Chapter 14 - Calculus of Vector-Valued Functions - 14.3 Arc Length and Speed - Exercises - Page 725: 4

$\frac{8}{27}((1+(9/2)\pi)^{3/2}-1)$

Work Step by Step

We have $r(t) =\lt \cos t,\sin t, t^{3/2}\gt$, and so the vector derivative is $r'(t) =\lt -\sin t,\cos t, \frac{3}{2} t^{1/2}\gt$. We calculate the length by integration: $$length=\int_{0}^{2\pi}\sqrt{\sin^2 t+\cos^2t+(9/4)t} \ dt=\int_{0}^{2\pi}\sqrt{1+(9/4)t} \ dt=\frac{4}{9(3/2)}(1+(9/4)t)^{3/2}|_{0}^{2\pi}=\frac{8}{27}((1+(9/2)\pi)^{3/2}-1).$$

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