Question

Biologists stocked lake with 800 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 9800. The number of fish grew to 910 in the first year.

a)Find an equation for the number of fish P(t) after t years P(t)
b) How long will it take for the population to increase to 4900 (half of the carrying capacity)? It will take years_ You may enter the exact value or round to 2 decimal places.

Answer 1

The equation for the number of fish P(t) after t years is P(t) = 800 * e^(11t). It will take approximately 0.258 years for the population to increase to 4900 (half of the carrying capacity).

Step-by-step explanation:

To find an equation for the number of fish P(t) after t years:

1. First, determine the initial population of fish, which is 800.
2. Next, find the growth rate by subtracting the initial population from the carrying capacity and dividing it by the initial population. In this case, the growth rate is (9800-800)/800 = 11.
3. Now, use the formula P(t) = P(0) * e^(rt), where P(t) is the population after t years, P(0) is the initial population, e is the base of natural logarithms (approximately 2.71828), r is the growth rate, and t is the number of years. Therefore, the equation for the number of fish P(t) after t years is P(t) = 800 * e^(11t).

To find how long it will take for the population to increase to 4900 (half of the carrying capacity), we can set up the equation:

4900 = 800 * e^(11t)

Dividing both sides of the equation by 800 gives:

e^(11t) = 6.125

Taking the natural logarithm of both sides:

ln(e^(11t)) = ln(6.125)

Using the property of logarithms, ln(e^(11t)) simplifies to 11t:

11t = ln(6.125)

Finally, divide both sides by 11:

t = ln(6.125)/11 ≈ 0.258 years (rounded to 3 decimal places).