Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.8 Length of Curves - 11.8 Exercises - Page 838: 12

Answer

$$L = 8\pi $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {4\cos 3t,4\sin 3t} \right\rangle ,\,\,\,\,\,{\text{for }}0 \leqslant t \leqslant 2\pi /3 \cr & {\text{find }}{\bf{r}}'\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {4\cos 3t,4\sin 3t} \right\rangle \cr & {\text{differentiating}} \cr & {\bf{r}}'\left( t \right) = \left\langle { - 12\sin 3t,12cos3t} \right\rangle \cr & {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr & {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr & L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr & {\text{then}} \cr & \,{\text{for }}0 \leqslant t \leqslant 2\pi /3 \to a = 0{\text{ and }}b = 2\pi /3.{\text{ then}} \cr & L = \int_0^{2\pi /3} {\left| {\left\langle { - 12\sin 3t,12cos3t} \right\rangle } \right|} dt \cr & L = \int_0^{2\pi /3} {\sqrt {{{\left( { - 12\sin 3t} \right)}^2} + {{\left( {12cos3t} \right)}^2}} } dt \cr & {\text{simplifying}} \cr & L = \int_0^{2\pi /3} {\sqrt {144{{\sin }^2}3t + 144{{\cos }^2}3t} } dt \cr & L = \int_0^{2\pi /3} {\sqrt {144\left( {{{\sin }^2}3t + {{\cos }^2}3t} \right)} } dt \cr & L = \int_0^{2\pi /3} {\sqrt {144\left( 1 \right)} } dt \cr & L = \int_0^{2\pi /3} {12} dt \cr & {\text{integrating}} \cr & L = \left( {12t} \right)_0^{2\pi /3} \cr & L = 12\left( {\frac{{2\pi }}{3} - 0} \right) \cr & L = 8\pi \cr} $$
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