Consider a lake of constant volume 12200 km^3, which at time t contains an amount y(t) tons of pollutant evenly distributed throughout the lake with a concentration y(t)/12200 tons/km^3.
assume that fresh water enters the lake at a rate of 67.1 km^3/yr, and that water leaves the lake at the same rate. suppose that pollutants are added directly to the lake at a constant rate of 550 tons/yr.
A) write a differential equation for y(t).
B) solve the differential equation for initial condition y(0)=200000 to get an expression for y(t). Use your solution to y(t) to describe in practical terms what happens to the amount of pollutants in the lake as t goes from 0 to infinity.
** i believe A is dy/dt = -67.1(y/12200) + 550 = -0.055y + 550
I believe A is correct. So, you just have a linear DE
y' + 0.055y = 550
which has the soution
y = c e^-0.55t + 10000
Since y(0) = 200,000
c = 190,000
A) Yes, you are correct. The differential equation for y(t) can be written as:
dy/dt = -67.1(y/12200) + 550
where dy/dt represents the rate of change of the amount of pollutant in the lake over time.
B) To solve the differential equation, we can rearrange it as follows:
dy/dt + (0.055)y = 550
This is a linear first-order ordinary differential equation with integrating factor e^(∫0.055 dt) = e^(0.055t).
Multiplying both sides of the equation by the integrating factor, we get:
e^(0.055t) * dy/dt + (0.055)e^(0.055t) * y = 550e^(0.055t)
Applying the product rule on the left side, we have:
d/dt (e^(0.055t) * y) = 550e^(0.055t)
Integrating both sides with respect to t, we get:
∫ d/dt (e^(0.055t) * y) dt = ∫ 550e^(0.055t) dt
e^(0.055t) * y = 10000e^(0.055t) + C
where C is the constant of integration.
Dividing both sides by e^(0.055t), we obtain:
y = 10000 + Ce^(-0.055t)
We can use the initial condition y(0) = 200000 to determine the value of C:
200000 = 10000 + Ce^(0)
C = 190000
Thus, the expression for y(t) is:
y(t) = 10000 + 190000e^(-0.055t)
Now, let's interpret the solution in practical terms.
As t goes from 0 to infinity, the term e^(-0.055t) approaches zero, which means that it decreases exponentially over time. Therefore, the amount of pollutants in the lake, y(t), approaches a constant value of 10000 tons.
In practical terms, this means that over time, the concentration of pollutants in the lake becomes relatively stable, with a constant level of 10000 tons. This suggests that the lake reaches a state of equilibrium, where the rate of pollutants entering the lake is balanced by the rate of pollutants leaving the lake, resulting in a constant concentration.
To answer part A of the question, let's break it down step by step.
The rate of change of the amount of pollutants in the lake, y(t), is given by the difference between the rate at which pollutants enter the lake and the rate at which pollutants leave the lake.
The rate at which pollutants enter the lake is the constant rate at which pollutants are added directly to the lake, which is 550 tons/yr.
The rate at which pollutants leave the lake is determined by the rate at which water leaves the lake, which is 67.1 km^3/yr, and the concentration of pollutants in the lake, which is y(t)/12200 tons/km^3. Thus, the rate at which pollutants leave the lake is (67.1 km^3/yr) * (y(t)/12200 tons/km^3).
Therefore, the differential equation for y(t) is:
dy/dt = pollutant input rate - pollutant output rate
dy/dt = 550 - (67.1/12200) * y(t)
dy/dt = -0.0055 * y(t) + 550
So, the correct differential equation for y(t) is indeed:
dy/dt = -0.0055 * y(t) + 550
Now, let's move on to part B.
To solve the differential equation for y(t), we need to separate the variables and integrate both sides.
dy / (-0.0055 * y + 550) = dt
Integrating both sides, we get:
- (1/0.0055) * ln(abs(-0.0055 * y + 550)) = t + C
where ln denotes the natural logarithm and C is the constant of integration.
Rearranging the equation, we have:
ln(abs(-0.0055 * y + 550)) = -0.0055 * t + C'
where C' = -0.0055 * C.
Taking the exponential of both sides, we get:
abs(-0.0055 * y + 550) = e^(-0.0055 * t + C')
Using the absolute value, we have two cases:
Case 1: -0.0055 * y + 550 = e^(-0.0055 * t + C')
Case 2: -0.0055 * y + 550 = -e^(-0.0055 * t + C')
To solve for y(t), we need to consider the initial condition y(0) = 200,000.
Plugging in t = 0 and y = 200,000 into the equation, we can find the value of C'.
From Case 1, we get:
-0.0055 * 200,000 + 550 = e^(C')
990 = e^(C')
Taking the natural logarithm of both sides, we find:
C' = ln(990)
Using this value of C', we can substitute it into the equations for both cases to find the expression for y(t).
However, the solution to the differential equation involves complex numbers, so we cannot directly describe the behavior of y(t) in practical terms as t goes from 0 to infinity.
Instead, we can analyze the equation to determine some general trends. The exponential term in the equation will cause the value of y(t) to approach zero asymptotically as t goes to infinity. This implies that the amount of pollutants in the lake will decrease over time, but it will never reach zero.
Please note that the solution provided should be verified independently as mathematical errors may have occurred during the process.