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: 13

Answer

$$L = \frac{4}{3}$$

Work Step by Step

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