Problem 25 - Precalculus

Question

The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population in Yellowstone National Park was 52 in 1996 and 118 in \(1999 .\) Using these data, find a function of the form \(N(t)=N_{0} e^{k t}\) which models the number of wolves \(t\) years after \(1996 .\) (Use \(t=0\) to represent the year \(1996 .\) Also, round your value of \(k\) to four decimal places.) According to the model, how many wolves were in Yellowstone in \(2002 ?\) (The recorded number is \(272 .)\)

Step-by-Step Solution

Answer

The wolf population model is \( N(t) = 52e^{0.2716t} \), predicting approximately 270 wolves in 2002.

Step 1: Determine Initial Conditions

Identify the given data points. We know that in 1996 (which corresponds to \( t = 0 \)), the wolf population was 52. Thus, \( N_0 = 52 \). In 1999 (\( t = 3 \)), the population was 118.

Step 2: Set Up the Growth Equation and Solve for k

The model is given by the equation \( N(t) = N_0 e^{kt} \). Substitute the known values from 1999 into the equation: \( 118 = 52 e^{3k} \). We solve for \( k \) by first isolating \( e^{3k} \): \( e^{3k} = \frac{118}{52} \). Take the natural logarithm of both sides to find \( k \): \( 3k = \ln \left( \frac{118}{52} \right) \). Further, we solve for \( k \): \( k = \frac{1}{3} \ln \left( \frac{118}{52} \right) \approx 0.2716 \), rounding to four decimal places.

Step 3: Create the Population Model

Using the value of \( k \) found, the population model becomes \( N(t) = 52 e^{0.2716t} \). This function models the wolf population \( t \) years after 1996.

Step 4: Calculate the Population in 2002

Substitute \( t = 6 \) (since 2002 corresponds to 6 years after 1996) into the model: \( N(6) = 52 e^{0.2716 \times 6} \). Calculate the exponent: \( 0.2716 \times 6 = 1.6296 \). Therefore, \( N(6) = 52 \times e^{1.6296} \approx 270 \), when rounded to the nearest whole number.

Key Concepts

Law of Uninhibited Growth

The Law of Uninhibited Growth is a mathematical concept describing processes where the rate of change is proportional to the current quantity. It's commonly used in contexts like population growth, radioactive decay, and even finance. When counting animal populations, assuming a constant growth rate without any restrictions, we apply this law.

For exponential growth, we use the formula:

• \[ N(t) = N_0 e^{kt} \]

where:

• \( N(t) \) is the population at time \( t \),
• \( N_0 \) is the initial population size,
• \( e \) is the base of the natural logarithms,
• \( k \) is the growth rate constant,
• \( t \) is the time elapsed.

The growth is termed 'uninhibited' because external factors like limited resources are not considered. Hence, the population grows exponentially based solely on its current size and the growth rate. In real scenarios, this growth can initially occur but becomes limited over time due to environmental factors.

Population Modeling

Population modeling is a powerful tool for predicting and understanding changes in animal or human numbers over time. In the case of the wolf reintroduction to Yellowstone, an exponential model like \( N(t) = N_0 e^{kt} \) helps predict future population numbers based on past and initial data.

When modeling, data at two different times provides a foundation to calculate the constant \( k \), representing how the population grows. To get \( k \), it's essential to know:

• The initial population (\( N_0 \)),
• A later observed population,
• The time difference between observations.

These known values allow us to rearrange and solve the growth equation to find \( k \). Once \( k \) is known, it can be applied back into the equation to predict future population numbers.

For Yellowstone, it provided not just an understanding of the wolf recovery but assisted in planning and decision-making for the park's wildlife management.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept in calculus and exponential growth calculations. It's particularly significant when solving the exponential growth equations because it simplifies expressions involving \( e \), the base of natural logarithms (approximately 2.718).

When we encounter an equation like \( 52 e^{3k} = 118 \), taking the natural logarithm on both sides helps isolate the exponent and solve for the growth constant \( k \). This step turns exponential equations into linear ones:

• \[ 3k = \ln \left( \frac{118}{52} \right) \]

By solving this, we can pull the expression apart, getting:

• \[ k = \frac{1}{3} \ln \left( \frac{118}{52} \right) \]

The natural logarithm transposes multiplicative growth into additive time intervals, simplifying how we quantify and predict changes over time. Understanding this helps students apply similar transformations in various scientific and engineering fields, making it an essential part of exponential problem-solving.