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: 39

Answer

The decomposition of ${\bf{a}}$ at $t=4$: ${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$ $\left( {0,1,4} \right) = 4{\bf{T}} + {\bf{N}}$, where ${\bf{T}} = \left( {\frac{1}{9},\frac{4}{9},\frac{8}{9}} \right)$ and ${\bf{N}} = \left( { - \frac{4}{9}, - \frac{7}{9},\frac{4}{9}} \right)$.

Work Step by Step

We have ${\bf{r}}\left( t \right) = \left( {t,\frac{1}{2}{t^2},\frac{1}{6}{t^3}} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {1,t,\frac{1}{2}{t^2}} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {0,1,t} \right)$, respectively. At $t=4$, we get ${\bf{v}}\left( 4 \right) = \left( {1,4,8} \right)$ and ${\bf{a}}\left( 4 \right) = \left( {0,1,4} \right)$. Thus, the unit tangent vector is ${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {1,4,8} \right)}}{{\sqrt {\left( {1,4,8} \right)\cdot\left( {1,4,8} \right)} }} = \left( {\frac{1}{9},\frac{4}{9},\frac{8}{9}} \right)$ By Eq. (2) of Theorem 1 we have ${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( {0,1,4} \right)\cdot\left( {\frac{1}{9},\frac{4}{9},\frac{8}{9}} \right)$ ${a_{\bf{T}}} = 4$ Next, we use Eq. (3) of Theorem 1 to find ${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$ ${a_{\bf{N}}}{\bf{N}} = \left( {0,1,4} \right) - 4\left( {\frac{1}{9},\frac{4}{9},\frac{8}{9}} \right)$ ${a_{\bf{N}}}{\bf{N}} = \left( { - \frac{4}{9}, - \frac{7}{9},\frac{4}{9}} \right)$ Since ${\bf{N}}$ is an unit vector, so ${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = \sqrt {{{\left( { - \frac{4}{9}} \right)}^2} + {{\left( { - \frac{7}{9}} \right)}^2} + {{\left( {\frac{4}{9}} \right)}^2}} $ ${a_{\bf{N}}} = 1$ To find ${\bf{N}}$ we use the equation ${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \left( { - \frac{4}{9}, - \frac{7}{9},\frac{4}{9}} \right)$ Thus, we obtain the decomposition of ${\bf{a}}$ at $t=4$: ${\bf{a}} = {a_{\bf{T}}}{\bf{T}} + {a_{\bf{N}}}{\bf{N}}$ $\left( {0,1,4} \right) = 4{\bf{T}} + {\bf{N}}$, where ${\bf{T}} = \left( {\frac{1}{9},\frac{4}{9},\frac{8}{9}} \right)$ and ${\bf{N}} = \left( { - \frac{4}{9}, - \frac{7}{9},\frac{4}{9}} \right)$.
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