Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.5 Length of Curves - 6.5 Exercises - Page 450: 25

Answer

$$\left. a \right)L = \int_1^{10} {\sqrt {1 + \frac{1}{{{x^4}}}} } dx,\,\,\,\,\left. b \right)\,\,L \approx 9.15$$

Work Step by Step

$$\eqalign{ & y = \frac{1}{x}{\text{ on the interval }}\left[ {1,10} \right] \cr & {\text{Definition of Arc Length for }}y = f\left( x \right): \cr & {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ }} \cr & {\text{The length of the curve }} \cr & {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr & {\text{Notice that }}y = f\left( x \right) = \frac{1}{x}{\text{ and }} \cr & \left[ {1,10} \right] \to a = 1{\text{ and }}b = 10.{\text{ Then}} \cr & \cr & \left. a \right)\,\,f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr & f'\left( x \right) = - \frac{1}{{{x^2}}} \cr & {\text{Using the arc length formula}}{\text{, we have}} \cr & L = \int_1^{10} {\sqrt {1 + {{\left( { - \frac{1}{{{x^2}}}} \right)}^2}} } dx \cr & L = \int_1^{10} {\sqrt {1 + \frac{1}{{{x^4}}}} } dx \cr & \cr & \left. b \right){\text{ Use technology to evaluate or approximate the integral}} \cr & L \approx 9.15 \cr} $$
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