Question

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

Answer 1

Hi dichalao,

Do you still need a solution?

Answer 2

Yes! Or just give me some clues first :)

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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.

Answer 7

Yea i see the solution is diluting

Answer 8

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\[\delta m = m( t + \delta t) - m(t)\]

Answer 9

Hello

Answer 10

Hi Lok!!

Answer 11

Yea

Answer 12

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).

Answer 13

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.

Answer 14

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\[-\frac{QC(t)}{V}=\frac{C(t+\delta t)-C(t)}{\delta t}\]

Answer 15

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

Answer 16

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

Answer 17

Yea

Answer 18

Sorry, my computer's acting up.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

Thanks i can take it from here on

Answer 23

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

Answer 24

ok

Answer 25

I have one more question

Answer 26

ok

Answer 27

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

Answer 28

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?

Answer 29

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.

Answer 30

Sure :) Have a good one :)

Answer 31

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

Answer 32

yes

Answer 33

ok

Answer 34

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).

Answer 35

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

Answer 36

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\[-QC(t)/V \approx m(t+ \delta t)-m(t)\]

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

This is a good explanation, i understand everything you did

Answer 41

-Q is -80L/min?

Answer 42

No, Q=-80L/min

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

1/4 kg/L, I mean.

Answer 48

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?

Answer 49

right side*

Answer 50

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

Answer 51

got it~

Answer 52

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

Answer 53

lol it will be super complicated?

Answer 54

Kind of...

Answer 55

where are you from btw

Answer 56

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.

Answer 57

UK citizen in Sydney.

Answer 58

Where are you from? Are you male or female?

Answer 59

I am from US, a kid

Answer 60

:P

Answer 61

So is this school stuff?

Answer 62

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

Answer 63

You are really good at problem solving

Answer 64

Is this for the international baccalaureate?

Answer 65

haha, thanks

Answer 66

international baccalaureate?

Answer 67

hey, I just found the answer to your other problem

Answer 68

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

Answer 69

AMAZING !!

Answer 70

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.

Answer 71

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

Answer 72

No it is just normal school work

Answer 73

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.

Answer 74

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

Answer 75

No, what i posted is everything given

Answer 76

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

Answer 77

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

Answer 78

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.

Answer 79

experience helps too

Answer 80

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

Answer 81

How old are you ~

Answer 82

if you dont mind :)

Answer 83

20s

Answer 84

You are a genius i guess ..

Answer 85

like I said, experience helps

Answer 86

You should pass some experiences to me haha

Answer 87

I have been! =D

Answer 88

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

Answer 89

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

Answer 90

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

Answer 91

ok

Answer 92

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

Answer 93

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.

Answer 94

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.

Answer 95

HAHA got it !!!!

Answer 96

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

Answer 97

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.

Answer 98

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.