## Calculus 8th Edition

Published by Cengage

# Chapter 10 - Parametric Equations and Polar Coordinates - 10.3 Polar Coordinates - 10.3 Exercises - Page 706: 13

#### Answer

$$2\sqrt7$$

#### Work Step by Step

Firstly we will have to convert polar points to Cartesian $(2, \pi/3)$ Thus, $x = rcosθ = 2\times cos(\pi/3) = 2\times \frac{1}{2} = 1$ $y = rsinθ = 2\times sin(\pi/3) = 2\times \frac{\sqrt 3}{2}= \sqrt 3$ Therefore we have First point $(x_1,y_1)=(1, \sqrt 3)$ $x = rcosθ = 4\times cos(2\pi/3) = 4\times \frac{-1}{2} = -2$ $y = rsinθ = 4\times sin(2\pi/3) = 4\times \sqrt 3/2 = 2\sqrt 3$ Thus, $(x_2,y_2)=(-1, 2\sqrt 3)$ Distance Formula, D = $\sqrt{ (x_2-x_1)^{2} + (y_2-y_1)^{2}}$ $=\sqrt {(-1-1)^{2} + (2\sqrt3-\sqrt3)^{2}}$ Hence, $$D=2 \sqrt 7$$

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