Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 7 - Applications of Integration - 7.4 Exercises - Page 473: 10

Answer

$\approx28.794$

Work Step by Step

The formula for the arc length is $s=\int_{a}^b \sqrt{1 + (y')^2} dy$ Differentiate $y = \frac{3}{2}x^\frac{3}{2} + 4$ with respect to x. $y' = \frac{1}{\sqrt[3]{x}}$ Substitute the value of y' in $1 + (y')^2$ $1+(y')^2=1+(\frac{1}{\sqrt[3]{x}})^2$ Substitute the value of $1 + (y')^2$ in the formula for the arc length $s=\int_{a}^b \sqrt{1 + (y')^2} dy$ Solve for the distance s $s=\int_{1}^{27}\sqrt{1+(\frac{1}{\sqrt[3]{x}})^2} dx$ $=\int_{1}^{27}(\frac{1}{x^\frac{2}{3}}+1)^\frac{3}{2}xdx$ $\approx28.794$
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