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

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Feb 24, 2012
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?

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ans: 40.5 days
I have posted a link there to this thread so the OP can find my work.
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Feb 24, 2012
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:

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

The ODE may be simplified to:

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

Writing this ODE in standard linear form, we have:

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

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

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

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

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

Integrating, there results:

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

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

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

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

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


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

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

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

\(\displaystyle 5=1+9e^{-\frac{t}{50}}\)

\(\displaystyle \frac{9}{4}=e^{\frac{t}{50}}\)

Convert from exponential to logarithmic form:

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

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

Thus, we find it will take about 40.5 days for the pollution concentration to reach the desired level.