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: 6


$\frac{136 \sqrt{17}-128 \sqrt{2}}{27}\approx 14.1$

Work Step by Step

We are given the vector: $r(t)=\lt 2t^2+1,2t^2-1,t^3 \gt$ Recall that $(x^n)'=nx^{n-1}$ Thus, we get the vector derivative: $r'(t) =\lt 4t, 4t, 3t^2\gt$. We calculate the length by integration: $$length=\int_{0}^{2}\|r'(t)\|dt=\int_{0}^{2}\sqrt{16t^2+16t^2+9t^4} dt\\=\frac{1}{18}\int_{0}^{2}18t\sqrt{32 +9t^2} dt=\frac{2}{18(3)}(32+9t^2)^{3/2}|_{0}^{2}\\=\frac{1}{27}((68)^{3/2}-(32)^{3/2})=\frac{136 \sqrt{17}-128 \sqrt{2}}{27}\approx 14.1.$$
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