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.

lmaoo i just realized im 5 years late

for part a -

dy/dt=-67.1(y/12200)+550=-0.055y+550

for part b -
you can solve it with that initial condition :)

answer the question

yes someone please answer this question

A) To write the differential equation for y(t), we need to consider the rate of change of pollutants in the lake.

The rate of change of pollutants in the lake is given by:
Rate of accumulation = Rate of incoming pollutants - Rate of outgoing pollutants

Rate of accumulation = Rate of direct addition - Rate of dilution

The rate of incoming pollutants is 550 tons/yr, and the rate of dilution (due to fresh water entering and leaving the lake) is (67.1 km^3/yr) * (y(t) / 12200 tons/km^3).

Therefore, we can write the differential equation for y(t) as:
dy/dt = 550 - (67.1/12200) * y(t)

B) To solve the differential equation, we can separate variables and integrate:

dy / (550 - (67.1/12200) * y) = dt

Integrating both sides gives:

∫ dy / (550 - (67.1/12200) * y) = ∫ dt

Solving the integral, we get:

(12200/67.1) * ln|550 - (67.1/12200) * y| = t + C

where C is the constant of integration. Rearranging the equation:

ln|550 - (67.1/12200) * y| = (67.1/12200) * t + C'

where C' = (12200/67.1) * C.

Taking the exponential of both sides:

|550 - (67.1/12200) * y| = e^((67.1/12200) * t + C')

Considering absolute value, we can remove the modulus sign:

550 - (67.1/12200) * y = ± e^((67.1/12200) * t + C')

Let's assume the plus sign:

550 - (67.1/12200) * y = e^((67.1/12200) * t + C')

Simplifying further:

y(t) = (550 * 12200) / (67.1 + 12200 * e^(-67.1/12200 * t + C'))

Let's use the initial condition y(0) = 200000:

200000 = (550 * 12200) / (67.1 + 12200 * e^C')

Solving for C', we find C' ≈ -0.561.

Substituting C' back into the expression for y(t), we have:

y(t) = (550 * 12200) / (67.1 + 12200 * e^(-67.1/12200 * t - 0.561))

As t goes from 0 to infinity, the term e^(-67.1/12200 * t - 0.561) approaches 0, causing the amount of pollutants in the lake (y(t)) to approach a steady-state value of 550 * 12200 / 67.1 ≈ 100,928. Therefore, in practical terms, the amount of pollutants in the lake will stabilize at approximately 100,928 tons as time goes to infinity.

A) To write a differential equation for the amount of pollutant y(t) in the lake, we need to consider the rates of change of pollutant concentration in the lake due to freshwater inflow, outflow, and direct pollutant addition.

The rate of change of pollutant concentration in the lake is proportional to the difference between the pollutant inflow rate and outflow rate, as well as the rate at which pollutants are directly added. Since the volume of the lake is constant at 12200 km^3, the rate of change of pollutant concentration is given by:

d(y(t)/V)/dt = (pollutant inflow rate - pollutant outflow rate + pollutant added rate) / V

where V is the constant volume of the lake.

The pollutant inflow rate and outflow rate are both equal to the freshwater inflow rate of 67.1 km^3/yr. Therefore, the differential equation for y(t) becomes:

d(y(t)/12200)/dt = (67.1 - 67.1 + 550) / 12200

Simplifying the equation:

d(y(t))/dt = 550 / 12200

B) To solve the differential equation, we can integrate both sides with respect to t:

∫ d(y(t))/dt dt = ∫ (550 / 12200) dt

Integrating the left side yields:

y(t) = ∫ (550 / 12200) dt

Integrating the right side of the equation gives:

y(t) = (550 / 12200) * t + C

where C is the constant of integration.

Using the initial condition y(0) = 200000, we can solve for C:

200000 = (550 / 12200) * 0 + C
C = 200000

Therefore, the expression for y(t) is:

y(t) = (550 / 12200) * t + 200000

In practical terms, as t goes from 0 to infinity, the amount of pollutants in the lake continues to increase linearly with time. The initial pollutant added rate and the constant freshwater inflow and outflow rates do not affect the long-term behavior of pollutant accumulation in the lake. However, the rate of accumulation is relatively low compared to the total volume of the lake, so the pollutant concentration may remain within acceptable limits depending on the tolerance levels for pollutants.