Polar Equation of a CircleDate: 04/12/97 at 16:26:52 From: Frank P. Subject: Polar Equation of a Circle I was wondering if you could provide for me a description of the following polar equations for a circle. r = a*sin (theta) and r = a*cos (theta). What influence does a have on the graph? What influence does theta have on the graph? What's the difference in using sin or cos, and how do they change the graph? Date: 04/12/97 at 17:34:07 From: Doctor Mitteldorf Subject: Re: Polar Equation of a Circle Dear Frank, You are familiar with writing an equation in x and y and then plotting the curve that represents all points obeying that equation. Well, you can do the same thing using r and theta instead of x and y. In fact, r and theta are related to x and y by x = r*cos(theta) and y = r*sin(theta). Equivalently, r^2 = x^2 + y^2 and tan(theta) = y/x. Another way to write this second equation is sin(theta) = y/sqrt(x^2+y^2). The obvious way to write the equation for a circle in polar coordinates is r = a. Then theta can be anything. This makes a circle of radius a around the origin as center. The equation you've written, r = a sin(theta), is also a circle, but it's not so obvious, and it's not centered about the origin. It's centered around another point. In a few minutes, you should be prepared to answer what that point is, and what the radius of the circle is. >what influence does a have on the graph? We should be able to answer this in a minute >what influence does theta have on the graph? This isn't a meaningful question. Theta is one of the variables in the equation, like x and y were in the cartesian equation. To see that you have a circle, and to see where the center and the radius are, just substitute into your equation r=a sin(theta). Let r = sqrt(x^2+y^2) and let sin(theta) = y/sqrt(x^2+y^2). Rearrange the equation and you'll have the form x^2 + (y-a/2)^2 = (a/2)^2. When you see the equation in that form, it should be clear what the radius is and where the center is. Let me know if this is not clear, please. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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