A newly constructed fish pond contains 2000 liters of water....

A newly constructed fish pond contains 2000 liters of water. Unfortunately the pond has been contaminated with 5 kg of a toxic chemical during the construction process. The pond's filtering system removes water from the pond at a rate of 200 liters/minute, removes 70% of the chemical, and returns the same volume of (the now somewhat less contaminated) water to the pond. Write a differential equation for the time (measured in minutes) evolution of: (a) The total mass (in kilograms) of the chemical in the pond: dm/dt = -7/100 m (b) The concentration (in kg/liter) of the chemical in the pond: dc/dt = -7c/100 (c) Let C denote the concentration of the chemical in the pond measured in grams/liter. The differential equation satisfied by C is: dC/dt = -1/25 C (d) The differential equation for the concentration C of the chemical in grams/liter, but with time measured in hours (let s denote time measured in hours) is: dC/ds =

Transcript

00:01 All right, so a differential equation governing both the amount and the concentration of a toxic chemical in a pond.

00:13 It looks like rather than solving the differential equation, we're more interested in changing units and changing the interpretation of the result.

00:24 So we'll get right to that.

00:25 But usually what i like to do is set up a differential equation for a rate.

00:35 Here, the amount of the chemical, the mass of the chemical in the pond is equal to the difference between the rate in and the rate out.

00:54 And here we've got a situation where the stuff coming out goes right back in, but with some subtraction of the chemical.

01:05 So the rate in is 0 .3 times the mass per unit volume.

01:13 And then the flow rate, 200 liters per minute.

01:21 And the rate out is the full m, sorry, m per unit v times 200 liters per minute.

01:40 And what i see is because the flow rates are the same, the v is, of course, the volume inside of the pond.

01:52 And so we can see that the volume remains constant, 2 ,000 liters.

02:04 So not too surprising, you can just replace the v with 2 ,000 liters in each of these.

02:14 And that unit will stay fixed.

02:18 We could change that unit, but we'll leave it as liters.

02:24 So your rate in is 0 .3 times 200 over 2000, which is 100.

02:34 10, sorry, not 100.

02:38 So 0 .3 over 10 times m minus 1 m.

02:48 So that is minus 0 .7 over 10 times m.

02:54 Okay, and if we wanted to rewrite in terms of the concentration, what we would do is simply divide both sides of the equation by the volume.

03:15 And since the volume is not a function of time, we can just scoot that right in with the mass.

03:24 Okay, so multiply by one over volume.

03:28 And so we get the mass per unit volume.

03:32 Idt is equal to minus 0 .7 mass per unit volume over 10.

03:43 Or deconcentration idt is minus 0 .7 over 10 times concentration.

03:55 Okay.

03:56 Now, it does pay before you solve a differential equation is to kind of go through and are the units the way you'd like them.

04:05 So the first question is, what if we wanted concentration in terms of grams per liter instead of kilograms per liter? and i do want to point out that nowhere in this derivation have we ever used the mass in kilograms.

04:42 So nowhere did we specify units for mass.

04:59 Okay, the only place it appears is in the initial condition up here.

05:05 And we never used that initial condition.

05:07 We did specify the volume in leader.

05:11 So we're okay...

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