 We are at lecture 11 This week, I don't we didn't I didn't mention this before we started, but we'll probably meet all Four days, and then the test is on Friday. So technically all five days this week We're going to do section 6.3 today. Hopefully it will fit in with kind of the line of thought of looking at Individual pieces skinny little rectangles Whatever those pieces happen to happen to represent and then describe one of those slices Solid disk washer today. We're going to look at the length of a curve or How can we use integral calculus to help us find the length of a curve and Then another application another couple of applications in 6.4 Tomorrow probably in the following day review on Thursday And take a test our first test in here on Friday. So Let's go into this application From chapter 6 section 3 the length of a curve or Arc length so we have and let's try to develop this. This is one of the easier developments. I think Let's say we have a curve in the plane and right now. I want the curve to be a Let's start off with it being a one-to-one function. So No repetition of x values or y values That will matter as we go through this but let's say we have a curve and we know the function that describes this set of points what we would really like to do is take this thing that's curved and Stretch it out straight put a ruler to it and see the length of it Okay, that probably would not be all that accurate if we put a ruler to it. We're going to have some error in measurement What we could do to approximate it is we could take the line segment That goes from here to here and say that Kind of the length of the curve, but we know the length of the curve is actually different from that So that one line segment is not going to do the job very well Just like one Rectangle under the curve is not going to do the job very well or one trapezoid or whatever the case may be So we take this curve and instead of having one of these let's go from here to here and Then once again from here to here. Do you think that'd be a better approximation? Probably get closer to the actual length of the curve if instead of just doing one Big old line segment that we somehow subdivide it and we can do a whole lot better and thus the reason for it being in this section if We have this curve And instead of having one line segment we have many I Don't know what many is and you're probably not going to be able to see the difference here because of the width of the Instrument that I'm using to draw it But all of these little line segments and if we could have a hundred of them or a thousand of them or six hundred trillion of them Isn't it true that this thing although it's a curve clearly a curve It's not very curved from here to here, right? In fact, you can I've heard a circle Which is obviously not a straight line in any way is the union of an infinite number of Tiny little line segments. Well, you know if you take a triangle and quadrilateral and Pentagon and hexagon and so on eventually You take this thing that was not very circular in shape and you make it more circular in shape That's the same rationale here taking line segment from this point to this point. It doesn't match the curve exactly But it's pretty darn close because the distance is small So let's go back up here to this diagram. Let's suppose we wanted the length of this This one red line segment, which is rough, but approximately the length of the curve from starting point to ending point So in going from here to here That's how much we've changed in terms of x in going from point a to point b so that would be some increment of x and Going from this point to this point. We've changed this much in terms of y so the hypotenuse Which is what we really want would be the square root of Delta x squared Plus delta y squared everybody in agreement with that and if that looks Reasonable to you just Pythagorean theorem then that's basically all you need to get you started on This particular line of reasoning would it change much if we moved over here to this diagram wouldn't this still be a Delta x some increment of x This would still be some increment of y that would approximate this line segment We've got another line segment here. That's delta x and That's delta y and then what would we want to do with these we'd want to add them together, right? Well, if we had an infinite number of these Representations I'm trying to avoid using the word hypotenuses because I don't really know if that's a word to begin with So if we had an infinite number of these red line segments Each of which is the hypotenuse of the right triangle. How did that go for avoiding it? What could we use to add an infinite number of these together? What have we used to this point integral, right? So this hypotenuse to this one this one this one and so on we can have an infinite number of these Hypotenuses and we want to add them together. Okay, so we're going to be able to do that We want the sum of an infinite number of these So a sum of an infinite number of these says we're going to be able to do that now That's a very basic kind of almost a crude Way of denoting what it is we're going to be adding together, but each one of these is a Hypotenuse we want to add them all together and if we have an infinite number of them then we should get the exact length of the curve from point a to point b Now typically when we do an integrand delta x's become what? The x's in delta y's then in like manner would be d y's so we're going to have to do some Adjustment here of this basic premise first of all we don't have a dx or a dy or a dt So we're going to somehow get that involved in the integrand outside of the integrand so we're going to add together an infinite number of Pieces each piece in this case is an individual hypotenuse. So let's start with this Let's finish with that Let's multiply by delta x And divide by delta x is that legal okay, which kind of seems odd So let's leave this delta x alone. That's going to become our dx eventually This one over delta x. I'm going to make that the square root of one over Delta x Squared is that the same thing any idea where I'm doing that shaking your head like I don't know you do all Kind of stupid stuff in here We got a square root here and a square root here. Is it okay to multiply the square roots? Yeah, that's the goal So I've got under this square root one over Delta x squared, so I'm going to put the square roots together So I'm going to have a delta x squared over a delta x squared I'll go ahead and write it this time, but pretty clear what that's going to be I'm going to have a delta y squared over a delta x squared and then outside of the radical I'm going to have this extra delta x, which is kind of what I wanted to Get in there anyway So a way of getting that there is to kind of compensate for it Luckily, we can put the two radical terms together Delta x squared over itself Might be Monday morning, but I think we can all agree that that's one What are we going to call delta y squared over delta x squared? dy over dx Squared wouldn't that be delta y over delta x the whole thing squared, right? And this delta x we're going to call dx so there's our first Version of one that we can actually use That's going to approximate arc length or the length of a plane curve So this would be from our initial x value since we're integrating with respect to x To some later x value now one thing we do want to be true about the curve is we do want it to be a Function in a sense that we're not going to have any repetition of Hypotenuse going from one x value to another x value so When we need that type of curve, which this one will suffice for a function When we don't have a function I'll make a diagram of that show you why this is Confounded by that it won't work and we'll have to kind of switch horses to another version but Regardless of the one that we use We can kind of always start with this and Develop the one that we think is going to be needed based on what the curve itself looks like if it's a function This is the one that we want Why don't we go ahead and develop all three of them and then we'll use the one that's appropriate for the problem that we're Confronted with so there's our first one length of a plane curve or arc length So we'll start with that. Why would we need something different? So let's say we have a curve that does something like this well, if we go from here to here and Then we're going to backtrack Let's say from here to here and then go from here to here The problem is that as we go from this x value To this x value then we have to backtrack and come back to this x value and we've got some repetition going on So we're not going to capture each hypotenuse the way we want to So that we can add them together. They're they're overlapping So what we would want to do instead of going in terms of x? Let's reorient and go in terms of y So do I have to do any of that? Doubling back or backtracking if I go from here to here in terms of y and Then pick up where I left off there and go from here to here that looks pretty good looks better to me and Then go from this last point Over to here did we do any doubling back or backtracking? We did none in terms of y So with respect to x we've got problems with respect to y We're flowing very nicely from here to here where this one ends the next one starts Same thing here So sometimes we want to be able to integrate with respect to y because with respect to x is going to fail So what can we do to this? Well very similarly Before we wanted a dx in the integrand now we want a dy in the integrand. What's your recommendation for? adapting our basic little hypotenuse Version so that we now get a dy in the integrand multiplied by delta y Divide by delta y We multiplied by delta x divided by delta x in the other one we can shorten the steps But let's at least get this step in here Delta y we're going to leave alone one over delta y We'll rewrite it so that it's under a radical so it's one over delta y squared now We can multiply the two square roots Put a couple of steps together since we've already done this once what are we going to have in the first position? dx over dy squared This times this delta y squared over itself is going to be one and this delta y is Going to be a dy we can do the same problem With respect to y we're going to differentiate the function that we have Derivative of x so if we have it x in terms of y that's wonderful set up perfectly Derive it with respect to y square it add one to it take the square root of it integrate with respect to y From some initial y value to some Terminal y value again that is because we need to do so Because trying to find the length of the curve in the fact that it's not a function We're not going to be able to do it with respect to x So there's our second version of this Well, we've got one with respect to x and one with respect to y about the third Sometimes we have x in terms of t and y in terms of t So we could have one in terms of t or theta or whatever the parameter is So that's what we're doing. We're adding an infinite number of these things together these line segments each of which comprises a hypotenuse of A right triangle So we want a d Something else other than x or y. Let's say t one over delta t we can put that in Another form one over the square root of delta t squared Now if we multiply the two radicals together What's going to be the first term under the radical? DX over dt squared next one dy over dt squared In a dt So if we wanted if we have a function we want to find the arc length or length of the curve in terms of x We've got it. It's not a function It would help to reorient it with respect to y We've got one of those and if we've got a two parametric equations X in terms of t and y in terms of t differentiate them Separately square them add them together put them under the radical integrate From one of the t values to another t value. So there's our third arc length formula They're basically all the same kind of just depends on what we're given in the problem and what the nature of that Picture tells us to do. All right. Let's do some examples. I don't think any of these are examples in your book There are some good examples in your book But let's take a look at these and then you have these that we're going to do plus those that are in your book The first one I was going to do is actually the hardest one. So let's let me do this a little different order So there's our curve kind of a strange function We want to go from the length of this curve from 1 13 12th to 2 7 over 6 Now there's probably a big hint Based on the a and the b that are given It's probably going to be a whole lot easier to do this problem in terms of x Because wouldn't the limits be a whole lot sweeter in terms of x than they are in terms of y So I would like for this to be a function it in fact is a function So we can do this problem in terms of x or with respect to x. So there's our format It's going to get us a solution in terms of x which this will In fact what happens on this problem You might say well when we're done you might say well, that's never going to happen on another problem Actually, this happens on a lot of these problems But what happens as we develop this problem? Just as many other times you're going to come up with something that is not integrable at all You're going to have some ugly function under a radical Substitution won't work any kind of change a variable trig substitution. Nothing's going to work if that's the case Then just go to some other approximating technique table of integrals something But this actually It's not that uncommon what's going to happen in this problem. We do need dy over dx or y prime What is that here x squared over four? I think I heard that is that right is that the derivative of 1 12th X cubed right three times one 12th x to one degree left last so x squared Over four derivative of one over x Okay, so there's our derivative So we want to square that derivative We're going to integrate with respect to x from our initial x value one to our final x value To are by okay to that point Anybody know what's going to happen a key or a clue to it is that there's an x squared up here And there's another x squared down here So the middle term is going to be kind of key to what happens all right when we've got a binomial squared So we need to square the first term the middle term and this is where Kind of this process starts that it's going to allow this to work the way it does The middle term ought to be twice their product right what is twice their product? Twice their product One half minus one half right So if you take twice the product of the first and last terms of the binomial you're going to get minus a half and the last term is That so can we combine any terms under the radical? Well, we've got a one and we've got a minus a half Very conveniently one minus a half is plus a half now. I said very conveniently Why does that make this problem workable or doable? This was already something squared right is this Something squared shouldn't it be the same two things with a different sign in the middle? if this is x squared over four Minus one over x squared squared. What should this be? x squared over four plus One over x squared the quantity squared. So the fact that this was minus a half It's being combined with one which now gives us the same number with a different sign instead of minus a half now It's plus a half so this thing under the radical which you should always look for this because it happens More frequently than you think it would happen is this mess that's now under the radical is it a perfect square? it is and What's the square root of that perfect square? What's in the parentheses right the square root of this quantity squared is? That quantity so we can integrate that there was a pretty ugly integrand until we got rid of the square root now We're rid of the square root now. It's integrating each piece What's the integral of x squared over four or one fourth x squared? It's gonna kind of almost get it to take us back to this original function x cubed over 12 That ought to look familiar and what's the integral of one over x squared? negative one over x Is that pretty darn close to the original function? It differs but where does it differ in the sign between these and doesn't that make sense based on how we got to this point in the problem? We kind of change the sign of the middle term therefore when we Take the square root of this perfect square Antidifferentiate we ought to be back, but we ought to have a sign change in between these two terms So we integrated now let's evaluate at two. What do we get? eight twelfths Minus one over two And at one one twelfth Minus one so eight twelfths minus one twelfth would be seven twelfths And we've got minus a half and then we've got plus one which is plus a half so our final answer would be what? 13 twelfths and that would be the exact length of that curve From one comma thirteen twelfths to two comma seven over six So when you get to this point when you've taken the first derivative and you've squared it and you combine things Check it to see if we have in fact a perfect square under the radical that would really help the cause if that is Is the case and that is the case with this problem Not going to be the case with all problems Questions on this one before we move to another one. Maybe we need a little Musical interlude When we go to the next problem express settings Yes, let's try that. Let's try it again. Did you hear that? I didn't hear it Hmm media players not working. That's disappointing a lot just to do that Got to have some sound effects from time to time. That's it. What is that? 24 best show on TV Okay, another example now that we're all fired up about 24 Jack Bauer Okay, this one's this one's a little stubborn and I had this written down as my first example But I didn't want to scare you away with this one X to the two-thirds So it really doesn't matter in this case if we go in terms of X. We're going to go from 8 to 27 That's going to be pretty easy. That's usually not the issue. Anyway, the y values are okay to 2 to 17 Let's see if we can since we have y in terms of X. Let's see if we can use the same format 1 plus dy over dx squared Integrate everything with respect to X and we're going to go then from 8 to 27 dy over dx What's the derivative of 3x to the two-thirds? Got agreement on that? Okay, and derivative of negative 10 I might as well do my part that would be zero, right? Go ahead and feel like I'm doing my part here. All right one plus first derivative squared What's the first derivative squared? Square of the two we're going to get four. How about squaring X to the negative one-third? Negative two-thirds, right? We have something raised to a power itself raised to a power you multiply the exponents It's probably worth a few seconds to check to see if what's under the radical is In fact a perfect square Not so lucky, right? This is not a perfect square under the radical But it's not integrable the way it is We could say let you equal the stuff that's under this square root symbol We're not going to have du right in the integrand. So regular substitution is not going to work I don't think we saw a trig substitution problem that was quite this awful in its original form. So What we could do is try to get it into another form Well, we've got That's really the same thing as 4 over X to the two-thirds Where am I headed with this? Let me do a little musical interlude again While we're thinking kind of like the Jeopardy background music anybody One's a fraction and one's not a fraction Put them together right get a common denominator see what happens Somehow I've got to change the form because the form it's in right now. It's we can't integrate it. So the first term If I'm going to give it the same denominator is going to be x to the two-thirds over X to the two-thirds And if we put those together into a single fraction X to the two-thirds plus four over X to the two-thirds Anything gained by that little manipulation Probably or I wouldn't have done it right what is what's gained by that? Yeah, we've we've now that we've got a fraction and we've got the square root of the fraction We can take the square root of the top Which isn't able to be found at this moment, but we can take the square root of the bottom Which is able to be found so since this is the first time we've encountered this We want to write it and there's no guarantee this is going to work There's no guarantee that it's going to get us where we want to but in fact it does on this problem So it is just another technique that you can try Now this in the bottom is the square root of x to the two-thirds The square root is raising something to the one-half, right? What's x to the two-thirds to the one-half? X to the one-third and that x to the one-third is in the denominator So if I brought it up to the numerator it'd be x to the negative one-third. Why is this now a doable problem? Now you can use you substitution exactly right So let you equal this ugly thing that's under the radical and That's going to work if we have du or at least a correctable version of du in the integrand. What is the? derivative of this choice for you So didn't we absolutely have to have x to the negative one-third dx to do the problem this way? Right doesn't this have to be in the integrand and this is going to be in the integrand because we're doing the problem with respect to x So x to the negative one-third. It's there now. It's a doable problem dx is there What do we lack in the integrand that we can correct for? We lack a two-thirds So let's multiply by two-thirds and multiply by three halfs I'm not going to write in the limits because those are limits in terms of x and we're changing the problem right now to a u problem What is this square root? What's it now become? Good a lot easier right you to the one half And how about all this other junk here at the end? That's all du. That's a much easier-looking problem Not an easier problem. It's the same problem can't be easier than itself, but it's easier looking Everybody okay with how we got from this integrand to this one. What's the integral of u to the one half? Two-thirds u To the three halves This is a definite integral. I'm not going to change the limits of integration I'll just take it back in terms of x and then bring back the 8 and the 27 So we kind of luck out here three halves that was the Compensatory term out in front and we created our own two-thirds when we integrated so u to the three halves becomes X to the two-thirds Plus four to the three halves is that right and we want to evaluate that from 8 To 27 so at 8 8 To the two-thirds plus four and at 27 27 To the two-thirds Plus four what's eight to the two-thirds? Sorry, thank you Getting Jumping ahead. I saw these we're gonna be nice things to raise to the two-thirds So we've got to raise this to the three halves right is that where I'm missing and we've got to raise this to the three halves Thank you Oh gosh Do you want to take over? How about this that work there we go Anxious to put that eight in there because I knew I could take it to the two-thirds and the same with 27 So this should be first upper limit of integration your instructor had any sense he would have put that first and this should be second So what's 27 to the two-thirds? Be the cube root of 27, which is 3 Squared which is 9 plus the 4 is 13. That looks like a nice value 13 to the three halves Okay, eight to the two-thirds and be the cube root of eight, which is two Two squared, which is four and four plus four is eight I would say it would be time to pull out a calculator on this one, right because we've got the square root of 13 I don't think I need to go any further. We're gonna take that and cube it the square root of eight We could probably deal with that, but then we turn around and have to cube it So I would abandon this exact value and then approximate each of these I don't know if I wrote this down or not. I did not write down the solution. I don't remember it either I was thinking 24 24 maybe that's why I chose this example 24 point Point 24 That's just that's like the awesome. Yes, we got to do that again I wish you could see what I'm seeing here. That's pretty cool 245 they're out of kind of ruins it Yeah, we could truncate it. That's what Jack always does right somebody's in his way just truncate some Questions on that process Well, all problems work this way where you end up with something that's not integrable You put the two fractions together. No, they won't so at some point in time That point in time in this problem would be right here At some point in time you have to abandon. I've tried everything I could can do I looked for the square root of a perfect square that wasn't there I Put two fractions together got a common denominator took the square root of the numerator and denominator If this didn't happen to kind of luck out being the big piece of DU that we needed We would want to abandon that process at this point in time on this problem All right, let's get one in terms of t and we will call it a day Let's say we have parametric equations I think this is one of the problems in your book, but it's not one of the examples in the section How does this version look different? What's under the radical when we've got parametric equations in terms of t we still want derivatives, right? It's either 1 plus dy over dx squared or 1 plus dx over dy squared But in this case we're differentiating with respect to t So it's each of the individuals derivatives Square right added together underneath this radical why the radical where'd that come from comes from the fact that we're adding together a bunch of little hypotenuses dx over dt is What dy over dt? Maybe it doesn't have a derivative. I think it does well four comes along for the ride Derivative of e to the t over two would be e to the t over two times one half So four times one half would be two e to the t over two that work so we want to take each of the Derivatives in terms of t square them I don't know what's going to happen But I like the fact that I'm squaring this second one that has this ugly exponent of t over two and when I square it It's no longer t over two because now it's what how do you take something to a power to a power? Multiply the exponents t over two times two is just t so that's going to help the cause So let's square everything e to the t minus one. It's a binomial squared first term would be e to the t squared what's that e to the 2t Middle term ought to be twice their product and the last term is the last term squared That's going to turn out pretty nice We square this next derivative dy over dt squared we get Four e to the t so tell me what's going to happen on this problem Based on where we are at this point It's gonna be perfect. It's gonna be a perfect square under the radical wasn't this something squared right With the minus 2t as the middle term now when we add 4t to that for e to the t What's this going to be it's going to be plus 2 e to the t? Which is again going to be a perfect square the same thing we squared which was this right? Except it's going to have a different sign in the middle So the minus 2 e to the t and the plus 4 e to the t and as Nicole said That's going to be a perfect square That's going to be e to the t plus one quantity squared The square root of that perfect square we get e to the t plus one we're going to integrate it What's the integral of e to the t? Integrated with respect to t would just be e to the t integral of one with respect to t Time so I'm just going to say what would we do we put in three We had any smarts at all and like me on the last problem we put in the upper limit of integration first We subtract from that what we get when we put in the lower limit of integration Okay, I will see you tomorrow