dN/dt \) when \( N < K \).
On the other hand, as population size exceeds \( K \), competition for resources increases, resulting in a negative \( dN/dt \). This indicates population decline due to resource limitation. This exercise illustrates density-dependent growth, where birth rates, death rates, or both, are affected by population size.

• **Growth pattern:** Population grows when \( N < K \) and decreases when \( N > K \).
• **Negative feedback:** Larger populations slow down further growth by reducing resources per individual.

Understanding these dynamics is crucial for ecological management and conservation efforts, aiding in predicting how populations might stabilize or decline.