Question

a 300gal capacity tank contains a solution of 200 gals of water and 50 lbs of salt. a solution containing 3 lbs of salt pergallon is allowed to flow into the tank at teh rate of 4 gpm. the mixture flows from the tank at the rate of 2 gal/min. how many pounds of salt are in the tank at the end of 30 min?

Answer 1

is that important?

Answer 2

well it's not given...it's also not in the formula...

Answer 3

aww

Answer 4

\[\frac{dQ}{dt} = 4(3) - 2[\frac{Q}{200+ (4-2)t}]\]
\[\frac{dQ}{dt} = 12 - \frac{2Q}{200 + 2t}\]
\[\frac{dQ}{dt} + (\frac{1}{100 + t})Q = 12\]
then linear DE thingy i end up with
\[Q(100+t) = 6(100+t)^2 + c\]

Answer 5

you are told that there are 200 gallons of water with 50lb of salt in it.
that mean 50/200 = 1/4 lb of salt per gallon.

mixture flows out at a rate of 2 gallons/minute, which means at the end of 30 minutes you would have lost 2*30 = 60 gallons of the original mixture.
so we now have 200 - 60 = 140 gallons of 1/4 lb per gallon of salt left from the original mixture. so amount of original salt left = 140 * 1/4 = 35lb

mixture flows in at rate of 4 gallons per minute, which means at the end of 30 minutes you would have gained 4*30 = 120 gallons of the new mixture.
so we now have 120 gallons of 3lb per gallon of salt added to the original. so amount of new salt added = 120 * 3 = 360lb.

Answer 6

so at the end of 30 minutes you have 35lb + 360lb = 395lb of salt in the tank

Answer 7

wait...can you explain this part sir


so we now have 200 - 60 = 140 gallons of 1/4 lb per gallon of salt left from the original mixture. so amount of original salt left = 140 * 1/4 = 35lb

Answer 8

you agree that mixture flows out at rate of 2 gal/min?

Answer 9

btw it says the answer should be 356.92 lbs...

Answer 10

maybe I made a algebra mistake somewhere

Answer 11

but see if you can follow my reasoning

Answer 12

there are some parts i dont get :(

Answer 13

BTW: I am assuming the mixture flowing in does not /mix/ with the mixture already in the tank. i.e. it just settles at the top.

Answer 14

it mixes

Answer 15

it is being stirred

Answer 16

oh - have they given you a formula for how it mixes?

Answer 17

\[\huge \frac{dQ}{dt} = r_1 c_1 - r_2 [\frac{Q}{V_o + (r_1 - r_2)t}]\]

Answer 18

it's the one i was using a while ago

Answer 19

what do all the symbols represent?

Answer 20

r1 is the rate of the thingy going in
c1 is the lg/gal thingy
r2 is the rate of the thingy going out
V_o is the volume of water in the tank
Q is the amt of salt in the tank

Answer 21

t is time of course

Answer 22

ok, so from this (and BTW you really should have stated all this with the question), we get:\[r_1=4gal/min\]\[r_2=2gal/min\]\[c_1=0.25lb/gal\]\[V_0=200gal\]

Answer 23

shouldn't there also be a \(c_2\) somewhere?

Answer 24

I am assuming \(c_1\) represents the concentration of salt in the original mixture, and \(c_2\) would be the concentration of salt being mixed in?

Answer 25

oh - hang on - looks \(Q\) represents the initial concentration of salt in the tank (i.e. at time zero)

Answer 26

so \(c_1=3lb/gal\) and \(Q=50lb\) at \(t=0\)

Answer 27

c1 is the salt coming in

Answer 28

the amt of salt coming in

Answer 29

and btw..that formula was a standard for these problems so i thought not to put it anymore

Answer 30

yes, so then you end up with:\[\frac{dQ}{dt}=12-\frac{Q}{100+t}\]

Answer 31

and we need to integrate this from t=0 to t=30

Answer 32

hmm yep

Answer 33

what i was thinking is to solve for c in that last equation i had
\[Q(100 +t) = 6(100+t)^2 + c\]
however i dont know how

Answer 34

oh - this is a linear DE of the form:\[Q'+(\frac{1}{100+t})Q=12\]

Answer 35

because if i have a value for c then i can just make t = 30 to get Q..idk

Answer 36

yep

Answer 37

have you solved these types of DE's? e.g. using an integrating factor?

Answer 38

yup...i solved the DE already... i got \[Q(100 + t) = 6(100+t)^2 + c\]

Answer 39

ok - let me check that

Answer 40

okay :)

Answer 41

I get a different answer...

Answer 42

but how :O

Answer 43

integrating factor = 100 + t

Answer 44

yes

Answer 45

which leads to:\[(100+t)Q=\int12(100+t)dt=1200t+6t^2\]giving:\[Q=\frac{6t(200+t)}{100+t}\]

Answer 46

and, for t=30, I get Q=318.462

Answer 47

oh hang on

Answer 48

forgot the constant!

Answer 49

\[Q=\frac{6t(200+t)}{100+t}+k\]

Answer 50

and we know Q at time t=0

Answer 51

Q=50lb at t=0

Answer 52

wait...

Answer 53

therefore:\[50=0+k\]therefore k=50

Answer 54

\[\int 12(100 + t)dt\]
\[12\int(100+t)dt\]
\[12[\frac{(100+t)^2}{2}]\]
right?

Answer 55

sorry, it should have been:\[Q=\frac{6t(200+t)+k}{100+t}\]

Answer 56

no:\[\int(100+t )dt=\int100dt+\int tdt=100t+\frac{t^2}{2}+k\]

Answer 57

hah so my teacher was wrong

Answer 58

so we have:\[Q=\frac{6t(200+t)+k}{100+t}\]and Q=50 at t=0, therefore:\[50=\frac{k}{100}\]\[\therefore k=50000\]

Answer 59

therefore:\[Q=\frac{6t(200+t)+50000}{100+t}\]

Answer 60

mmhmm

Answer 61

so now t = 30

Answer 62

sorry - one too many zeros in there!

Answer 63

k=5000

Answer 64

\[Q=\frac{6t(200+t)+5000}{100+t}\]therefore, at t=30:\[Q=356.92\]

Answer 65

weirdd...it gave the same answer as when 12(100+t) was used

Answer 66

12(100+t)^2 i mean

Answer 67

i mean 6(100+t)^2

Answer 68

really?

Answer 69

yep

Answer 70

are you sure - even AFTER computing the constant by using Q=50 at t=0?

Answer 71

well the constant in that one was 55000

Answer 72

what was the formula you ended up with?

Answer 73

lol 4th time im gonna type this :))
\[Q(100+t) = 6(100+t)^2 + c\]

Answer 74

so at t = 0
100(50) = 60,000 + c
5000 = 60,000 + c
c = -55,000

Answer 75

then at t=30
Q(130) = 6(130)^2 - 55,000

Answer 76

that is really wierd

Answer 77

let me see if there is a reasonable explanation for that

Answer 78

do you think it's a fluke?

Answer 79

or there really is a calculus reason...

Answer 80

both formulas are identical after simplification!

Answer 81

but (100+t)^2/2 = 5,000 + 100 + t^2/2

Answer 82

oh lol i see

Answer 83

thanks for the help sir :D this topic is the only one im having difficulty in our diff eq :/ i get all the other applications except this

Answer 84

yw :)

Answer 85

I still am amazed that the answers came out the same - because I am sure your integration was not correct :/

Answer 86

ah! I see how it happened!

Answer 87

i was following what my teacher did...but i think it's somehow right if i let u = 100+t
du = dt so

it's int udu

Answer 88

look at this: http://www.wolframalpha.com/input/?i=integrate+12%28100%2Bx%29

Answer 89

\[\int 12(100 + t)dt=6t^2+1200t+c\]but, you worked it out as:\[\int 12(100 + t)dt=12\frac{(100+t)^2}{2}+c=6(100+t)^2+c=6(100^2+200t+t^2)+c\]\[=6*100^2+1200t+6t^2+c\]\[=6t^2+1200t+c+6*100^2\]

Answer 90

so the only difference here is the value of the constant.
my constant is just your constant plus 6*100^2

Answer 91

your constant came out to be -55000
so -55000 + 6*100^2 = -55000 + 60000 = 5000 = my constant

Answer 92

mystery solved!

Answer 93

wow cool

Answer 94

so that proves that in diffeq constants dont matter much

Answer 95

yes - and more importantly, your teacher WAS right after all :)