Problem 6 - Algebra and Trigonometry

Question

The frog population in a small pond grows exponentially. The current population is 85 frogs, and the relative growth rate is 18% per year. (a) Find a function that models the population after \(t\) years. (b) Find the projected population after 3 years. (c) Find the number of years required for the frog population to reach 600.

Step-by-Step Solution

Answer

(a) \( P(t) = 85e^{0.18t} \); (b) 145 frogs; (c) approximately 11 years.

Step 1: Understanding Exponential Growth

Exponential growth can be modeled using the formula \( P(t) = P_0 e^{rt} \), where \( P(t) \) represents the population at time \( t \), \( P_0 \) is the initial population, \( r \) is the growth rate, and \( t \) is time in years. The current population \( P_0 \) is 85 frogs, and the growth rate \( r \) is 18%, or 0.18 as a decimal.

Step 2: Modeling the Function

To model the function, we substitute the known values into the formula: \( P(t) = 85 e^{0.18t} \). This function will represent the frog population after \( t \) years.

Step 3: Calculating the Population After 3 Years

Substitute \( t = 3 \) into the model \( P(t) = 85 e^{0.18t} \) to find \( P(3) = 85 e^{0.54} \). Using a calculator, compute \( e^{0.54} \) and then multiply by 85 to find the population after 3 years.

Step 4: Solving for the Exponential

First, calculate \( e^{0.54} \) which approximates to 1.718. Therefore, \( P(3) \approx 85 \times 1.718 \approx 145.03 \). Hence, the population after 3 years is approximately 145 frogs.

Step 5: Solving for Time to Reach 600 Frogs

To find when the population reaches 600, set the equation to 600: \( 85e^{0.18t} = 600 \). Divide both sides by 85 to isolate \( e^{0.18t} \): \( e^{0.18t} \approx 7.059 \).

Step 6: Using Natural Logarithm to Solve for Time

Take the natural logarithm (ln) on both sides: \( \, \ln(e^{0.18t}) = \ln(7.059) \, \) which simplifies to \( 0.18t = \ln(7.059) \, \). Calculate \( \ln(7.059) \), which approximates to 1.957, and solve for \( t \): \( t \approx \frac{1.957}{0.18} \approx 10.87 \).

Step 7: Rounding to the Nearest Year

Round 10.87 up to the nearest whole number since time in years must be a whole number. Thus, it will take approximately 11 years for the population to reach 600 frogs.

Key Concepts

Frog Population Modeling

Frogs are known for their incredible ability to adapt and reproduce in environments like small ponds. When modeling frog populations, it is common to use mathematical functions that predict future numbers based on a variety of factors. One popular method is to utilize exponential growth models. The key is identifying the initial population, the growth rate, and then applying these to the model. This helps predict how many frogs will exist in the future.
Frogs in a pond typically follow exponential growth patterns when there's adequate space, food, and no overwhelming predators. This type of growth signifies that as the population becomes larger, it grows at an increasingly faster pace. This assumption enables scientists and ecologists to predict population trends and plan for potential environmental impacts.

Exponential Functions

Exponential functions are a powerful mathematical tool, used in various scientific fields to model growth processes. The general form of an exponential growth function is:

• \( P(t) = P_0 e^{rt} \)

where:

• \( P(t) \) is the population at time \( t \)
• \( P_0 \) is the initial population
• \( e \) is the base of the natural logarithm, approximately equal to 2.718
• \( r \) is the rate of growth
• \( t \) is time in years

Plugging a known initial population and growth rate into this equation allows us to model and understand how a population will change over time. In our example, the initial population \( P_0 \) is 85 frogs, and the rate of growth \( r \) is 18% per year, or 0.18. Substituting these values gives us the exponential function that models the frog population: \( P(t) = 85 e^{0.18t} \). This equation is essential for predicting future population sizes.

Relative Growth Rate

The relative growth rate is crucial in exponential growth models. It describes how fast the population is growing compared to its current size. This rate is usually expressed as a percentage. In our scenario, the frog population's relative growth rate is 18%. This means that each year, the population increases by 18% of its current size.
Understanding this concept is beneficial when considering population dynamics because it allows us to estimate how quickly a population, such as these frogs, could reach significant numbers. This can be particularly important when managing wildlife populations or planning conservation efforts. It can help set expectations and inform decisions in population management strategies.

Natural Logarithm Calculations

Natural logarithms are used in various scientific calculations, especially when dealing with growth and decay processes. The natural logarithm function, denoted as \( \ln \), is the inverse of the exponential function. When solving equations where the unknown variable is in the exponent, taking the natural logarithm can help isolate that variable.
For instance, if you need to determine when a population will reach a certain size, such as 600 frogs, you might end up with an equation like \( e^{0.18t} = 7.059 \). To solve for \( t \), you take the natural logarithm of both sides: \( \ln(e^{0.18t}) = \ln(7.059) \). This simplifies to \( 0.18t = \ln(7.059) \), allowing you to solve for \( t \). The calculation of a natural logarithm transforms complex exponential equations into manageable linear ones, making it an indispensable tool in math and science. In our example, calculating \( \ln(7.059) \) yields an approximate value of 1.957, which helps determine that the frog population will reach 600 in about 11 years.