Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.7 Exercises - Page 381: 56

Answer

$$y = 2{\sin ^{ - 1}}\left( {\frac{x}{5}} \right)$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{2}{{\sqrt {25 - {x^2}} }} \cr & {\text{Separate the variables}} \cr & dy = \frac{2}{{\sqrt {25 - {x^2}} }}dx \cr & {\text{Integrate both sides}} \cr & \int {dy} = \int {\frac{2}{{\sqrt {25 - {x^2}} }}} dx \cr & y = 2{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) + C{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}\left( {5,\pi } \right) \cr & \pi = 2{\sin ^{ - 1}}\left( {\frac{5}{5}} \right) + C \cr & \pi = 2\left( {\frac{\pi }{2}} \right) + C \cr & C = 0 \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & y = 2{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) \cr & \cr & {\text{Graph}} \cr} $$
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