Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.3 Exercises - Page 688: 58

Answer

The slope of the tangent line to the given polar curve when $\theta = \pi$ is equal to: $$-\sqrt 3$$

Work Step by Step

1. Write the formula for the slope of the tangent line: $$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}sin(\theta) + rcos(\theta)}{\frac{dr}{d\theta}cos(\theta) - rsin(\theta)} $$ 2. Calculate $\frac{dr}{d\theta}$ and substitute the equation into the formula: $\frac{dr}{d\theta} = \frac{d(cos(\theta/3))}{d\theta} = -\frac{1}{3}sin(\theta/3)$ $\frac{dy}{dx} = \frac{(-\frac{1}{3}sin(\theta/3))sin(\theta) + (cos(\theta/3))cos(\theta)}{(-\frac{1}{3}sin(\theta/3))cos(\theta) - (cos(\theta/3))sin(\theta)} $ 3. Calculate the slope when $\theta = \pi$ $\frac{dy}{dx} = \frac{(-\frac{1}{3}sin(\pi/3))sin(\pi) + (cos(\pi/3))cos(\pi)}{(-\frac{1}{3}sin(\pi/3))cos(\pi) - (cos(\pi/3))sin(\pi)} = \frac{(-\frac{1}{3}\frac{\sqrt 3}{2})(0) + (\frac 1 2)(-1)}{(-\frac{1}{3}\frac{\sqrt 3}{2})(-1) - (\frac 12)(0)}$ $\frac {dy}{dx} = \frac{-\frac 12}{\frac {\sqrt 3}6} = - \frac {3}{\sqrt 3} = -\sqrt 3$
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