 So the midterm scores are up on the web page if you want to look at them. You put in your last name, which may not be working to keep your last name in, but it's what registers for your last name is. So for some people who have two names in their last names, sometimes it comes out weird. You put in your last name, your ID number, and you can see what grade you got and what letter it's available into is. People did worse on the exam than I expected. I thought it was a relatively easy exam, but clearly you don't agree. If you've got below an 80, which means you've got an F, a D, or a C minus, that doesn't mean you're going to fail this class. So one thing that confused people at first is a C minus is not a grade that you want because it means that while this course counts towards graduation credit, it doesn't allow you to take the next course in the sequence. If you need this for your major, it also doesn't count. I mean it counts, but it doesn't fulfill the requirement. So if you're an engineer and you need to take this class, you need to get a C or better. If you were a math major, you need to get a C or better because it's a very, very good calculus, whatever. So whatever program you're in, if you're taking this course because it's required, you need to get a C or better. If you've got a C minus on this exam or even a D, it doesn't mean your life is over, your family, your life is died. What it means is either you have to change what you're doing. So the next exam will actually be harder than this one. Sorry, it just is harder material. So the next exam will be harder. I just said that. If you've got below an 80 on this exam, you can recover from that by not necessarily working harder but working more intelligently. So I know that some people, for example, when they do their homework, they do it all together in a group. And what answer did you get? Oh, I got that answer too. Okay, that's the answer. That is a way to figure out stuff but that isn't necessarily a way to make sure that you know it rather than your friend knows. There's also the possibility of moving down from Math 132 to Math 126 or even Math 131 or even Math 103 if you want. But that needs to be done by a week from Friday. There's a form on the class web page also on the registrar's website which you fill out. If you're a freshman, you get your advisor to sign it and you also get either someone in the math department or the professor in the new course to sign it. You take it to the registrar's office and then have her cadaver and you get in that course since the beginning of the semester. So there's no W on your transcript. You should not withdraw from this class. If you're one of the two people who got a zero on the exam then you should not withdraw from this class. What you should do is get that form, figure out what math class you belong in, take that form to the math department, you get a sign, take it to the registrar and then be in that other class. If you move to Math 126, Math 131 or Math 125, it's the same textbook. So you just look back at some amount of space in your container. I'm not telling you to get out of this class. I'm just telling you that's an option. One thing you should be aware of, if you've always got A's in high school you can get an A on this test. That doesn't mean you should get out of this class. I only get, usually in most of the math classes, only about at most 20% of the students get an A or an A minus. So on this exam, another A is about including A minus if we're about 1,000%. It's just the way it is. A's are hard to get. A's mean something. At least they mean something here. There are other places where the average grade is an A. That's not true. Getting an A is a significant issue. A B? Well, B is another, I mean I gave a lot of B's and I gave a lot of C's. So unfortunately, I gave a lot of C's. A lot of people did like one or two things. Okay, any questions about the exam? You get the exam back at your recitation either today, tomorrow, or Wednesday depending on when you have a recitation. Any questions? No? Okay, so everybody knows how to check. You probably have already done it, right? No? Okay, so if you go to the classroom today you can check this. All right. Other questions? No? No? All right, so we need to move along then. So what we've been doing and what we will continue doing for the next class or two is this idea about sort of the core idea of this section of the class is that an integral, well, not the troubles here. It's the same as adding up. So we saw already that an integral we see that we add up lots of skinny lines within an area. We see that an integral of the cross section and we're adding up, we have some object in the area of lots of slices and I get a volume. Another thing that we did is we saw that the average value is just an area. So we take something like that and we average by the width that gives us the average value and the last thing that we saw last time was that arc length. So the length of the curve, we have a curve and we approximate it by it's like 9.3 degrees. So if I just pass out of here it's 9.4. Then you can leave. Okay, so the length of the curve which we also did last time came from a bunch of lines and these short lines and so what we did last time is this is dx and this is dy and this is the square root of dx squared plus dy squared which is the same thing as 1 plus derivative. So if we divide through by dx back to dx out so we did this last time. Does anyone need to see another example of arc length? You do. So the problem with arc length is almost always the integrals that you get are horrible and you have to do them numerically because there are very few functions for whom 1 plus the derivative squared take the square root is something you can integrate. So I'm just going to write down an integral and I'm not going to do the integral as in the other one. And plan 1 let's just say so just to remind you of this example so what is so as you can see this would be a quicker question but the computer's still working so just pretend it's it, okay? So this is the kind of question you should be able to do. What I'd like you all to do is I don't know how to do this question. So if I call on one of you with random you can just tell me the right answer, right? So I don't need you to do the integral I just want to write an integral. This is something that should be trivial and the formula is right there. So it's like really a stupid question but given what I saw in the exam even if the formula is right there there are lots of people that can't do the problem. Okay so somebody want to tell me what this integral is? Nobody wants to tell me, okay I'll start it here put another part in here we go here I'll put it in the next part okay somebody tell me what's next please now this integral this is 1 plus 1 over x to the 4 x to the 4 plus 1 does not look clear to everyone that goes with the same thing is not and maybe you should drop down to any of these so this is 1 over x to the 4 we get where we get x to the 4 okay so now this is not it's not an obvious substitution you can do you could maybe let x square be a tangent so let's move on so I can ask the computer to do that I don't know maybe you could do this by trace substitution move on to the next topic so we have a bunch of these little ideas that slice a thing up in different ways then we get something we want so one application of this that comes up in for example physics is this idea of work one does in physics calibration and the amount of work involved in exerting a force how much you need to observe that force so for example the work to observe a force over a certain distance the amount of force you need to observe is the distance that you need to exert it so for example I don't know how much it weighs let's call it 5 pounds the work involved in lifting this book let's say it weighs 3 pounds the work involved in lifting this book 1 foot because there's the force of gravity involved so the force of gravity is I want it to be 36 36 32 is 3 times the force of gravity there's this confusion about whether we're talking about forces whether we're talking about weight or mass so this 3 pound book this is the math all the force involved but if we're talking about a force then we just there's the newtons and jewels and all of that junk if you prefer I don't care this is not physics class depends on how the question is phrased so that's fine that just says we exert a force a constant force over a given distance and over the force of 2 feet the force of gravity is not very very much so we can call it constant but if the force is not constant so if the force is not constant then we have to do what so if we don't know if the force moves a little if the force changes so at this point it's harder to do than it is to do how can we solve this so for example so let me give you an example of this stretching a spring we're stretching a spring and Hooke's law says that the force proportional to the distance stretch stretch in other words the force is some constant times x so if I have a spring which I don't have one here okay so if I have a spring so imagine a spring here and I just pick it up and it's this long then I can stretch it out like that and I need to know that the hold it here takes no force the hold it stretched out here takes some amount of effort to hold it stretched at that length and the amount of effort I need is proportional to how far I stretched it beyond the place where it just wants to hang out so I don't know if you can even think in this case so I have a spring here which has some difficult natural length and the force to stretch it here is proportional to how far it's gone so to stretch it from say my spring is should I put it in meters now no as long as I put it in meters I can't stretch it pretty far but it has a natural length of one half meter and now if I want to stretch it and let's just say it has some constant so I'll just give you a spring constant I wouldn't tell you what that means so then how would we figure out how much work because to stretch it from a half meter from 0.5 meters to 0.6 meters is a whole lot easier than to stretch it from a nine meters to a whole lot harder pulling a lot harder when I'm out here near one meter then it is when I'm out here near one meter then that's good when I'm out here from one meter is when I'm in here right? I can't tell if you're following this or you're just dying of heat yeah both of those both is okay is this clear so how would we figure it out what would we do I want to integrate the force I want to integrate the force here dx and in this case the force is times the distance to be stretched past one half in this case okay so it's going to be 2 times x so I can write this in two ways you need to write this and x is going from 0.5 to 2x dx or if you prefer from one half to one to minus half the reason I wrote this in these two different ways is to try and make a formula in the same so here if this is my spring and this is one half I could either let x be or I could use either way whichever way seems to make more sense to you is fine so because I wrote the formula 2 x minus one half so force either 2 times the distance I stretched minus the distance I stretched to 2 minus its natural length which is the same thing where x at the distance is stretched beyond its natural length so these are not hard is it clear to everyone why this isn't integral is it clear to anyone why this isn't integral so force is integral somebody want to tell me why this isn't integral so why if not a constant force should I integrate after taking any way I want to so I mean why possibly would I integrate if it's a constant force would I integrate it would it make sense if the force is constant to use an integral I see a lot of nodes why not it works perfectly well if I use an integral when it's a constant force I want to force of 18 and I want to apply that force over a distance of 2 meters and I integrate from 0 to 2 of 18 dx 18x evaluated from 0 to 2 is 36 it's exactly the same thing so it makes perfect sense to integrate if it's constant it's just stupid it's like it makes perfect sense to drive your car to your mailbox in a 10 meter driveway except it's just stupid by the time you start up the car and roll down to the end of the driveway you've gone up and down 14 times but you can do it and it's exactly the same thing so the idea here the reason we're integrating is over a tiny little interval of time or a tiny little interval of distance the force is approximately constant so this integral here when we integrate force dx this means add up the work tiny interval so this f dx here represents the amount of work I need to do to move my object dx units so this dx is not just the stupid thing that means this I'm right it actually represents the distance the infinitesimal distance that I'm applying this force over this means I add up the amount of work that I do to stretch my spring from here to here and then the amount of work I do to stretch my spring from here to here and then from here to here and then from here to here this alright so we can do other examples of this kind of thing like the amount of work required to say pump some water out of the tank or any of those kinds of things maybe I should do an example like that or should I do another spring example where we can grab the concept it seems to me in the context of figuring out the constant on the exam there's a lot of trouble for a lot of people so let me do a slightly different example in the vein of this before I do say water from the tank or something so say that it takes so say I have a spring 40 newtons of course so maybe 40 newtons but whatever that's enough so then you stretch it from one meter beyond because newtons are naturally measurable so the work will be in joules but newtons over meters so how do I do this this is enough information to do this problem even though I didn't tell you what the constant is so how do I figure out what the constant is what is the constant so let's make a change okay that's pretty hard so here the constant since it takes 40 newtons to hold 2 meters beyond its natural length that means that the constant is so the force is the constant by the distance beyond the natural length and the force here is 40 so the constant is 1 okay that's pretty easy now that I have the constant this problem becomes the same as the last problem force involved in work involved I have to apply it from 1 to 3 to the constant of 20 so I just integrate 1 to 3 of the force so to read the question it takes me 40 newtons of force this is 2 meters beyond its natural length now my question is still I want to stretch it from 1 meter to 3 meters how much work so that's the integral 1 to 3 of 20 which is nx squared from 1 to 3 which is 90 minus 20 which is quite a constant aren't you supposed to integrate force instead of the constant the force is 20 the force is 20x where x is the distance stretched beyond its natural length so I did integrate the force the constant is the force is constant times x if your force is more complicated and you have more complicated but usually the force is not so complicated so I guess let's do one more example of this but again the idea here is that work is just integral of force over so say I have a cylindrical tank filled with water and I want to pump let's make it a little harder let's say that the tank has a radius which is like 20 meters so this cylindrical tank is filled with water so I want to pump this tank and how much work is involved notice that it's more work to pump it from the bottom I'm going to pump it this way so how do I figure out what is this what do I do again the point here is not to memorize formula the point here is to attack problems and figure out so what is it that I need to do to calculate how much oh I guess I should say how much the water weighs so the mass let's say the mass of water the mass of water is 1 kilogram per square meter so water and everything is in meters so we can just forget about everything is in the right units we don't have to think about the units so what do I do sorry someone is telling me what to do what do I do calculate the volume the volume of water so you're saying if I slice it you're saying that volume so the volume let's call it h of a width v h times 100 right if I slice if I take a little slab of water which is v h all this is this limit all this 10 meters across then the volume of that height that little slab of water is v h I mean this 100 pi v h so the mass that was right multiply by the height so why do I multiply by the height but I'm not draining the tank and pumping it out at the top so what's wrong with that so this is almost wrong what's wrong with this approach your h goes to the wrong way so if I'm draining the tank your answer is perfectly right because it's going this way so there's a lot of energy here and very little energy here but I'm not draining the tank I'm picking it up from the bottom and taking it out at the top so this stuff is a lot of work to get out and this stuff is not much work to get out and there's a space so what I have to do is let h one thing I need to do is let instead of h I can call it well d's are bad if I call it d then it's e d so what's another word for depth I'm just calling it h so h is the distance this way and h go that way then I find it's the height I have to lift the watch yeah so if you want h minus h not yeah no 100 so the volume of the slide is i r squared times the h's times the height that's the volume of the little cylindrical slab in this case h is dh it's a skinny little slab r is 10 so r squared is that the volume is 100 pi dh the mass of the slide is easy because it's 1 kilogram per cubic meter so it equals the volume so this is 100 pi times dh kilograms and now this little discussion that we have is now the work involved 30 meters high if I have a little slab of water somehow just working in the sky here the amount of work is the mass times the distance I have to move it so the work is in this case the mass is there 105 times this distance here from here to there so if this this distance is h and it's 30 minus h this distance is not this slab is a high h then this is 30 minus h then it has to go so now the work is a pretty straightforward integral it's the integral the water goes up to a depth of 30 meters and so I want to figure out the work that it is see the water from the bottom all the way up to the 20 meter line so it's 100 pi I'm going to integrate 30 minus h that's the force needed that's that's the anyway that's what I need to get from here to there I just integrate that from 0 to 1 so I mean we can make these problems a little harder we can make the tank be conical or spherical or whatever but the idea is the same the idea of of pressure is the same and all of these ideas are the same idea in fact the whole thing is kind of the same idea you want to figure out a big thing figure out what happens if it has to be an input so it's too hot here to move on so I'm going to stop