Final answer:
To find the time at which the fish populations in the two lakes will be identical, solve the two differential equations representing the growth and decline of the populations. The time at which the populations will be identical is given by t = ln(10) / (k + m).
Step-by-step explanation:
To find the time at which the fish populations in the two lakes will be identical, we need to solve the two differential equations. The first equation represents the growth of the fish population in the first lake:
dP/dt = kP
where P is the population of fish and k is a constant. We are given that at time t = 0, there were 700 fish in the first lake, so we have the initial condition
P(0) = 700
The second equation represents the population decline in the second lake due to pollution:
dQ/dt = -mQ
where Q is the population of fish and m is a constant. We are given that at time t = 0, there were 7000 fish in the second lake, so we have the initial condition
Q(0) = 7000
To find the time at which the populations will be identical, we need to find the value of t for which P(t) = Q(t). We can solve these differential equations using separation of variables and integrating:
For the first equation:
dP/P = k dt
ln|P| = kt + C1
P = Ce^(kt)
Using the initial condition, we can find the value of C:
700 = Ce^(k * 0)
700 = C
So, the equation for the first lake becomes:
P = 700e^(kt)
For the second equation:
dQ/Q = -m dt
ln|Q| = -mt + C2
Q = De^(-mt)
Using the initial condition, we can find the value of D:
7000 = De^(-m * 0)
7000 = D
So, the equation for the second lake becomes:
Q = 7000e^(-mt)
Now, we can set P(t) equal to Q(t) and solve for t:
700e^(kt) = 7000e^(-mt)
e^(kt+mt) = 10
e^((k+m)t) = 10
Take the natural logarithm of both sides:
kt + mt = ln(10)
t(k + m) = ln(10)
t = ln(10) / (k + m)
Thus, the fish populations in the two lakes will be identical at a time of t = ln(10) / (k + m).