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Numerade Educator



Problem 33 Hard Difficulty

In Section 9.3 we looked at mixing problems in which the volume of fluid remained constant and saw that such problems give rise to separable differentiable equations. (See Example 6 in that section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable.
A tank contains 100 L of water. A solution with a salt concentration of 0.4 kg/L is added at a rate of 5 L/min. The solution is kept mixed and is drained from the tank at a rate of 3 L/min. If $ y(t) $ is the amount of salt (in kilograms) after $ t $ minutes, show that $ y $ satisfies the differential equation
$ \frac {dy}{dt} = 2 - \frac {3y}{100 + 2t} $
Solve this equation and find the concentration after 20 minutes.


$$\approx 31.85 \mathrm{kg}$$


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Video Transcript

Okay, so this one, because the rate of change eyes equal Teoh the ray in minus the rate out we can We can show that this equation matches what? That would be eso The rate ends fairly easy. Um, we're only putting in 0.4 kilograms per leader. Uh, times five liters per minute. So our ray is gonna be 0.4 times five of her minute. How much is equal to two? The way out is a little bit different, since it's a different rate, we're removing three per minute times, whatever the concentration is of the fully mixed solution. So the concentration of the fully makes solution is gonna be why? Which is the total amount of salt in the ah tank divided by, um, the amount in the tank that's the amount in the tank is 100 plus two times T because every second we're gaining two leaders or every minute we're getting two leaders. So that's why the rate out is equal to three y time over 100 minus two t her plus tuti sari because 100 plus two t is the total amount of solution and three is the total amount that we're taking out the three per minute and the wise the total amount of salt in the tank. Okay, so that's why this equation works. Okay, So in order to solve this this we can write it, Um, in the form of d y d t Uh, plus p x times lying equals Q pecs. Okay, so in this case, P vivax is just equal to three. Divided by 100 plus two x her to team his pft. Thank u of t has just equal to two. Okay, so I'm gonna let my Ivax equals e to the integral of p of that pip T d t. Yes, we need to integrate pft. So the integral of three over 100 plus two t we can write that with a U substitution as three halves, uh, integral of whenever you do you where you is 100 plus two t and D you is to deal eyes that this gives us three halves times the natural log of 100 plus two t. And then if we, uh we can, uh, simplify this s o that Ivax. You need to eat to that. We're gonna get 100 plus two. T plan to the three halfs power and three house power just comes from the cowfish on the front of the natural log. Yes, Another we have. I've t We can say that our rate, her amount of salt in the tank is as a function of time is equal to one over. I said, I I have x times the inner girl of p of x No, not your backside. Relax. Times Cuban X. The X changed all the access to tease. I'm sorry again. Like you've exit just twos. That should make this fairly straightforward. Um, so we're gonna have for one of her Eye of X. We can write this as 100 plus two t to the negative three. House you were multiplying it by the inter grow of to times 100. Those two teams to the three house duty again a U. Substitution. Um, where you is 100 plus two t and do you is to, uh DT um we can simplify. Let's just say the integral of you to the three halves to you, which is equal to to you to the five halves in revenge. Okay, so so far we have. Um the amount of salt in our tank is going to be equal to 100. Let's to t the nature of three halves. Times two. It was 100 plus two t 25 house over five. Oh, and I forgot the plus c. So this is gonna be plus C so we can come down here and say us, see, because the first part of the negative three halves times the 100 plus two t to the five house just gives us 100 plus two t two. The first that's that's just gonna be 2/5 times 100 plus two team, we're gonna have some see value. Ah, 100. Let's to t to the three house so we can use the fact that we know that at time zero our concepts are are y value, um, is gonna be zero so we can solve for C doing that. So 2/5 times 100 is 40. Okay, We're gonna have plus R c value divided by 100 to the three halves. Okay, So well, use a calculator and I should just be 1000 eso 100 to the three. He has? Yeah, just 2000. And so we're gonna see over 1000. So and we know that that has to come out to be zeros. To see over 1000 has to be negative. 40. The negative. 40 times 1000 means that c is equal. Teoh. Negative. 40,000. Okay, so we just put that back in. That's our original equation. No, I must know how much is in there after 20 minutes. So we've seen to plug in 22 That. So we're gonna have 2/5 times. 100 US, two times 20 which is 40 minus 40,000. Divided by 100 plus 40 23 house power. Okay, so the so the amount of salt after 20 minutes is going to be around. I would That cannot be 31.85 kilograms. Okay. And because now there are after 20 minutes, there are 140 leaders. We just divide that by 140. That's gonna give us a final concentration of 0.228 Uh, so I guess the concentration down 0.228 All right. Kilograms per leader. No salt. Okay. Thank you very much.