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About this Lesson
 Type: Video Tutorial
 Length: 15:04
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 Posted: 06/26/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Parametric Equations, Polar Coordinates (18 lessons, $27.72)
Calculus: Calculus and Parametric Equations (4 lessons, $8.91)
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
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Parametric Equations and Polar Coordinates
Understanding Parametric Equations
Finding Arc Lengths of Curves Given by Parametric Equations Page [1 of 3]
So to find arc length, the way we can think about this if I'm given a parameterized curve is just to integrate from the one point we want to find the distance length of all the way to the other point  so integrate from there to there  the square root of the derivative of x with respect to t^2 plus the derivative of y with respect to t^2. The idea is that I'm breaking this very complicated, prettylooking curve into a whole bunch of little teeny line segments where each of those line segments actually is the hypotenuse of some tiny triangle and the length of the hypotenuse, in fact, is that.
Another way of thinking about this, by the way, just as a formality in terms of notation, is instead of thinking about this as , I could think about it as f^1 of t if I wanted in here. So, in fact, an alternate way of actually stating that integral would be to write it as the integral from a to be of the square root of...and instead of , I could put down f^1(t)2 + g^1(t)2 dt . So these two things, of course, are the same, just different notations. Some people like one notation more than the other, but the point is there is a formula for arc length  the same formula, just a different way of looking at it.
So let's look at an example where we actually use the formula to find the length of a particular arc. Let's take a look at the following. Suppose that I have the following parametric equations: e(t) = e^t cost and y(t) = e^t sint. Let's now first of all ask the question: What does the graph of this look like? Well, the graph of this, in fact, is a spiral, and it looks something like this. It spirals out, and you see that it starts at one  and it keeps going, of course  but here at one, we're at zero. Then it goes out and spirals around and very tightly coils around the origin. In fact, if we had an even thinner font size and if our printers were even better, you would see this thing spinning around, spinning around, and spinning around. But no matter what your printer thing is, it would never see that sort of infinite spinning through you. In fact, what we've shown here is actually what happens when t goes from zero out to 2. So for example, if t is zero, what is x? If I put in zero for t, I see e^0, so that's one, and cos0 is one, so I see that x is one. And what is y if I plug in zero? e^0, which is still one; and then sin0, which is zero. So that's why we're right at this point  that's (0,1). And if you spiral in and you go to 2, you get over here.
Now, the question is: What is the length of this arc? So it's sort of a prettylooking arc. In fact, you can sort of make a guess. Let's make a guess, shall we? So I'm going to just line out a red curve, so I use red rope. It's always good if you're doing a question or something to see if your answer is reasonable enough. Is your answer reasonable? I mean, if we got like 400 for the length, would that make sense? Well, from here to here is one; four hundred would probably not be a good answer. Yeah, we made a mistake somewhere. So let's just make a guess here. It's hard to get that spiraling, but let's suppose that that's it. So where my finger is, that seems to capture the spirit of the spiral. If we now just sort of stretch that out here, it certainly looks bigger than one. This is .5 right here, so it seems like it's less than 1.5 but certainly bigger than one  so pretty close to 1.5 but less. So, in fact, let's write that down as a guess, not quite 1.5, so maybe 1.4 is a guess. So that's our guess, and we'll see how close we are.
Now, let's actually use calculus to find the exact dot. So what do we do? Well, to find the arc length, which is the question, I use this formula, that it's just an integral. Now, I'm trying to find the arc length from t ranging from zero to 2. So let's see where we are. We have to set up this integral, so I have the integral of zero to 2   because this a and b represent the endpoints of the region I'm trying to figure out the length for. Then I'm going to have square root, and then I'm going to need to find the derivative of x with respect to t, so I can write that as ^2 + ^2 dt. So that's the integral I have to figure out for arc length.
That requires me to find a couple of derivatives first. Not a big deal, but a little bit involved so let's work through these things because to find , we're going to have to use a product rule  this is a product. So we have first  the e^t  times the derivative of the second, and the derivative of cosine is minus sine  I'll put the minus out here  and then sint. Then, plus the second, which is plus cost, times the derivative of the first. The derivative of that is the little p chain rule, e^blop  the derivative of e^blop  and then times the derivative of the blop is just a 1, so in fact, I have the negative sign way out here, and we can note that with a 1 way out there. So it actually changes the sign of this to negative. So, in fact, I see that the thing works out to be e^t sint, and then e^t cost. Or if I factor out, I could factor out that e^t, and if I factor it  which sort of might be handy to do  then I'd see et x sint + cost. So I just factored out the e^t, and then I'm left with sint + cost. So that's this first piece right there, but don't forget, I've got to square it.
Okay, so let's just record that for now, so this equals integral from zero to 2   and then I'm going to have this big squarerootlooking thing, and then here I'm going to put this thing squared. This is not for the meek, by the way. Not hard, but certainly not for the meek. It's going to be a long one. So I square all of that...square all of that. That was just this piece right here.
Now I've got to put down what I do when I take the derivative of this. This is going to be a similarlooking kind of thing. Let's see if we can do that really fast. So what's ? That's the first times the derivative of the second, so that's e^t x cos(t) plus the second times the derivative of the first, and the derivative of this would be a e^t  so e^t, and if I factor out the e^t, I see cost  sint. So that's the derivative for this thing in a similar way. So I have to put that in  oh, boy, this is a long one! So plus...so I have e^t x cost  sint, all squared, and that is the entire square root . Wow! So that integral is looking pretty threatening, pretty threatening. We can simplify this a little bit, so let's think about how to simplify this. One thing I can do is sort of square this number  this is the product of two things I'm squaring  so I can square this and multiply it by squaring this. Then I can square this thing because I'm multiplying it by this, and then square this. So I can square this term, square that term, and then add it to squaring this term and squaring that term. So let's do that. I'm just trying to simplify this as best as I can, and it's looking pretty awful. So let's see if we can do that. If I square this out, then I have a 1^2. 1^2 is just 1, so that actually goes away when I square it. I'm left with then e^2t. I'm just doing the inside piece here now. Then when I square this out, I've got a foil. So I've got to foil this, and so I see sin^2t + 2sintcost  those are the middle terms. And the last times the last is a +cos^t2. So let's just recap here. If I square this term, then I see a 1^2 is just one; and then I see this thing squared, and I e^2t. Then when I square this thing out, it's a little binomial. I use the foil technique. I have the first squared, I have the last squared, and the inside terms give me 2sin. Ah! But sin^2t plus cos^2t, that's one, so that's actually finally a slight simplification. You can rewrite this as e^2t times  and now that's all just one  so I have 2sintcost + 1. So this whole thing becomes that. So that is a good simplification. It still doesn't look that great, but at least it's progress. A big square root, and then I have e^2t x 2sintcost + 1, so that's this piece right here. Now I've got to do the same kind of thing here, which we can do it a little bit faster because it's the same idea. If I square this out, this thing squared just gives me an e^2t, and then if I foil this out, I see cost x cost  that's cos2t. Then my inside terms and outside terms will combine to give me a 2sint x cost. And the last times the last would be a +sin^2t. Again, I see ^cos2t + sin^2t, that's one. So again, this simplifies to e^2t x 2sintcost + 1. So, in fact, all of that becomes this. So let's put that in. Still, it's not clear that this is progress. Boy, is there any hope in sight? I don't know.
So I have e^2 x 2sint+1. You can almost, barely not read it  et  and then all of that under the square root. So, that looks pretty threatening until you look at it and realize that after I distribute everything, I see an e^2t times this stuff, and then here I see an e^2t with a negative that stuff. So, in fact, they add to give zero  they cancel out. You see, I've got this thing, and then I've got minus that thing. They actually drop out.
So what am I left with? I'm just left with this term, which is e^2t, and I'm left with this term, which is another e^2t, and so I have two of them. So, in fact, this now really is going to simplify dramatically. This is going to be a dramatic simplification, and the dramatic simplification is going to be this  the integral from zero to 2 of the square root  and this, after canceling, is just 2e^2t dt. So that looks like a much, much, much easier integral. So let's see if we can evaluate that really fast. Well, that's just the, which I'll pull out in front...it's a constant. And I have the integral from zero to 2, and now I'm looking at the square root of e^2t dt. But what's the square root? That's raising something to the power. So what I see here is e^2t divided by 2. That's taking the square root, and what does that equal? Well, the two's cancel here, so I'm just looking at this integral: the square root of 2, integral from 0 to 2 of e^t dt. And what's that integral? Well, it's just e^t.
Now what I've got to do is evaluate that from zero to 2. So when I plug in 2, I see e^t, and then I subtract off what I get when I plug in zero, which is just the negative square root of two. So what this equals is this is just +  ^e2^^^^^^. If you want to write that nicely, it's x 1  . Phew! That is the exact answer; that is the precise length. Now, what does that length work out to be? Well, if you use a calculator, what do you see? What you see is...I'll show you what the readout says  it says 1.41157. So the length is that. Now if you remember, a long time ago we made a guess, and there was our length just based on what we saw with the thing, and you can see that we were actually really, really close. But this is the exact value, and that just follows by computing the integral  that integral of the square root of the derivative squared added up.
Okay, that's arc length. I'll see you at the next lecture.
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