The growth of a certain species​ (in millions) since

1970

closely fits the following exponential function where t is the number of years since

1970

.

Upper A left parenthesis t right parenthesis equals 3500 e Superscript 0.0166 t

a. The population of the species was about

3552

million in

1980

.

How closely does the function approximate this​ value?

b. Use the function to approximate the population of the species in 2000.​ (The actual population in 2000 was about

5046

​million)

c. Estimate the population of the species in the year 2015.

We are given the exponential growth function for the population of a species, where:

\[
A(t) = 3500 e^{0.0166 t}
\]

Here, \( t \) is the number of years since 1970, and \( A(t) \) gives the population (in millions) at time \( t \).

### Part (a): Approximation for 1980

We are asked to compare the population given by the function to the actual population in 1980. The actual population is given as **3552 million**. Since 1980 is 10 years after 1970, we substitute \( t = 10 \) into the equation:

\[
A(10) = 3500 e^{0.0166 \times 10}
\]

Let's calculate this value.

### Part (b): Approximation for 2000

For the year 2000, \( t = 30 \) since it's 30 years after 1970. We substitute \( t = 30 \) into the equation:

\[
A(30) = 3500 e^{0.0166 \times 30}
\]

This gives us the approximate population for the year 2000, which we will compare with the actual population of **5046 million**.

### Part (c): Estimate for 2015

For 2015, \( t = 45 \) since it's 45 years after 1970. We substitute \( t = 45 \) into the equation:

\[
A(45) = 3500 e^{0.0166 \times 45}
\]

Let's now calculate the values for each part.Here are the results for each part:

### Part (a): Approximation for 1980
The function estimates the population in 1980 to be approximately **4132 million**, while the actual population was **3552 million**. The function overestimates the population by about **580 million**.

### Part (b): Approximation for 2000
The function estimates the population in 2000 to be approximately **5759 million**, while the actual population was **5046 million**. The function overestimates the population by about **713 million**.

### Part (c): Estimate for 2015
The estimated population in 2015 using the function is approximately **7387 million**.

Would you like further details or clarifications? Here are 5 questions to expand on this topic:

1. How can we calculate the percent error between the actual and predicted values?
2. What factors might cause a real population to differ from the model?
3. How would changing the growth rate \(0.0166\) affect the predictions?
4. What other models (besides exponential) can describe population growth?
5. How would we modify the function if population growth slowed over time?

**Tip:** Exponential growth models are best for short-term predictions; for long-term predictions, other factors like resource limitations often need to be considered.

The growth of a certain species (in millions) since 1970 closely fits the following exponential function where t is the number of years since 1970. A(t) = 3500 e^0.0166t. a. The population of the species was about 3552 million in 1980. How closely does the function approximate this value? b. Use the function to approximate the population of the species in 2000. (The actual population in 2000 was about 5046 million) c. Estimate the population of the species in the year 2015.

Explore how to use the exponential growth function A(t) = 3500 e^0.0166t to approximate the population of a species from 1970 to 2015. This solution compares predicted and actual populations for 1980 and 2000, and estimates the population for 2015, covering exponential growth and algebra concepts. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation