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 - 14.5 Motion in 3-Space - Exercises - Page 745: 36

Answer

The decomposition of ${\bf{a}}$ at $t=0$: ${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$ $\left( {1,0} \right) = \frac{1}{{\sqrt 2 }}{\bf{T}} + \frac{1}{{\sqrt 2 }}{\bf{N}}$, where ${\bf{T}} = \left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ and ${\bf{N}} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$.

Work Step by Step

We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},1 - t} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}, - 1} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {{{\rm{e}}^t},0} \right)$, respectively. At $t=0$, we get ${\bf{v}}\left( 1 \right) = \left( {1, - 1} \right)$ and ${\bf{a}}\left( 1 \right) = \left( {1,0} \right)$. Thus, the unit tangent vector is ${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {1, - 1} \right)}}{{\sqrt {\left( {1, - 1} \right)\cdot\left( {1, - 1} \right)} }} = \left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ By Eq. (2) of Theorem 1 we have ${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( {1,0} \right)\cdot\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ ${a_{\bf{T}}} = \frac{1}{{\sqrt 2 }}$ Next, we use Eq. (3) of Theorem 1 to find ${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$ ${a_{\bf{N}}}{\bf{N}} = \left( {1,0} \right) - \left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ ${a_{\bf{N}}}{\bf{N}} = \left( {1,0} \right) - \left( {\frac{1}{2}, - \frac{1}{2}} \right) = \left( {\frac{1}{2},\frac{1}{2}} \right)$ Since ${\bf{N}}$ is an unit vector, so ${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} $ ${a_{\bf{N}}} = \frac{1}{{\sqrt 2 }}$ To find ${\bf{N}}$ we use the equation ${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \frac{{\sqrt 2 }}{1}\left( {\frac{1}{2},\frac{1}{2}} \right)$ ${\bf{N}} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ Thus, we obtain the decomposition of ${\bf{a}}$ at $t=0$: ${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$ $\left( {1,0} \right) = \frac{1}{{\sqrt 2 }}{\bf{T}} + \frac{1}{{\sqrt 2 }}{\bf{N}}$, where ${\bf{T}} = \left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$ and ${\bf{N}} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$.
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