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Calculus: Finding Tangent Line Slope in Polar Form


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About this Lesson

  • Type: Video Tutorial
  • Length: 7:35
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 81 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: Polar Functions and Slope (2 lessons, $3.96)

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.

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Parametric Equations and Polar Coordinates
Polar Functions and Slope
Finding the Slopes of Tangent Lines in Polar Form Page [1 of 2]
When you're thinking about these polar functions that have these beautiful sweeping-type curves, one natural question that we tend to ask in calculus is how would you find the slopes of tangent lines? Well, that requires us to find - change in y over change in x - . So what does mean when you're thinking about 's and r's? Well, the answer is not much. What we have to do is remember the conversion - how to go from polar form back to Cartesian form. So what I'd like to do is show you how to find when you're given a polar form.
Let's just pretend that we're given some function: r = f(). So this is a polar form. What I'd like to do now is figure out - find - because represents slope always - the slope of a line. How do I do that? What I've got to do is somehow bring this back to y's and x's with the conversion method. Let me remind you that in the conversion method, what we see is that the y is always r sin , and the x is defined to be r cos by a little application of a right triangle - right triangle action.
So these two things allow us to find the y and the x. What I want to find is . So how do I do that? Well, the answer is, "It's too hard." So when something's too hard, don't do it! Instead, let's do something easier. Now what would be easy to do? Well, since I know that r is really a function of in disguise, I could actually insert that right into r. And if I were to do that, I would see the following: I would see y = f() sin , and I would see x = f() cos . Just plugging this fact in to this general formula.
Now if I do this, this is great news because what I see here is that y now just depends on upon . It doesn't matter what anything else in life is, if I just know , then I can plug it in and find out what y is. Similarly with x, if I just know , I can plug it in and find out what x is. That means that it would be really easy for me to take the derivative of y with respect to . I can just differentiate. And it would be really easy for me to take the derivative of x, with respect to , I could just differentiate. So could I somehow use those two pieces of information to get what I want, and the answer is yes, by using the chain rule. If I use the chain rule, what would I see?
Let's take a look at the chain rule here. Let's take a look at something like because that's an easy thing to compute - I just take a derivative here. In fact, all I would have to do is use the products rule. How could I think about taking ? What I could do is say, "Well, first what I'll do is I'll take , and now I'm going to do some fantasy math." This is something you should never do at home. I'm going to say, "Okay, what would I have to write in here in order to make it ?" Well, I already have the dy on top, so that's looking great. I want a don the bottom. I don't have that, so I'd better put a d right here. And I certainly don't want that in here, so I'd better get rid of it by putting it on top. Well, look, this is fantastic because this says take the derivative of y with respect to , and we just determined that's pretty easy. And this says take the derivative of x with respect to , and we've already said that's really easy. So, in fact, these both are easy to compute, and I can solve for that very easily and find what I'm after. So, in fact, what I see here is that this equals what? Well, it equals divided by . So if I just take the derivative of y with respect to and divide it by the derivative of x with respect to , I actually have , which is what I was after.
Okay, now what would that actually work out to be? Let's actually figure it out. If I'm given the function f(), I plug it in here, I just have to take some derivatives. So let's take some derivatives right now and see what we get. Where should I take the derivatives, over here or over here? All right, so let's now compute the derivative. What is ? I have to use the product rule. So it's the first times the derivative of the second. So that's the first times the derivative of the second, plus the second times the derivative of the first. So that would be f prime () sin. So that's . There it is. Now what is ? Well, first times derivative of second, that's going to give me a -sin - that's the first -- then plus the second times the derivative of the first, so that's going to be f prime times cos.
So, if you put all of this together, what I see is the following fact: is going to equal , which is up here, f() cos + f prime () sin divided by , which is -f() sin + f prime () cos . So, in fact, there is the formula for finding . And in fact if you try it - if you actually work this out with the example of the Rose curve here - what you would actually see is that if you were to plug in = , this number would turn out to be tan because this number, if you work this out, will just become , just by taking the derivative and plugging in. So what you see here is that this slope should be tan, and that's exactly right. Since this angle is , that means that its slope - -- is tangent. is tangent, so the slope should be , which would be tan. So this actually verifies once again the fact that right here this is going to graze the origin along this tangent line. So this, indeed, is the tangent line of this function. The function, remember, is r = cos(3), and it braves it right along there. Similarly, if you plug in or , you'll see the same effect. So in fact this will always give you the slope of a tangent line of any polar form of this type just by knowing the and the function.
See you at the next lesson.

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