Explanation Given information: Volume and outflow in Great Lakes: V (km 3 ) R (km 3 /year) Superior 12,200 65.2 Michigan 4900 158 Erie 460 175 Ontario 1600 209 Calculation: The differential equation for Q changing with time, Rate of Q changes = − ( Rate of pollutants leave in outflow ) where the negative sign represents the fact that Q is decreasing. At time t, the concentration of pollutants is Q / V and water containing this concentration is leaving at rate r. Thus, Rate pollutants leave in outflow = Rate of outflow × Concentration = r ⋅ Q V . So, the differential equation is d Q d t = − r V Q and its general solution is Q = Q 0 e − r t / V For Lake Michigan, r V = 158 4900 = 0.032 , so, at time t we have Q = Q 0 e − 0.032 t . When 90% of the pollution has been removed, 10% remains, so Q = 0.1 Q 0 . Substituting gives 0.1 Q 0 = Q 0 e − 0.032 t Cancelling Q 0 and solving for t gives ln ( 0

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ISBN: 9781119399353

Author: DA

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Chapter 9.4, Problem 11P

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The time it would take for 90% of the pollution to be removed from Lake Michigan and from Lake Ontario and which lake will take more time to be purified by using the data and also give explanation.

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The time it would take for 90% of the pollution to be removed from Lake Michigan and from Lake Ontario and which lake will take more time to be purified by using the data and also give explanation.

Solution Summary: The author calculates the time it would take for 90% of the pollution to be removed from Lake Michigan and Lake Ontario.

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To find: The time it would take for 90% of the pollution to be removed from Lake Michigan and from Lake Ontario and which lake will take more time to be purified by using the data and also give explanation.

Chapter 9 Solutions

To find:

The time it would take for 90% of the pollution to be removed from Lake Michigan and from Lake Ontario and which lake will take more time to be purified by using the data and also give explanation.