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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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