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3 (2 points) A tank initially contains 50 gallons of water and 100 ounces of salt: Water containing a salt concentration of 0.1 ounces per gallon flows into the tan...

Question

3 (2 points) A tank initially contains 50 gallons of water and 100 ounces of salt: Water containing a salt concentration of 0.1 ounces per gallon flows into the tankata rate of 2 gallons per minute, and the well-stirred mixture leaves the tankata rate of 1 gallon per minute. Let Q(t) be the amount of salt (in ounces) at t minutes after this process started. Set up (but don't solve) an initial value problem that describes this situation: ie , write down a differential equation (involving onl

3 (2 points) A tank initially contains 50 gallons of water and 100 ounces of salt: Water containing a salt concentration of 0.1 ounces per gallon flows into the tankata rate of 2 gallons per minute, and the well-stirred mixture leaves the tankata rate of 1 gallon per minute. Let Q(t) be the amount of salt (in ounces) at t minutes after this process started. Set up (but don't solve) an initial value problem that describes this situation: ie , write down a differential equation (involving only the variables Q and t) and an initial condition.



Answers

Salt water of concentration 0.1 pound of salt per gallon flows into a large tank that initially contains 50 gallons of pure water. (a) If the flow rate of salt water into the tank is 5 gal/min, find the volume $V(t)$ of water and the amount $A(t)$ of salt in the tank after $t$ minutes. (b) Find a formula for the salt concentration $c(t)$ (in Ib/gal) after $t$ minutes. (c) Discuss the variation of $c(t)$ as $t \rightarrow \infty$

Hello, everyone. Today we're going to solve problem about 32. Here's one like the Cube I d. P equals two Cuban are one minus are two into curiosity so we can have a person like you off t equal toe capital Q divided by be off key. So the off key equal to the not plus are one minus are too in do Okay, So que off equal toe capital Q developed by the not plus are one minus R two in duty so de que developed by they take equals two Cuban are one minus You are to hold developed by we not plus are one minus are two into the full that we can do like r one equal toe are to equal toe So the cube I DT plus Q are developed by the not equal to q one are Thank you

Okay, So for this problem, we're told that at time equaling zero a tank contains 25 ounces of salt dissolved in 50 gallons of water. So for our why of zero, we have 25 pounds. And then, of course, we can kind of say that this isn't 50 gallons of H 20 So then it says the then Brian containing four ounces of salt per gallon is allowed to enter the tank at a rate of two gallons per minutes. So we have entering four ounces per gallon with a rate of two gallons per minute. Okay, so what I want to do is I want to Duthie rate of change. So it's gonna be d y over DT where why is the mount? And obviously T is time. And so it's going to be the rate in, ah minus the rate out or rate out minus rate in depending on the situation. So for this one, what I want to dio because I want to say the rate in So I'm starting off its force. What's going in in four Alice is per gallon at two gallons per minute and in minus my the amounts over 50 cause it's a 50 gallon tank times the rate So d y over DT is going to equal eight minus Why over 25. So now what? I want to dio because I want to set this up for linear set up solving using a linear solution. So what I want to dio is I want to add 1/25 y equals eight. Then what I want to do is I want to find my integrating factor. So I'm gonna have e to the integral of 1/25 DT. So that's going to equal e to the 1/25 t. So then what I know is that my d over d d t is going to be the integrating factor times the variable which, um, is going to be Why equals eight DT. Then I have to add put my integrating factor. So now if I'm integrating each of these, I know that this is going to cancel. Its gonna leave e to the 1/25 tea. Why? And so now I'm gonna have my eight and then eats of the 1 2050 and then I have to do the reciprocal of what the constant is in front of the tea, which is gonna be 25 and then plus my constant I'm gonna divide by E to the 1 2015 So we'll be left with y equals 200 plus c e to the negative one 25th t. So, since I have an initial condition, which we said is why zero equals 25 then I can go through and plug this in to find out my constant. So, you know, eat of the zero is one, so it looks like, see is going to equal negative 1 75. So my equation is going to be why equals 200 minus 1 75 e to the negative t over 25. So we were asked to first, um, find the equation and find it for an arbitrary T, then were asked to find what it is after 25 minutes. So then what I want to dio is I want to find why of 25 so it's gonna equal 200 minus 1 75 e to the negative 25/25. So putting this into a calculator, it's gonna be approximately 100 35.62 and then we're talking about ounces

Okay, we have this thes tanks of salt water. This first one has 30 gallons of water, and at any time, t it has Q one ounces of salt in it at time. Zero has 25 ounces over here. Then we have this other tank that initially as 20 gallons of water in it. And it has que tu t of t ounces of salt. So that changes the amount of salt changes as the water comes in. Initially, it has 15 ounces, so those two things are given there. Yeah, Okay. This one has salt water coming in from somewhere else. Saltwater coming in from this tank over here. And then it's also sending saltwater out can. Then this tank has water coming in. Hopes are coming in up there. And then it's sending water over to the other tank, and it's sending water out onto the ground or in another somewhere else. And the other tank is sending water to it. Okay, so we want to make some differential equations that show the rate of change of salt do per time. Okay, so that's what she's black, D. Q one D t. Okay, so let's start here at the top. We got 1.5 gallons per minute gallons in one minute times, one ounce per gallon. So that's going to give us ounces per minute. Because then we have this coming in. It's coming in at 1.5 gallons in one minute, and it has cute, too, of T ounces of salt for 20 gallons. And then it's sending out this mix right here. So minus three gallons per minute and how much salt is in it. Kyu won t ounces for 30 gallons. Okay, so that that's sort of the strength of it. Okay, so let's get all the junk out of there. So we got 1.5 ounces per minute plus 1.5 Q 2/20 minus three Q 1/30. So 1.5 plus 15 over 200. So 3/40 que two minus 1/10 Q 1. So that's our first differential equation. Okay, Now we have to write one for the other one. I need the picture, though. I'll screwed it just a little bit there. Yeah, we'll just have to move it up now. Okay. D Q two d t. Okay, we have the fresh stuff coming in one gallon per minute. Three ounces per gallon, one gallon for one minute times three ounces per gallon. Plus we have the water coming from the other tank. Three gallons per minute Q. One ounces per 30 gallons, three gallons per minute times Q. One ounces for 30 gallons minus. It's sending water out. I'm going to do them separately. Worse so we have. There we go. So we have the one going this way. Hopes up, and it's going at 1.5 gallons per minute. And that one's strength is Q T ounces per 20 gallons. And it's sending some this way. 2.5 gallons per minute times cute to over 20. Or you could just send it out as four gallons per minute. It's gonna it's gonna be the same. So this one we get three. Just forget per minute plus 1/10 Q 1 minus 3/40 Q 2 minus 2.5 over 20. So 25 over 200 518 Find us 18 Q 2. So three plus 1/10 Kyu Wan That's 5 40. So that's 8/40 minus 1/5 que two take on DT de que two d t. So those two equations Oops. This one really is just tea this year and this year, okay, with these initial conditions, which are given in the book, Okay, then the next question is find the values for Q one and Q two, for which the system is eat in equilibrium. So it means where the rate of change of salt is equal to zero. Okay, so we're solving this system. Zero equals 1.5. You know what I'm gonna write? That is three halves, three halves plus 3/40 que tu minus 1/10 Q 1 and zero equals three plus 1/10 Q 1 minus 1/5 Q 2. Okay, so I'm trying to solve this system. So first thing, I mean, I do get everybody in order, get all the fractions out of here. I'm gonna multiply the first one by 40 so I'm gonna have okay, four Q one minus three, Q two. So I've moved them over here equals 2060 and this one. I'm gonna go ahead and multiply it by 40 also, So I get four Q one minus eight. Q two equals well when I move them this way, minus four. Q. One plus a Q two equals 120. Well, im going sideways here. Okay, so I'm gonna add those together, so I get five. Q two equals 180. So Q two equals five goes into that 36 times. And then if I plugged that in, I get four. Q one minus three times 36 equals 60 for Q one minus 108. Equal 60 for Q one equals 168 so Q one equals 42. Okay, so when we get 36 ounces in the second one and 42 ounces in the first one, then the systems will be a new equilibrium and they won't change. The amount will change. Okay, Now it says called those que on e and Q two e and then make a new system of equations X one and x two. Alright, so part c X one equals Q one minus Q one e, which is this'll and X two equals Q two minus Q two e, which is this? Okay, so then we wanna make a differential equation, and so we're gonna take some derivatives here. All right, So, uh, x one prime is Q one prime minus zero. Because that's just a constant. Okay, Q one, prime. Don't forget Waas 1.5 minus 1/10 Q 1 plus 3/40. You too. Okay, now X one Waas Kyu Wan minus 42. And x two is Que tu minus 36. So here I get three halves minus 1/10 of Q one, which is from here X one plus 42 plus 3 40 sq, too, Which will be X two plus 36. Okay, so we get three halves minus 42 tents minus 1/10 x one plus 3/40 X two plus 108. Oh, let's reduce their nine will go in their 40 no, no. Before we go in there, 10. Oh, just leave 100 and 8/40. Let's get a common denominator of 40. So this one will be 60/40 minus 31 times 41 68 over 40. Plus 108 over 40 minus 1/10 x one plus 3. 40 s x two. All right, So this one turns out to be x one Prime or DX one d t. If you wanna make it look like that, Okay, those all canceled minus 1/10. Next one plus 3. 40 s x two. Okay, then we'll do the same thing over here. X two Prime will be cute to prime, which, if you'll remember, is three plus 1/10 Q one minus 1/5 Q 2 So three plus 1/10 x one plus 42 minus 1/5 x two My plus 36 three plus 42 tents minus 36 5th plus 1/10 x one minus 1/5 x two Get a common denominator here of 10 30 plus 40 to minus 72/10 plus 1/10 x one minus 1/5 x two so x two Prime or D X two DT is 1/10 x one minus 1/5 x two So notice this new system that we have is homogeneous doesn't have any Constance with it, as opposed to the one with Q, which did. There you go


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