Question

A lake is stocked with 95 fish at time - The fish have a birth rate (b) = 0.15 day-1 and a death rate (d) = 0.02 day-1. The carrying capacity of the lake is 7800 fish. Assume that no fish swim into or out of the lake. Calculate the number of fish at t = 30 days and then find how many days until the carrying capacity is reached. Round to the nearest whole number of fish.

Answer 1

To calculate the number of fish at t=30 days, use the exponential growth formula. The number of fish at t=30 days is approximately 300. To find how many days until the carrying capacity is reached, rearrange the formula and solve for t. It will take around 51 days for the carrying capacity to be reached.

Step-by-step explanation:

To calculate the number of fish at t=30 days, we can use the exponential growth formula:

N(t) = N(0) * e^(r*t)

Where N(t) is the number of fish at time t, N(0) is the initial number of fish, r is the net growth rate (birth rate minus death rate), and t is the time interval. In this case, N(0) = 95, r = (b-d) = (0.15 - 0.02) = 0.13, and t = 30 days. Plugging in these values, we get:

N(30) = 95 * e^(0.13 * 30) ≈ 300.42 ≈ 300 fish

To find how many days until the carrying capacity is reached, we need to find the time it takes for N(t) to reach the carrying capacity of 7800 fish. We can rearrange the exponential growth formula to solve for t:

t = ln(N(t)/N(0)) / r

Plugging in N(t) = 7800, N(0) = 95, and r = 0.13, we get:

t = ln(7800/95) / 0.13 ≈ 51.41 ≈ 51 days

Therefore, it will take approximately 51 days for the carrying capacity to be reached, and the number of fish at t=30 days is around 300.