Final answer:
To find the time when two differently behaving fish populations are equal, we solve their respective exponential growth and decay equations. The populations meet when 500e^{0.2t} equals 4000e^{-0.1t}.
Step-by-step explanation:
The student is asking about the time at which two fish populations in different lakes will be identical. The populations are changing over time, one growing and one decreasing, described by differential equations. To solve this problem, we'll integrate the differential equations and solve for the time when the populations are equal.
For the first lake with an initial population of 500 fish and a growth rate of 0.2 per year, the equation becomes y(t) = 500e^{0.2t}. For the second lake with an initial population of 4000 fish and a decline rate of 0.1 per year, the equation is y(t) = 4000e^{-0.1t}. We set these equations equal to each other and solve for t.
So, 500e^{0.2t} = 4000e^{-0.1t}. Dividing both sides by 500 gives e^{0.2t} = 8e^{-0.1t}. Solving for t gives us the exact time when both fish populations will be equal.
The fish population in the first lake is growing and satisfies the differential equation dy/dt = 0.2y where t is time in years. At time t = 0, there were 500 fish in this lake. The fish population in the second lake is dying due to pollution and satisfies the differential equation dy/dt = -0.1y, with an initial population of 4000 fish. To find the time when the fish populations in the two lakes are identical, we need to set the two equations equal to each other and solve for t:
0.2y = -0.1y
0.2y + 0.1y = 0
0.3y = 0
y = 0
Since y represents the fish population, a population of 0 does not make sense in this context. Therefore, there is no time when the fish populations in the two lakes are identical.