Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.1 - The Unit Circle - 5.1 Exercises - Page 408: 61

Answer

$$P(x,y) = \Bigg(\frac{\sqrt 3}{2}, \frac 12\Bigg)$$

Work Step by Step

Going from Q to R using the arc of the circle, we have a total of $$\frac{\pi}{2} + \frac{\pi}{6} = \frac{4\pi}{6}$$ Thus, the middle point in this arc is $\frac{4\pi}{6} \div 2 = \frac{\pi}{3}$ away from Q and R. The segment with the red arrows has a total length of $$\frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$$. Therefore, $P$ is the midpoint of the arc between $R$ and $Q$. And since the unit circle is symmetrical, the distances PQ and PR are the same. $$PQ = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2}$$ $$PQ = \sqrt{(x - x)^2 + (y - (-y))^2}$$ $$PQ = \sqrt{(0)^2 + (2y)^2} = \sqrt {(2y)^2}$$ $$PQ = 2y$$ $$PR = \sqrt{(x_P - x_R)^2 + (y_P - y_R)^2}$$ $$PR = \sqrt{(x - 0)^2 + (y - 1)^2}$$ $$PR = \sqrt{x^2 + (y - 1)^2}$$ Therefore: $$2y = \sqrt{x^2 + (y-1)^2}$$ $$2y = \sqrt{x^2 + y^2 - 2y + 1}$$ $$2y = \sqrt{2 - 2y}$$ $$(2y)^2 = (\sqrt{2 - 2y})^2$$ $$4y^2 = 2 - 2y$$ $$4y^2 + 2y - 2 =0$$ $$2y^2 + y - 1 = 0$$ $$y = \frac{-(1) \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}$$ Since $P$ is in Quadrant I, the y-coordinate value must be positive. $$y = \frac{-(1) + \sqrt{9}}{4} = \frac{-1 + 3}{4} = \frac{1}{2}$$ $$x^2 + y^2 = 1$$ $$x^2 + (\frac 12)^2 = 1$$ $$x^2 = 1- \frac 14 = \frac 34$$ $$x = \pm \sqrt{\frac{3}4}$$ Since $P$ is in Quadrant I, the x-coordinate value must be positive. $$x = \frac{\sqrt 3}{2}$$
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