Problem 27 - Applied Calculus

Question

Find a formula for the number of zebra mussels in a bay as a function of the number of years since 2010 , given that there were 2700 at the start of 2010 and 3186 at the start of 2011 (a) Assume that the number of zebra mussels is growing linearly. Give units for the slope of the line and interpret it in terms of zebra mussels. (b) Assume that the number of zebra mussels is growing exponentially. What is the annual percent growth rate of the zebra mussel population?

Step-by-Step Solution

Answer

(a) Linear growth: 486 mussels per year; (b) Exponential growth rate: 16.55% annually.

Step 1: Understanding Linear Growth

To assume that the number of zebra mussels is growing linearly, we need to define a linear function of the form \( N(t) = mt + b \), where \( N(t) \) is the number of zebra mussels, \( t \) is the number of years since 2010, \( m \) is the slope, and \( b \) is the initial quantity of zebra mussels.

Step 2: Calculating the Slope for Linear Growth

Given that there were 2700 mussels at the start of 2010 and 3186 at the start of 2011, we calculate the slope \( m \) using the formula \( m = \frac{N(1) - N(0)}{1 - 0} = \frac{3186 - 2700}{1} = 486 \). This slope, in the context of zebra mussels, means that the population increases by 486 mussels per year.

Step 3: Writing the Linear Growth Formula

Now that we know \( m = 486 \) and \( b = 2700 \), the linear growth function becomes \( N(t) = 486t + 2700 \). This formula gives us the number of mussels as a function of time in years since 2010.

Step 4: Understanding Exponential Growth

To assume exponential growth, we use the formula \( N(t) = N_0 e^{rt} \), where \( N_0 \) is the initial quantity, \( r \) is the growth rate, and \( t \) is time. Initially, \( N_0 = 2700 \).

Step 5: Solving for the Exponential Growth Rate

From \( N(1) = 3186 \), we solve \( 3186 = 2700 \cdot e^r \). By dividing both sides by 2700: \( e^r = \frac{3186}{2700} = 1.18 \). Taking the natural logarithm: \( r = \ln(1.18) \approx 0.1655 \).

Step 6: Calculating the Annual Percent Growth Rate

The annual percent growth rate is \( r \times 100 \), so \( 0.1655 \times 100 \approx 16.55\% \). Thus, the mussel population grows by approximately 16.55\% every year.

Key Concepts

Linear Growth

Linear growth is a simple model that describes how a population or quantity increases by a constant amount in a given time period. Think of it like climbing a staircase, where each step represents a consistent increase. In our example, we look at the zebra mussel population growth as a straight line to estimate future numbers.

For the zebra mussels, we started with 2,700 in 2010 and increased to 3,186 by 2011. We represent this scenario by a linear function: \[ N(t) = mt + b \] where:

• \(N(t)\) is the number of mussels after \(t\) years
• \(m\) is the slope, representing the yearly increase
• \(b\) is the initial number of zebra mussels in 2010, which is 2,700

To find the slope \(m\), use the formula: \[ m = \frac{N(1) - N(0)}{1 - 0} \] Plugging in the numbers, the slope \(m = 486\) mussels per year. It shows that every year we're adding 486 more zebra mussels. So our growth equation is: \[ N(t) = 486t + 2700 \] This formula tells us how their numbers grow. Every year, just add the number of years times 486 to the original 2,700.

Exponential Growth

Exponential growth models a situation where a quantity grows by a fixed percentage rather than a fixed number each time. Imagine this like compounding interest in a savings account, where the more you have, the more you gain. Instead of stepping up one stair at a time, it's as if you're doubling steps every so often, leading to rapid increases.

To examine exponential growth for the zebra mussels, we use the formula: \[ N(t) = N_0 \cdot e^{rt} \] where:

• \(N_0\) is the initial number of mussels, 2,700 in 2010
• \(r\) is the rate of growth
• \(t\) is time in years

Using the population of 3,186 in 2011, the calculations show:\[ 3186 = 2700 \cdot e^r \] Solving for \(r\), gives us:\[ e^r \approx 1.18, \quad r \approx \ln(1.18) \approx 0.1655 \] This means each year, the number of mussels grows by about 16.55%. Exponential growth predicts a much faster increase as the population itself increases, unlike linear growth, because the growth accelerates over time as the population base gets larger.

Growth Rate Calculation

Understanding how to calculate growth rates is essential in distinguishing between linear and exponential growth. Growth rates reflect how much a population grows over a specific period of time, and it's expressed either as a constant number of additions (linear) or as a percentage (exponential).

For linear growth, as in our example with the mussels initially, the growth rate is simply the slope \(m\) of our linear function, obtained via:

• \( m = \frac{3186 - 2700}{1} = 486 \)

This tells us the population increases by an exact number, 486 mussels, every year, reflecting a stable, consistent growth without acceleration.

In exponential growth, the calculation involves a bit more effort, as the rate is a percentage increase. We calculate using the equations:

• \( e^r = \frac{N(1)}{N_0} \)
• \( r = \ln\left(\frac{3186}{2700}\right) \approx 0.1655 \)

Converting that to a percent gives a dramatic annual increase of 16.55%. Recognizing how these values shape population dynamics helps us predict future scenarios and choose appropriate models for growth.