Question

The population of a species is given by the function \(P(t)=\frac{K t^{2}}{t^{2}+b},\) where \(t \geq 0\) is measured in years and \(K\) and \(b\) are positive real numbers. a. With \(K=300\) and \(b=30,\) what is \(\lim _{t \rightarrow \infty} P(t),\) the carrying capacity of the population? b. With \(K=300\) and \(b=30,\) when does the maximum growth rate occur? c. For arbitrary positive values of \(K\) and \(b\), when does the maximum growth rate occur (in terms of \(K\) and \(b\) )?

Short answer

Answer: The carrying capacity of the population is 300. The maximum growth rate occurs at t = 1 year.

Step-by-step solution

a. Finding Carrying Capacity

To find the carrying capacity, we need to evaluate the \(\lim_{t \rightarrow \infty} P(t)\) with the given function $P(t)=\frac{K t^{2}}{t^{2}+b}\(. We have \)K = 300\( and \)b = 30$. Therefore, the carrying capacity is: $$\lim_{t \rightarrow \infty} \frac{300t^{2}}{t^{2}+30}.$$ To evaluate this limit, we can divide both the numerator and the denominator by \(t^2\), which gives: $$\lim_{t \rightarrow \infty} \frac{300}{1+\frac{30}{t^2}}.$$ Since \(\lim_{t \rightarrow \infty} \frac{30}{t^2} = 0\), we simplify the expression and get: $$\lim_{t \rightarrow \infty} \frac{300}{1} = 300$$ So, the carrying capacity of the population is 300.

b. Finding the Time of Maximum Growth Rate

To find when the maximum growth rate occurs, we need to find the critical points of the function \(P(t)\). First, we calculate the derivative of \(P(t)\) with respect to \(t\): $$\frac{dP(t)}{dt} = \frac{K(2t(t^2+b)-t^2(2t))}{(t^2+b)^2} = \frac{2Kb}{(t^2+b)^2}.$$ Now, to find the maximum growth rate, we look for when the derivative is at its highest point, which is when the derivative of the function with respect to \(t\) is equal to zero: $$\frac{d^2P(t)}{dt^2} = 0.$$ Finding the second derivative of \(P(t)\): $$\frac{d^{2}P(t)}{dt^{2}} = \frac{-8Ktb}{(t^2+b)^3}.$$ We are given \(K = 300\) and \(b = 30\). Now, we set the second derivative to zero and solve for \(t\): $$0 = \frac{-8\cdot 300\cdot 30t}{(t^2+30)^3}.$$ \(t=1\) (solve for t): The maximum growth rate occurs at \(t = 1\) year.

c. Maximum Growth Rate with Respect to \(K\) and \(b\)

For arbitrary values of \(K\) and \(b\), let's follow the same steps as in part (b). We already have the second derivative calculated as: $$\frac{d^{2}P(t)}{dt^{2}} = \frac{-8Ktb}{(t^2+b)^3}.$$ Set the second derivative to zero and solve for \(t\): $$0 = \frac{-8Ktb}{(t^2+b)^3}.$$ We can see that this equation holds true for t = 1, regardless of the values of \(K\) and \(b\). Therefore, for arbitrary positive values of \(K\) and \(b\): The maximum growth rate occurs at \(t = 1\) year.

Key concepts

Carrying Capacity

In the study of population dynamics, the concept of carrying capacity is crucial. It represents the maximum population size that an environment can sustain indefinitely. This is directly influenced by resources and ecological conditions. For any population, the limiting factor is often represented mathematically as a constant, denoting the highest number that can be supported.

In the provided function, where population over time is modeled by:

• \( P(t) = \frac{K t^{2}}{t^{2} + b} \)

The carrying capacity can be determined by evaluating the behavior of this function as time \(t\) approaches infinity. By simplifying the expression and understanding the impact of each component, we determine it through limit evaluation:

• \( \lim_{t \rightarrow \infty} P(t) = \lim_{t \rightarrow \infty} \frac{K t^{2}}{t^{2} + b} \)

With given values of \(K = 300\) and \(b = 30\), the limit evaluates to 300. This indicates that the population's carrying capacity, or the maximum sustainable size, is 300 individuals based on the resource limits simulated by \(K\) and \(b\). Understanding this concept aids in recognizing the effects of resource limits on population growth.

Derivative Calculations

For analyzing changes in population growth, derivative calculations play a fundamental role. By differentiating a function with respect to time, we identify how its rate changes at any point, which is essential for determining the maximum growth rate. In our case, the function is:

• \( P(t) = \frac{K t^{2}}{t^{2} + b} \)

The first derivative, \( \frac{dP(t)}{dt} \), tells us how the population is changing per unit of time:

• \( \frac{dP(t)}{dt} = \frac{2Kbt}{(t^{2} + b)^{2}} \)

The process involves applying differentiation rules for quotients, and solving this reveals crucial points in the population function. Setting the second derivative \( \frac{d^2P(t)}{dt^2} \) to zero helps find when the growth rate is maximized:

• \( \frac{d^{2}P(t)}{dt^{2}} = \frac{-8Ktb}{(t^{2} + b)^{3}} \)

By solving for zero, it’s established that maximum growth occurs at \(t = 1\) year, no matter the values of \(K\) and \(b\), showcasing how factor derivatives help predict growth patterns.

Limit Evaluation

Limit evaluation provides insight into the behavior of functions as inputs grow large or approach certain critical points. In population models, limits help predict long-term outcomes like carrying capacity. By applying limits to:

• \( P(t) = \frac{K t^{2}}{t^{2} + b} \)

We can simplify complex expressions for easier analysis. As \(t\) approaches infinity, the function simplifies by focusing on dominant terms. Dividing each term by \(t^2\):

• \( \lim_{t \rightarrow \infty} \frac{K}{1 + \frac{b}{t^2}} \) simplifies to \( \lim_{t \rightarrow \infty} \frac{K}{1} = K \)

This evaluation is critical, showing that the function approaches a real number as \(t\) increases. In our scenario, this result helps verify that the carrying capacity is indeed \(K\), contributing to clearer, more insightful predictions of population dynamics. Through limits, we effectively capture and comprehend the stabilizing trends of the modeled population growth over time.