Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.3 Arc Length and Speed - Exercises - Page 725: 5



Work Step by Step

We have $r(t) =\lt t,4 t^{3/2}, 2t^{3/2}\gt$, and so the vector derivative is $r'(t) =\lt 1, 6 t^{1/2}, 3t^{1/2}\gt$. We calculate the length by integration: $$length=\int_{0}^{3}\sqrt{1+36t+9t} dt=\int_{0}^{3}\sqrt{1+45t} dt=\frac{2}{45(3)}(1+45t)^{3/2}|_{0}^{3}=\frac{2}{135}((136)^{3/2}-1).$$
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