Question

Hi everyone! Can someone check my work on the differential equation mixing word problem? I will post the question and my work. Thanks! :o)

Answer 1

@ganeshie8 @Kainui

Answer 2

It looks right, I'm just having trouble focusing right now so I might have missed some thing but your process seems correct. The setup is good, the concentrations for R1 and R2 are set up perfectly, like the only error you could have is if there is some kind of arithmetic error like messing up a sign or accidentally messing up a number or something like that is what would be where you went wrong that I didn't check. =)

Answer 3

I think the main source of error would have been just at the start, but you nailed it pretty explicitly when you wrote it out. 400+(3-4)t gal is what I was looking for right off the bat, everything else is just really algebra imo.

Answer 4

ok I don't want to say too much, but I am just so glad to see someone who has done this problem with units. Too many people don't use units on this type of problem and that's almost always why they get this wrong. It's like they don't realize the point of differential equations is to model physical phenomena and they think "Oh it's just math, so I can forget other things I learned" lol

Answer 5

@Kainui
Okay...regarding the question about the concentration after 60 minutes... when I plugged and chugged, I got 7.89 The thing I am confused about is this: When I first setup the problem, I turned 6% and 3% into .06 and .03 when I got my final answer, it was simply 7.89 I was thinking since I got rid of the % at the beginning by moving the decimal place, that perhaps I should move the decimal place to the right in my final answer, but 789% would not make sense! Not sure why I am confused here...any ideas?

does my confusion make any sense or am I just confusing myself: :o/

Answer 6

Wait a second I am starting to see some problems now. At t=0 we have 400 gallons in the tank right? So if 3 gallons flow in and 4 gallons flow out per minute, we have 1 gallon per minute leaving. That means: \[\Large V(t) = 400 - t\] and the tank is empty after 400 minutes. So let me see here, I'll check that other thing now.

Answer 7

I didn't really recognize that (400+(3-4)t is the same as v(t)=400-t
my teacher did something weird like turn that v(t) into its own little DE but I didn't understand but when I started to do the homework, I was getting correct answers without doing what he did so I just ignored it as confusing and never thought about it again...did I make a mistake by not considering his method?

Answer 8

I also have some info on the tensor calculus online course, but we can discuss that later I guess :o)

Answer 9

You can do it by a second little mini differential equation but it's pretty clear to figure it out without calculus.

Answer 10

I almost feel like my answer after 60 minutes should have been .0789...that way I could move the decimal over to the right and confidently say its 7.89% but it wasn't, therefore I am confused

Answer 11

It really isn't necessary but it might help you to understand better if you can do it both ways. I wouldn't worry about it too much, as long as you understand my reasoning for why v(t)=400-t. You can start out with the change in volume over time as dV/dt being volume in - volume out which is \[\Large \frac{dV}{dt} = 3 \frac{gal}{\min} - 4 \frac{gal}{\min} = -1 \frac{gal}{\min}\]
Simple little differential equation, the rate of volume change doesn't depend on time and we're not even worrying about how much alcohol is in it! super simple, so integrate to get: \[\Large V(t) = -1 \frac{gal}{\min} t +C\] We solve for constant of integration C since we know at t=0 the volume is 400 \[\Large V(0)=400 gal = -1 \frac{gal}{\min} 0 +C \\ \Large 400 gal = C\] \[\Large V(t) = -1 t + 400\] There you have it, the "differential equations way" lol.

Answer 12

Yeah sorry I went on a tangent there, I don't know about the percentage thing I'm pretty distracted at the moment so I'm kinda slowwww sorry

Answer 13

sorry...I couldn't connect to openstudy grrr

Answer 14

but anyway...wow...so that's what he was doing! omg
okay but that seems way more confusing than just
(amount in tank+(ratein-rateout))
no wonder I and a lot of the class was confused!

Answer 15

Yeah, it's pretty simple now but what if the rate of flow in and out weren't constants? Then you'd be screwed since that trick wouldn't work anymore. =P

Answer 16

oh yeah...very true! excellent point!
guess I better start doing it the DE way then!

Answer 17

so do you have any insight about the percentage thingy?

Answer 18

so does it even make sense for me to think that I almost feel like my answer after 60 minutes should have been .0789...that way I could move the decimal over to the right and confidently say its 7.89% but it wasn't, therefore I am confused

Answer 19

You're right about that, what you have must be wrong since it ends up being 789%. I'm checking now

Answer 20

thank you sooo much!

Answer 21

I see what the problem is. You're interpreting A(t) to be the percent alcohol and the question itself is saying A(t) is the gallons of alcohol. So the difference is you have to keep that in mind and maybe use a different letter when you work the problem, Q to mean concentration maybe. The relationship being: \[\Large \frac{A(t)}{V(t)} = Q(t)\] A is the amount of pure alcohol in gallons, V is the volume of the total liquid which includes water and alcohol in gallons and Q is the concentration, which is the ratio of gallons of pure alcohol to the total content of liquid.

Kind of confusing, I just figured this out myself because I wasn't paying attention haha.

Answer 22

hmmm...so did I set up the problem wrong then?

Answer 23

Nope, it's just you're interpreting the results wrong. It's a quick fix. You found out there were 7.89 gallons of alcohol in the vat at t=60. But you don't want the total amount of alcohol, you want the concentration of alcohol to the whole container since it's mixed with water or other stuff. So just divide it by V(60) which is just 400-60=340.

That comes to 2.19%

Answer 24

if I did, then that must mean that I need to figure out what 6% of a gallon is and 3% of a gallon is so that I can get the amount of alcohol right?

Answer 25

What I mean is \[\Large \frac{A(t)}{V(t)} = Q(t) \\ \Large \frac{A(60)}{V(60)} = \frac{7.89}{340} = .0219 = 2.19 \% \frac{alcohol}{gallons}\] See how the units line up nicely?

Answer 26

i posted a little late after your comment...I guess what I am saying is that if I can find what "amount" 6% of a gallon is and do the same with the 3% thingy, then I wouldn't have to make a correction at the end of my problem, I would probably get my .0219 and then be able to shift the decimal place to 2.19% right?

Answer 27

no

Answer 28

wait

Answer 29

6% of a gallon is .06
lemme think about this for a sec

Answer 30

Sure, yeah it's sort of weird you'll have to rethink the question even though you set it up correctly.

Answer 31

okay...i think i get it...one sec

Answer 32

@kainui
okay, here is what I am missing i think...
in all the other problems I have done, the question asks like "how many pounds of salt is left after t minutes" and stuff like that...
if this question asked:
"The amount of alcohol that flows into a tank at 3gal/min is
.06 alcohol bits/gal...
bla bla bla...find the amount of alcohol bits per gallon after 60 minutes"
Then I would have known I would have had an amount...

But wouldn't it be 7.89/400 which 400 being the total volume of the tank?
shouldn't the final answer be .019725 or 1.97% ?

Answer 33

Nope, cause it's not amount of alcohol to the volume of the tank, it's amount of alcohol to the volume of liquid.

For instance, if I open a bottle of beer right now, pour a little of it into a shot glass and the rest into a large cup, they have the same alcohol/gallons since that is just concentration but the shot glass contains less alcohol

Answer 34

@Kainui
Nevermind, I think I get it for reals this time! During the process of filling and draining, the tank was draining faster than it was filling... since I need to know how many alcohol bits are remaining after the 60 minutes is over, there obviously wouldn't be 400 gallons still in the tank, there would be less... and less by exactly whatever v(t) gives us after 60 minutes which is: v(60)=400-60=340 yeah so 7.89 alcohol bits/340 gallons = .0232 = 2.32%alcohol/gallons

Answer 35

right? please say yes or I'm the one going to be in need of 789% alcohol! :o~

Answer 36

Yeah that looks good I guess I just did the division wrong and was slightly off, 2.32%

Answer 37

yay! :o)
that's a tough question to get your head around...
there are so many rates of change happening in the same problem!
but it does show the importance of understanding the whole picture!

so really, the initial setup should really look like this:

Answer 38

Yeah at no point do we know the volume of the container by the way. =P

Answer 39

We are assuming it is just at least 400 gallons that's it.

Answer 40

does that make sense?

Answer 41

Nope, cause dV/dt = 3-4

Answer 42

hmmm...grrr...one sec
lemme fix

Answer 43

Also, that would end up giving you the formula for the _change_ in amount of alcohol per gallons of total liquid. Not the same. You wand A/V not dA/dV