Calculus (3rd Edition)

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

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.2 Arc Length and Speed - Exercises - Page 611: 21


$$ \frac{d s}{d t}\bigg|_{ \text{min}}=\sqrt{4.89} $$

Work Step by Step

Since \begin{aligned} \frac{d s}{d t} &=\sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}} \\ &=\sqrt{9 t^{4}-24 t^{2}+16+4 t^{2}} \\ &=\sqrt{9 t^{4}-20 t^{2}+16} \end{aligned} Now, we will find the minimal speed, $$f(t) =\sqrt{9 t^{4}-20 t^{2}+16} $$ Then $$ f'(t) =\frac{36 t^{3}-40 t }{2\sqrt{9 t^{4}-20 t^{2}+16} }$$ Hence $ f'(t) =0 $ at $t=0 $ and $t= \sqrt{\frac{10}{9}}$ then $$\frac{d s}{d t}\bigg|_{t=0}= 4,\ \ \frac{d s}{d t}\bigg|_{t=0}=\sqrt{4.89}$$ Hence $$ \frac{d s}{d t}\bigg|_{ \text{min}}=\sqrt{4.89} $$
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