Ta C.
asked • 08/19/20Write the point (3, pi/6) with the following conditions: a. r < 0, theta > 0 b. r > 0, theta < 0 c. r > 0, theta > 0
Write the point (3, pi/6) with the following conditions:
a. r < 0, theta > 0
b. r > 0, theta < 0
c. r > 0, theta > 0
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1 Expert Answer
Mike D. answered • 08/19/20
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Ta
(c) is obvious - r=3 theta = pi/6
(b) r = 3. For a negative theta we go clockwise starting from the x axis , as a whole revolution is 2 pi, we need to go 2 pi - pi / 6. So r = 3, theta = - (2 pi - pi /6)
(a) r<0. This means theta = pi + pi/6, and r = -3. BECAUSE -3 units down the line theta = pi + pi/6 is the same as +3 units up the line theta = pi/6
Mike
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Joseph C.
I find this question confusing, because the point (3,pi/6) is already implies that r>0 and theta>0. This is because the (3,pi/6) -> (r,pi/6). In fact, in polar coordinates r<0 is not a valid value.08/19/20