Polar EquationsDate: 05/05/2003 at 11:21:18 From: Derek Subject: Polar equations If you have r^2 = 9*cos(theta) and r = 1 + cos(theta), how many points of intersection are there? If I attempt to use simultaneous equations, by splitting the first equation into two: r = -(9*cos(theta))^0.5 and r = (9*cos(theta))^0.5, I then have the following: 1 + cos(theta) = -(9*cos(theta))^0.5 and 1 + cos(theta) = (9*cos(theta))^0.5 I cannot square both sides to get rid of the root, because I end up with two identical equations and only two intercepts, 1.424 and (2pi - 1.424). When I draw the two graphs using a software package called Autograph it clearly shows 4 intercepts. Please help. Date: 05/05/2003 at 12:40:54 From: Doctor Luis Subject: Re: Polar equations Hi Derek, You weren't doing anything wrong by ending up with a single equation. You want to solve for values of t that satisfy the original two equations. As you noted, by squaring both sides you end up with the same equation as if you had substituted for r in the first equation. (1 + cos(t))^2 = 9*cos(t) For the purposes of solving this problem, let x = cos(t). We must satisfy the quadratic equation (1 + x)^2 = 9*x 1 + 2*x + x^2 = 9x or x^2 - 7*x + 1 = 0 This gives you two values of x using the quadratic equation (i.e. two distinct real roots x_1, x_2). However, one of roots lies outside the range [-1, 1]. If you recall, the cosine of a real number cannnot exceed 1, so the arccos is not defined (well, technically, it's an imaginary number). So one of the roots (say x_2 was the one with x_2 > 1) is bad. The good root x_1 that's in the range [-1, 1] satisfies x_1 = cos(t) or t = arccos(x_1) There are two values of t that satisfy that equation. You can see this from the identity cos(2pi - t) = cos(t). So, the only two values of t are t = arccos(x_1), 2pi - arccos(x_1) And that's where they intersect (to find r, find r(t) for these t). I'm not sure how you did your graph, but here's mine, done in Maple 8: As you can see, there are only two intersections. The origin looks like a solution, but it doesn't satisfy the equations 0^2 = 9*cos(t), 0 = 1 + cos(t). (One says cos(t) = 0, the other cos(t) = -1.) I hope this helped. Let us know if you have any other questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ Date: 05/05/2003 at 15:43:02 From: Derek Subject: Polar equations Many thanks for your prompt reply. I believe I was misled by a lesser software package than Maple. My software showed a dashed shape on the left-hand side of the graph, symmetrical to the right-hand side of your graph (r^2 = 9cos(theta)). I assume, now, the dashed lines meant that the values for r were negative(?) and in reality lay on top of the right-hand side. In that case am I right in saying that the two equations: 1 + cos (theta) = -(9*cos(theta))^0.5 and 1 + cos (theta) = +(9*cos(theta))^0.5 have the same real solutions and no others? You mentioned that values outside -1 to 1 are not real. Can I do anything with these using complex numbers? Again many thanks for your help. Derek Date: 05/06/2003 at 18:49:52 From: Doctor Luis Subject: Re: Polar equations Hi Derek, When the value of r becomes negative you add pi radians to the angle theta before graphing. That's why you saw the dashed lines in your program. In general, you want to avoid taking square roots. You'll be introducing extraneous solutions by doing that. It's better to square and then just use the quadratic formula. r^2 = 9*cos(theta) r = 1 + cos(theta) The REAL problem (pun intended, heh) with r^2 = 9*cos(theta) is that theta can only range from -pi to pi. It is forbidden from ever being in the range from pi to 3pi/2. As soon as theta enters that range, cos(theta) becomes negative, which would mean that r^2 is negative, or that r is imaginary. By taking the square root, you are making a reflected image on the (until now) forbidden left half of the plane. If you want those "solutions" too, simply add or subtract pi radians from the two original solutions. You can see in this diagram (or from your program) that the solutions are indeed diametrically opposite (180 degrees apart): To find the respective radii, just use r = 1 + cos(theta). As for the complex numbers, I don't think they'll be of much use in your diagram. They usually represent a *lack* of intersections. For example, when graphing the function y = x^2 + 1, you can see that it doesn't cross the x-axis at all. When you try to find an x-intercept by setting y = 0, you'll find imaginary roots x = +i, -i. Therefore, imaginary roots represent the non-existence of an intersection. I hope this helped you understand the problem better. Let us know if you have any other questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
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