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: 56

Answer

$$\eqalign{ & \left( a \right)L\left( x \right) = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}\left( {x - \frac{\pi }{4}} \right) \cr & \left( b \right){\text{graph}} \cr & \left( c \right)0.696781 \cr & \left( d \right)0.0001\% {\text{ error}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \cos x;{\text{ }}a = \frac{\pi }{4};{\text{ }}\cos \left( {0.8} \right) \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos x} \right] \cr & f'\left( x \right) = - \sin x \cr & {\text{Evaluate }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ at }}a = \frac{\pi }{4} \cr & f\left( 0 \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \cr & f'\left( 0 \right) = - \sin \left( {\frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2} \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{Substitute }}f\left( a \right){\text{ and }}f'\left( a \right){\text{ into }}\left( {\bf{1}} \right) \cr & L\left( x \right) = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}\left( {x - \frac{\pi }{4}} \right) \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = 0{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}\cos \left( {0.8} \right) \cr & L\left( x \right) = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}\left( {x - \frac{\pi }{4}} \right) \cr & L\left( {0.8} \right) = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}\left( {0.8 - \frac{\pi }{4}} \right) \cr & L\left( {0.8} \right) \approx 0.696781 \cr & \cr & {\text{Therefore}}{\text{,}} \cr & \cos \left( {0.8} \right) \approx L\left( {0.8} \right) \cr & \cos \left( {0.8} \right) \approx 0.696781 \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 & \cos \left( {0.8} \right) = 0.69670670 \cr & \cr & \frac{{\left| {0.696781 - 0.69670670} \right|}}{{0.69670670}} \times 100\% = 0.0001\% {\text{ error}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.