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: 11

Answer

$$L = 168$$

Work Step by Step

$$\eqalign{ & y = \frac{1}{3}{x^{3/2}}{\text{ on }}\left[ {0,60} \right] \cr & {\text{Definition of Arc Length for }}y = f\left( x \right): \cr & {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr & {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr & {\text{Notice that }}y = f\left( x \right) = \frac{1}{3}{x^{3/2}}{\text{ and }}\left[ {0,60} \right] \to a = 0{\text{ and }}b = 60.{\text{ then}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{3}{x^{3/2}}} \right] \cr & f'\left( x \right) = \frac{1}{3}\left( {\frac{3}{2}{x^{1/2}}} \right) \cr & f'\left( x \right) = \frac{1}{2}{x^{1/2}} \cr & {\text{Using the arc length formula}}{\text{, we have}} \cr & L = \int_0^{60} {\sqrt {1 + {{\left( {\frac{1}{2}{x^{1/2}}} \right)}^2}} } dx \cr & {\text{simplifying}} \cr & L = \int_0^{60} {\sqrt {1 + \frac{1}{4}x} } dx \cr & L = \int_0^{60} {\sqrt {\frac{{4 + x}}{4}} } dx \cr & L = \frac{1}{2}\int_0^{60} {\sqrt {x + 4} } dx \cr & L = \frac{1}{2}\int_0^{60} {{{\left( {x + 4} \right)}^{1/2}}} dx \cr & {\text{integrate and evaluate}} \cr & L = \frac{1}{2}\left( {\frac{{{{\left( {x + 4} \right)}^{3/2}}}}{{3/2}}} \right)_0^{60} \cr & L = \frac{1}{3}\left( {{{\left( {x + 4} \right)}^{3/2}}} \right)_0^{60} \cr & L = \frac{1}{3}\left( {{{\left( {60 + 4} \right)}^{3/2}} - {{\left( {0 + 4} \right)}^{3/2}}} \right) \cr & L = \frac{1}{3}\left( {512 - 8} \right) \cr & L = 168 \cr} $$
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