Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Exercises - Page 676: 41

Answer

$L = 4 \sqrt 2 - 2 $

Work Step by Step

1. Write the formula for the length "$L$"of a curve: $$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$ 2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ $\frac{dx}{dt} = \frac{d(1 + 3t^2)}{dt} = 6t$ $\frac{dy}{dt} = \frac{d(4 + 2t^3)}{dt} = 6t^2$ 3. Substitute the values into the definite integral: $L = \int_{\alpha}^{\beta} \sqrt{(6t)^2 + (6t^2)^2} \space dt$ $L = \int_{\alpha}^{\beta} \sqrt{36t^2 + 36t^4} \space dt$ $L = \int_{\alpha}^{\beta} \sqrt{36t^2( 1 + t^2)} \space dt$ $L = \int_{\alpha}^{\beta} 6t \sqrt{( 1 + t^2)} \space dt$ $u = 1 + t^2$ $\frac{du}{dt} = 2t \longrightarrow \frac {du}{2 \space dt} =t$ $L = \int_{\alpha}^{\beta} 6(\frac{du}{2 \space dt}) \sqrt u \space dt$ $L = \int_{\alpha}^{\beta} 3 \sqrt u \space du$ $L = 3 \int_{\alpha}^{\beta} \sqrt u \space du$ - Since $t$ goes from 0 to 1: $u_1 = 1 + 0^2 = 1$ $u_2 = 1 + 1^2 = 2$ $u$ goes from 1 to 2 $L = 3 \int_{1}^{2} \sqrt u \space du$ $L = 3 (\frac 2 3 u^{3/2})]^2_1$ $L = (2 u^{3/2})]^2_1$ $L = ( (2 \times 2^{3/2}) - (2 \times 1^{3/2})) $ $L = 4 \sqrt 2 - 2 $
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