University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 27



Work Step by Step

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Here, $\dfrac{dx}{dt}=t$ and $\dfrac{dy}{dt}=(2t+1)^{1/2}$ Thus, $S=\int_{0}^{4} \sqrt{(t)^2+((2t+1)^{1/2})^2} dt=\int_{0}^{4} (t+1) dt$ Thus, $S=[\dfrac{t^{2}}{2}+t]_{0}^{4} =\dfrac{16}{2}+4=12$
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