Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 49: 89

Answer

$$\eqalign{ & {\text{amplitude: 3}} \cr & {\text{period: 6}}\pi \cr} $$

Work Step by Step

$$\eqalign{ & g\left( \theta \right) = 3\cos \left( {\frac{\theta }{3}} \right) \cr & {\text{The functions of the form }}y = A\cos \left( {B\left( {\theta - C} \right)} \right) + D{\text{ have a vertical}} \cr & {\text{stretch }}\left( {{\text{or }}{\bf{amplitude}}} \right){\text{ of }}\left| A \right|,{\text{ and a period of }}\frac{{2\pi }}{{\left| B \right|}} \cr & {\text{Therefore, rewriting the function }}g\left( \theta \right) = 3\cos \left( {\frac{\theta }{3}} \right){\text{ we obtain}} \cr & \underbrace {g\left( \theta \right) = 3\cos \left( {\frac{1}{3}\left( {\theta - 0} \right)} \right) + 0}_{y = A\cos \left( {B\left( {\theta - C} \right)} \right) + D} \cr & A = 3,{\text{ }}B = \frac{1}{3},{\text{ }}C = 0,{\text{ }}D = 0 \cr & {\text{The amplitude is: }}\left| A \right| = 3 \cr & {\text{Period: }}\frac{{2\pi }}{{\left| B \right|}} = \frac{{2\pi }}{{\left| {1/3} \right|}} = 6\pi \cr & \cr & {\text{amplitude: 3}} \cr & {\text{period: 6}}\pi \cr} $$
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