 Five years keep that keep the string going Welcome back. We just had a test in this class and By the way, I haven't started to grade those but when as soon as I get this stuff turned in for next fall Then I'll start to grade those today. What I usually do is I'll return them as I grade them So I don't wait till I get them all graded So if I have half the class graded tomorrow then I don't see any reason to hold those back till the other half are Graded, I'll just hand those out with a solution sheet once everybody gets their test back. We'll talk about it I will tell you a couple of you at the end of class Friday I mentioned this to every problem that was on that test was in your textbook So I don't know if you've seen it by going through the chapter reviews or a you know a problem actually from web assign But every problem on that test and I'll point it out to you if you have a question about that when you get it back Was a problem from your textbook a lot of times I look in the chapter review Sometimes I don't find problems that I think are kind of Middle-of-the-road test type questions, but in this case I found them and some of you asked me questions during the test First of all, I would never give you a partial fractions decomposition problem that had like negative 11 17th's For the a or the b or the c value. I just I mean I'm not the nicest person on the planet, but I'm not mean No, it wasn't now see dog on it. You missed that right? But I mean they were integers fact what I do is I'll go from the decomposed part back to the single fractions so that That way I'm guaranteed that they're integers So probably says that something's gone haywire when when that happens Another problem that kind of I really thought this was a gift and many of you rejected my gift Which happens to me all the time The first time I saw the problem in this I know this was a problem from the book It had the e to the x here So the way I started to do the problem I Was thinking so I kind of started this out Let you equal e to the x and in my mind. I'm thinking that's you Because it kind of looks like a you right if you is e to the x Knowing that the real reason I chose you was that this was you squared right? e to the 2x is the same thing as e to the x squared So there's my u squared, which is kind of what prompted me to make this choice when I looked at this problem to see if I thought It was a fair test question And then I'm saying okay. Well, there's a you well Do you if you is e to the x do you is also e to the x? So then that e to the x is needed. So actually when I wrote it on your test Not that this is any big major hint, but I actually wrote it with the e to the x after So that that you would not be tempted or be less tempted to call that you and Then when you discovered this out here at the end you'd say well, okay, there it is Do you is all at the end of the problem? so this is One minus u squared In fact that was on the table of integrals. That was the first one, right? I don't know what number it was 30 maybe 30 Okay, was this was the very first formula because it was a squared minus u squared So I tried to kind of take that from the form It was in the book and send it out there to the right and maybe maybe you think it look a little bit more like D u as opposed to just a u that's out here floating around by itself But I saw a couple of you asked me some questions during the test that You had it going Way different than that and it was way more complicated than that So I really don't necessarily try to do that to you on a test The syllabus let's take a look at that and where we are in kind of some options that we have So section 6.5 is applications dealing with physics Problems and engineering problems It probably to do justice to this section in the textbook. It's probably going to take us all week So we have some options 6.6 or 6 7 Notice what they say in parentheses or More 6.5 if we have time probably not 6.6 Maybe some 6.7 although. I've never gotten to that before in this course 6.7 deals with something that I'm sure all of you are very familiar with is the normal distribution the bell shape curve and When you deal with how many how much of the area is within one standard deviation of the mean and how much is it Within two standard deviations of the mean I mean that's the tip kind of basic statistics glass stuff Isn't it? Are you taking statistics now and then how much is within three standard deviations anybody? if this is a bell shaped curve and we're going one standard deviation here and here how much of the Kind of the population of this normal distribution does that capture somebody recall that 68 68% Okay, and we these are some things we could validate with calculus if you go to standard deviations from the mean it captures How much of the area under the skirt? 95 and if you go three standard deviations from the mean 90 99 okay right around 99% so these are some things that you know in a statistic class you've been told this but in if we get to this in a math class we can actually take the equation and Find out the area under the curve and you know how sure can we be? Based on what one standard deviation is for that particular distribution That we could be within so many standard deviations of the mean so that's actually 6.7 I Don't know if we'll get there or not. We'll kind of see how the week goes there are a lot of Applications in six point five and let's start six point five and you might just kind of as we get to new Types of applications. It's almost like a completely new section in the book So you might want to kind of what you normally do when you go from six point three to six point four Kind of make a mark in your notes that this is a new type of problem We have several new types of problems even though they all fall under this big umbrella of work so some things that We can either see for the first time or hopefully it's review for most of you What force is Force which we're going to need because we're going to take an examination of little I know it sounds Stupid in a way little pieces of work and then we'll use integral calculus I mean you've seen somebody that's a piece of work, right? That's not what I'm talking about. I'm talking about actual descriptions of work and We'll analyze that little incremental piece of work and then use integral calculus to tell us how much work is done So in doing so we're going to need force and distance So let's talk about what force could be in some of the units So force can be mass times acceleration. We've seen that and we've used that Now the different units that we could have The ones that are the most stubborn and I'm by no means I'm not Fluent in these units, but I've come across them enough where I recognize them I don't actually have a wonderful handle on what what a Newton is and what a Newton meter is other than it's a jewel Okay, I don't really have a great handle on that Like I do with a foot pound. I mean a foot pound is you know the force it Takes to move something that's one pound one foot that just have a better handle on that but When we when we take force And if it's mass times acceleration well, we know what acceleration is it's the second derivative of The distance height or position function the s of t Function with respect to t both times So mass could be in kilograms And if our mass is in kilograms then we multiply that by an acceleration constant That then is in another unit, which is meters per second squared and we end up with this strange unit kilogram meters per second squared which is a Newton so somebody that has a really good handle on what a Newton really is Enlightened me this morning on this handle that you have on what a Newton is No, no enlightenment this morning. Hopefully you have a better handle on what a Newton is than I do then if we Convert work, which is what we really want into force Times distance and this this is what we look at incremental pieces of I Agree, it's it's really hot Chandler if we can just prop the door open see if we can get something that we're not all going to be just Rolling and sweat here at the end of class I mean it is work right something we should have a little bit of sweat going because we're talking about work But we don't you know might get a little stinky in here So if work is force times distance, we'll see how much force is required to Stretch or compress the spring or move this particular slice of water to this height And then we'll figure out Incrementally what the distance is that it's going to move little delta x's at a time or little delta y's at a time Then we'll convert it to an integral calculus problem, but basically we're going to need force Well force might be mass times acceleration, especially if we're given something in kilograms We're going to have to convert it to that and then we want to analyze how far that Object is being moved incrementally Little dx's or little dy's at a time. So this could be feet Sorry, this could be pounds And this could be feet and if it's that way it's pretty darn simple then our work required would be foot pounds Okay, if our force happens to be like we have up here Kilogram meters per second squared and let's say this is meters Then we've got an even stranger sounding unit kilogram Meters per second squared times meters or this is we decided was newtons, right? So this is newtons times meters. So a newton meter is a Jewel, okay, and if you have a really good handle on what a jewel is then Please enlighten me. There is a nice conversion. What is it a jewel is one point? Three six foot pounds. I think is the conversion that helps me but as far as So if we have newton meters, then we have jewels So we do have to kind of pay attention to the nature of what we're given If it's a force required to move something or stretch or compress something that is in terms of pounds We're good. Just let it go We want feet probably you can have inch pounds. I guess too, but normally it's foot pounds and pay attention if we've got Newtons and meters then we'll convert it to jewels. All right. So how simple can work be without integral calculus? first example and We don't need calculus to do it. So we have a book That is 1.2 kilograms. It's on the floor We want to The amount of work required to bring that book to a height of point seven meters So work being force times distance Nothing real Mysterious or complicated about this problem. We've got a Let's get agreement here to move a book That Is 1.2 kilograms? I guess it's a little tougher to think about that you think of it Does it how much force does it require to move that book? Well? We've got to convert that to a force first of all So we know that it requires that much force to move that book or lift that book So if it's kilograms, which it is we've got to convert that to Something here that's going to occupy this capital F position the force position So that it actually is converted to a force rather than just a mass So if you've got mass we need to multiply it by acceleration and in terms of meters per second squared What is acceleration? Gravitational constant associated with acceleration 9.8 So that would be a handy thing to hang on to if you haven't come across that in a physics class Acceleration due to gravity 9.8 meters per second per second What would that look like if we needed to use that acceleration constant which in pounds we don't But if we needed it somewhere down the road, what's it look like in feet? 32 feet per second per second right So now we've converted that to a force So the force required to move that book is this and then how far are we going to move it? to a height of 0.7 meters So that's the distance. We're going to move it. We could set up an integral calculus problem to do this, but it's It's just way too simple to Rely on calculus to do it. So the unit here we're going to have kilogram meters per second squared Newton's times meters, which would be Newton Newton meters, which is Joules and It's the product of this number times this number times this number right Force times distance So that should be my arithmetic is right here 8.2. Somebody correct me if that's not What you get when you multiply those three units? 8.2 Joules is the work. All right, let's take another simple non-calculus example Let's say we have something that weighs 20 pounds and we want to move it six feet So the force required to move 20 pounds is 20 pounds of force And if we want to use move it six feet, there's our distance So the work done in this situation required to move something that weighs 20 pounds Six feet would be 120 foot-pounds. No calculus now let's take this one and I don't know how much this is worth, but it's probably worth a minute or two of class time. I suppose we wanted to make a Real simple integral calculus problem out of this one So we're going to move 20 pounds. We're not going to be able to like if it's let's say it's a Dumbbell and we want to move 20 pounds where you're not going to be able to like Partition it up and move it one or two or five or seven pounds at a time. It's 20 pounds We've got to move the whole thing, but we can look at the distance. We're going to move it Incrementally if you're going to move it six feet total, can you move it one inch and then another inch and then another inch? When you get done won't won't you have done the same work as if you just picked it up and placed it up here six feet So we can't incrementally look at the force required, but we can incrementally look at the distance. You're going to move it so There we go, okay 20 is the force Let's say we're going to move it and let's look at this vertical movement We're going to lift it up a total of six feet So we want to lift it up little Increments at a time little delta y's at a time and then we want to add together all of those Moving 20 pounds little increments of y each time From wherever it started we'll call that y equals zero to a position six feet above y equals zero So we could take this problem don't do this because it doesn't make any sense to make the problem more difficult But we're converting it so that it could be an integral calculus problem So I know it's not 20 delta y we would call that 20 dy But I guess the purpose of spending a couple minutes is that here's the force That dy is really a delta y Which is really the distance So 20 is in pounds zero to six those are y values in there and feet well if you integrate 20 with respect to y What do you get? 20 y and if you evaluate that from zero to six You get the same answer again no reason to use integral calculus Just take 20 multiply it by six hundred and twenty foot-pounds, but integral calculus does work on this problem So what we're going to do is we're going to try to analyze the force needed And it could be a variety of things stretch or compress a spring to actually physically move something Look at the incremental pieces of distance And then see if we can decide where to start the process and where to end the process I don't think this is from your book These things are mentioned in your book, but this is in essence what we're doing in this first box is This is the force required to Move this or stretch this or compress this So the work done in moving an object from the point with coordinate a to the point with coordinate b is this These little delta x's or delta y's are the incremental breakdowns of the distance This is I don't think this notation is used in this book That Capital P with that looks like double absolute value Brack's Brackets is is actually called the norm of the partition And if the norm of the partition is approaching zero that kind of is the fattest sub-interval the widest delta x Then if the fattest one or widest one is approaching zero then certainly the ones that aren't so fatter Aren't so wide. They're also approaching zero. So this just says that Whatever we've incrementalized if that's a word They're all practically zero Little we're going to move it little delta y's We're in this case little delta x's that are just little bitty distances each time we move it So what does that become? What does that look like that pretty much looks like the definition of? What a definite integral actually is and that's what we're going to be doing. Here's the force. Here's the distance incrementalized Where does it start? Where does it stop? Now we're going to do a spring problem. So let's take a look at Hook's law Before we do the spring problem. I Know this is in your reading, but I don't think it's kind of it's framed out exactly this way There is a description in sentence form of what hooks law is So let's read it and then we'll try to put that into our own terms the force f of x So we are going to deal with force because it's going to be in the integrand So this force f of x required to stretch and you could put or compress Depending on the nature of the spring problem if the spring is able to be compressed It takes just as much work to compress it as it does to stretch it From its natural length so the force f of x required to stretch or compress a spring x units beyond its natural length is given by this so it's a Pretty simple linear relationship So the force is some constant Times x well, what is x x is the number of units it has been stretched or compressed from its natural length Okay, it's called the spring constant x we can call it's I don't know. I've seen it described as its elongation So the force required to stretch or compress a spring is Some constant times its elongation So think about a spring for a second If you've already stretched a spring out, so we're already beyond its natural length Isn't it more difficult to stretch it even further? Right so the force it's going to take more force to stretch it even further Because it's already elongated so many inches or centimeters or meters beyond its natural length So the k is the same the reason it requires more force is because it's already stretched Beyond its natural length So the force required to stretch or compress a spring is Directly proportional to its elongation is how we'll analyze that in terms of a spring problem all right, let's look at one in terms of Newtons Is this example in your book if it's not there's one similar to it This is in your book and then we'll do one that's not in your book So this is actually on page 472 says a force of 40 newtons is Required to hold a spring That has been stretched from its natural length of 10 centimeters centimeters are not good If we put them together with newtons. We don't want newton centimeters. We want newton Meters, so we're going to have to convert that to meters Force of 40 newtons is required to hold a spring that has been stretched from its natural length of 10 centimeters To a length of 15 centimeters. Why do they give us that statement? Why do they tell us how much force is required to hold this spring in its stretched position? Five centimeters beyond its natural length So we can find the spring constant so that's in the first sense, okay So the force what is the force required in this first sentence 40? Newtons so we don't know okay, that's what we're searching for and A force of 40 newtons is required to hold a spring that has been stretched from its natural length of 10 centimeters To a length of 15 centimeters. So what's the elongation? Five centimeters, what are we going to call that? centimeter is a One one hundredth of a meter right with the prefix C E M T I So that'd be point oh five Right meters now the constants in this case has a funny unit But let's just keep track of the fact that when we're done. We're dealing with meters and newtons, so We're probably going to have jewels for an answer So K is equal to what? So this is what we're solving for 40 divided by point oh five 800 Is that right? So we have our K so the work That's what we want to do we want to analyze the work done I guess I should read the rest of the problem So we know what we're actually searching for how much work is done in stretching the spring From 15 centimeters to 18 centimeters So here's what the work is kind of generically Well f of x in this case is K times x and we know what that is There's K There's x So our little increments of distance Kind of we're stretching this spring little delta x's at a time From what starting position to what ending position so it's all related to the natural length of the spring So we're going to stretch it from a length of 15 centimeters. What's our starting position? Related to the natural length point zero five is that right so we're Starting point zero five meters beyond the natural length and we're ending how much beyond the natural length With your point zero eight That work so this is Units beyond the natural length if that if we're starting from natural length that's position zero kind of its equilibrium But we're not there. We're starting point oh five meters beyond the natural length ending at point oh eight Meters beyond the natural length the calculus problems are not going to be difficult. So we'll just kind of once we're there We've got 800. We're integrating x. So that's X squared over two Point oh five to point oh eight. Let me see if I have my arithmetic down here. I think I do If you analyze this at point oh eight Square divided by two multiplied by 800 same thing here and subtract them I Have written down one point five six and this should be Jules. Does that work? Because the force required to stretch or compress a spring is Is Directly proportional to the elongation. So here's our force in this problem on a spring problem That's what the force is going to look like So I'm spring constant times x this linear Relationship so the further you stretch it The more force is required and how much more for how much additional force is required. It's a linear relationship That spring constant times x All right, let's look at one That is not and I thought I had one written down. I don't so I'll just make one up So let's say a spring has a natural length of 13 inches And let's say that we have a Description let's say that That it takes a force of 20 pounds Actually, let's make this a little different. Okay, let's not kind of stretch it like we're doing the stretching So let's put a 20 pound weight Use your imagination. That's a spring. Okay, so there's the spring At its natural length. Okay This is 13 inches Again, use your imagination So we put a 20 pound weight At the end of that spring And it stretches the spring. I don't know four inches Four inches beyond its natural length So the fact that we put a 20 pound weight on here and it stretches the spring four inches beyond its natural length That gives us our equation from Hook's law Force required to stretch or compress the spring is directly proportional to the elongation. So the force is 20 pounds Okay, we're searching for and x is the elongation which was actually given in that format, right? We didn't have to figure it out. It stretched it four inches Do we want inches? I don't know if I want inches because then we're going to have a final answer of Inch pounds I don't know if I want that. So let's convert this. Okay So we've still got 20 pounds And four inches can be what? A third of a foot That way when we're done with the problem, we have a more convenient unit which would be foot pounds So what is k? In this problem 60 60 So regardless of what we do in this problem, we've got our spring constant We know the force is some k times x We know we're going to move this little delta x's at a time Let's say how much work is required to stretch it from its natural length to a length of 21 inches So what would we call natural length? Zero call it zero Zero feet. So that's its equilibrium. So we put this weight on the end. It stretches it. We figured out the spring constant Now we want to be able to stretch it. How much force is it going to take to stretch it? From its natural length to a length of 21 inches. Well, what are we going to call 21 inches? So it's 21 Twelfths right foot and what is that? Is it one and three quarters? Yeah three be seven fourths, right? Let's just call it seven fourths. So from zero feet to a position that is one and three quarters feet now it's All related to its natural length So there's our integral calculus problem. Yes, sir. We're going to stretch it to a length of 21 from the natural 13 Right the difference would only be oh, that's what I want. Okay. Sorry That's what I want I told you that I kind of lose my train of thought today. We just lost it. Let's regain it here Natural length. So that's going to be position Zero because that's 13 inches already. That's its equilibrium To a length of 21 inches How much beyond the natural length is that? That's what we need. Thank you So beyond the natural length that is Eight So what is eight inches in terms of feet? It is two-thirds So thank you for catching that. Hopefully we would have caught that before we left it, but 13 inches is where it already is. That's equilibrium position zero We want to go to 21 inches. How much beyond the natural length is that? Two-thirds of a foot Now we're good, right? So we would integrate 60 x squared over two Evaluate from zero to two-thirds. I just made up the problem. So I don't have the answer but Doesn't take a whole lot to get this answer put in two-thirds square it Right four ninths multiply it by 30. What is that? 13 and a third and the units Foot pounds. Okay. I think we can start this one. I don't know that we can finish it. All right. Here's our diagram See now. I don't know your names yet. No, I'm in trouble I just have to call the four people I know See that's trouble that you're back here again Here is a conical tank I don't think this example is in your book one similar to it is in your book This one is not in your book a right circular conical tank of altitude 20 feet and base radius five feet Has its vertex at ground level That kind of helps it hold water if the vertex at the ground because if it were the other way It probably would struggle holding water And its axis is vertical now probably we don't have a lot of Right circular cone tanks. Okay, but we can take this Procedure and I think it'll help us solve a lot of other problems that maybe aren't conical in nature If the tank is full of water And we'll also look at problems when it's not full when it's full to a certain point But the first example it's probably easiest to look at it when it's full of water Find the work done in pumping the water Out over the top of the tank So we've got this cone Base radius five feet. It's 20 feet tall. It's full of water And we want to analyze the work done in pumping the water out to the top of the tank Now this already has the slice of water pictured Why do we need to analyze different slices of water Different from some other slice of water Well, this particular slice that's here To take this slice to the top of the tank. Don't we have to take it this far, right? Whereas another slice of water Down here Even though there's less water in that particular slice Doesn't it have to travel further to get to the top of the tank? So we've got some variable Weights if you want to call them that because there's the weight of this slice is different from the weight of this slice Not only are the weights different the distances they have to travel are also different So we have to describe those in terms of the variables that we've been given in this problem All right, so we've got feet 20 feet tall in a base radius of five feet The tank is full of water find the work done in pumping the water over the top of the tank So, um, what about the since we've got Feet and feet What about the weight Or density if you want to call it that of water? Anybody remember what that is? I've actually seen two different units 62.4 Pounds per cubic foot and 62.5 Pounds per cubic foot. So let me see what they want us to use here Try to be consistent with this book and I'm not seeing it right away 62.5. Do you see it in this book? Yeah, what's 74 on the side? Left side Small script. Okay. There we go page 474. So we'll use their version 62.5 Pounds per cubic foot Now if we're careful with our units too, we can actually kind of See what the answer is going to be By talking about, you know, here's feet. Here's another so we have foot feet times feet that's square feet And then we have another incremental distance involved linear distance. It's feet again So we can actually keep track of the units So do you feel Confident that if we describe How much water Is in this slice Not just the volume of that water that's going to play a part but the weight of that water And then we figure out how far we have to move this amount of water Wouldn't that same thing work for this also? As long as we describe it Generically and then we'll add them all up here. We are back to that little incremental model again So we're going to add up all of the work Plurals works involved in moving each slice of water to the top of the tank So let's look at the one that was actually pictured for us How much water is there in that little slice? What is that little slice? If you had to describe that What is that? It's a cylinder right short little squatty cylinder. Well, what's the volume of a cylinder? So the slice is really a cylinder volume of a cylinder Pi r squared H we've already used that right earlier in chapter six Okay, I think we can do that Can you describe The radius? Actually, it probably would be better down here Because then we'd be going to the curve They've got it Marked out here either way We've got this thing looking pretty thick or pretty tall It's actually a whole lot thinner than that right if we do the process right So this distance and this distance We're talking about the same thing. This doesn't actually go all the way to the edge of the Cone but in reality there's no difference because it's so thin So what do you call the distance from the y-axis over to a point on the curve which in this case is a line? That's its x value right so the radius is x Can be described by x. Does that work down here also? Is the radius of that slice also described as x? Okay, so every slice that we have the radius could be described as x How about the height or thickness of each slice? A little delta y little incremental Y and you can see that right here. There's the delta y there's the x sub i so depending on where we are it's an x value for that particular slice So eventually in the integrand that's going to become a dy So let's finish because we're about out of time but let's finish with How much does that slice of water or any slice of water actually weigh Well, we know how much water is there pi r squared h that's the volume What do we do to the volume to get converted so that it's actually how much force it's going to take to move that slice Well, what are our units thus far? We said the radius was x The height x is in feet, right? So there's feet squared Delta y is also in feet So we've got feet squared times feet which is feet cubed What do we need to multiply by so that we know how much force it takes to move this slice? something that has Feet cubed we've got to get rid of cubic feet. We've generated feet squared times feet We've generated cubic feet. We got to get rid of that unit Right and we want it to be a force. So we want it to be what pounds How many pounds of force are we going to take is it going to take to move that slice of water? so feet cubed Reduces with feet cubed And we're left with pounds does that sound like a force to you sounds like a force to me So this is part of our integrand. I said we'd stop there So we will what else do we need in the integrand which we're not going to get until tomorrow We need distance right and the distance of each slice is different So we have to describe that in terms of what it takes to move each slice, but we'll pick up from this point tomorrow