 We are at lecture 28 today. We talked about concepts that the test will cover. The test is tomorrow for this group of students in this classroom this semester. So pay attention if you're doing this course on DVD or cable TV that you pay attention to your particular pacing guide for when your test is, but these students are taking a test tomorrow on 6.5 up to 2 and including 7.5, the logistic model. Now there are some sections we deleted in there. The end of chapter 6, a couple of sections that are not part of this course. So we deleted those, obviously not included on this test. I handed out an old test. Actually this old test is posted under the Distance Ed Math 241 thing on my web page. But there is quite a bit of overlap with the test that you're going to be taking tomorrow. There's the center of mass. Question number 4, how much work is done in stretching or compressing a spring? Number 5, pressure, hydrostatic pressure? Number 6, average value of a function is not for this particular test for you. So you might just put a line through number 7. Differential equation is given. Initial conditions are given in number 8. Use Euler's method. That is something you're responsible for. And then actually solve it exactly, get an exact solution, and see how close the approximation was. I actually like that kind of line of questioning because then you have an idea if your answer to A is even in the ballpark because you have the exact solution in part B for the y of 2. Number 9 was mentioned right before the cameras were rolling about just regular exponential uninhibited growth if you're given two data points. So that's fair game for tomorrow's test. And number 10, orthogonal trajectories is also part of this. Now there's some things that didn't make it to this that are also included but there's quite a bit of overlap with this test in your test. So we talked about the topics yesterday. So I think we can pretty much set up problems and work problems that you want to make sure we look at today in class before tomorrow's test. Briefly, what did not make it to this test tank problems? So we might want to look at a tank problem today, set it up, see what the initial conditions are, and go about at least getting to where we have a solution. Newton's law of cooling did not make it to this test but has a chance of making it to your test. Number 9 is an exponential growth problem. It's a population problem but continuously compounded interest would be the same exact mathematical model. So we've really kind of got one on this previous test but it doesn't say continuously compounded interest. And then we did not get a logistic growth problem on this previous test that you need to be responsible for for your test. So you do the prioritization and tell me which ones we should look at. What do you want to make sure you look at, Nicole? You look at one like number 10 and one like number 6. Okay. All right. Do you want to take a look at 10? I mean you have the solution to it here. Can we do a different one if you want to? All right, so let's take a look at what this one has first since we can kind of just browse through the solution just to look at the process. So we are given, what is that that we're handed here in number 10? What is that original set of points? x squared plus 3 y squared equals c. An ellipse. An ellipse. A family of ellipses, right, based on the value we put in for c. So we've started with this family of ellipses. We want to find out what are the slopes, right, of the tangent lines on the family of ellipses. So we take the derivative implicitly. Solve for dy over dx. If there is by chance a c or k value in our derivative, then we want to replace it with what c or k is in the original x and y relationship. That is not the case here because the derivative of c disappeared. So dy over dx is negative y, excuse me, negative x over 3 y. So we're not going to write the whole thing down, but we have this equation. Then we find what are the slopes of the tangent lines to any point on this, any of the ellipses. So there's the slopes of the tangent lines on that. What do we do to find the orthogonal trajectories? Negative, reciprocal. That's what the word orthogonal means, is that they're mutually perpendicular. So on the orthogonal trajectories, we want the slope to be what, 3 y over x. Then what's next? What does it look like happen next on this solution? Separate the x's and dx's from the y's and dy's. So it's kind of a separable differential equation problem as well. Integrate both sides and it looks like, actually that's not the one that resulted. I think the previous equation is one we should look at. 3 natural log x equals natural log of y plus k. So after we separate, integrate both sides, we come up with that. The 3 that's out in front can be replaced in that position, right? If we can take it from that position and bring it out in front, we can certainly take it from the out front position and put it back in the exponent position. What happens from there? I write that down right? Yes. Exponentiate, right? And you can get it down to the point where you have x cubed equals y times a constant. And you could solve for y by dividing both sides by a, but if you divide by a constant it just becomes another constant so you can see the final result is y equals b times x cubed. So this is a family of cubic polynomials that all go through the origin. Kind of picture how those would look. A typical cubic polynomial through the origin would look like this and they must be mutually perpendicular to the set of ellipses that we started with so you can kind of picture them and the plane is where they meet. They're going to form right angles. So that's the process. Let's try to find an example in the book. Let's try to find one that has a k in it so we have to make a substitution for, which would be slightly different. All right, solutions of differential equations, initial value problems. Here we go. 23 through 26. Find the orthogonal trajectories of the family of curves. 23 is a lot like the example we just did. I'm trying to remember the examples we did in class. Don't want to do the same problem again. You want to try 26? I may abandon it if it doesn't have a k in it in the derivative because that's what I want to get for the next example. But let's see what this looks like. Any ideas what that is? What is this family of curves? What would they look like? Just try to picture some different numbers in here for k. x over 1 plus 3x, x over 1 minus 5x. What would that look like? It's the thing with the asymptotes. It's got some asymptotes. It's got a vertical asymptote at negative one-third, right? Would all of these have some kind of vertical asymptote? Right? Based on k. And how about a horizontal asymptote? One-third. Degree is the same. x to the 1 over x to the 1. So isn't it the quotient of the lead coefficients? 1 over 3. So it's got horizontal asymptotes as well. And if you pictured this, let's see if x were zero, we get zero. So that means we've got this one. Probably this one. So some hyperbolas, right? In the plane. Family of those. Let's see what dy over dx is. How do we go about finding dy over dx? Let me go back. I don't think I sufficiently read the directions. Find the orthogonal trajectories of the family of curves using a graphing device. Draw several. We're not going to do that. Okay. So let's just find them. Tell me what to write down. 1 plus kx times 1. So quotient rule, right? Numerator times derivative of denominator. K. All over original denominator squared. Square. Wow. We've got to get rid of k, don't we? Bad example, Katie. Sorry. Bad. Bad choice. Let's see if we can do it somewhat painlessly. You just throw it away. Multiply, add, and divide by x. Yeah. Maybe it's not quite as bad as it looks up here because if we take off to the side, y, and call it y over 1, we cross multiply, and if you wanted to solve it for 1 plus kx, it's what? x over y. So we've got a 1 plus kx here, right? So we could call that x over y the quantity squared. Let's see how bad this is. This may not be kind of a fair test question. We've got a 1 plus kx here. So what is 1 plus kx? It's x over y times 1. Here's the one we probably have to work at and solve for k. So if we solve for k, divide everything by x. So if we divide everything by x, what do we get? So we'd really multiply this side by 1 over x. 1 over y minus 1 over x. That would go... I have a question. The 1 plus kx minus kx on the top, wouldn't that just be 1? Yeah, wouldn't that just be 1? Oh, do we just drop them? That would be even better. I think we're going to see that right now also. So if... Because we'd have x over y minus x over y, so they're going to drop out. So it would be cleaner to see it up here. And if your teacher had any sense, he would have seen that earlier, but he doesn't. But either way, we're going to get them to drop out. So we've got a plus kx and a minus kx, so they're going to drop out. They would also do that right here. So we have a 1x times 1 over x, so we have a 1. And... x over y quantity squared in the denominator, which is... what? y squared over x squared? Any other obvious reductions that I'm not seeing here? So what's the next task? That we got a derivative, marked out some like terms, got rid of the k, which I wanted to make sure we had one of those. This is the slope of the tangent lines at any point on this curve. I don't know, is that realistic? For this curve, for the slopes, what about the slopes of the tangent lines here? Aren't they all positive? Right? Because this curve is increasing everywhere. So is that true about these? Seems to kind of match up in that respect that at least the slopes are always going to be positive. So we want the orthogonal trajectory. So the slopes of the tangent lines should be what? Negative. Just so we make sure we get to several examples today. What would be the next step? Right there. Actually, you can cross multiply here, right? Get the x's and the dx's and the y's and dy's together. Integrate both sides. We're going to have a y squared dy, negative x squared dx, both pretty easily integrable, right? So I think we can get it to that point. I just wanted to make sure we looked at one that had a k in it so we could eliminate the k. That suffices for a second problem. And then we'd have this family of curves that have a c or a k or some a or b letter in there based on the fact that we had to do a general anti-differentiation. All right, what else did you say, Nicole? Number six. Actually, let's just set this one up because I think that's the work to the problem anyway once we get it set up. Water trough is ten meters long. That really doesn't have much bearing on the problem with triangular ends that are one meter high and at the top two meters wide. So I guess this is one. And this is one. The density of water is given. The gravitational constant, 9.8 meters per second per second. Find the hydrostatic pressure on one end of the trough. So our trough is actually here. Setup only do not integrate. Now that's a possibility for your test too is get it to the point where it's set up and then we've done different types of integration. We've already been tested over that. So that's a way to kind of save time to make sure you have plenty of time to set each problem up. What do we need for hydrostatic pressure? Depth of what? Not the trough itself because different parts of the trough are different depths, right? So don't we need the depth of some generic horizontal strip, right? And then each horizontal strip is different depth which we have to describe. So the depth of one of the strips and what the area also of one of these horizontal strips. And it does say full, okay? I was looking for that very last statement. It's full so the depth, there's our water level up here. Let's kind of go to the end of this first. How about the area of this horizontal strip? Wouldn't it be this distance doubled, right? Mm-hmm. 2x by dy, delta y. I'm going to go ahead and call that dy since we're already in the integrand. Why is it delta y? That's how thick it is or tall. There's a little increment of y, sorry. So it's 2x wide and delta y tall. So that's delta y becomes dy in the integrand. So if we were doing a segment notation setup, we could have a 2x and a delta y and when that becomes the other s, then that can become our dy. So that means we're going to be integrating with respect to y. Let's go there. So from y equals, where do we start these little strips? Y equals 0 and our last one is up here at y equals 1. So we've got that taken care of. See a problem with what we have thus far? 2x dy, we're going to integrate with respect to y and we have an x here. Don't we have to get it some way to describe what x is in terms of y and don't we need to use this line right here to do that, right? Don't we need the equation of that line? So we have, it goes through the origin and what is this point right here? Over 1 and up 1, right? So that's a pretty easy line. What's that line? Y equals x. So we can replace x with y, which we'll do. Alright, so that looks pretty good. How about the depth of this particular horizontal strip? Here's how deep it is. How do we describe that? This is 1 and what is this from the x-axis up to this point? That's y. So what we want is 1 minus y for the depth of that horizontal strip. So here's 1 and there's y, so what we want is 1 minus y. And density, don't we just throw in these constants, right? We've got meters, so 1,000 kilograms. So are we going to be able to get rid of cubic meters here in the integrand? What is 1 minus y? Isn't that meters? 2x was also meters and dy or delta y was also meters. So there's meters cubed, so we're going to get rid of this. And then we also need the gravitational constant, right? In this case, because we're working in meters, 9.8 meters per second per second. If it's in pounds, like if it's 62.5 pounds per cubic foot, that's kind of got the gravitational constant already built into it, but if it's anything in terms of meters and kilograms in seconds, we need to throw that in. Now, how does this match the setup? What problem is this? Problem 6, I think I threw a G in there and then I put that gravitational constant in the integrand the second time through. So you can bring the 9.8 out in front, you can bring the 1,000 out in front and really, I guess you can bring the 2 out in front as well. Not a whole lot left, you've got a y and a 1 minus y. So the integration is by far the easiest part of this problem. Distribute the y, so you've got an integral of y minus y squared, which is y squared over 2 minus y cubed over 3. Evaluate from 0 to 1. So the setup is the important part. And really, that's, I mean, we did, I think we did a trapezoid. The end of a water trough was a trapezoid. If it's something like a rectangle, that you can't really get a whole lot easier than that because the x value wouldn't vary as you work your way up this line that serves as the edge, the x value would always be the same, right? If it were rectangular, it would be 2 or 1.5 or whatever that distance is. What would change if the tank is only half full? Okay, that's a good question. Tank is now half full of water. What changes in this problem? So from the depth standpoint, from where the water is, we'd have to get our horizontal strip picture down here, but if it's half full in a sense that it's up to a half instead of all the way up to the top, which is up to 1, then this would be a half minus y. That's how deep beneath the surface that horizontal strip is. The area of the strip stays the same, and then where do we form our horizontal strips and where do they stop? They would go from zero to one-half, right? I mean, how much of the tank is above where the water is, that becomes irrelevant, that there's no pressure created by the fact that the water level is here and you've got some more tank up here, or water trough in this case. Anything else before we leave this one? All right, what else? What else should we look at today? Orthogonal trajectory, pressure, tank, okay? And there isn't one on this test, so let's find one. We've done several as examples in class. I don't think we did... No, we did the one with pure water, didn't we, because we had zero salt coming in. That's probably the simplest case. Let's see if there's any in the chapter review that we could pick off as an example. I don't know, the best one on the review page is here. We can address that after we set this up. This is on page 553, tank problem. Now, when I say tank problem, I don't mean any problem that's talking about a tank. We just did a water trough, which if you want to think of that as a tank, but I don't classify that one as a tank problem, that would be a pressure problem. So a tank problem is a mixing problem or a change in dilution problem. So this is problem 18, looks like pretty standard tank problem, testable problem. A tank contains 100 liters of pure water, brine that contains 0.1 kilogram of salt per liter enters the tank at a rate of 10 liters per minute. So we've got the rate coming in. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after six minutes? So I think the WebAssign type question was how much is in the tank after T minutes so you get kind of the formula and then plug in a specific value. All right, so salt. So let's let S equal the amount of salt in the tank at time T. So the rate of change of salt is based on two highly scientific things. The rate that salt is coming in, kind of wrap your mind around that, and the rate at which salt is going out. Not a tricky beginning for tank problems. Now, to address chandlers, there may be two different spickets coming in. You just analyze them separately, and it better be true that the sum of the flow of the liquid here is equal to the liquid that's going out. Otherwise, this type of problem falls into a different category and you're not responsible for it at this point in time. You can mix them if they're different densities. Yes, so they might have different concentrations of... one might just be pure water, the other might be some concentration of brine. They're both coming in. But the sum of the liquid flow coming in for us at this point in time better equal the liquid flow going out. So the tank is going to remain constant at 100 liters. Now, can you envision what's going to happen if they're different? When we start to do the concentration of the stuff coming out, it's not going to be S over 100. It's going to be S over 100 either plus a certain amount because it's growing because we got more coming in than going out or 100 minus a certain amount because we've got more... it's a mess. It's quite a bit messier when the flow in and the flow out are not the same. Somebody said something. Yes, Shayna? So like if one's coming in at five and the other's coming in at 10, the rate going out has to be 15. Yes. All right. So what do we have here? Brine that contains 0.1 kilogram of salt per liter. So here's what's coming in. It enters the tank at a rate of 10 liters per minute. So I usually take a couple seconds to say, does that sound like to me the rate at which salt is coming in? So many kilograms per minute. Kind of sounds like the rate at which salt could be coming in. The rate at which salt is going out, well, it better be leaving at 10 liters per minute. And then you ask yourself the question, how much salt do I have in the tank at any point in time? Well, it's already written down on this sheet. How much salt do we have in the tank at any time, too? S, that's what S is. And it's... so that'd be S kilograms out of 100 liters. So that might look like a five. That's an S. So our... So 0.1 times 10 would be 1 minus, what? S over 10. Pretty simple differential equation. I always kind of cringe when I see the minus here because some kind of minus problems. I like all multiplication and division and I kind of get a little uneasy when I see subtraction, but I think we'll be okay. Before we separate what recommendation did we make the first time around? Pull out the coefficient of S. So we're going to pull out a negative 1-tenth. Don't have to do it, but I think it makes it easier from this point. And that would be negative 1-tenth S minus 10. Is that right? This is probably not one of those problems where you're just going to set it up and not finish the problem because there are some conditions that are given in the problem. So we want a specific amount in the tank at the end of six minutes. So this will be a complete completion type problem. So that's gone. That's gone. So we integrate both sides. This is where I think it's advantageous to have factored out the negative 1-tenth that's in front of the S. This left side becomes what? Natural log of S minus 10. We did the absolute value route and we see at some point in time later in the problem we can dispense with that. So let's just dispense with it right now. Next, exponentiate both sides. Left side is S minus 10. The right side, we've done this enough. Let's just skip a couple steps. B. Does that work? Right. It's going to be E to this, times E to the C, E to the C is a number, so we'll just call it B. All right, so we've got it solved for S. Let's go back to the problem. A tank contains 100 liters of pure water. What does that tell us? Initial conditions. At time zero, what is S? Zero. No salt at all in the tank. So there's E to the zero, which is one. So that's just B. So B must be what? Negative 10. So there's our equation in terms of T. And as long as everything was given in terms of minutes, which it was, as our original unit time, we want to know the amount of salt in the tank at time six. We just plug in a six. Is that correct? Yeah. So for T, we'll plug in six. Do the arithmetic button pushing, whatever you want to call it on the right side, and it should give us the amount of salt in the tank at the six-minute mark. Is that all right? The very first statement in the problem, a tank contains 100 liters of pure water. Right. So keep in mind what we're talking about in the problem. Are we talking about a pesticide? If it's pure, there's zero pesticide. Are we talking about some concentration of salt? If it's pure water, there is zero salt. Okay, what else should we look at today? Can we do something out of a tank? Pump something out of a tank? The work? Yeah. A work problem? Okay. Center of mass. Okay. There is a center of mass on here. Let's look at this real quickly, and then we'll go to a work problem, and that's probably all we have time for. Center of mass on here is number four. Off to the side on the center of mass problem, you will probably want to find the area under that curve, because X-bar, I think the formula in the book says one over A, or just remember to divide through by the area under this curve. What do we have for X-bar? X, F of X, DX for Y-bar, one-half F of X squared. So we do need the area. So in this problem, let's save a few moments, we need the area. So the area under this curve, it is the square root of X curve from zero to four. I don't think that's an issue at this point in time in this course to integrate this and evaluate it from zero to four. It's on the sheet, and that is 16 thirds. So that's going to go here and here for the centroid or center of mass. So for the numerator, zero to four, we want X, F of X up here. What is F of X, the square root of X, right? So I'm going to divide that by 16 thirds, so I'm going to multiply it. Now let's just put it down here. So for Y-bar, one-half F of X squared integrated with respect to X divided by the area, which is still 16 thirds. That ends up being really nice and simple. Square root of X squared is just X. That's a pretty easy integration and evaluation problem. For the numerator there, you're going to get 16. Is that right? I'm having trouble picking out the arithmetic there. No. Actually, we're going to get eight. So that ends up being three-fourths. No. I've got my arithmetic wrong here somewhere. So X squared over two from zero to four. Now it would probably be four, right? Yes. So there's Y-bar. This X times the square root of X is just X to the three-halves integrated would be X to the five-halves over five-halves. Evaluate it from zero to four. And that ends up being two-fifths of... So we'll bring that up three-sixteenths. We're going to have a two-fifths from the division by five-halves. And so four to the five-halves. Yeah, square root of four or two to the fifth, 32. So there's the arithmetic on that one. Which X-bar is 12-fifths. Might be helpful on a center of mass problem. Here, it's just X, and we're going to integrate with respect to X. So that just stays the same. I put four in. Yeah, four to the five-halves. So it'd be the square root of four, which is two, two to the fifth, which is 32. So there's our region underneath this curve from zero to four. So X-bar should be 12-fifths. This curve looks a little bit heavy on the right side. So it should be larger than two, right? Which it is, two and two-fifths. So it's over here somewhere. And Y-bar is three-quarters. What is this right here? Two, right? So probably right in here. So it seems like a reasonable answer for that region. And I think that's certainly a fair test question. A good indicator that I thought that was a fair test question is because I put it on a test. Therefore, I must have thought it was fair. Actually, it's probably on the easy end of fair. All right, a work problem. Let's say that we have a... So let's say we have this trough, or lack of a better word, and it ends our trapezoids. And let's say that this, I'm just making up numbers. Let's say this is two. And up here, this is five. And it is six feet deep. It's full of water. So we want to know how much work is involved in pumping the water that's in this particular water trough out to the top of the tank or trough. So work is going to be force times distance, so we're in feet here. So we need to take a... And not just a strip of area, because we don't just want pressure on the end. We want kind of how much volume is in that particular slice. And then not just the volume, because it's not just volume that's going to be able to move it. We want to know how much that actually weighs. That's going to take that amount of force to actually move it. So that distance is x. And so we want 2x by delta y. So I'm going to off to the side do a little volume. 2x by delta y. Don't we also need to know this dimension? Which we don't. So let's put in a number. Let's say it's 10. That's going to be given to you. So that's the volume of water in that particular slice. So that is feet, that is feet, and that is feet. So we've got cubic feet. Now that's not the force required to move that. So what is the force required to move it? That times 62.5 pounds per cubic foot. So we've got a 2x, we've got a 10, and we've got a dy. That's the volume. 62.5 pounds per cubic foot. So that's the force now that's going to move that. And how far do we have to move that particular slice to get it to the top of the tank? Don't we have to move it this far? So 6 minus y is how far we have to move it. We've already got our incremental value, which is delta y or dy. So we would start with the strip at the bottom of this tank, which is what? y equals 0 and go all the way up to the strip or slice at the top of the tank, which is y equals 6, right? And then what work is left? Okay, all the y's are okay because we've got dy's. We've got to get rid of x. Don't we need the equation of this line? Do we have two points on that line? What's this point? 2, 0. And this point is 5, 6. So you would use that point and that point to find the equation of that line. Once you have it, you can solve it for x. It'll be x in terms of y and replace that quantity for x. So there's a work problem and we're out of time. I don't want to keep you from your next class. Tomorrow is the test. We can start it at 5 minutes before the normal time and end it 5 minutes after the normal time. So that way you get it out.