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: 16

Answer

$$L = 30\pi $$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {4\cos t,4\sin t,3t} \right\rangle ,\,\,\,\,\,{\text{for 0}} \leqslant t \leqslant 6\pi \cr & {\text{find }}{\bf{r}}'\left( t \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {4\cos t,4\sin t,3t} \right\rangle \cr & {\text{differentiating}} \cr & {\bf{r}}'\left( t \right) = \left\langle { - 4\sin t,4\cos t,3} \right\rangle \cr & {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr & {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr & L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr & {\text{then}} \cr & \,{\text{for 0}} \leqslant t \leqslant 6\pi \to a = 0{\text{ and }}b = 6\pi .{\text{ then}} \cr & L = \int_0^{6\pi } {\left| {\left\langle { - 4\sin t,4\cos t,3} \right\rangle } \right|} dt \cr & L = \int_0^{6\pi } {\sqrt {{{\left( { - 4\sin t} \right)}^2} + {{\left( {4\cos t} \right)}^2} + {{\left( 3 \right)}^2}} } dt \cr & {\text{simplifying}} \cr & L = \int_0^{6\pi } {\sqrt {16{{\sin }^2}t + 16{{\cos }^2}t + 9} } dt \cr & L = \int_0^{6\pi } {\sqrt {16\left( {{{\sin }^2}t + {{\cos }^2}t} \right) + 9} } dt \cr & L = \int_0^{6\pi } {\sqrt {16 + 9} } dt \cr & L = \int_0^{6\pi } 5 dt \cr & {\text{integrating}} \cr & L = 5\left( t \right)_0^{6\pi } \cr & L = 5\left( {6\pi - 0} \right) \cr & L = 30\pi \cr} $$
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