Preview
Buy lesson
Buy lesson
(only $1.98) 
You Might Also Like

Calculus: Using Integrating Factors 
Calculus: Differentiability 
Calculus: 1stOrder Linear Differential Equations 
Calculus: Solving Separable Differential Equations 
Calculus: Power Series  Integration 
Calculus: Power Series  Differentiation 
Calculus: Power Series DifferentiationIntegration 
Calculus: The Polar Coordinate System 
Calculus: Fibonacci Numbers 
Calculus: Applying Implicit Differentiation 
College Algebra: Solving for x in Log Equations 
College Algebra: Finding Log Function Values 
College Algebra: Exponential to Log Functions 
College Algebra: Using Exponent Properties 
College Algebra: Finding the Inverse of a Function 
College Algebra: Graphing Polynomial Functions 
College Algebra: Polynomial Zeros & Multiplicities 
College Algebra: PiecewiseDefined Functions 
College Algebra: Decoding the Circle Formula 
College Algebra: Rationalizing Denominators

Calculus: Applying Implicit Differentiation 
Calculus: Fibonacci Numbers 
Calculus: The Polar Coordinate System 
Calculus: Power Series DifferentiationIntegration 
Calculus: Power Series  Differentiation 
Calculus: Power Series  Integration 
Calculus: Solving Separable Differential Equations 
Calculus: 1stOrder Linear Differential Equations 
Calculus: Differentiability 
Calculus: Using Integrating Factors
About this Lesson
 Type: Video Tutorial
 Length: 9:37
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 103 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: Understanding Polar Coordinates (4 lessons, $6.93)
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.
About this Author
 Thinkwell
 2174 lessons
 Joined:
11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
This lesson has not been reviewed.
Please purchase the lesson to review.
This lesson has not been reviewed.
Please purchase the lesson to review.
Parametric Equations and Polar Coordinates
Understanding Polar Coordinates
Converting Between Polar and Cartesian Forms Page [1 of 2]
Now we've been thinking about these polar coordinates and how to plot points and some very, very simple functions, but now what I want to think about is going back and forth between the Cartesian coordinates  the usual rectangular ones  and the polar ones. It's sort of like thinking about Curious George. What you've got is the polar coordinates here, and then you've got the Cartesian coordinates here. There should be a way of getting from one to the other, and then back again. So the question is: How do you convert back and forth?
Let's see if Curious George or someone can figure this out. Now anyone can do this, by the way  anyone. Here's how. Let's take a look and think about how to convert back and forth. Let's go back and forth.
If you think about it, if we plot everything on one [???] axis, it would look like this. If I put the x here  so I go over x  and I go up y, then in Cartesian form I call this (x,y) because it's the rectangular thing  over x, up y. Now what would this be in polar form? Now in polar form, I would sort of think of it this way: I would think about this angle   and then this direction here  this that somebody thought of as a radius  r. So now I want to find a connection that will go. So in polar form, this is (r,). So this is polar form, and this is rectangular form.
Now how do we go back and forth? Let's just see. What would x equal. Well, to figure out what x equals, let's think about that. What is x? If I take the cosq, that's going to be , so that's going to be . So I can actually solve for x, and if I solve for x, I see that x = r cos q. So if I'm given this thing in polar form, I can figure out what the xvalue should be in rectangular form by looking at r cos q. And what about the y? Well, it's opposite it, so let's consider it a sin function. Sinq would be opposite over hypotenuse, which is . So if I now crossmultiply, I see that y = r sin q. So now I see that if I'm given q and r  so polar form for a particular point  I can find that exact location in the Cartesian or rector linear way by just using this conversion thing. Put r cos q, and that's how far to go in the xdirection, and r sin q is how far up and down to go in the ydirection. So this is a way to convert from one to the other.
Now, if you also look at this, you can get another relationship that's useful in converting back and forth, too, which is the Pythagorean theorem. So that's worth noting, that x^2 + y^2 = r^2. So there's a relationship that takes the x and the y, and then produces an r.
So, for example, if you were given (x,y) in Cartesian coordinates and you wanted to find the polar form, one way to find r is to use this formula and to realize that if I take x^2 and add it to y^2, that will give me r^2, and then I can find the angle by using a trig thing. Let's take a look at just a couple of examples so we can see this in action.
Let's take a look at (r,q), and suppose it equals. First of all, let's just plot it so we can see what it looks like so we have a visual sense of what we're trying to figure out. The question, by the way, is to convert this back to good, oldfashioned Cartesian coordinates  (x,y)'s. Let's see, so the angle is  it's a positive direction  I always start here. is 45 degrees, so I go right up to here, and then I go over two units. So I go two units out, so I go 45 degrees, and then two units out. So that is the polar form .
If I want to figure out how far over and up, what do I do? Well, we've already seen this: x = r cos q and y = r sin q. So what's the x in this case? It's going to be r, which is two, and then cosq, so that would be . What does that equal? That's 2 x , so this just produces . So I must go units over in this direction; and if I compute this for y, I'll see y = , but is the same as , so it's still , so with the two in front, I see the exact same value, . That actually makes a lot of sense that these two numbers should be the same. Why should they be the same? Because if this is actually pitched at an angle of 45 degrees, then this is along the line y = x. So, in fact, whatever x equals, y should be the same. What's that exact value? Well, it turns out to be . Does that make sense? Let's use the Pythagorean theorem. Is this squared plus this squared equal to this squared? Well, this squared was , that's two. Two, plus this squared, which is another two, that's four, and that does equal 2^2. So, in fact, this checks. This actually checks. So the answer is that if you want to write this now in good, old Cartesian coordinates, it's (,). So there's a nice example going from this side to that side.
Let's now go the other direction. Suppose I gave you the point (x,y)  so this is in Cartesian form  and it equals (1,2). Well, what's the polar version of that? Well, again, let's draw a picture just to give us a sense of what's going on. These are x's and y's. So we go over one unit, and then up two. So, in fact, that's the point  (1,2)  and we now want to write that in terms of a pitch  so I should draw this line in  a and an r. Well, the r is pretty easy to figure out because I can just think of this as being a right triangle and use the Pythagorean theorem. So when I use the Pythagorean theorem, I see that 1^2 + 2^2 = r^2, or 1 + 4 = r^2, or r = . Why is it positive? Because, I'm going out in this direction, and that's a positive length. So that's . Okay, so r = , so I'm halfway home. Now I need to figure out that angle. How can I figure out that angle? There are a lot of ways of doing it. One way is just to be very slick and say, "Well, what is the angle?" Notice that the tangent of this angle has a particularly nice form because the tangent, remember, is . So the tangent is , or just 2. So a very slick way of writing is to use the inverse function, arctan, because remember if tan = 2, that means that equals the angle whose tangent is two  in particular, = arctan(2). So that's . That's some number. So (, arctan(2)) is the polar coordinate for the point (1,2). So you might say, "Well, gee, I'm not satisfied with arctan(2)." Well, I think you should be. That's completely fine. In fact, that's the most accurate way of saying it; but if you want, I'll actually tell you what arctan(2) is: 1.107... radians. So what is that? Well, remember that this is radians. So what's that? This is 1.5 something. So 1.1 something should be a little bit shy of that, and you'll notice that that's exactly what we have here. So if you don't like this  which I think you really should like  this is 1.107 radians.
So anyway, you can see that there are ways of easily converting from polar form to Cartesian coordinates and back and forth, just as you would see Curious George on a unicycle, and it's going back and forth, back and forth. It could cost people some money, and that's how you make money in math, by converting back and forth. I'll see you at the next lecture.
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet: