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A vat with 500 gallons of beer contains $ 4% $ al…

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Problem 46 Easy Difficulty

The air in a room with volume 180 $ m^3 $ contains $ 0.15% $ carbon dioxide initially. Fresher air with only $ 0.05% $ carbon dioxide flows into the room at a rate of 2 $ m^3/min $ and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run?

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

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Video Transcript

So to start these mixing problems you have to actually write the differential equation before we do that. It might be helpful to clarify what the question would mean by percent as you can see if they're written as decimals. So basically what percent would be is the proportion, so the pe is gonna be the proportion of um uh sio two Um carbon dioxide um among the whole 181 80 volume of the of the room. So we can call that C over 1/8 the volume of the room. So if that's the case then that would mean that uh C is equal to one ADP and therefore uh C. Prime or rather the derivative of C Um respect he is equal to 180 times. Sorry that's a 1 81 80 times the derivative of P with respect to T. D. P. By DT. And now we can actually start writing the equation first week we'll write it in terms of C just to make things clear. So basically basically the amount of them, the rate of um ah carbon dioxide entering in the room is going to be equal to the rate of the rate in which is going to be that um Well it's two m cube coming per minute and 0.5 times that is the is the actual amount of C. 02 coming. So 0.5 three, times 2 is going to operate in. Yeah. Mhm. And then we're gonna have minus the rate out. So the rate out is going to be, well once again it's the it's a percent. Um but specifically it's the percent of since it's all coming uniformly out of the room, it's going to be the percent that's already in there, which is P and then times a Times two It says the same rate coming out for in terms of volume. And now we as I said, uh this is just Equal to 180 DP by D. T. So to be clear, what we have is um 1 80 DP by DT is equal to um 0.05 or 0.1 -2 p. and from here it could be helpful to ah make sure that there's not like a coefficient on the p by dividing both sides by by two Or even rather by -2. Um It's up to you, you don't necessarily have to do that, but what we would get then is negative 90. DP by DT is equal to um zero, sorry? Uh P -0.05. And uh now notice this is a separable equation. Um we have since everything in terms of P What we have to do is divide both sides by P -0.05. And by -90. Yeah, to get um Uh p zero. Sorry? No, yes, they were dividing by that. So Man over P -0.05 and then dP by DT It's equal to negative 1/90 and then um what we can do is sort of multiply both sides by DT and then integrate both sides. Now integrating that left side with respect to D. P. You're gonna get um Negative Lawn Absolute Value P -0.05. And on the right side here um that would of course just be negative 1/90 t. And then you can put arbitrary constant, which I'm just going to call them. See not too don't get that confused you with this one that was capital C. For amount of carbon dioxide here, this is just a lowercase C arbitrary constant. Um And you can put it on either side since you're integrating either way it doesn't matter now. What we're gonna do is we're gonna isolate P. So first of all we can Uh huh. Yeah so there was a mistake here that should not be, there should be no negative sign, no negative on the p. We made sure that earlier. So uh you can ignore that part. But yeah. Now what we can do is of course to isolate P. We can take E. To the power of each side. Um like this And basically what that gives us is absolute value B -0.05 is equal to you to the Um C -1/90 T. And then um we can do is this trick that we do is um so that when you get rid, getting rid of the absolute value side, you have to put plus or minus on the other side. We're gonna do is we're gonna change this part using X. We're gonna rewrite this part using exponent rules and we have E. To the sea and then you to the negative But 1/90. T. or t. over 90. And basically this part we're just gonna treat it as a whole other one constant of its own which is called K. And we get um yeah p minus 0.52 K. E. To the negative t minus 90. And once we get to this point we can use our initial value that were given. Um Which is that P. of zero is equal to ah 0.15. Because that's what the percent was initially At T. equals zero. So something that in we get um P. or 0.15 -0.05. It's equal to K. Times E. To the negative 0/90. That's just E. To the zero which is just one. Get rid of that. And they get that Kay Is equal to 0.1. So that means what we had was um T -0.05 Is equal to 0.1. Uh sorry about jumping on his K. um need to the negativity over 90 And finally we have p equals 0.1. It's negative slash 90 plus 0.5. And we can rearrange this how we want. It's used to be some factoring and we get that P. Um we did we weren't writing of teeth throughout but you can do that and PFT is gonna be equal to uh we can factor out a 0.05 or one out of 20. And then inside we can the order doesn't matter. We can take this term first and we get um Well if you factor one out of 20 from 0.25 that's gonna be one. And then we're going to have one factor out 1/20 from 0.1. You're going to have to two. I get in to get to that uh sort of painted tee 28 to the negative T. Over 90. That's our final answer for what the function is. And then we also asked what happens in the long run. So you're just gonna take the yes limit as puberty. Sorry. As T approaches infinity of um Of the this over 20. Uh huh. 1 40. This um what we can do is uh you can take out first of all, we can easily take out this constant for the limit and the then we take this we can also rearrange it. So basically we have since this is just a constant as well and and this is a constant where we basically just have is um 1/20 um 1-plus 2 times. And now we can take will limit as T approaches infinity of E. To the Negative T over 90. And as you should know from limits um since there's a negative sign here, so basically coefficient is negative one or 90 it's negative. Then as this S. T approaches infinity, this whole thing is going to approach zero. So basically that term will be gone in terms of that limit and we are left with 1/20 or 0.05. So that's the answer to the second point. That's what happens in the long run. The% 0.05. Mhm.

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals