Question

The population of fish in a lake grew exponentially. In 2000, a researched estimated

the population to be 4780 fish. By 2010, the population had grown to 8200 fish. Let P(t) be
the population of fish t years after 2000.
(a) Assuming exponential growth (so P(t) = P0e
kt), find an expression for P(t).
(b) What was the estimated population in 1990? Round your answer to the nearest thousand.
(c) When will the population 24,000? Round your answer to the nearest year?

Answer 1

a) P(t) = P₀ e⁰•⁰⁵⁴ᵗ

b) 2786 fishes

c) The estimated population will be 24000 in the year 2030.

Explanation:

The function representing the population of fish in a lake as it grows exponentially

P(t) = P₀eᵏᵗ

If the base year is taken to be 2000

P₀ = 4780 fishes

k = constant

In 2010, P(t=10) = 8200 fishes, we can now solve for the constant

kt = k × 10 = 10k

8200 = 4780 e¹⁰ᵏ

e¹⁰ᵏ = (8200/4780) = 1.7155

In e¹⁰ᵏ = In 1.7155 = 0.5397

10k = 0.5397

k = 0.054 to 3 d.p

P(t) = P₀ e⁰•⁰⁵⁴ᵗ

b) What was the estimated population in 1990?

1990 is 10 years before 2000, hence, t = -10

P(t) = P₀ e⁰•⁰⁵⁴ᵗ

0.054t = 0.054 × -10 = -0.54

P₀ = 4780

P(t) = 4780 e⁻⁰•⁵⁴ = 4780 × 0.5829 = 2,786.4 = 2786 fishes to the nearest thousand

c) When will the population 24,000?

P(t) = 24000

P₀ = 4780

t = ?

24000 = 4780 e⁰•⁰⁵⁴ᵗ

e⁰•⁰⁵⁴ᵗ = (24000/4780) = 5.021

In e⁰•⁰⁵⁴ᵗ = In 5.021 = 1.6136

0.054t = 1.6136

t = (1.6136/0.054) = 29.88 years = 30 years to the nearest whole number.

Since the base year is 2000, 30 years after that is 2000 + 30 = 2030.

Hope this Helps!!!