Which Quadrant in the Unit Circle?Date: 11/30/98 at 21:57:23 From: Rosalynn Subject: The Unit Circle Hello, I'm trying to determine how to do this problem: "for each value of s, find the quadrant in which C(s) is found." I've studied the examples in my textbook. One example was: Find the quadrant in which C(s) is located s = 14pi/3 = 2pi/3 + 4pi = 2pi/3 + 2(2pi) C(14pi/3)= C(2pi/3) Thus, C(14pi/3) is in quadrant II I tried to apply this example to one of my problems, s=15pi/3, but I still cannot determine the answer... please help. Date: 12/01/98 at 13:30:35 From: Doctor Peterson Subject: Re: The Unit Circle Hi, Rosalynn. I assume your "C(s)" is the point on the unit circle at angle (or arc length) s from the point (1,0). Here's what you need to remember to do this: Each quadrant takes up pi/2 radians; that is, for every pi/2 that s increases, you go into a new quadrant. After four quadrants (4 * pi/2 = 2 pi), you come back where you started, since you've gone all the way around the circle. So what I would do is to divide s by 2 pi to get the number of times you went around the whole circle, and keep the remainder; then multiply the result by 4 to find the quadrant number. For example, for 14pi/3, we divide by 2 pi: 14 pi ----- 3 14 pi 1 7 ------- = ----- * --- = --- = 2 1/3 circles 2 pi 3 2pi 3 Two full circles takes us back to the starting point; how many quadrants does 1/3 circle take us through? Multiply by 4, and we get 4/3 = 1 1/3 quadrants. So we've gone entirely through the first quadrant and into the second. In terms of the explanation in your book, we can rewrite our division as a multiplication: 14 pi ----- = (2 1/3)(2 pi) = 2 (2 pi) + (1/3) (2 pi) 3 = 2 (2 pi) + 2/3 pi = 2 (2 pi) + (4/3) (pi/2) = 2 circles + 1 quadrant + a little more Your problem, 15/3 pi = 5 pi, will be a little special: 5 pi 5 ------ = --- = 2 1/2 circles 2 pi 2 This takes us through exactly 2 full circles and 2 more quadrants, so we are right on the line between quadrants II and III. We're not in a quadrant at all! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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