Calculus (3rd Edition)

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

Chapter 14 - Calculus of Vector-Valued Functions - Chapter Review Exercises - Page 754: 38

Answer

(a) $\begin{array}{*{20}{c}} {{\bf{T}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{\sqrt 5 }},0,\frac{2}{{\sqrt 5 }}} \right),}\\ {{\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right),}\\ {{\bf{B}}\left( {\frac{\pi }{2}} \right) = \left( {\frac{2}{{\sqrt 5 }},0,\frac{1}{{\sqrt 5 }}} \right)} \end{array}$ (b) the equation of the osculating plane at the point corresponding to $t = \frac{\pi }{2}$ is $2x + z = \pi $

Work Step by Step

(a) We have ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,2t} \right)$. So, the tangent vector is ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t,2} \right)$. The unit tangent vector is ${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( { - \sin t,\cos t,2} \right)}}{{||\left( { - \sin t,\cos t,2} \right)||}}$ ${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {\left( { - \sin t,\cos t,2} \right)\cdot\left( { - \sin t,\cos t,2} \right)} }}\left( { - \sin t,\cos t,2} \right)$ ${\bf{T}}\left( t \right) = \frac{1}{{\sqrt {{{\sin }^2}t + {{\cos }^2}t + 4} }}\left( { - \sin t,\cos t,2} \right) = \left( { - \frac{{\sin t}}{{\sqrt 5 }},\frac{{\cos t}}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)$ So, ${\bf{T}}'\left( t \right) = \left( { - \frac{{\cos t}}{{\sqrt 5 }}, - \frac{{\sin t}}{{\sqrt 5 }},0} \right)$ At $t = \frac{\pi }{2}$, we get ${\bf{T}}'\left( {\frac{\pi }{2}} \right) = \left( {0, - \frac{1}{{\sqrt 5 }},0} \right)$. Thus, ${\bf{N}}\left( {\frac{\pi }{2}} \right) = \frac{{{\bf{T}}'\left( {\frac{\pi }{2}} \right)}}{{||{\bf{T}}'\left( {\frac{\pi }{2}} \right)||}} = \sqrt 5 \left( {0, - \frac{1}{{\sqrt 5 }},0} \right)$ ${\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right)$ The binormal vector is defined by ${\bf{B}} = {\bf{T}} \times {\bf{N}}$. So, ${\bf{B}}\left( {\frac{\pi }{2}} \right) = {\bf{T}}\left( {\frac{\pi }{2}} \right) \times {\bf{N}}\left( {\frac{\pi }{2}} \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ { - \frac{1}{{\sqrt 5 }}}&0&{\frac{2}{{\sqrt 5 }}}\\ 0&{ - 1}&0 \end{array}} \right|$ ${\bf{B}}\left( {\frac{\pi }{2}} \right) = \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{k}}$ In summary, we obtain $\begin{array}{*{20}{c}} {{\bf{T}}\left( {\frac{\pi }{2}} \right) = \left( { - \frac{1}{{\sqrt 5 }},0,\frac{2}{{\sqrt 5 }}} \right),}\\ {{\bf{N}}\left( {\frac{\pi }{2}} \right) = \left( {0, - 1,0} \right),}\\ {{\bf{B}}\left( {\frac{\pi }{2}} \right) = \left( {\frac{2}{{\sqrt 5 }},0,\frac{1}{{\sqrt 5 }}} \right)} \end{array}$ (b) In part (a) we obtain ${\bf{B}}\left( {\frac{\pi }{2}} \right) = \left( {\frac{2}{{\sqrt 5 }},0,\frac{1}{{\sqrt 5 }}} \right)$. The point corresponding to $t = \frac{\pi }{2}$ is ${\bf{r}}\left( {\frac{\pi }{2}} \right) = \left( {0,1,\pi } \right)$. Thus, by Theorem 1 of Section 13.5, the equation of the osculating plane at the point corresponding to $t = \frac{\pi }{2}$ is $\frac{2}{{\sqrt 5 }}\left( {x - 0} \right) + 0\left( {y - 1} \right) + \frac{1}{{\sqrt 5 }}\left( {z - \pi } \right) = 0$ $2\left( {x - 0} \right) + \left( {z - \pi } \right) = 0$ $2x + z = \pi $
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