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: 16

Answer

$$L = \frac{{55}}{3}$$

Work Step by Step

$$\eqalign{ & y = \frac{2}{3}{x^{3/2}} - \frac{1}{2}{x^{1/2}}{\text{ on }}\left[ {1,9} \right] \cr & \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{ 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{2}{3}{x^{3/2}} - \frac{1}{2}{x^{1/2}}{\text{ and }}\left[ {1,9} \right] \to a = 1{\text{ and }}b = 9.{\text{ then}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{2}{3}{x^{3/2}} - \frac{1}{2}{x^{1/2}}} \right] \cr & f'\left( x \right) = \frac{2}{3}\left( {\frac{3}{2}{x^{1/2}}} \right) - \frac{1}{2}\left( {\frac{1}{2}{x^{ - 1/2}}} \right) \cr & f'\left( x \right) = {x^{1/2}} - \frac{1}{4}{x^{ - 1/2}} \cr & {\text{Using the arc length formula}}{\text{, we have}} \cr & L = \int_1^9 {\sqrt {1 + {{\left( {{x^{1/2}} - \frac{1}{4}{x^{ - 1/2}}} \right)}^2}} } dx \cr & {\text{simplifying}} \cr & L = \int_1^9 {\sqrt {1 + {{\left( {{x^{1/2}}} \right)}^2} - 2\left( {{x^{1/2}}} \right)\left( {\frac{1}{4}{x^{ - 1/2}}} \right) + {{\left( {\frac{1}{4}{x^{ - 1/2}}} \right)}^2}} } dx \cr & L = \int_1^9 {\sqrt {1 + {{\left( {{x^{1/2}}} \right)}^2} - \frac{1}{2} + {{\left( {\frac{1}{4}{x^{ - 1/2}}} \right)}^2}} } dx \cr & L = \int_1^9 {\sqrt {{{\left( {{x^{1/2}}} \right)}^2} + \frac{1}{2} + {{\left( {\frac{1}{4}{x^{ - 1/2}}} \right)}^2}} } dx \cr & {\text{factoring }} \cr & L = \int_1^9 {\sqrt {{{\left( {{x^{1/2}} + \frac{1}{4}{x^{ - 1/2}}} \right)}^2}} } dx \cr & L = \int_1^9 {\left( {{x^{1/2}} + \frac{1}{4}{x^{ - 1/2}}} \right)} dx \cr & {\text{integrate and evaluate}} \cr & L = \left( {\frac{{{x^{3/2}}}}{{3/2}} + \frac{1}{4}\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right)} \right)_1^9 \cr & L = \left( {\frac{{2{x^{3/2}}}}{3} + \frac{{{x^{1/2}}}}{2}} \right)_1^9 \cr & L = \left( {\frac{{2{{\left( 9 \right)}^{3/2}}}}{3} + \frac{{{{\left( 9 \right)}^{1/2}}}}{2}} \right) - \left( {\frac{{2{{\left( 1 \right)}^{3/2}}}}{3} + \frac{{{{\left( 1 \right)}^{1/2}}}}{2}} \right) \cr & L = \frac{{39}}{2} - \frac{7}{6} \cr & L = \frac{{55}}{3} \cr} $$
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