 We are rolling I should have looked at that lecture 12. I think okay. Good guess that that one right first guess today I Came across a problem that I thought would be worth our while to look at on arc length We saw a couple of situations that happened yesterday arc length length of a curve But this is an equation that's not solved for x Nor is it solved for y and we kind of have to make that choice whether we solve it for x and do dx over dy squared Under the radical or do we solve it for y and go dy over dx the quantity squared? so Let's take a look at this a little bit different Situation than we looked at yesterday So we've got this equation of this curve We want to go from one comma two-thirds to eight comma eight-thirds So we don't have parametric equations so we can throw out that third type so we know we're going to use either one plus dy over dx quantity squared integrated with respect to x or or One plus dx over dy Squared integrated with respect to what so the question is do we solve this equation for x and then we differentiate x with respect to y and throw it in this one and Put in y values What would that be solve for x and I'll stop when we get to the point where I think we have to Kind of make a decision which way would be easier or better Okay, so if we just did the number stuff separately we'd have 27 over 8 right We okay to keep just positives right principal square roots because of where we are in the plane between these two points and We'd have what y to the three-halves Because we're going to take square root of both sides Right, so that's if we solve for x if we solve for y What would we get? numerically, we'd have what cube root of 8 over 27 and never let the numerical stuff Persuade you or dissuade you in one way or the other because that's just a matter of pushing buttons anyway and then we would have x to the Two-thirds all right, so let's take a little hesitation here and see if we can make a decision Which of these two methods we want to use x equals this in terms of y Okay, that would be my choice now. Let's see if we agree on the reason why Okay Let's Amplify that just a tad What are we going to have to do with this? What's the next thing we do with this or with this? Take the derivative we've got I don't worry about the numbers Just look at the y to the three-halves and x to the two-thirds When you take the derivative of something to the three-halves. What do you get? To the one-half and then what are you going to do with that? You're going to square it wouldn't that be nice As opposed to when you take the derivative of something to the two-thirds. What do you get? Something to the negative one-third Squared you're going to get it to the negative two-thirds probably not as nice This one is probably not going to be as Convenient once we take the first derivative and square it as this one is So if you've got something to the three-halves That's going to be a good one to choose take the derivative of something to the three-halves. It's to the one-half Square it. It's to the first. That's always going to be a lot easier to work with and Something to some other power other than one so the derivative we bring the Coefficient along for the ride and we add to it what? three-halves Why to the one-half? So what are we going to do with that? We're going to take that Put it into here. We're going to square it When you square a square root actually, let me just go ahead and write that down. We've got this whole thing We're going to go from two-thirds to eight-thirds Now let's go ahead and do what it says to do which we think is going to be easier than the other Squaring of something to the negative one-third So one plus this square root squared is just 27 over 8 This three-halves squared is nine over four and y to the one-half squared Which is kind of why this method is going to work a little bit better is y so one plus Any reductions anywhere? Don't have any right? What's 27 times 9? 243 is that sound right? Is that right? 243 over 32 Again the fact that that's an ugly coefficient we can deal with that What's the method going to be whether you do the method in your head or you actually write it down What's the substitution? Okay, let you equal what's under the radical? Which means that du since this is to the first du is just going to be what 243 over 32 dy So do we need to make any corrections or compensations in the existing integrand? Okay, we need a 243 over 32 to get that We multiply by its reciprocal to make that legal So we multiply by one the purpose is that that becomes what do you? And this term right here is Square root of u U to the one half So whether you write it down or do that part in your head We've made that substitution now. We're integrating u to the one-half du What's the integral of u to the one-half du you've got this? Outfront we're integrating ignore the limits right now because those go for why not you Integral of u to the one-half du is u to the three halves over three halves and what is U equal to a lot of ugly fractions, but at least the integration part the substitution part the actual integration part is Not bad at all and when it comes to coefficients and all that stuff We're eventually going to form that out to our calculator anyway, so I think we have a doable problem, right? so the reason I thought it was worth looking at in class is that you can kind of anticipate if You've got an equation that's not solve for x. It's not solve for y. You can solve for either one Anticipate what you're going to do with it after you take the derivative, which is square it Kind of try to make it easier on yourself once you square that first derivative that ends up under the radical Question about how to take this to a solution from this point Okay, I thought they'd be worth a couple minutes You might think differently at this point, but I still think it's probably worth it to look at things that can happen differently in different examples all right last section that will cover prior to the test which is on Friday has to do with Average value of a function value of a function isn't that a y value? right The value of the function is y value. So we want the average Y value and if you think what a y value is on these pictures a Y value is From the x-axis up to the curve So it's kind of like a height of that region. So another way to think of this is an average an average height So we've got this region got this function and we're going from a to b How many different y values are there? Average value of a function average y value. Well, there's a y value right here on the other side of a There's another one right on the other side of that actually I'm kind of leaving a little bit of a space But there's one a whole lot closer than that There's another y value right here another y value right here How many different y values do we have in going from a to b? We have an infinite number of y values, but aren't every one of these line segments Then I'm drawing aren't they all y values So if we have an infinite number of these little line segments and going from a to b we want to know What's the average y value? What's the average height if you want to think of these as heights? This is a height This is a height. These are all heights Or y values or values of a function. So in going from here All these different heights or y values to here. What's the average of those? So since there's an infinite number of these We can think back to what we've done as far as representing one of the skinny little rectangles or one of the trapezoids or one of the parabolic regions one of the solid discs one of the washers Whatever it is, we're representing one of the little hypotenuses and then we add them all up We're in a sense going to try to undo that process So we've already added them up. We want to work our way back to what constitutes the average Y value or the average height. So how how do we add all these up? We've done that in this process by finding the area under the curve from a to b That gives us Back to this picture We want a bunch of skinny little rectangles. How skinny do we want the rectangles to be? Practically a line segment in thickness. How thick is a line segment? Practically zero right a zero thickness So didn't we add up all these kind of skinny little rectangles with little increments of x along the way from a to b by finding the area under the curve So if that gives us the area under the curve, let's divide the area by the width in fact the entire width What is the width in going from a to b? It's b minus a so we'll take this which is the area and divide it by B minus a it'll give us exactly the number that we want. It'll give us the average height or the average Y value because these y values are really heights of the skinny little rectangles That's how thin they are. That's how skinny they are So if it is possible for us to find the area or even easier than that if we're given the area All we have to do is divide it by the width Does that seem logical to you that if we take area and divide it by width? What should we get? We should get height right and we should in a sense get an average height So back to our picture the other one's kind of cluttered So I don't know what the average height is but let's say that Let's say it's this distance right here There's our average height so if we take the average height So this I'll just call it f of c because that's what it's going to become in a couple minutes anyway If we take this average height times the width the width is b minus a don't we get The same area in this rectangle as is bounded under the curve Does that look okay on the diagram? The area of this rectangle would be the same as the area under the curve Now if you divide both sides by b minus a Aren't we back up here to this? This point by the way would be c so that f of c would be the average Height or the average y value So if the area of the rectangle which is right here Equals the area under the curve then we're in business And that average height on this picture is the f of some c value somewhere between a big obviously It's got to be a continuous curve for us to even talk about this But that's a point that's well taken by the authors in this section So if you divide both sides by b minus a you're right back up here. So from The tax I don't think you're going to see anything a whole lot different here But just to make it official. It's not just me writing some letters down This is The definition of the average value of a function on the interval from a to b will take the area Under the function and we'll divide it by the width area divided by width ought to give you height And a height is nothing more than a y value. That's pretty easy Because we already know how to integrate and evaluate all we have to do is divide it by b minus a and we're in business So let's see what a problem like that Looks like So let's take this function and we want the average height on this interval from x equals zero to x equals four We could draw a picture. You don't need a picture to do the problem Sometimes it's helpful especially in one of the first problems that you do What's this picture look like What is this function? Brabola tell me something about it opens down from what point? eight right so we come up here to eight and Which is where we want to start anyway when x is zero y is eight and we want to stop it when x is four What's the y value when x is four? Would be zero right so we're going Those are all the same. I think we got those answers in there. So we're going from zero eight right over to four Zero is that our consensus? Okay, so we're here at four zero. So we've got this Parabola that opens down or really kind of half of that because of our interval from zero to four a Lot of different y values in there y values from eight And then they reduce to the right here right in front of four They're practically zero and at four the y value is zero. So we want the average y value So let's find the area under the curve from zero to four and Do what with the area? divide by Four right so we want one over b minus a one over four minus zero So we're dividing it by the width it ought to kick out an average height, which is an average y value one fourth Go ahead and integrate the integral of eight with respect to x The integral of minus one half x squared still negative x to the Third over six I heard that a couple times Evaluate that from zero to four at zero. We're going to get zero right You put zero in in both places you're going to get zero so it just matters what we get when we put in four so 32 minus four cubed 64 over six So if we change well, we could I guess reduce that to what 32 thirds So 32 minus 32 thirds 64 thirds is that right and we'll take a fourth of that because we want to divide by the width 16 thirds That makes sense Our lowest our highest y value is eight Our lowest y value is just shy of zero So is our average y value? Five and one third does that seem to mesh with the diagram? We have if this were diagrammed accurately don't we have a lot more Larger y values before this thing starts to taper off So even though we're going from eight to zero The average one is five and a third, so I don't know where five and a third is let's say it's right there So the area of that rectangle what rectangle the rectangle that is five and one third units tall and How wide would this rectangle be four? Is the area of that rectangle the same as the area under the curve? What was the area under the curve 64 thirds, right? We take their product do we get the area under the curve we do Now the let the next topic in this section is now that we know how to find the average value of the function Which is not that difficult find the area divided by the width Can we find? The C value somewhere between a and B Where the height is in this case five and one third Okay, and we find the number between zero and four where this occurs that actually is Called something different it's called the mean value theorem for integrals the prefix mean means kind of Middle we're finding that value in the middle not necessarily the arithmetic mean there are lots of means Arithmetic mean is the average. What's a geometric mean? You might remember what a geometric mean that'd be a good bonus question Just you know how to start doing that more often just put stuff from your Mathematical paths that are you know middle-of-the-road kind of problems. What's the geometric mean of four and Nine it's a number between four and nine. It's not their average. That'd be the arithmetic mean anybody geometric mean That's uncaring of me to ask that Six years ago, I might have been able to answer Anybody geometric means what if I told you it's six So you knew that how did you know it was six I guess you guess Well one way is to set up a proportion Four is to x is x is tonight And then you end up with x squared equals 36 x equals plus actually plus or minus six But the only one that makes sense in this problem is six because we want a number between Four and nine so six is the geometric mean of four and nine another way. It's the square root of their product, right? It still is a number between them, but it's not directly in the middle So don't think every time you hear the word mean it's going to be you know right in the middle. It's somewhere between them So we have a mean sorry. I've digressed there with that but You know every don't think mean means right in the middle every time it means somewhere between a and b So the mean value theorem for integrals First of all the function has to be continuous Talked about that briefly on the interval. We can't find the area under the curve if it's not continuous So we've already done this part. We found the area. We've divided by b minus a we know That's the average value of the function now. We're going to call it the f of c and we're going to search for c So if that's the f of c, here's the picture that goes along with it. There's the f of c that distance Well f of c is the average height or the average Y value So if we take f of c The average height now we're dealing with the rectangle the f of c and we multiply it by b minus a We wanted the area of the rectangle to be exactly the same as the area under the curve It's kind of hard to separate these two ideas But we just go one step further with the mean value theorem for integrals. We go ahead and find c Such that the f of c is this average y value or average height So there's average height There's the width and with that the area of the rectangle should be the same as the area under the curve interesting little caption this is taken from your book Page 468. I don't always read the captions, but It says on page 468 underneath this picture You can always chop off the top of the Parenthesis two-dimensional mountain at a certain height and use it to fill in the valleys So that the mountain becomes completely flat kind of an interesting little picture, right? We're going to chop this off Got a little deficiency going on here. We'll just take the chopped off part pour it in here, and we've got a flat surface Actually did that I was a part of that one time. That's why I thought it was kind of funny I went on a mission trip We went to eastern Kentucky a little town called Saliersville anybody ever been to Saliersville? No We were part of a crew that was helping to build a church huge church so there were about a hundred and forty of us there for a week and The church had purchased land down Across the street from the existing church and it went down toward a creek and of course this happened after they purchased the land They purchased the land then it was determined that they couldn't build on the land because it was They just couldn't do it. They had to fill it in so they were going to have to buy here. We are here's the church property Okay So they purchased. I don't know how many acres of land couldn't even build their new church on it So across the street, this is very mountainous part of Kentucky Across the street they were widening the road on Where the existing church was well across the street from the church is this humongous mountain So guess what happened in order to widen the road? What did they have to do with the mountain? they kind of had to use some dynamite and Get pieces large chunks of the mountain out of the way to what widen the road guess where the mountain went So they took the mountain that they cleared the road This was actually pretty cool that all this happened so they widen the road the state says what are we going to do with all this? Stuff. Oh, why don't we take it right across the street and we'll fill in that land Forget what it was a hundred and thirty three thousand cubic yards. That's a lot of Stuff so they go in they get all this free Fill in the lot and they build this humongous church on this site in Salyersville, Kentucky It's actually the largest building in the county But they were kind of stuck for a while they didn't they couldn't build so we get there about a hundred and forty of us and Basically in a week's time constructed the their church on this land. So can you really move mountains? Yes, you can move mountains. In fact, they had church shirts printed with all this, you know With enough faith you can move mountains because it actually happened Sorry, but I read that caption yesterday, and I said, you know what that that actually happened But anyway, so you can move mountains So that's the next piece of the problem is we're going to find f of C. We're going to find C Such that the f of C is the average height Salyersville, Kentucky We haven't had any trig for a while So let's put a little trig in here Used to have a daily dose of trig in 141 But it's simple trig sign of X So we're going to find the C value such that the f of C Generates this average y-value or average height on 0 to pi Wouldn't be very interesting from 0 to 2 pi Because the average height would be what? 0 because you've got as much positive as you do negative. So this is probably a little more interesting from 0 to pi So we've got this sine curve this much of it anyway from 0 to pi We've got a whole bunch of different y-values Starting with just barely larger than 0 up to the largest y-value, which is 1 and then they start decreasing again So we need to find the area under the sine curve from 0 to pi, excuse me What do we want to do with that to find the average height? divide it by 1 over or divide it by b minus a or multiply by 1 over b minus a which is 1 over pi So we're going to get the average value and we know that that is going to be the f of the C value and in fact We might have two C values right in this problem Two values between 0 and pi such that when we put those values into the function They will generate the average height because of the symmetry that we have on either side of pi over 2 All right integrate sine of x. What do we get? What's the anti derivative of sine of x? negative cosine So we want to evaluate negative cosine of x at pi and then subtract from that what we get at 0 so negative cosine pi Now that I know to evaluate at the upper limit of integration, which I learned yesterday in here Some of you were not very very nice about that and correcting me which I wrote that down after class negative cosine of pi I Wrote down your names. I know who you are. I do I figure your grades, too Negative cosine. What is cosine of pi? negative one right and then we wanted to get that one and And cosine of zero is one so we're going to subtract The negative of one so it looks like we get two over pi Somebody that has a calculator out in working. What is two? Divided by pi. That's the exact average y value, but let's get a decimal approximation six three Seven seven so we got a bunch of y values as we go from zero to pi But it looks like the average y value is a little more than six tenths point six three seven So we want the f of C. I'm going to go ahead and go back to the exact value We want the f of C To be two over pi. What is the f that we're dealing with on this particular problem? Sign of x so we want the sign of x and we're searching From zero to pi So we want to use our calculator as much as possible Not a common value So this becomes a sign problem or an inverse sign problem, and it's an inverse sign problem. So x is The inverse sign of two over pi now. It's not going to give you both answers But we can figure out what the other answer is from the answer that the calculator is going to give us By the way when your calculator is doing an inverse sign problem, it's choosing an answer Anybody remember the restricted domain for inverse sign? That's why it knows to give you one answer when sometimes there are multiple answers It's going to choose an answer from negative pi over two to positive pi over two Depending on what value you enter in so we're entering in two over pi and we're punching the inverse sign key So this is going to give us the angle From negative pi over two to pi over two that has this value for its sign That's too much information So what is the value that your calculator wants to give you? Such that when we take the sign of that value you might want to have your calculator working in radians to if in doubt on a calculus class Degrees or radians always choose radians kind of a unitless description Whereas degrees carries that little Small elevated circle up there radians. You don't need anything. It's just pi or pi over six or whatever You don't have to attach a unit to it What angle has pi over two for its sign? 0.690 and that's radians right I'll put that in Parentheses you don't need to say that you don't need to write it if you say 0.69 and it doesn't have a unit it's radians So back to our picture Pi over two or pi divided by two is about what 1.57 So this solution says that we need to go to about 0.7 radians 0.69 radians And we wanted the average height to be What should that average height be for this problem? 0.6372 over pi right so that rectangle That is 0.69 approximately units tall and Pi units wide has the same exact area as Is bounded under this sine curve? Is that true? So here's one of our C values. I'll call it C1 We could say C the C that we're searching for such that the f of C is The area divided by the width Where's the other one? Exactly right so this value Right here is pi minus 0.69 right because of the Symmetry of this sine graph and because the fact that the sine is positive in the first and second quadrant and don't these both have a reference angle of 0.69 right if you drew Obviously 0.69 Has a reference angle reference angle is between the terminal side and the kind of the nearest ray of the x-axis so here the reference angle is 0.69 if you came over here to This angle which is pi minus 0.69 so you go over pi and come back 0.69 isn't the reference angle Also 0.69 radians, so we want those two Angles 0.69 and pi minus 0.69 because the sine si in sign of Both of these angles is The same first and second quadrant and they both have a reference angle of 0.69 radians So find C actually there's two possible C values C1 is 0.69 radians C2 pi minus 0.69 either one will suffice or it's probably Good to go ahead and give both if there are two like there are in this case Questions on that problem back to the original picture Guess we can use this one. What is that distance? two over pi and two over pi times pi Pi being the width should give us the area under the curve, and I don't know that we ever found that by itself But because we'd already divided it by pi, but the area under this curve is two square units, right? Area under a sine curve One branch of the sine curve is exactly two square units Believe it or not, we usually go right to the buzzer in this class But believe it or not we're done with everything that we need to do today. How about that better write that one down? That doesn't happen very often. So I will see you tomorrow