Question

Because of the absence of predators, the number of rabbits on a small island increases at a rate of \(11 \%\) per month. If \(y(t)\) equals the number of rabbits on the island \(t\) months from now, find the rate constant \(k\) for the growth function \(y(t)=y_{0} e^{k t}\)

Short answer

Answer: The rate constant k is approximately 0.1044.

Step-by-step solution

Calculate the growth after the first month

Since the growth rate is \(11\%\), the number of rabbits after the first month will be \(1.11\) times the initial number of rabbits. If the initial number of rabbits is represented by \(y_0\), then the number of rabbits after one month is \(1.11y_0\).

Compare the growth after the first month to the exponential growth function

Now, let's compare the growth after the first month to the given exponential growth function \(y(t)=y_{0} e^{k t}\). Plug the values: \(t = 1\) (for one month) \(y(1) = 1.11y_0\) (number of rabbits after the first month) Use in the growth function: \(1.11y_0 = y_{0} e^{k(1)}\)

Solve for k

To solve for the rate constant \(k\), divide both sides of the equation by \(y_0\): \(1.11 = e^k\) Now, we can take the natural logarithm (ln) of both sides to isolate k: \(\ln(1.11) = k\) Now we can calculate the value of \(k\): \(k = \ln(1.11) \approx 0.1044\)

Final answer

The rate constant \(k\) for the growth function \(y(t)=y_{0} e^{k t}\) is approximately \(0.1044\).

Key concepts

Rate Constant

The rate constant, often denoted by the letter \( k \), is a fundamental concept in exponential growth models. It essentially measures how quickly the population grows. In the context of our rabbit population problem, the rate constant reflects the consistent percentage increase each month.

In this specific problem, the growth rate is given as \( 11\% \) monthly. To determine \( k \), we followed a process to find out by how much this growth rate corresponds to a constant rate. By comparing the growth after one month to the formula \( y(t) = y_0 e^{kt} \), we found \( e^k = 1.11 \). This equation expresses that the exponential factor \( e^k \) applied for one month must equal the growth factor \( 1.11 \).

To solve \( e^k = 1.11 \) for \( k \), we use the natural logarithm, which inverted the exponential function to expose the rate constant. This approach unlocked k's value as approximately \( 0.1044 \), explicating the rate at which the population would expand under these conditions.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a valuable mathematical tool when working with exponential functions. It's specifically useful in this problem to pull down the exponent of the base \( e \).

Why the natural logarithm? Because it's perfectly configured to work with the constant \( e \) (approximately equal to 2.71828), a fundamental component of continuous growth models. By using \( \ln \), we can reverse an expression like \( e^k \) into a linear form where \( k \) can be easily isolated and determined. This is exactly what we did to uncover the rate constant in our rabbit population task.

• Taking Natural Logarithms: When faced with an equation like \( 1.11 = e^k \), employ \( \ln \) to both sides to access the exponent, resulting in \( \ln(1.11) = k \).
• Importance: This method is crucial for solving exponential growth equations, providing the means to translate percentage increase into a clear numerical constant.

When we applied \( \ln \) correctly on 1.11, it visually helped in approximating \( \ln(1.11) \approx 0.1044 \), delivering the required \( k \).

Rabbit Population Dynamics

Rabbit population dynamics offer a fascinating insight into exponential growth. In the absence of natural predators, these populations can increase rapidly, which is precisely why this problem focuses on such a scenario.

Understanding rabbit population dynamics involves recognizing specific factors that contribute to their growth. Primarily, exponential growth functions like \( y(t) = y_0 e^{kt} \) describe the situation where populations grow by a fixed percentage over regular intervals, such as 11% per month in our case. It characterizes a consistent multiplication factor over each period.

• No Predators: Without predators, rabbits reproduce quickly, causing their numbers to rise sharply.
• Fixed Growth Rate: Set at 11% can succinctly predict future population sizes.

By accurately calculating the rate constant \( k \), we can model future rabbit populations effectively and anticipate changes over time. For instance, without intervention, a small number of initial rabbits can lead to millions within a short amount of time because of their ability to multiply exponentially.
These dynamics are critical for ecological planning and understanding how ecosystems might manage certain populations without achieving overpopulation.