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Calculus: Derivatives of Parametric Equations


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About this Lesson

  • Type: Video Tutorial
  • Length: 12:43
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 136 MB
  • 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 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 UT-Austin 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, p-adic 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.

About this Author

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Parametric Equations and Polar Coordinates
Understanding Parametric Equations
Derivatives of Parametric Equations Page [1 of 3]
Suppose you're given a path in the plane that's described by some parametric equations. Is that path going to be nice and smooth to the touch so that if you touch it it feels very, very nice? Or will it be sort of bumpy and sharp? For example, in the cycloid we saw that every time that that marked spot on the tire hit the ground, we get a spiky point like that. In fact, we even saw in some graphs, for example the graph of y = x^2/3, there's this spiky point here, too.
So when, in fact, do we not get spiky points? Well, in fact, when we don't get spiky points, then we say that the curve is smooth. Now what does that mean from a mathematical point of view? A smooth curve given by parametric equations - the curve is called smooth - but suppose that you're given parametric equations x and y. So you have x, maybe this is some function of t, and you have y, depends upon t in a different way. Suppose that these describe now some sort of curve. We say that that curve is smooth if the derivative of this and derivative of this - so the derivative of x with respect to time - and the derivative of y with respect to time both are continuous. So if you take the derivative of each of these functions, these are continuous. And we have to think about one thing: They both cannot be zero at the same time. So they are continuous, and they are not simultaneously equal to zero.
Now why do we put on that extra weird condition not simultaneously equal to zero? Let's actually take a look back at this example. In this example, if I take the derivative of x with respect to t, the answer is 3t^2. When is that zero? 3t^2 is zero when t = 0. What about here with y? If I take the derivative of y with respect to t, I see 2t. And when is that zero? Well, that's also zero when t = 0. Notice that when t = 0, both of these derivatives produce zero. What happens in the picture? Notice in the picture I've got that sharp point. So, in fact, this curve is not smooth there, it's actually very sharp. If you touched it, your finger would start to bleed because it's like a needle there. So it's not smooth there. Notice everywhere else, though, if you look at the derivatives everywhere else, 3t^2 and 2t, if you plug in any other point, in fact those functions are continuous. Right? There are no breaks in those functions at all. 3t^2 is a very nice, continuous function, as is 2t. So, in fact, everywhere else it's nice and smooth, and the picture actually depicts that. It's nice and smooth everywhere else except there. So, in fact, this is the definition of smoothness that seems to make sense, especially looking at this example - this picture.
In fact thinking about this, these are the parameters and this is the actual graph of which this parameterization lives along. If there were a bug actually climbing around here, how would the bug actually move around? Well, when t = 0, you notice the bug is right here - (0,0). Then when you go forward in time, as t gets bigger, the bug starts to move out in this direction. And if you could somehow go backwards - if you could go into the time machine and go backwards - where would the bug have been? The bug would actually have been in this direction. That is to say that the bug, like a negative affinity, would be way out there; and then as you move forward with negative time, you come in. When time is zero, it's here; and then when time is positive, it's out here. You can actually see how the bug is moving. So it's like if I put on sort of like the bug line, the bug line will sort of look like this. It will come down, touch there, and then start again. So those are sort of the bug lines for this, and it lives along this path.
Let's look at another equation. In this equation, you saw x = e^t and y = 3e^2t. It somehow travels along this path. Let's see exactly how it's actually traveling along that path. If I plug in t = 0, I have e^0, which is 1, and then here I see 3 x e^0, so that's 3 x 1, which is 3. So actually when t = 0 - time equals zero - the bug is sort of located here at (1,3). Then as time moves forward, the bug goes out like this.
And what if you go backwards in time? Then if you think about it, the bug is going to keep going backwards and backwards and backwards. But will it actually get up to this wing and actually see this part of the world? The answer is no because think about it: As time gets smaller and smaller and smaller - so more negative, negative, negative going out to negative infinity - what's this thing approaching? Well, e-something very, very tiny, like -1,000 or -1,000,000 and so forth. This whole thing is actually shrinking to zero as is this. So as the bug goes backwards in time, it's coming from the origin, although it never actually gets there. So if you go way back - infinitely back - it's going to get really intimately close to the origin and start to climb there. So the actual path is sort of along there. You can actually see these paths, and now you can see whether the things are smooth or not by looking at this little condition.
Actually, how would you find the derivatives? Suppose you actually want to find the derivatives of these paths. Okay, they're smooth. Smooth paths, by the way, means that we can actually find the derivatives. So let's actually now see if we can figure out what calculus would look like along these parametric equations. So what would calculus look like? Suppose that I'm given x is the function of t and y is some other function of t. These are parametric equations - you give me time, I find out my location. And, of course, the mission always at hand is to find . How do I do that? All I can find here is . That's really easy. Just differentiate this with respect to t. I already have x in terms of t, and I can differentiate this with respect to t and see . So what's easier - so here's easy - is to find and . Look at me, I said t. Now watch this, I'm going to use y. That's actually a t. It might not look like a t to you, but there it is - t! Hah! You thought I couldn't do it. I know what you're saying, you're saying, "Well, if he breathes really hard, it will blow away." Well, that's true, but even my breath cannot withstand this Scotch tape.
So these are really easy to find. You just take the derivative of this with respect to t. So in this other example that we saw here, taking the derivative with respect to t, we did that in our heads. It's 3^t2. So that's really easy. Similarly here, you can find those derivatives. But, what about ? How can you figure out what is? I mean do you figure out what is? That represents slope, that represents the derivative of the curve. Well, the way to think about this is to think about the chain rule. Everything in life boils down to the chain rule. You see, if you want to chain together , one way to do that is first of all pretend you had , and then multiply that by some derivative, and multiply it by some derivative so that the answer is this. Now all you've got to do it do fantasy math. If you do fantasy math, in fact you get the right answer.
So let's do fantasy math and pretend these aren't derivatives but just these funny-looking fractions. If you want to get this fraction, I already have the dy on top, but I need the dt on the bottom. Why don't we put a dt on the bottom here. But I want to get rid of that, so to cancel that, I'll put the x on top. Notice that these sort of fantasy cancel, and I'm left with . Well, it turns out the chain rule makes the fantasy a reality. So, in fact, this is just the chain rule. So if I want to find , all I've got to do is take and divide it by . So, in fact, this gives us a really neat formula. It tells us that - the thing I was after - is nothing more than divided by , and those are both easy to find. Of course, whenever you divide by something, you want to make sure that thing is not zero. So I have to put on the extra condition that we can't let equal zero, otherwise we're dividing by zero, which is, of course, not at all allowed. So in fact, this is a formula for finding the derivative of y with respect to x if you're given the formula parametrically. If you just take and divide it by , and that quotient is going to give you the derivative.
So, in fact, let's try this with a little example. Let's actually return back to this example right here, and in this example right here, I want to find the derivative. Now, of course, we know what the answer is. The answer is just a derivative of y with respect to x. Since we already found this curve, we see it's going to be the derivative is 6x...6x. But let's see if we can verify that using this particular formula. So let's use this formula and see if we can get the same answer. Let's see how that would go. In fact, let me just take this whole stuff. It's already over there anyway, so I don't need to have that there.
So let's now find . Well, , if I take the derivative of y with respect to t, I've got to use a little chain rule, so I have 3e^2t, so that's going to be e^2t, and the derivative of that is e^2t, and then times the derivative of 2^t would produce a 2 out in front. So I get 3 times 2, which is 6e^2t. That's the derivative. And , the derivative of d^t is just d^t. So I have ; I have ; and so should just equal the quotient of those two things. So let's divide one to the other and see what we get. If we divide one to the other, we should get the answer. Let's see if we're close. Supposedly, is going to equal this thing so that 6e^2t divided by this - e^t. Notice that I'm just applying this formula, taking this derivative, which is up here, and dividing it by that derivative, which is right here. Now, let's see, what happens? Well, you can cancel this away a little bit, and if you cancel this away, I just am left with six. Here, I have e^2t, here I have e^t, so if I subtract the exponents, I'm just left with one t upstairs, and that seems to be the answer. Well, that doesn't sound like the answer that we said it should be. We said it should be 6x, so that sort of...oh, oh, maybe there's a mistake. Oh! Wait a minute, though! - e^t is actually x. So, in fact, this is just a camouflaged way of saying 6x. And that, in fact, is the right derivative. So you can see this formula really is consistent and is right, and will provide rates of change and slopes - whatever you want. When you're given parametric equations, you can always find .

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