MHB ZZZZZzzz's question at Yahoo Answers regarding a mixing problem

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Mixing
AI Thread Summary
The problem involves calculating the time required to reduce pollution in a lake from 2 grams per cubic meter to 1 gram per cubic meter, given a constant inflow of cleaner water. The differential equation modeling the pollution level is derived and simplified, leading to a solution that incorporates an integrating factor. By applying initial conditions, the pollution concentration over time is expressed as a function. The final calculation shows that it will take approximately 40.5 days to achieve the target pollution level. The mathematical approach effectively demonstrates the physical application of calculus in environmental science.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Physical application of calculus?

A lake has a volume of 10^6 m^3 and an initial pollution of 2 grams per cubic metre. Every day a river flows in 2 x 10^4 m^3 of a lower pollution of 0.2 gram per cubic metre, and the same amount of water flows out to another river. Assume the water is perfectly mixed at any time. How long will it take to reduce the pollution to 1 gram per cubic metre?
thanks

Additional Details:
ans: 40.5 days

I have posted a link there to this thread so the OP can find my work.
 
Mathematics news on Phys.org
Hello ZZZZZzzz,

We have an initial amount of pollution in the lake, and we have pollution flowing in and pollution flowing out. So we may model the amount of pollution $P$ in the lake at time $t$ with the initial value problem:

$$\frac{dP}{dt}=0.2\cdot2\cdot10^4-\frac{P(t)}{10^6}\cdot2\cdot10^4$$ where $$P_0=2\cdot10^6$$

The ODE may be simplified to:

$$\frac{dP}{dt}=4000-\frac{P(t)}{50}$$

Writing this ODE in standard linear form, we have:

$$\frac{dP}{dt}+\frac{1}{50}P(t)=4000$$

Multiplying through by an integrating factor of $$e^{\frac{t}{50}}$$, we obtain:

$$e^{\frac{t}{50}}\frac{dP}{dt}+\frac{1}{50}e^{\frac{t}{50}}P(t)=4000e^{\frac{t}{50}}$$

Observing the left side is the differentiation of a product, we may write:

$$\frac{d}{dt}\left(e^{\frac{t}{50}}P(t) \right)=4000e^{\frac{t}{50}}$$

Integrating, there results:

$$e^{\frac{t}{50}}P(t)=200000e^{\frac{t}{50}}+C$$

Solving for $P(t)$, we have:

$$P(t)=200000+Ce^{-\frac{t}{50}}$$

To determine the parameter $C$, we may use the initial value:

$$P(0)=200000+C=2000000\implies C=1800000$$

Hence:

$$P(t)=200000+1800000e^{-\frac{t}{50}}=200000\left(1+9e^{-\frac{t}{50}} \right)$$

Now, to determine when the pollution concentration is down to 1 gram per million cubic meters, we set $P(t)=1000000$ and solve for $t$:

$$1000000=200000\left(1+9e^{-\frac{t}{50}} \right)$$

$$5=1+9e^{-\frac{t}{50}}$$

$$\frac{9}{4}=e^{\frac{t}{50}}$$

Convert from exponential to logarithmic form:

$$\frac{t}{50}=\ln\left(\frac{9}{4} \right)$$

$$t=50\ln\left(\frac{9}{4} \right)\approx40.5465108108164$$

Thus, we find it will take about 40.5 days for the pollution concentration to reach the desired level.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top