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&#039; + [(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&#039; + [(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 /> No, you don&#039;t know that because there is no &quot;s&quot; 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&#039;s why you should have specificed the &quot;s&quot; 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'.