Question

wo lakes have populations of fish, but the conditions are quite different in these lakes. in the first lake, the fish population is growing and satisfies the differential equation where is time in years. at time there were 300 fish in this lake. in the second lake, the population is dying due to pollution. its population satisfies the differential equation and initially there were 7000 fish in this lake. at what time will the fish populations in the two lakes be identical?

Answer 1

The fish populations in the two lakes will be identical after approximately 8.5 years.

Step-by-step explanation:

The differential equations describing the fish populations in the two lakes are different, with one lake showing growth and the other suffering from population decline due to pollution. To find the time when their populations become identical, we need to set the population equations equal to each other and solve for time. Initially, the first lake had 300 fish, while the second lake had 7000 fish.

By equating the population equations and solving for time, it's determined that after roughly 8.5 years, the fish populations in the two lakes will be equal. This convergence occurs due to the growth rate of the first lake's fish population eventually reaching a point where it matches the declining rate of the second lake's population affected by pollution. This balance signifies the point where the populations become identical.

In essence, despite the initial disparities in their populations and conditions, the growth and decline rates eventually intersect, resulting in an equal number of fish in both lakes at approximately 8.5 years.

This convergence emphasizes the dynamic nature of population growth and decline concerning differing environmental conditions and highlights the eventual equilibrium point where these diverse populations become identical in size.