Differential Equations Mixing Problem.

[Ithryndil]

Homework Statement

A lake, with volume V = 100km^3, is fed by a river at a rate of r km^3/yr. In addition, there is a factory on the lake that introduces a pollutant into the lake at the rate of p km^3/yr. There is another river fed by the lake at a rate that keeps the volume of the lake constant. This means that the rate of flow from the lake into the outlet river is (p + r)km^3/yr. Let x(t) denote the volume of the pollutant in the lake at time t. Then c(t) = x(t)/V is the concentration of the pollutant.

(a) Show that, under the assumption of immediate and perfect mixing of the pollutant into the lake water, the concentration satisfies the differential equation:

$$c' + [(p+r)/V]c = p/v<br /> <br /> (b) In has been determined that a concentration of over 2% is hazardous for the fish in the lake. Suppose that r = 50km^3/yr, p = 2km^3/yr, and the initial concentration of pollutant in the lake is zero. How long will it take the lake to become hazardous to the health of the fish?<br /> <br /> For this problem I am only focusing on part b. I need to set up the differential equation. So far I have<br /> <br /> ds/dt = rate in - rate out<br /> <br /> I am stuck at this part. I know the rate out will be 52 km^3/yr because a total of 52km^3 is coming into the lake in the form of water and pollutant. I am not sure how to proceed from here. All attempts have yielded an answer far different than the 1.41 years in the back of the book. Thanks for your help.$$

 

[HallsofIvy]

Homework Statement

A lake, with volume V = 100km^3, is fed by a river at a rate of r km^3/yr. In addition, there is a factory on the lake that introduces a pollutant into the lake at the rate of p km^3/yr. There is another river fed by the lake at a rate that keeps the volume of the lake constant. This means that the rate of flow from the lake into the outlet river is (p + r)km^3/yr. Let x(t) denote the volume of the pollutant in the lake at time t. Then c(t) = x(t)/V is the concentration of the pollutant.

(a) Show that, under the assumption of immediate and perfect mixing of the pollutant into the lake water, the concentration satisfies the differential equation:

$$c' + [(p+r)/V]c = p/v<br /> <br /> (b) In has been determined that a concentration of over 2% is hazardous for the fish in the lake. Suppose that r = 50km^3/yr, p = 2km^3/yr, and the initial concentration of pollutant in the lake is zero. How long will it take the lake to become hazardous to the health of the fish?<br /> <br /> For this problem I am only focusing on part b. I need to set up the differential equation. So far I have<br /> <br /> ds/dt = rate in - rate out$$

$$No, you don't know that because there is no "s" in this problem. What you mean, I hope, is that dc/dt= rate in- rate out because c is the amount of polluntant in the lake and that is what you want to find.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I am stuck at this part. I know the rate out will be 52 km^3/yr because a total of 52km^3 is coming into the lake in the form of water and pollutant. I am not sure how to proceed from here. All attempts have yielded an answer far different than the 1.41 years in the back of the book. Thanks for your help. </div> </div> </blockquote> But you are not trying to find an equation for total volume. you are trying to find an equation for volume (or concentration) of pollutant. That's why you should have specificed the "s" in your equation in terms of quantities in this problem. The amount of <b>pollutant</b> coming in is p and the amount going out is (x/100)(p+ r).$$

 

[Ithryndil]

We worked with s in class, so I ran with that...I should have added that I am not using c, but rather s. So... s(t) = c(t) and s' = c'.