differential equations - mixing problem

A room containing 1000 cubic feet of air is originally free of carbon monoxide. Beginning at time t=0 cigarette smoke containing 4 percent carbon monoxide is blown into the room at 0.1 ft^3/min, and the well-circulated mixture leaves the room at hte same rate. Find the time when the concentration of carbon monoxide in the room reaches 0.012 percent.

rate = rate in - rate out ?

the 4% carbon monoxide part is really throwing me off. i don't exactly know what to do with this number. but other than that, i should be alright.

CO enters : 0.04 x 0.1 OR 0.1, i'm not sure whether to factor in the 4% CO
CO leaves : 0.04 x 0.1 S(t)/1000 OR 0.1 s(t)/1000

i need to figure that part out.. after that, everything i can handle. can someone please help me with the 4% carbon monoxide part
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 okay so i tried working it out with the 0.04 factored in as shown above... and at the end i get a ln -#, so i guess that is wrong. so now i am doing it without the 0.04. if i do it without the 0.04, then i do not understand the concentration part. it says to find the time when conc is 0.012 %, so set c(t)=0.00012 OR shoudl i be factoring in 4% CO somewhere still. i am really confused
 Recognitions: Homework Help Here's a hint: make M(t) be the quantity of carbon monoxide in the room at any time and then the concentration is given by C(t) = M(t)/1000.

differential equations - mixing problem

i have used, s(t) = amount of CO
c(t)=s(t)/1000

so i have rate in as: 0.04 x 0.1
rate out: 0.1 s(t)/1000

i have modified my rate in and rate out from when i first posted the eq'n as i don't believe i have to add the 4% CO factor twice. tell me what you think now.
i have solved this ALL the way to plugging in C(t)

i plugged in c(t) as 1.2 x 10^-4
my eq'n is
c(t) = 4x 10^-10 (1-e^(1.0x10^-4)t)
so pluggin in c(t)
1.2 x 10^-4 = 4x 10^-10 (1-e^(1.0x10^-4)t)
solving for t
300000 = 1-e^(1.0x10^-4)t
299999 = -e^(1.0x10^-4)t

so.. once again. i am stuck.
 here is all my work. rate in: 0.04 x 0.01 rate out: 0.1 x s(t)/1000 s(0) = 0 s'(t) = 0.004 - 1.0x10^-4 s(t) s'(t) + 1.0x10^-4 s(t) = 0.004 solving diff eq'n: a(t) = 1.0x10^-4, b(t)=0.004 using formula: u(t) = exp(integ(a(t)dt)) u(t) = exp(integ(1.0x10^-4 dt)) u(t)=exp(1.0x10^-4 t) using formula: d/dt (u(t) s(t) ) = u(t)b(t) d/dt ( (e^(1.0x10^-4 t )) s(t) ) = (e^(1.0x10^-4 t)) x 0.004 (e^(1.0x10^-4 t )) s(t) = integ (0.004(e^(1.0x10^-4 t))dt) (e^(1.0x10^-4 t )) s(t) = 0.004(1.0x10^-4)(e^(1.0x10^-4 t) + C s(t) = [(4x10^-7) (e^(1.0x10^-4 t)) + C]/(e^(1.0x10^-4 t))] s(t) = 4x10^-7 + Ce^(1.0x10^-4 t) sub s(0)=0 0 = 4x10^-7 + C C= -4x10^-7 so, s(t)=(4x10^-7)(1-e^(1x10^-4 t)) c(t) = s(t)/1000 c(t) = (4x10^-10)(1-e^(1x10^-4 t)) find t when c(t) = 1.2 x 10^-4 1.2x10^-4 = (4x10^-10)(1-e^(1x10^-4 t)) 300000 = 1-e^(1x10^-4 t) 299999 = -e^(1x10^-4 t) stuck. any suggestions or any wrong steps???
 Recognitions: Homework Help Use natural logarithm.
 i end up ln-ing a negative..
 can anyone find the mistake? i have some help from the instructor as he said the percentage is percentage of the volume, so i guess that is the part that is wrong??? i'm not sure what he means by that.

Recognitions:
Homework Help
 Quote by braindead101 can anyone find the mistake? i have some help from the instructor as he said the percentage is percentage of the volume, so i guess that is the part that is wrong??? i'm not sure what he means by that.

$$\frac{dm(t)}{dt} = 0.1 * 0.04 - 0.1 \frac{m(t)}{1000}$$

Maybe you will have less mistakes in your calculations if you solved it like this.

$$c(t) = \frac{m(t)}{1000}$$

$$1000\frac{dc(t)}{dt} = 0.1 * 0.04 - 0.1 \frac{1000c(t)}{1000}$$

Recognitions:
Homework Help
 Quote by braindead101 here is all my work. (e^(1.0x10^-4 t )) s(t) = 0.004(1.0x10^-4)(e^(1.0x10^-4 t) + C s(t) = [(4x10^-7) (e^(1.0x10^-4 t)) + C]/(e^(1.0x10^-4 t))] s(t) = 4x10^-7 + Ce^(-1.0x10^-4 t)
The mistake lies in these steps. You forgot the negative.
 i'm working it out your way now, but the negative i 4got to type in, but i had it on paper and it doesnt really make a difference as when i am plugging in s(0)=0, c is still the same regardless of that.

Recognitions:
Homework Help
 Quote by braindead101 i'm working it out your way now, but the negative i 4got to type in, but i had it on paper and it doesnt really make a difference as when i am plugging in s(0)=0, c is still the same regardless of that.
It DOES make a difference for the logarithm.
 okay so here is my work: starting from pluggin in the concentration: s(t) = (4x10^-7) - (4x10^-7)e^(-1.0x10^-4 t) C(t) = s(t)/1000 c(t) = (4x10^-7)(1-e^(-1.0x10^-4 t)) / 1000 pluggin in c(t) = 1.2x10^-4 1.2x10^-4 = 4x10^-10(1-e^(-1.0x10^-4 t)) 300000 = 1-e^(-1.0x10^-4 t) 299999 = -e^(-1.0x10^-4 t) ln 299999 = ln -e^(-1.0x10^-4 t) i am stuck agian.. at the same place
 Recognitions: Homework Help Well i decided to do the problem to see why you weren't getting it right, anyway, i see another mistake, your integration is wrong. Btw, i get 30.05 as the answer, is t in minutes?, looks awfully fast if it was in seconds. $$\int e^{kx} dx = \frac{1}{k} e^{kx} + C$$
 oh my god , thank you so much. I have checked over my work so many timse and still did not catch that! reworking my solution as we speak and hopefully i get the same answer as you.
 i got the same answer as you, thanks very much.