Properties of vector addition commutative property

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Unformatted text preview: he graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us. 1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover, we know from our work there that as 0 ≤ θ ≤ π , we are tracing out the ‘leaf’ of the rose 2 which lies in the first quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through 0, π . Hence, the region we seek is the leaf itself. 2 y r 5 π 4 π 2 3π 4 π θ x −5 (r, θ) : 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π , so 6 the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π . In other words, we are 6 looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since r ≤...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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