Question

Rate Problem: Diff Eq, Intitally, 5 pounds of salt are dissolved in 20 gallons of water. A salt solution with a concentration of 2 lb/gal is added to the tank at a rate of 3 gal/min. The well stirred mixture is drained from the tank at the same rate. How long should this mixture be added in order to raise the amount of salt in the tank to 25 lb?

Answer 1

So the general equation I have to go off of is

\[\frac{dQ}{dt}= Rate_{IN}-Rate_{Out}\]

\[Rate=Concentration(\frac{lb}{gal})*Flow Rate(\frac{gal}{\min})\]

with the resultating units being

\[\frac{lb}{\min}\]

Answer 2

Also according to the book the answer is

\[t=\frac{20}{3}Ln(\frac{7}{3})=5.6 \min\]

Answer 3

Seems like you have pretty much everything you need to figure this out. Where are you getting stuck?

Answer 4

I can't seem to figure out how to set up the differential equation. I've tried a few things now for example,

\[dq/dy+\frac{3/t}{20}q=\frac{20}{t}\]

Answer 5

I just can't seem to get it into the form of the answer

Answer 6

I guess my major difficulty right now is figuring out where to assign the variables.

Answer 7

So whenever I come across these I like to list out all my variables with units, since units are what will save you. There's a little taught thing called "dimensional analysis" and it's your best friend in solving these types of problems.

Very quickly we have:

5 lb = Q initial
20 gal = V initial
3 gal/min = rate in = rate out
2 lb/gal = Concentration in

and you want to eventually find out how long you have to pour until you have 25lb of salt.

So first off, what're the units of this equation? \[\frac{ dQ }{ dt }=rate_{IN}-rate_{OUT}\] The units are: (lb/min) which is crucial to understand. If you don't have pounds per minute as the units on both sides of the equation it would be exactly like somehow saying some quantity of oranges is equivalent to apples. You can't do that, so make sure the units agree ALWAYS, or else you know it's wrong.

So dQ/dt is fine, but the right side rate in/out must also have that lb/min unit. So for the first one, the rate in is: 3 gal/min. This isn't lb/min, we need to divide out gallons some how and multiply by pounds, so likely we will multiply by something that has units of lb/gal to make this make sense. Luckily, looking above we're given it. It's a little more complicated for the other term, but see if you can make a guess and I'll help you out!

Answer 8

well times that by the concentrations and you end up with 6 lb/min

Answer 9

oh you want lb/gal?

Answer 10

Exactly it is 6 lb/min, so that's the rate in, now how about the rate out?

Answer 11

this is where I get jammed up I think it is something along the lines of

\[\frac{3}{t}\]

Answer 12

sorry 6/t

Answer 13

or rather 6/t*q

Answer 14

Don't forget the units, and I think you're on the right track, but I don't know where you're getting your numbers from.

Q has units of lbs
your rate out is 3 gal/min

You're getting there, but remember, you need to end up with lb/min and if you multiply those together you'll have:

\[Qlb*3\frac{gal}{ \min }=3*Q \frac{ lb*gal }{ \min }\]

You can't use 2lb/gal since that's the concentration of salt pouring in, not out.

You already have time in the denominator, so you don't need to include "t" in here.

Answer 15

hint: Since the rate of flow in is the same as the rate of flow out, the volume of the tank will be constant. =)

Answer 16

This isn't always true, and you might have to solve another differential equation to get just a formula for volume with respect to time, but luckily this isn't that difficult.

Answer 17

so for rate out it's just 3Q?

Answer 18

No, since 3Q doesn't have units of lb/min it's not.

Answer 19

\[rate=(\frac{Q lb}{gal}) (\frac{3 gal}{\min})= \frac{3Q}{\min}\]

Answer 20

You only have:

Q lbs
and
3 gal/min

you don't have anything that's Qlb/gal. You had the concentration as a constant given to you earlier, but now the salt concentration is different flowing out and changes over time since the salt is mixed in the tank when it flows out. So when the salt comes in, it enters the 20 gal tank of water.

\[\frac{ Q lbs }{ 20 gal }*\frac{ 3 gal }{ \min }\] is what it should look like, and now you can start doing the real calculus stuff haha.

Answer 21

So instead of doing a indefinite integral can I do a defendant from 5 to 25 to account for the difference of pounds of salt?

Answer 22

Now you should solve the differential equation and use the initial conditions to solve for your constant of integration. Then you'll get a general formula Q(t) and then you can just plug in 25lb for Q and solve for t.

Answer 23

right on it seems that I made that problem much harder than it needed to be. I was working down this line earlier but gave up on it because I didn't think it was going to work out. Thanks for the help!

Answer 24

Yeah sure, try to spend a little time understanding what all is going on, like why should rate of salt flowing out depends on the volume of the tank?