Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.4 Length Of A Plane Curve - Exercises Set 5.4 - Page 375: 3

Answer

$$L = \frac{{85\sqrt {85} - 8}}{{243}}$$

Work Step by Step

$$\eqalign{ & y = 3{x^{3/2}} - 1{\text{ from }}x = 0{\text{ to }}x = 1 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {3{x^{3/2}} - 1} \right] \cr & \frac{{dy}}{{dx}} = \frac{9}{2}{x^{1/2}} \cr & {\text{Use }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr & L = \int_0^1 {\sqrt {1 + {{\left( {\frac{9}{2}{x^{1/2}}} \right)}^2}} dx} \cr & L = \int_0^1 {\sqrt {1 + \frac{{81}}{4}x} dx} \cr & L = \frac{1}{2}\int_0^1 {\sqrt {4 + 81x} dx} \cr & {\text{Integrating}} \cr & L = \frac{1}{{2\left( {81} \right)}}\left[ {\frac{{2{{\left( {4 + 81x} \right)}^{3/2}}}}{3}} \right]_0^1 \cr & L = \frac{1}{{243}}\left[ {{{\left( {4 + 81} \right)}^{3/2}} - {{\left( {4 + 0} \right)}^{3/2}}} \right] \cr & L = \frac{{{{85}^{3/2}} - 8}}{{243}} \cr & L = \frac{{85\sqrt {85} - 8}}{{243}} \cr} $$
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.