 Welcome back to episode five of Math 1050, College Algebra. I'm Dennis Allison here in the Math Department at Utah Valley State College. And today we're going to talk about two big topics. The first one is using functions as models. And the other one is on direct and inverse variation. So let's begin by looking at functions and models. And let's go to the first graphic that says the graph of a function can tell a story. In this case, it becomes a, quote, model for the application. Now let's take an example and go to the next graphic here. We actually saw this same problem several episodes back. But I just want to remind you how this goes. This is a graph of a salesman's distance from home. So you see it starts off at 8 AM. He's at home. And by 9 o'clock, he's a certain distance away from home. And it looks like between 9 and 10, he stopped. Maybe for breakfast. Maybe he stopped at his office. But then he travels on between 10 and noon. He travels in a linear fashion, we might point out. And he stops between 12 and 1. Perhaps he had lunch. Then he's traveling again between noon and 5 o'clock. He gets a little closer to home. Then he travels further away. And then he stopped between 6 and 7, perhaps, for dinner. And then he comes home in a hurry. So you see, here's a graph of a function that is a model for a story. And when you look at the graph, you can interpret the story. We don't know that we have all the details right. We don't know if he stopped for dinner from 6 to 7, but it looks like he was stopped anyway. Let's go to the next graphic. Here's a new one. This one says, a homeowner mows the lawn every Wednesday afternoon, sketch the graph of the height of the grass as a function of time over a four week period, beginning with Sunday. Well, you see, beginning on Sunday, the grass has some height to it, but it keeps growing until Wednesday. And that's when the homeowner mows the grass. And then the height drops back down. And then the grass starts growing until the next Wednesday. Then it gets mowed again. So you see, this graph is chopped up into pieces. Does that graph, would that graph be a function? Does it represent a function? Well, let's see, it has to pass the vertical line test. Does it pass the vertical line test? Yes, every vertical line seems to intersect it no more than once. Why is it that the graphs, those little branches don't start off at zero? What would be your explanation for that, Jenny? You don't shave your grass off completely. You have to leave about a half an inch there. Yeah, exactly. So you have to leave a little bit of grass. You don't cut it right down to the ground. So the graph starts off, those branches start off a little bit higher than zero. But then it seems to grow in a linear fashion. Now, of course, when I say a function as a model, this is not an exact model of the grass because I doubt if grass grows in a perfectly linear fashion. If you get rainfall, grass grows faster. If you're in a drought, grass may grow slower. So this model is supposed to represent, but not perfectly, the growth of the grass. By the way, it's said to grow this or to represent this for a four-week interval. And actually, it looks like I've only drawn about a three and a half-week interval. So I should put an extra short branch coming out of the right-hand Wednesday if I were actually going to show four weeks. Okay, let's go to the next graphic. Here's a quite different problem. This time, we have an illustration, but we don't have a graph. It says a property owner uses 120 feet of fencing to make a rectangular garden beside a garage. So you see the garage is represented. We're sort of a bird's eye view. The garage is on the left. We have the rectangular garden. We come out X yards on each side. And the perimeter of the three sides is 120 feet. Now, the questions say, first of all, find the four-sided perimeter. We'll call that function P of X, P for perimeter. And then find the area, A of X. So we're gonna choose the function, name the function A for area. So let's try solving this on the greenboard. I'm gonna draw that illustration up here. Here's the garage. We're going to come out, down, and back. And we know that we have 120 yards all together of fencing. And I'm gonna call this X, call this X. So if I put X yards here and X yards here, how many yards are left over for this side? Stephen? 120 minus 2X? 120 minus 2X, yes. And I think I'm running out of room there, so I'll put it, I'll write it over here. 120 minus 2X. So part A asks us to find the function P of X, which is the perimeter of the garden. Now that's for all four sides. So the perimeter, I'd have to add up X and 120 minus 2X and X, and another 120 minus 2X, even though there's no fencing here, it's part of the perimeter. So that's gonna be X. Actually, I'll have two of those X's. Why don't I put 2X there? And I also have two of the 120 minus 2X's. Two times 120 minus 2X. Now that is the function P of X, but it's just begging us to simplify it, so let's go ahead and do that. So that'll be 2X plus 240 minus 4X. And that's a total of 240 minus 2X. And what units would I put on that? Yards. Yards, okay. So that would be 240 minus 2X yards. So that is the function rule for the perimeter. Now you see what this means is, if you tell me how much X is, like if you tell me X is 10, if I plug in a 10 here, that'll tell me what the perimeter of the garden is. Let's just see. P at 10 would be 240 minus 2X 10 minus 20 is 220 yards. That's the total perimeter. Now you remember, we only have 120 feet of fencing, but the perimeter of the garden is 220 yards. That would be 10 and 10. How much would this side be? 100. 100, and this would be 100. So 10 and 10 and 100 and 100 is 220 yards. On the other hand, what if we chose X to be 50? Well, then this would be 240 minus 2X. That'll be 100. And I get 140 yards for the total perimeter. That would be 50 and 50. And what would this be? 20. 20 and 20, 50 and 50 and 20 and 20. Sure enough, that's 140 yards. And you might say, well, Dennis, look, if you can figure out those answers just by pointing at it, 50 and 50 and 20 and 20, what's the purpose of this? Well, now I have a general representation for all possible, for all perimeters, for all possible Xs. By the way, there is a restriction on the domain of this function, because X can't just be any number. One restriction, for example, is that X can't be negative. So we know that the Xs here have to be positive numbers, because you can't have a negative dimension on a garden. On the other hand, what's the biggest X could be? That's kind of a tricky one. Let's see, we have 120 feet of fencing. So if I go straight out and straight back, how much would X be? 60. It'd be 60, so X can't be bigger than 60. If I go out 70, I don't have enough fencing to even come back. So the biggest X could be, would be 60. And as a matter of fact, we might even argue that 60 isn't allowable, because that's not really a rectangle. If you go out 60 and back 60, you just have a line segment with zero area enclosed. But we'll include those as sort of the extreme possibilities. And I would say that the domain of this function is restricted, and that is the domain is zero to 60. The reason I point that out is we mentioned in the first episode, I think it was, that if there's no mention of a restriction on the domain, we assume the domain is every allowable number. Well, you see, this function allows every possible number to give you an answer. It just, it doesn't make sense in this physical application. So we have this smaller domain. Part B of the same problem says, find the area of that garden. So let's do that now. We call this function A of X. Once again, the garden looked like this. We come out X, and we said that this side was, oh, this is X down here. We said this side is 120 minus 2X. So what would be the area of the garden? Susan, what would you say is the area? X times 120 minus 2X. Exactly, X times 120 minus 2X. Now, we could leave the answer like that, but why don't we multiply it out? And this says that A of X will be 120X minus 2X squared. If you wanna turn that around and put the X squared first, you could say minus 2X squared plus 120X. It doesn't really matter which way you write it. But in either case, this is not a linear function because it's not in the form F of X equals MX plus B. It has this X squared term in it. So this sort of a function is called a quadratic function. You remember you've solved quadratic equations before. Well, this is a quadratic function. Can anyone tell me what would be the area of the garden if we chose X to be 15? That was not my calculator. Well, let's see. Let's see how we'd do that. Now, we could plug in a 15 here and here. We'd have to square the 15. We'd have to multiply 15 times 120. We could plug in the 15 here and here. That would be essentially the same thing. Or we could plug in the 15 up there. So why don't we go to that first version because I think it'll be simpler to multiply. If I put in a 15 for X and then if I put in a 15 here, that'll be 120 minus 30. How much is that? 90. 90, okay. So what I need to do is take 90 times 15. Let's see, 15 times 90 is going to be 0, 5, 1, 3. So that'll be 1350, 1,350. And what are the units on that answer? Yard squared. Yard squared, exactly, because we're talking about area. So this gives me yard squared because it's an area function. You notice that the name of the function is not f of X because f is sort of a generic name for function. In this case, the area has, or the function has a certain meaning. So what I choose is a letter to allow me to remember what this function is. P for perimeter, A for area, 1350 yards in this case. Okay, let's go to the next graphic. This next problem is actually a rather famous one. It was first solved by Galileo. Let me write his name up here, Galileo. Galileo was born in 1564 and he died in 1642. And Galileo was the first mathematician to notice that the height of a projectile, if you throw it up into the air, would obey a quadratic function. Here's what he discovered. If we go back to that graphic, you see if we have a building, let's say we're standing on top of that building, and let's say the building is H sub zero tall. There's a H sub zero in that equation that you see there. That's how tall the building is. And let's say you throw the object up in the air with an initial speed, which I call V sub zero, then the height of the ball can be calculated by the formula you see on the screen. Let me just write that up here. Capital H of T. Why do you think I'm calling the function capital H? For height. For height, okay. Is equal to minus 16 T squared plus V sub zero T plus H sub zero. Now, the number H sub zero is your initial height. That's why it's H sub zero. It's your height that times zero when you first release the object. V sub zero is the initial speed. If you throw the object up in the air, the initial speed is positive, and if you throw the object down, I'd call this negative, in which case it's actually velocity. And the minus 16, well now that's the strangest part of all, this has to do with the acceleration of gravity. If I were to solve this problem, let's say on the moon, where the acceleration of gravity is different, because you're way less on the moon, I'd put a different number there. The minus 16 stays the same because we're doing all of our problems here on the earth, luckily, so we don't have to worry about other accelerations of gravity. Okay, let's go to the problem that follows the graphic of the building. It says that a ball is tossed upward from a 240 foot building at a speed of 32 feet per second. And there are three questions that we ask. First of all, find the function H of T. Well, we have the general form of the function here on our green board. Part B says, how high will the ball be after four seconds? Part C, when will the ball strike the ground? And part D, what is the domain of the function H? Okay, let's come back to the green screen and take a look at those one at a time. The first question was to find the function H of T for this particular problem. So this function is going to be a model of that problem. Let's see, so we start off minus 16 T squared plus the initial speed V sub zero. Do you remember what the initial speed was? 32 feet per second. 32 feet per second, exactly. And you remember what the height of the building was? 240. 240 feet, yes. By the way, before I go on, let me just mention, for those of you at home taking notes on this, all of these problems are worked out in detail on the website. So if I give you a value like the height of the building and you say, wait a minute, I don't remember what you said the height was. If you look on the website, you'll see a full discussion of this problem and all the others that we've done. And you can check the numbers and you can look at that more slowly than listening to me explain it here verbally. Okay, so this is the function that describes the height of the ball at time T. Now, the second question was, how high is the ball after four seconds? Well, T represents time in seconds. So I'm gonna substitute in a four for T and this answer is gonna be given in feet. That's gonna be the height of the ball in feet. Well, if I substitute in four, that's gonna be a minus 16 times four squared, that's 16 plus 32 times four plus 240. Now, we just have to simplify that. 16 times 16 is 256, that's a minus 256 plus four times 32 is 128, thank you, plus 240. Well, why don't we combine those two together? Those are real easy to combine. Negative 256 plus 240, that's a minus 16 plus 128. And so 128 subtract 16 is how much? 112. 112, right. So that's 112 feet. So the ball is 112 feet in the air. Now, let me just erase some of this and let's just look at the illustration of that building. I'm gonna draw the building here on the graphic. Let's keep in mind that after four seconds, the ball is 112 feet in the air. And so if this is the building and if this is where we're standing initially at h sub zero, we throw it up in the air and then it comes back down and it's on the way to the ground. So this altitude, if I make a vertical axis right here, this would be 240 because that's our initial height. Now, of course, the ball goes higher than that for a while but then it comes back down. After four seconds, it's 112 feet off the ground. 112 would be, I'll estimate it right about there. So after four seconds, the ball is right here. I'm gonna say t equals four, which means that it has already got up to its maximum. It's on the way back down, it's traveling downward but it hasn't reached the ground yet because it's height is in zero. So the ball's right about there. Now, if you were to actually carry out this experiment, if you went to the top of a building that was 240 feet high and if you could throw the ball up in the air at exactly 32 feet per second straight up and if you'll let it come down the side, it probably wouldn't be exactly there. Do you know what other effects would influence the progress of the projectile? Air resistance. There'd be air resistance. So you know what happens is the ball actually slows down. It probably wouldn't go quite that high because of air resistance and it would probably accelerate not quite at the acceleration that Galileo predicted because his formula is actually intended to work in a vacuum but we're not in a vacuum when you throw the object in the air. There's air resistance but this is a very close approximation. That's why I'm saying this function H of t is a model for the application. It's not the exact representation of it but it's close enough for our purposes say in a physics course. The next question we asked related to this problem was when does it hit the ground? Let's see, let me just write in our function for this particular situation. We said it was minus 16 t squared plus 32 t plus 240. Now how are we gonna find out when it hits the ground? That's question C. Any suggestions? Ginny? When H equals zero? Exactly, when the height is zero it's on the ground. So I'm gonna put a zero right here and what I have now is a quadratic equation. We had a quadratic function. Now we have a quadratic equation. There are several ways to solve this but if we can factor it I think that's the quickest way to do it. Trouble is there's a negative 16 coefficient. Why don't we divide that out? I think 16 divides all of these. I'm gonna divide by a negative 16. Of course even divide the zero by negative 16 and I get a zero. This would be a t squared, what will this be? Negative two. Negative two t, yep. And this one will be a negative, let's see, 16 into 24, one, 16 to 85. That's a minus 15. That looks a whole lot nicer. Now of course this equation is different than the previous equation. But what they have in common is they have the same solution. So I've actually made a transformation in the equation to a simpler form and now I'm gonna solve this one. Just like last time we made transformations in graphs, here I'm making a transformation in an equation. I can factor that into, let's see, t, t. This sign's a negative. That tells me something's positive, something's negative. Has anyone figured out what those numbers would be? Three and five. Three and five, yeah. I think we wanna put the five with the negative but the three with the positive. And if this product is zero, one of the factors is zero. So either t plus three is zero or t minus five is zero. So either t equals negative three, hey, what's that all about? Or t is equal to five seconds. Of course, this is the answer that we'll wanna choose. It hits the ground in five seconds. You remember it was 112 feet off the ground after four seconds. In one more second it should be on the ground. Now, why did we get a negative three? What does that mean? Well, you know, if I go back and draw the illustration of the building, I think we can see where that comes from. In this problem, the algebra doesn't know that the problem begins when t is zero. So we said we released the ball at time t equals zero, it goes up and then it comes down to the ground. Well, for all the algebra knows, we released it prior to time zero. We released it at maybe negative one second, maybe negative two seconds, maybe negative three seconds we were on the ground. So three seconds prior to the release of the ball on the top of the building, we could have been on the ground and thrown it much faster and it would have followed the same path and it would have gone up to that peak and come back down. So the algebra gives us every possible answer and it asks us to figure out what answer we want. So we're gonna take the answer t equals five and we're going to reject the answer t equals negative three seconds because we know this problem began at the moment when t was zero. But the algebra doesn't know that we don't want that so it gives us every answer. Now, we have one more application I'd like to look at. This one involves an interview with an oral motorcycle policeman. So let's go to that tape. War class, we've been discussing how functions are used as models in our college algebra class and so I've called Officer Gaines from the Orem Police Department who's gonna show us how his radar gun works and we'll go back to the classroom and we'll talk about how we can make a mathematical function that models the radar gun. So Officer Gaines, thank you for coming. Thank you. Basically we use two different types of radars. One is one that we use in our car which uses radio frequencies. The other one is the one I have here which uses laser or light frequencies to figure out the speed. Basically, the difference in the two is with the radio frequency that's sending the radio frequency out. It's hitting the object and coming back into the radar. The radar's taking the change in that frequency or the change in that wavelength to figure out the speed in miles per hour. We can use that unit when we're either moving down the road or when we're stopped. If we're moving down the road the speed we're traveling plays an effect in the way it figures out that speed. It needs to take that speed out of there so it comes up with an accurate speed of the target vehicle. With the LiDAR, the big difference is, one, it's using a light wave and it's sending out pulses of light at about 600 pulses per second. It's hitting an object and returning back into the unit. When it returns, the unit uses that time it laps to calculate the speed and convert that into miles per hour. The one disadvantage to this unit is that we have to be stationary or stop to use it so we can't be driving down the road to use it. The other benefit of this one, obviously when you're using a radio frequency it's going out there and getting wider and wider. There's some other steps as an officer that you have to use in order to make sure you're getting an accurate reading. With the LiDAR, the beam width at 1,000 feet is only about 12 inches wide so we can place the LiDAR unit on a particular object whether it's a person, bike, car, whatever, and we know that's the reading we're getting is from that particular object. Okay, with this unit, it tells us both the speed and the distance the object is from us and then you'll see either a plus or minus next to that number which lets us know if the vehicle's coming towards us or going away from us. What I'm using inside this unit here is just a small red dot so that I can use it to point it at the object or what I want to get the speed from. And you can see as the speed changes in the unit it changes in the speed here as well as it giving us the distance and again there's the plus sign which is telling us that object was coming towards our location. Okay, now you notice you said there are actually two different types of radar guns. The one that's mounted in a police car you can actually be traveling while you're using that radar gun and it's tied to the speedometer of the car so it makes an adjustment for the speed of the police car and they can clock you whether you're going away or toward them. But with this radar gun that he has works on a different principle and he has to be sitting stationary and he has to be looking right at you you have to be coming right at him if he moves off to the side like at a 30 degree, 45 degree angle then your speed will be registered as too slow. So that's why they step out in the middle of the street and they look right at you to get an accurate reading and if you think well if they're on the side maybe I'm gonna be they're gonna think I'm going faster no actually they'll think you're going slower so there'll be a difference there and in fact that uses something he calls the cosine effect that we'll talk about in math 1060. Well let's take a look at the radar gun that the motorcycle officer is using and the principle is that it sends out pulses of light he said 600 per second and that light bounces off the car comes back to the radar gun so it takes two different readings first it sends out a pulse see how long it takes for it to come back then it takes another pulse sometime later and bounces off the car comes back and now it's not gonna take as long if the car's coming toward him and so by measuring the change in the time intervals of those two pulses he can determine what your speed is. Okay let's say the officer is standing right here where sort of a bird's eye view the officer standing along this line and your car is over here moving toward the officer so the officer points the radar gun at you sends out a radar beam that goes out hits the car and comes back in a total of T1 seconds and let's say at that moment you're a distance of D1 miles away and then in a short length of time you've moved closer to the officer you're still moving toward the officer and now the radar gun sends out a second pulse that goes out and hits the car comes back but this time this will be a little bit shorter so we'll call this T2 seconds and let's say at this moment you're D2 miles away. Now clearly T1 is bigger than T2 because it takes longer for the first pulse to go out and hit the car than the second pulse and also that D1 is bigger than D2 and let's say for the sake of argument that the second pulse is sent out let's say 1 tenth of a second after the force pulse is sent out so I'm gonna assume 0.1 seconds is the delay for this second pulse Okay, so let's see now why is D1 bigger than D2? Well, your car has moved a little bit closer in that time interval in the time interval of 1 tenth of a second so I'm gonna use the formula rate times time equals distance you've used this formula a number of times in other courses, algebra courses and maybe some science courses as well so I'm thinking that this distance could be measured by the rate of the car which is your velocity times the amount of time that's lapsed between this moment and that moment and that's the same as the difference between the two pulses pulse emanations and that would be 1 tenth of a second so that would equal the distance travel I'll call that delta, delta D Okay, so how can we put all of this information together? Oh, there's one more thing I wanna mention let's say that C is the speed of the light or the speed of the radar Now, if this is actually the speed of light then this would be 186,000 miles per second and that is, I would assume to be the same as the speed of the radar beam in the gun at the same time Okay, so what can we say about this? Well, I know that D1 is equal to D2 plus the distance traveled by the car and the distance traveled by the car was 0.1 times V V being your velocity in this case so if I solve for V I'll need to subtract D2 from both sides so let me turn this all around and say 1 tenth V is equal to D1 minus D2 Now, I know that D1 can also be computed by the formula rate times time is distance D1 would be the amount of time it takes for the radar beam to go from the officer to the car times the speed of the radar gun so if I substitute in here the amount of time it takes for the beam to go from the officer to the car would be T1 over 2 Now, why am I taking half of that? Well, you see T1 is the total round trip time so T1 over 2 is the time it takes to travel strictly to the car and I multiply it by the speed of the radar beam and we're assuming that's C down here minus D2, well, let's see, now D2 that would be distance times or that would be rate times time and the time would be D2 over 2 times C once again so if I simplify this a little bit I have 0.1 V is equal to C T1 over 2 minus C T2 over 2 and if I get a common denominator I can put it all over 2 and if I factor out the C then I have T1 minus T2 C times T1 minus T2 all over T Now, the difference between those two time amounts I'll call delta T and so this is C times delta T over 2 Now, delta T is the difference in the readings between the first impulse coming back and the second impulse coming back we call that delta T Okay, and over on this side this is 1 tenth times V so if I multiply both sides by 10 then I have V is equal to 10 times this quantity and let's see the two will cancel out so I'll get 5 C times delta T now you see C is a constant it's the speed of the light and therefore my velocity is a function of only one variable and that's the difference between the two time readings on the radar gun so I assume what happens is when the radar gun first sends out a signal and it comes back it measures the distance or it measures the time the time it takes for the first radar beam to come back and then it measures the time interval for the second radar beam to come back and it takes a difference in those two times and that's what we're referring to here as delta T Okay, you know maybe we can distract the police officer next time you get pulled over with an argument like this but for now I think we better move on to a new topic let's look at variation as a matter of fact here's direct variation there are three types of variation there's direct variation which you see on the screen there's also inverse variation and then there's compound variation so let's discuss these one at a time two quantities X and Y if they're related by the equation Y equals K times X where K is a non-zero constant those two variables are said to vary directly with one another let me just write that on the screen if I have two variables X and Y and if I tell you that Y is just some constant multiple of X this is where K is not zero but it's a constant then I say that Y varies directly with X there's an abbreviation for that you may have seen this in some science books before if I put a little Wiggly sign in here Y is proportional to X then what that means is there's a constant I can multiply X by so that Y is equal to a constant times X and the constant K in this case K is the constant of proportionality the constant of proportionality let's go to the next graphic and I'll show you some examples of things that vary directly first of all suppose a car is driven at a constant speed then the distance D that it travels varies directly with the time T that you drive the car in other words T is equal to K times T D is equal to K times T now in this case K represents the speed of the car or the rate because distance equals rate times time you see if you drive at a constant speed then your speed is constant that's what the K is D equals K times T in another example this is called Hooke's law Hooke's law says that force is equal to K times X this has to do with two weights suspended from springs and I'm going to set up a model that demonstrates this that demonstrates Hooke's law from our physics lab I'm going to hang two springs from these hooks there's one and there's one and at the moment you'll if the camera can zoom in on this you'll see that there's a gradation I think maybe I better turn this one around so we can see those better you'll see that there's a scaling along here that goes from zero then point one point two point three down to two point oh I think it is and there's a little marker on this on this spring if you can see that right there going up and down and that comes up to zero same thing on the other one now I'm going to take two masses and hang them from these springs this first mass is a hundred grams and that pulls the spring down I think you have to look a little higher to see the scale it pulls it down to a mark just above one point zero right here there's a mark right here not quite one point zero the second mass I hang on the second spring has twice the mass it's two hundred grams and it pulls the spring down twice as far this time the marker comes down to just above two point oh that one was just above one point oh this one's just above two point oh now you see what's happened is when I put on twice the mass I get twice the stretch now the mass is equivalent to saying I put on twice the force or the weight over here and so twice the force stretches it twice as far now this relationship is called Hooke's Law let's go back to the green screen and I'll show you how to represent this mathematically as a matter of fact let's go to the next graphic I think we have an illustration of this that would be helpful uh... on the left there we see a spring that's unstretched and then on the right we see a circle that represents a force f suspended from that spring and it stretches it stretches at a distance x now the fact is that the force is proportional or direct or varies directly with the distance x so if I were to double the force I would double the stretch if I triple the stretch I would triple the force so I can make this an equation by saying f equals kx this is where k is a constant and so therefore f becomes a function of x it's equal to a constant multiple of x now the constant in this case in this application is referred to as the spring constant k is called the spring constant and every spring has a different constant to it uh... for example if you look at the springs in the back of your car so that if you go over a bump the springs absorb the shock there uh... those have a very high spring constant because it takes a huge force to stretch it just a little bit on the other hand the springs that we have right here it didn't take that much force or weight to stretch it fairly far so the spring constant on these springs is relatively small okay let's look at another uh... application of direct variation and we have this on a with an interview another interview this with one of our physics professors dr paul mills of the physics department let's let's show that interview well college algebra students uh... we're in the sacred halls of the science building here at uvse and uh... this is dr paul mills of the physics department and he's going to show us some uh... applications of our recent discussion of variation direct and inverse variation so paul i'm gonna turn it over to you uh... newton pointed out that uh... object that starts moving has an inertia as he called it or perseverance in trying to keep going and the mathematical way of describing that perseverance that the object has in wanting to keep moving is in the the perseverance of this thing is equal to the mass times the velocity or we call it the momentum is equal to the mass times the velocity in this kind of a relationship if you increase either one of these you have a lot of perseverance in wanting to keep going and we have a range for a little demonstration of this we're gonna start with a very small mass of ping pong ball and we're going to get it moving really fast and will look at its perseverance this is what i call a mosquito killer uh... it's got a cap in here that's going to ignite and it's going to expect produce hot gases which will then expand really fast and that will drive this ping pong ball at the end mosquito killer and it will take off moving at really high velocity okay this goes right it will not hit you okay you better get a little closer i'm not that good of a shot he's lined up and hope you don't fall into the plant are we ready okay hold your papers up you can see that that ping pong ball just wanted to keep going it didn't want to stop it coasted right through both pieces of paper and it kept going until it hit the wall over on the other side momentum is equal to mass times velocity okay now you know that's the same principle uh... that uh... you might have uh... heard a reference to say after a hurricane say if a hurricane goes through florida sometimes these they find straw embedded straight into a telephone pole you wonder how could you take a piece of straw and stick it into a telephone pole like that but the straws going so fast it has very little mass but it has such high velocity that if it hits the telephone pole straight on rather than shattering the straw it actually sticks into the telephone pole so they they see examples of these things now and then uh... the formula that he wrote was that perseverance he called it is equal to mass times a velocity now if we think of uh... the masses being the constant your member uh... we had uh... why equals k x so if we think of this is being the constant then we can say that the perseverance is directly uh... varies directly with the velocity so as the velocity gets bigger so does the perseverance now in this case when he fired the ping pong ball out of that little gun it had such a high velocity even though it didn't have that much mass it had such a high velocity it had a lot of perseverance and it persevered with the effect that it went right through the two sheets of paper it didn't split them from top to bottom it just split them open in one little little spot so that was his example for perseverance uh... okay now another type of variation is inverse variation let's go to the next graphic we have this for the screen instead of direct variation inverse variation says that if x and y are related by the equation uh... y is equal to x uh... y is equal to k over there should be x not a p y equals x over uh... y equals k over x where k not zero is a constant then y varies inversely with x i probably had p on my mind when i wrote that because i've just seen the perseverance uh... here's an example if total sales s let's say you have a company total sales s varies inversely with the unit price p if that happens then we say uh... s equals k over p now uh... as you might imagine what if you're selling uh... what if you're selling cds and what if you double the price of the cds wouldn't that make the sales probably go down now if it goes down in such a way that when you double the sales price you cut the sales in half if you triple the sales price you cut the sales down to one third in other words as one goes up the other one goes down and then that said to be inverse variation in this case i said that the sales would equal k over sales price now it's not always the case that just because the price doubles the sales prices cut in half but if that were to be the case then we would say that s varies inversely with p with p now we have a physical model that will demonstrate this for us so uh... let's just take a break while i set up this demonstration and we'll come right back okay uh... we had to take a break there so we can set up this apparatus but now i want to demonstrate a property of pendulums and uh... in this case the pendulums are basically two strings with masses hanging from the ends one of them is very short one of them is fairly long uh... actually when i set this up i should have gotten two masses that were exactly alike but uh... the mass appears a little bit smaller but it turns out the mass doesn't affect the period of the pendulum when it oscillates it's the length of the string that affects how long it takes the pendulum to go back and forth so uh... the first string is twenty centimeters long the first pendulum the other one is eighty centimeters long so in other words this one is four times longer than the other one and we're going to see how that affects the period now you notice when i displace this one it has a fairly short period it only takes a short time for it to make one complete oscillation that's what i mean by the period on the other hand this one when i just place it a small amount it has a longer period it takes a longer amount of time to make one complete oscillation now how are they related well it turns out that these two quantities very directly these very directly uh... okay so let's see what happens if i pull these out i'm going to release them at the same time uh... the one up above i believe will make two oscillations in the amount of time this one makes one oscillation so here we go uh... i didn't quite release it straight one two one two whoops i touch that one sorry about that but i think you are i think you will see that this one let me turn this like so and uh... i think this camera in front of us may be able to pick it up uh... okay so one two one two one two one two and uh... that should progress indefinitely so what we have here if we go to our green board is that the if i call t the period of a pendulum then uh... and if i let l be the length of the pendulum that would be the length of the string actually it's the distance from the pivot down to the center of gravity of the weights is not just to the end of the string but to the center of gravity so that would be the length of the pendulum then the period varies directly with the square root of the length so it's a bit more complicated because i have a square root now but it's still a direct variation so what that says is the period is equal to a constant multiple of the square root of the length this would be my constant proportionality and uh... so if i put a if i if i multiply the length of the pendulum by four if i put a four in here four l because see the longest the longer pendulum was four times as long then that should be the period of the new pendulum but when i reduce this that four comes out as a two two k square roots of l now if k square roots of l was the original period of the shorter pendulum then when i when i when i quadruple the length of the pendulum i get two times k square roots of l so that says the period has been doubled because i have two times the previous period so this is a direct variation but it involves a square root okay now the last type of variation involves both direct and inverse variation and uh... the example i'll use is the law of gravitation the law gravitation and this was first proposed by isaac newton by the way isaac newton was born in sixteen forty two that was the year that galileo died just a moment ago uh... and uh... he died in seventeen twenty seven now isaac newton said that if you have two masses for example if you have a small planet and if you have a big planet maybe let's say the sun and they have masses little m and big m and if there's centers of gravity which i'll just take to be the center of each object if the distance between them is called r probably r for radius is what he what he meant then there's a force of gravitation that attracts the small planet to the big planet and it also attracts the big planet to the small planet so there's a certain force uh... there's a certain uh... certain force of gravity that attracts these two uh... if these weren't planets this could be you and the earth there's a certain gravitational force attracting you to the earth and there's a gravitational force attracting the earth to you and uh... this would be your mass this would be the earth isn't the earth's mass and the r would be the distance from your center of gravity to the center of the Earth, so that'd be a little over, or right around 4,000 miles. Now here's what Isaac Newton concluded. This force is directly proportional to the mass of the first object, the mass of the first planet, the mass of you, whatever the two things we're considering. But you know also this force is directly proportional to the mass of the second object. It's inversely proportional to the square of the distance between the two. You know in the pendulum problem we had a direct variation with the square root of the length. This is an inverse variation with the square of the radius. Now if I put these three things together I can make one compound variation out of it. And I'll say little m big m over r squared. The force is directly proportional to, or it's proportional to, or varies with little m big m over r squared. Now the way I make that an equality, so now let's go back, the way I make that an equality is I put in a constant of variation. So f equals, I'm going to multiply by this, the sort of the name that's normally given to the constant is g little m big m over r squared. And this is called the gravitational constant. So let me just write that on the side here. G, I'm not using k, but g is the gravitational constant. And it's the constant that makes the variation into an equality. It's the gravitational, gravitational, let's say, let me get that right here, constant. And the way you normally see this law written is to have g in the numerator, so it's big g little m big m over r squared. Now I call this a compound variation because we have several things going on. f is varying directly with little m, f is varying directly with big m, and f is varying inversely with r squared. So what that means is when I say f varies with little m, if you hold big m constant, if you hold the radius constant so that everything is constant except little m, and if you double little m, you'll double the mass, you'll double the force. Same thing with big m. If you leave everything else constant, little m are, if you triple big m, you'll triple the force of gravitation. On the other hand, if you double the radius so that you move the objects further apart, the force will not get greater, the force gets smaller, and if you double the radius, the force will be one fourth of what it used to be. So if you want to lose weight, what you do is just get further away from the earth and you'll weight less. It doesn't mean your mass will change, but you just need to go further away from the earth, that's the problem. Go up to the top of the mountain, top of a mountain, and you may weight slightly, slightly less. Okay, that's called Newton's law of gravitation. Let's see, let me work one example of inverse variation because we really didn't get to, we didn't, we saw an example of direct variation, but I didn't work an example of inverse variation. Suppose I tell you that x and y vary inversely, very inversely. Now what that means is y is equal to k over x, or if you want you can say k times 1 over x. Now what if I told you a particular value that say when x is equal to 4, that y is equal to 10. So when x is 4, y is 10, the question I might ask is, what is k? Well I think if we substitute n 4 for x and 10 for y, we can solve for k. So if I put a 10 here, k over 4, then solving for k, I get k equals 40. So if I have one reading of an x and a y, I can figure out the constant of proportionality. So what this tells me is that the relationship is y is equal to 40 over x. Let's take a different one. Suppose we have a joint, rather a compound variation, where I have products and quotients. Suppose I have three variables, a, b, and c. And what if I tell you that a varies directly with b, and a varies inversely with c? Now if I put those two things together into one statement, I would say that a varies with b over c. And this is what I refer to as a compound variation. Compound variation. I can make that an equality if I just multiply by a constant of proportionality, which we could write as kb over c. Now if I have one reading of a corresponding a, b, and c, I can calculate the k. If we say for example that a is equal to 2, at the same time b is equal to 5, and c is equal to 20, then the question is, what would be the corresponding, the corresponding value of k? Well now that I have my relationship, I can plug in the numbers 2 equals k times 5 over 20. And that says that 40 is equal to 5k. Let's see, I'm running out of room. What is k going to equal? k is going to equal 8. k is going to equal 8, exactly. Now if I go back up here to this relationship, a is equal to kb over c, I can now say a is equal to 8b over c. Some of your homework problems are going to be such that they will describe the proportion by telling you how a is related to two other variables, let's say, and they'll ask you to find k. And to do that, you'll have to have one set of readings that corresponds to each other, so you can substitute those in. Well, if we summarize what we've done today, we started off by looking at functions as models for applications. We saw two examples of graphs that are used as models, and those graphs represented functions, although we didn't know the exact rule. There was the traveling salesman problem, and there was the, there was the graphs on the lawn growing example. And you see from those illustrations, those graphs told a story, from the story we could tell what was happening. Then we looked at some very specific functions that modeled the perimeter of a garden, the area of a garden, at the radar gun model. And then we looked at, at the very end of the episode today, we've looked at some variations, at various variations. We'll see you next time for episode six.