A large tank contains 100 liters of a salt solution which has a concentration of 0.25 kilograms per liter. Pure water is added to the solution at a rate of 80 liters per minute. At the same time, the well-mixed solution drains from the tank at 80 liters per minute.
Find the amount S of the salt in the solution as a function of t:
kilograms
At what time is there only 12 kilograms of salt remaining in the tank?
t= minutes

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- anonymous

Hi dichalao,
Do you still need a solution?

- anonymous

Yes! Or just give me some clues first :)

- anonymous

What i have tried is to set up the first differential equation as : dS/dt= 25/(100+80t)(1-80t)

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## More answers

- anonymous

You need to consider the physics of what's going on in a little bit of time,\[\delta t\]

- anonymous

and you need to consider what's actually changing. You're told 80 L/min enters the tank, and 80 L/min leaves the tank, so volume's the same.
What's changing is the concentration, or even better, the mass, of salt in the solution.

- anonymous

You need to come up with an expression for the change in mass of salt in the tank as a function of time, then use that to get an expression for concentration and solve the system.

- anonymous

Yea i see the solution is diluting

- anonymous

So, let's consider what happens in that little bit of time, delta_t. No salt is entering the system, since what's coming in is pure water. Salt *is* leaving the system in the outflow. I said we need to consider mass, so, let the mass at time t be\[m(t)\] and that at time t+delta_t be\[m(t+\delta t)\]The difference between the two will be\[\delta m = m( t + \delta t) - m(t)\]

- anonymous

Hello

- anonymous

Hi Lok!!

- anonymous

Yea

- anonymous

The change in mass is the negative of the amount leaving (because these delta quantities are always read (final)-(initial), and since you have less at the end, the change will be negative).
That is,\[\delta m = -Q \delta t C(t)\]where Q is the flow rate (liters/min) and C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).

- anonymous

Notice the units on the left-hand side and the right-hand side. LHS is (mass), and right-hand side is (volume)/(time) x (time) x (mass)/(volume). The units equate.

- anonymous

So what we have now is\[\delta m = m(t+ \delta t)-m(t)\]\[-Q \delta t C(t)=m(t+\delta t)-m(t)\]Divide both sides by the constant volume in the container, and by delta_t to get\[-\frac{QC(t)}{V}=\frac{C(t+\delta t)-C(t)}{\delta t}\]

- anonymous

because \[C(t) = \frac{m(t)}{V}, C(t+\delta t)=\frac{m(t+\delta t)}{V}\]

- anonymous

That expression above (above what I *just* wrote) should look familiar. Does it?

- anonymous

Yea

- anonymous

Sorry, my computer's acting up.

- anonymous

You can now take the limit as delta_t approached zero on both sides to get,

- anonymous

\[\lim_{\delta t \rightarrow 0}-\frac{QC(t)}{V}=\lim_{\delta t \rightarrow 0}\frac{C(t+ \delta t)-C(t)}{\delta t}\]\[\rightarrow -\frac{QC(t)}{V}=\frac{dC}{dt}\](the left-hand side is independent of delta_t, so it's just a constant in terms of taking the limit).

- anonymous

You now have a separable differential equation:\[\frac{dC}{C}=-\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{C}=\int\limits_{}{}-\frac{Q}{V}dt\]

- anonymous

Thanks i can take it from here on

- anonymous

\[\log C=-\frac{Q}{V}t+c\]

- anonymous

ok

- anonymous

I have one more question

- anonymous

ok

- anonymous

Suppose taht a_n>0, and lim n->infinity n^2a_n=0. Prove that sigma a_n converges

- anonymous

Since lim n^2a_n->0, so i assume n^2a_n behave like 1/n^p where p >0, if so a_n should be have like (1/n^(2+p), a_n is a p-series, with p >1, so a_n converges. Do u think this approach is reasonable?

- anonymous

You might have to leave it with me. I just got up and need to get ready for uni/work. I'll get back to you.

- anonymous

Sure :) Have a good one :)

- anonymous

Before I go, I want to confirm the sequence is\[n^2a_n\]

- anonymous

yes

- anonymous

ok

- anonymous

i have a question regarding C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).

- anonymous

Do you mean C(t)*delta_t wont be valid? Or just C(t)

- anonymous

Well, C(t) is the concentration at time t, but that's always changing. We're looking at times t and t+delta_t, but only using C(t) to approximate the amount of mass that's leaving. This is valid in the limit, but if you leave everything as a finite difference (which is what you have before taking the limit), what you're really saying is\[-QC(t)/V \approx m(t+ \delta t)-m(t)\]

- anonymous

But because our increment in time was small, the CHANGE in C(t) over that time is assumed very small too.

- anonymous

So then all you need to do to find the mass is multiply it by volume flow rate.

- anonymous

I don't know how well this is being explained. We could have used differentials instead of delta notation, but I think understanding the approximations before taking limits is useful.

- anonymous

This is a good explanation, i understand everything you did

- anonymous

-Q is -80L/min?

- anonymous

No, Q=-80L/min

- anonymous

I took Q as the magnitude of flow, that's why I introduced the minus sign when setting up the equation.

- anonymous

You should end up with\[C(t)=\frac{1}{4}e^{-\frac{4}{5}(\min^{-1})t}kg/L\]

- anonymous

which makes sense, since as t approaches infinity, C(t) approaches zero...the salt is being drained away.

- anonymous

and at time 0, C(0) = 1/4 kg, which is what you've stated as an initial condition.

- anonymous

1/4 kg/L, I mean.

- anonymous

if i neeed to find the amount S of the salt in the solution as a function of t: i just need to multiply both side by 100? since V is constant?

- anonymous

right side*

- anonymous

Yes, (concentration) x (volume) = (mass) here.

- anonymous

got it~

- anonymous

If the volume flow rates were different either side, you'd have a very different situation to look at.

- anonymous

lol it will be super complicated?

- anonymous

Kind of...

- anonymous

where are you from btw

- anonymous

So, in these types of problems, look at what's actually changing, and try to come up with an expression that focuses on that. Try to formulate something you can turn into a differential equation in the end.

- anonymous

UK citizen in Sydney.

- anonymous

Where are you from? Are you male or female?

- anonymous

I am from US, a kid

- anonymous

:P

- anonymous

So is this school stuff?

- anonymous

yea, i am learning sequence/series and my friend is learning 1st degree differential equation

- anonymous

You are really good at problem solving

- anonymous

Is this for the international baccalaureate?

- anonymous

haha, thanks

- anonymous

international baccalaureate?

- anonymous

hey, I just found the answer to your other problem

- anonymous

haha do you have tips on how to approach a solution?

- anonymous

AMAZING !!

- anonymous

the stuff you're studying is covered in the international baccalaureate - it's an international diploma that can be used to get into universities around the world.

- anonymous

actually, hang on, I'll need to check it. I was looking to apply the Dirichlet test with a series 1/n^2...but I want to see if I can get around one of the assumptions

- anonymous

No it is just normal school work

- anonymous

I'll think about tips on how to approach a solution and get back to you. Really, all you have to do is determine what you're being asked to find, put down the information you have before you and build a bridge.

- anonymous

are you told that \[a_n \ge a_{n+1}>0?\]

- anonymous

No, what i posted is everything given

- anonymous

i wouldn't have thought about setting -Qdelta_t(Ct)=m(delta_t+t)-m(t) haha

- anonymous

Did you ask yourself what was actually changing in the system?

- anonymous

Concentration's harder to come to terms with than mass, since concentration is a ratio of quantities, whereas mass is something we have an intuitive feel for.

- anonymous

experience helps too

- anonymous

yea i know that the amount of salt in the container is lessening and the amount of salts going out is also lessening

- anonymous

How old are you ~

- anonymous

if you dont mind :)

- anonymous

20s

- anonymous

You are a genius i guess ..

- anonymous

like I said, experience helps

- anonymous

You should pass some experiences to me haha

- anonymous

I have been! =D

- anonymous

LOL True.. Are you still trying to find the solution to the problem?

- anonymous

yes, but I should be showering...I have an idea, but not much time to check it out.

- anonymous

take your time buddy, i have to go to class right now.. Thank you for helping out :) see you

- anonymous

ok

- anonymous

Damn, I figured it out just after you left. You can use the limit comparison test.

- anonymous

The version we need says,
If \[\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=0\]then\[\sum_{n=0}^{\infty}b_n\]convergent implies\[\sum_{n=0}^{\infty}a_n\]is also convergent.

- anonymous

Now, you need to show that the sum of a_n is convergent, and we can consider for out 'b' series, \[\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}\frac{1}{n^2}\]which is convergent (p-series, p=2>1). Therefore, taking the ratio of a_n and b_n and forming the limit gives,\[\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=\lim_{n \rightarrow \infty}\frac{a_n}{1/n^2}=\lim_{n \rightarrow \infty}n^2a_n=0\]by construction (i.e. by the setup of the problem).
By the theorem above, your series is convergent.

- anonymous

HAHA got it !!!!

- anonymous

Yeah. Whenever I see a polynomial in the variable, I think ratio test, limit comparison or Dirichlet first.

- anonymous

A large tank contains 450 liters of a salt solution which has a concentration of 0.65 kilograms per liter. A salt solution which has a concentration of 0.8 kilograms per liter is added to the solution at a rate of 20 liters per minute. At the same time, the solution drains from the tank at 20 liters per minute.

- anonymous

It's going to be similar. Volume in the tank won't change again, which is useful. Try to form a mass balance like we did before.

- anonymous

I learned the limit comparison and ratio test just now haha

- anonymous

Ah..well, that may have been why you couldn't do it ;)

- anonymous

There are a couple of versions, depending on how the limit a_n/b_n behaves. I had to pick the right one.

- anonymous

So this time, there is outflow and inflow of salt.

- anonymous

Yes

- anonymous

What does your mass balance look like?

- anonymous

i am trying to figure it out

- anonymous

There's a general formula for changes of 'stuff' in a system (for further reference):\[\Delta = i-o+ \gamma - \delta\]

- anonymous

Change = in - out + generated - destroyed

- anonymous

This isn't a chemical or nuclear reaction, so nothing is generated or destroyed.

- anonymous

So gamma = delta = 0

- anonymous

inflow=Qdelta_t(C(t)) outflow=-Q(deltat)(C(t+deltat)?

- anonymous

Nearly...

- anonymous

Your inflow isn't changing with time - it's fixed at 0.8kg/L * 20 L/min

- anonymous

First, in a unit of time, delta t, how much salt is entering the system?

- anonymous

0.8(20)(delta t) (delta t in min of time)?

- anonymous

Yes :)

- anonymous

Just set it equal to some constant, a, or something.

- anonymous

You don't have to, but when you have numbers floating around, it can stuff the algebra if you're not careful.

- anonymous

so it will just be a-outflow=m(t+delta t)-m(t) and then everything goes the same way?

- anonymous

Yep

- anonymous

give a round of applause to loki haha

- anonymous

but what about the outflow...what's your expression?

- anonymous

LOL ouchie

- anonymous

might want to save the round of applause for a bit =D

- anonymous

You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.

- anonymous

outflow= 20(delta t)(Constantly changing concentration) :(

- anonymous

- anonymous

Your outflow is write. Don't be :(

- anonymous

right

- anonymous

not 'write'...

- anonymous

LOL you need more sleep

- anonymous

i know

- anonymous

concentration = mass over volume mass=(mass initial+mass inflow-mass outlfow)?

- anonymous

Now, what does\[m(t+ \delta t)-m(t)\]equal?

- anonymous

(constant inflow)a - (changing outflow), b delta t(C(t+delta t) ?

- anonymous

lol, you're giving me a headache...I'll write it out

- anonymous

sorry.. i am not floating with you :(

- anonymous

\[m(t+\delta t)-m(t)=a \delta t Q-C(t) \delta t Q=(a-C(t))Q \delta t\]

- anonymous

the change is just *in* - *out*, as per that Delta equation I put up.

- anonymous

give me some time to digest haha

- anonymous

a is constant concentration for inflow, while C(t) is changing concentration

- anonymous

Yep.

- anonymous

delta t is intended to cancel the time component in Q?

- anonymous

Yes :)

- anonymous

got it now i think .. haha

- anonymous

i got to go grab food now...otherwise i have to eat myself tonite .see you ~~thanks !!!

- anonymous

Ok. I'll finish it.

- anonymous

Divide both side by V, the volume of the system to obtain\[C(t+\delta t)-C(t)=\frac{Q}{V}(a-C(t))\delta t\]and divide both sides by delta t\[\frac{C(t+\delta t)-C(t)}{\delta t}=\frac{Q}{V}(a-C(t))\]

- anonymous

Take the limit as delta_t approaches zero on both sides to get\[\frac{dC}{dt}=\frac{Q}{V}(a-C(t)) \rightarrow \frac{dC}{a-C}=\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{a-C}=\int\limits_{}{}\frac{Q}{V}dt\]

- anonymous

You should get\[C(t)=a-C(0)e^{-\frac{Qt}{V}}\]

- anonymous

Oh wait...I forgot about the salt already in there! This is what happens when you do it on the fly. Go eat.

- anonymous

I'll write something later.

- anonymous

Okay, this is much easier when you do it on paper. This site needs something like a whiteboard. I'll write it up and scan.

- anonymous

##### 1 Attachment

- anonymous

Wow! you are the most amazing person in the world haha! So for the first question do i also need to add the original amount of salt to the equation?

- anonymous

You know, I've forgotten the first question already. I don't think so since if you put t=0 into the solution, you should get the initial amount in the tank. Just check.

- anonymous

Oh yea. it is right

- anonymous

Always check the initial and terminating conditions (t=0, t-> infinity) to see if your answer makes sense.

- anonymous

LOL i am too lazy to do so when there is a smart guy in the house :)

- anonymous

was*

- anonymous

Yeah, flattery will get you everywhere.

- anonymous

LOL

- anonymous

I am just fooling around ~~

- anonymous

I know :P

- anonymous

Is there a way that i can use to find square root super fast

- anonymous

What do you mean? By hand?

- anonymous

yea or better in mind

- anonymous

Newton-Raphson method is a numerical method for extracting solutions from equations. It has something called, "quadratic convergence" which makes it very fast. Plus, it's easy to use.

- anonymous

I explained it to someone last week. I'll see if I can find it.
If not, the result for square root is
\[x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\]

- anonymous

You start with an initial guess, x_0, and the number it spits out on the LHS will be x_1, which will be closer to the solution. But that means that this is a BETTER guess than before, so we can plug the result into the RHS to get an EVEN BETTER guess on the LHS and so on...you repeat for infinity, but in reality, it only takes a couple of cycles to get something close.

- anonymous

Oh, that's only for the square root of 2.

- anonymous

LHS= left hand sum?

- anonymous

For\[y=\sqrt{a}\]it would be\[x_{n+1}=\frac{x_n}{2}+\frac{a}{2x_n}=\frac{1}{2}\left( x_n+\frac{a}{x_n} \right)\]

- anonymous

LHS = left-hand side.

- anonymous

So, start with an initial guess for sqrt(2) of 1.5. Plug 1.5 into the RHIS of the equation and you should get an answer, 1.4166666...

- anonymous

How you got a/(2X_n)

- anonymous

Plug that answer into the RHS again and you get on the LHS: 1.414215686...

- anonymous

Already it's getting close.

- anonymous

is this something similar to newton's method?

- anonymous

Yes, it's the same.

- anonymous

You've used it?

- anonymous

My professor talked a little bit about it last quarter

- anonymous

Oh do you have any cool math tricks that you can show me haha

- anonymous

I'd have to think...
Some tricks, you can add zero or multiply by 1 and not change a number...so you can extend this idea when it comes to solving a lot of problems since 0 and 1 can be written in many convenient forms.
Example
\[\int\limits_{}{}\frac{u}{1+u}du=\int\limits_{}{}\frac{u+0}{1+u}du \int\limits_{}{}\frac{u+1-1}{1+u}du\]\[=\int\limits_{}{}1-\frac{1}{1+u}du=u-\log (1+u) + c\]

- anonymous

O.o magic~

- anonymous

Why multiplying two negatives gives a positive...

- anonymous

cuz ..mathematicians say so LOL

- anonymous

\[-2 \times -3 \]\[= -2 \times -3 +0 \]\[=-2 \times -3 + 2 \times 0 \]\[=-2 \times -3+2 \times (3+-3)\]\[=-2 \times -3 + 2 \times 3 + 2 \times -3\]\[=-3 \times (2+-2)+ 2 \times 3\]\[=-3 \times 0 + 2 \times 3\]\[=0+2\times 3\]\[=2 \times 3\]

- anonymous

Everyone takes it for granted. Hardly anyone ever checks or bothers to figure out why.

- anonymous

Yea this is how education been conducted these days

- anonymous

Were you math major? i hear that only math majors actually get into this kind of proofs, for example proof 1+1=2

- anonymous

yes. pure mathematics and physics.

- anonymous

COOL so you can help me with physics too haha!!

- anonymous

why do you have classes so late?

- anonymous

I had no choice, i picked my class on the last day of classes registration because i didn't have many credits when i entered this school.

- anonymous

Hassle.

- anonymous

did you go to University of Sydney ?

- anonymous

Partly.

- anonymous

Postgrad. stuff at Oxford

- anonymous

yea i had to fight for classes

- anonymous

Isn't it a school though?

- anonymous

LOL that explained why you have 129 fans here haha

- anonymous

Have you heard of University of California, Davis?

- anonymous

Ah, yes.

- anonymous

I don't get the American system.

- anonymous

yea i am attending Davis ..since i couldnt get into UC berkerley

- anonymous

It is annoying and difficult to deal with..

- anonymous

Can't you try and transfer, or do postgraduate stuff...or even try for another course?

- anonymous

I can transfer after my sophmore year , but it is really hard to transfer from one college to another college at the same level

- anonymous

It is much easier if you try to transfer from a city college to a University

- anonymous

Your system's quite weird. It seems almost punitive. In the Commonwealth, it's so much easier. We can transfer easily and get exchange placements.

- anonymous

brb

- anonymous

Yea, i had an internship at a company last summer. One of my colleagues is an immigrant from UK. He told me that the school system here is much more complicated than the one in UK

- anonymous

Hell yeah.

- anonymous

I think it has to do with the fact all universities in the Commonwealth (well, most) are public, and there are links between schools that act as 'feeder schools', although you can choose where you go.

- anonymous

The governments ensure as much as possible there is enough money to allow people places in courses they want (assuming the academics think they can handle the work).

- anonymous

So you actually go to lower division schools, elementary, middle and high school, that help you to get into the university you want?

- anonymous

We have nothing like that here. NOw they are evening cutting classes because of the recent budget cut

- anonymous

No, it's not like that. Because the government funds everything, the standards of all the universities have to be the same, which means most people are happy studying at a university that is close geographically to their home.
There's no (as far as I'm aware) discrimination in selection.

- anonymous

If you want to do research, then you have to get yourself into one of the major ones here. At Sydney, I was taught and worked with people who were at Oxford, Cambridge, Yale, etc.

- anonymous

I think there was a mathematician who worked at Stanford too.

- anonymous

Where did you do your undersgraduate

- anonymous

Sydney.

- anonymous

There's a guy in my old physics school there who worked at NASA after graduating. He's back now. I think he discovered something to do with black holes.

- anonymous

What are you going to do when you finish your studies? I get the impression with the US system that you guys study for a general degree and then use it (if you want) to get into a professional school.

- anonymous

OMG. You are surrounded by ...genius man so jealous

- anonymous

I dont know...what i am gonna do.. I am majoring in mechanical engineering.. i dont know what i can do with it

- anonymous

Well here, mechanical engineers make a lot of dosh.

- anonymous

All engineers do.

- anonymous

Maybe you could focus on making moolah?

- anonymous

LOL you mean i design and produce my own moolah?

- anonymous

I mean money ;)

- anonymous

yea that s what i meant

- anonymous

LITTERALLY producing moolah

- anonymous

Sorry, subtleties are lost online.

- anonymous

Why not, your Government does.

- anonymous

*?*

- anonymous

LOL

- anonymous

How about you .what you going to do after post grad

- anonymous

Well, I'm taking a break and teaching mathematics at a grammar school in Sydney. I'll get back into research a bit later.

- anonymous

I C

- anonymous

Do you think the earth's magnetic field is gonna turn 180% around

- anonymous

i asked about the international baccalaureate before because that's part of what I'm teaching.

- anonymous

Re. magnetic field, I think it's supposed to flip, and can do so anytime 'soon', but 'soon' is in astronomical terms, so it could be like, forever.

- anonymous

Oh

- anonymous

What kind of research are you working on

- anonymous

gtg i will ttyl man..Thanks for all the help~:)

- anonymous

No worries :D

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