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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)=3500e⁰.⁰¹⁶⁶ᵗ
a. The population of the species was about 4142 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 5869 ​million)
c. Estimate the population of the species in the year 2015.

User Julieann
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Final answer:

The exponential function approximates the 1980 population very closely as 4131.75 million against the actual 4142 million. The 2000 population estimate is slightly lower at 5760.95 million versus the actual 5869 million. For 2015, the estimated population is 7388.85 million based on the function.

Step-by-step explanation:

To assess the accuracy of a given exponential function describing the growth of a species population and to use it for population estimates at various years, we will apply the function A(t) = 3500e0.0166t with different values of t representing the number of years since 1970.

a. Population Function Approximation for 1980

For 1980, the year t is 10 since 1980 is 10 years after 1970. Thus, we calculate:

A(10) = 3500e(0.0166 \(\times\) 10) = 3500e0.166

Next, we find the value of this expression:

A(10) = 3500 \(\times\) e0.166 \(\approx\) 3500 \(\times\) 1.1805 = 4131.75 million

The approximation is very close to the actual population of 4142 million.

b. Use the Function to Approximate the Population in 2000

For the year 2000, t is 30. Calculate:

A(30) = 3500e(0.0166 \(\times\) 30) = 3500e0.498

The approximate value then is:

A(30) = 3500 \(\times\) e0.498 \(\approx\) 3500 \(\times\) 1.6457 = 5760.95 million

This approximation is slightly less than the actual population in 2000, which was about 5869 million.

c. Estimate the Population in 2015

For 2015, t is 45. We calculate:

A(45) = 3500e(0.0166 \(\times\) 45)

After calculation, we get:

A(45) = 3500 \(\times\) e0.747 \(\approx\) 3500 \(\times\) 2.1111 = 7388.85 million

The population estimate for the species in 2015 would be approximately 7388.85 million according to the given exponential function.

User Carli
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