Question

Population models 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 \(\lim P(t),\) 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 is 300. Question: When does the maximum growth rate occur for the given population function with \(K=300\) and \(b=30\)? Answer: The maximum growth rate occurs at time \(t=\sqrt{30}\). Question: When does the maximum growth rate occur with arbitrary values of \(K\) and \(b\)? Answer: The maximum growth rate occurs at time \(t=\sqrt{b}\).

Step-by-step solution

Calculate the Limit of the Population Function

Given \(P(t)=\frac{K t^2}{t^2 + b}\) with \(K=300\) and \(b=30\), the carrying capacity of the population is defined as the limit when \(t\rightarrow\infty\). Therefore, we need to calculate \(\lim_{t\rightarrow\infty}P(t)\): $$ \lim_{t\rightarrow\infty}P(t) = \lim_{t\rightarrow\infty} \frac{300t^2}{t^2 + 30} = \lim_{t\rightarrow\infty} \frac{300}{1 + \frac{30}{t^2}} = 300. $$ So, the carrying capacity of the population is 300.

Calculate the Derivative of the Population Function

To find the maximum growth rate, we need to calculate the derivative of the population function and find where the derivative is equal to zero. First, let's find the derivative of \(P(t)\): \(P'(t) = \frac{d}{dt}\left(\frac{Kt^2}{t^2 + b}\right) = \frac{2Kt(t^2 + b) - Kt^2(2t)}{(t^2 + b)^2} = \frac{2Kbt}{(t^2 + b)^2}\).

Solve for Maximum Growth Rate

Now that we have found the derivative of the population function, we need to solve for when it is equal to zero. However, since \(K\), \(b\), and \(t\) are all positive, the derivative \(P'(t)\) will never be equal to zero. Therefore, we must find where the derivative is undefined. This occurs when the denominator of the fraction \(P'(t)\) is equal to zero: \((t^2 + b)^2 = 0 \Rightarrow t^2 + b = 0\). Given \(K=300\) and \(b=30\), we can solve for when the maximum growth rate occurs: \(t^2 + 30 = 0 \Rightarrow t = \pm\sqrt{30}\). Since \(t\geq0\), we have that the maximum growth rate occurs at time \(t=\sqrt{30}\).

Solve for Arbitrary Maximum Growth Rate

Finally, we need to find the time for maximum growth with arbitrary values of \(K\) and \(b\). We can reuse the equation \(t^2 + b = 0\), and instead of substituting \(b=30\), we just leave the arbitrary value: $$ t^2 + b = 0 \Rightarrow t = \pm\sqrt{b}. $$ Again, since \(t\geq0\), the maximum growth rate occurs at time \(t=\sqrt{b}\).

Key concepts

Carrying Capacity

Understanding the concept of carrying capacity is crucial in population models. It refers to the maximum population size an environment can sustain indefinitely, given the available resources like food, habitat, water, and other necessities in the environment.

In the context of the given population function, \(P(t) = \frac{Kt^2}{t^2 + b}\), the carrying capacity is reached when time \(t\) approaches infinity. As time grows indefinitely, the population stabilizes and reaches its upper limit, which, in this scenario, is dictated by the parameter \(K\). With \(K=300\) and \(b=30\), the limit of this function as \(t\) goes to infinity is 300, indicating that the environment can support a population of up to 300 individuals of that species.

Maximum Growth Rate

The maximum growth rate in an environment occurs when the population growth is at its peak. This rate reflects the most rapid increase the population can experience and is a pivotal point in understanding population dynamics.

By calculating the derivative of the population function, we investigate how the population grows over time. Observing where the derivative \(P'(t)\), which symbolizes the rate of change of the population, reaches its highest positive value, we discover when the population's growth rate is maxed out. However, in our specific function, since all involved parameters are positive, the derivative will never mathematically equal zero. Instead, we look for when the derivative is undefined, revealing an essential behavior of the model. Accordingly, in our example with \(K=300\) and \(b=30\), the population's maximum growth rate occurs at time \(t=\sqrt{30}\) years.

Derivative of Population Function

To further interpret population models, it’s necessary to grasp the concept of the derivative of the population function, as it conveys the rate at which the population size is changing at any given time.

In calculus, the derivative of a function at any point gives us the slope of the tangent line to the function's curve at that point. For the population model \(P(t)\), the derivative \(P'(t) = \frac{2Kbt}{(t^2 + b)^2}\) represents the instantaneous growth rate of the population. It tells us how fast the population is increasing or decreasing at the time \(t\). Consequently, for a population model with arbitrary positive constants \(K\) and \(b\), the derivative helps us pinpoint the precise moment when the population experiences maximum growth, which, as calculated previously, occurs at \(t=\sqrt{b}\) years.