Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.4 - Curvature and Normal Vectors of a Curve - Exercises 13.4 - Page 765: 1

Answer

$${\bf{T}} = \cos t{\bf{i}} - \sin t{\bf{j}},\,\,\,\,\,\,\,\,\,{\bf{N}} = - \sin t{\bf{i}} - \cos t{\bf{j}}\,,\,\,\,\,\,\,\,\,\,\kappa = \cos t$$

Work Step by Step

$$\eqalign{ & {\bf{r}}\left( t \right) = t{\bf{i}} + \left( {\ln \cos t} \right){\bf{j}},\,\,\,\,\,\,\,\, - \pi /2 < t < \pi /2 \cr & {\text{Calculate }}{\bf{v}}\left( t \right).{\text{ Use }}{\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + \left( {\ln \cos t} \right){\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = {\bf{i}} + \frac{{ - \sin t}}{{\cos t}}{\bf{j}} \cr & {\bf{v}}\left( t \right) = {\bf{i}} - \tan t{\bf{j}} \cr & \cr & {\text{We calculate }}{\bf{T}}{\text{ from the velocity vector}} \cr & {\bf{v}}\left( t \right) = {\bf{i}} - \tan t{\bf{j}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - \tan t} \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {1 + {{\tan }^2}t} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\sec }^2}t} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sec t \cr & \cr & {\text{use }}{\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} \cr & {\bf{T}} = \frac{{{\bf{i}} - \tan t{\bf{j}}}}{{\sec t}} \cr & {\bf{T}} = \frac{1}{{\sec t}}{\bf{i}} - \frac{{\tan t}}{{\sec t}}{\bf{j}} \cr & {\bf{T}} = \cos t{\bf{i}} - \sin t{\bf{j}} \cr & \cr & {\text{Calculate }}{\bf{N}}\left( t \right){\text{ using the equation }}{\bf{N}} = \frac{{d{\bf{T}}/dt}}{{\left| {d{\bf{T}}/dt} \right|}}{\text{ }}\left( {{\text{see page 764}}} \right).{\text{ Then}} \cr & \frac{{d{\bf{T}}}}{{dt}} = \frac{d}{{dt}}\left( {\cos t{\bf{i}} - \sin t{\bf{j}}} \right) \cr & \frac{{d{\bf{T}}}}{{dt}} = - \sin t{\bf{i}} - \cos t{\bf{j}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {{{\left( { - \sin t} \right)}^2} + {{\left( { - \cos t} \right)}^2}} = 1 \cr & {\bf{N}} = \frac{{d{\bf{T}}/dt}}{{\left| {d{\bf{T}}/dt} \right|}} = \frac{{ - \sin t{\bf{i}} - \cos t{\bf{j}}}}{1} \cr & {\bf{N}} = - \sin t{\bf{i}} - \cos t{\bf{j}} \cr & \cr & {\text{Calculate }}\kappa {\text{ using the equation }}\kappa = \frac{1}{{\left| {\bf{v}} \right|}}\left| {\frac{{d{\bf{T}}}}{{dt}}} \right|{\text{ }}\left( {{\text{see page 764}}} \right).{\text{ Then}} \cr & \kappa = \frac{1}{{\sec t}}\left( 1 \right) \cr & \kappa = \cos t \cr} $$
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