Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 14

Answer

$$\eqalign{ & \left( a \right)L\left( x \right) = \frac{{\sqrt 2 }}{2}x - \frac{{\sqrt 2 }}{8}\pi + \frac{{\sqrt 2 }}{2} \cr & \left( b \right){\text{graph}} \cr & \left( c \right)0.6820 \cr & \left( d \right)0.053\% {\text{ error}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \sin x,{\text{ }}a = \frac{\pi }{4}{\text{ at }}f\left( {0.75} \right) \cr & \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x} \right] \cr & f'\left( x \right) = \cos x \cr & \cr & \left( a \right){\text{Use the linear approximation formula }}\left( {{\text{See page 287}}} \right) \cr & f\left( x \right) = L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Evaluate }}f\left( a \right){\text{ and }}f'\left( a \right) \cr & f\left( a \right) = f\left( {\frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \cr & f'\left( a \right) = f'\left( {\frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \cr & {\text{Substitute }}f\left( a \right){\text{ and }}f'\left( a \right){\text{ into }}\left( {\bf{1}} \right) \cr & f\left( x \right) = L\left( x \right) = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}\left( {x - \frac{\pi }{4}} \right) \cr & L\left( x \right) = \frac{{\sqrt 2 }}{2}x - \frac{{\sqrt 2 }}{8}\pi + \frac{{\sqrt 2 }}{2} \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = \frac{\pi }{4}{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}f\left( {0.75} \right) \cr & L\left( {0.75} \right) = \frac{{\sqrt 2 }}{2}\left( {0.75} \right) - \frac{{\sqrt 2 }}{8}\pi + \frac{{\sqrt 2 }}{2} \cr & L\left( {0.75} \right) \approx 0.6820 \cr & \cr & \left( d \right){\text{ The percent error is:}} \cr & \frac{{\left| {{\text{approximation}} - {\text{exact}}} \right|}}{{{\text{exact}}}} \times 100\% \cr & {\text{The exact value given by a calculator is }} \cr & f\left( {0.75} \right) = \sin \left( {0.75} \right) = 0.68163876 \cr & f\left( {0.75} \right) = 0.68163876,{\text{ then}} \cr & \frac{{\left| {0.6820 - 0.68163876} \right|}}{{0.68163876}} \times 100\% = 0.053\% {\text{ error}} \cr} $$
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