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: 44

Answer

${\bf{T}}\left( t \right) = \left( {\frac{t}{{\sqrt {{t^2} + 4} }},\frac{2}{{\sqrt {{t^2} + 4} }}} \right)$ and ${\bf{N}}\left( t \right) = \left( {\frac{2}{{\sqrt {{t^2} + 4} }}, - \frac{t}{{\sqrt {{t^2} + 4} }}} \right)$. The decomposition of ${\bf{a}}$: ${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}}\left( t \right) + {a_{\bf{N}}}{\bf{N}}\left( t \right)$ ${\bf{a}}\left( t \right) = \left( {\frac{{2t}}{{\sqrt {{t^2} + 4} }}} \right){\bf{T}}\left( t \right) + \left( {\frac{4}{{\sqrt {{t^2} + 4} }}} \right){\bf{N}}\left( t \right)$

Work Step by Step

We have ${\bf{r}}\left( t \right) = \left( {{t^2},4t - 3} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {2t,4} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {2,0} \right)$, respectively. Thus, the unit tangent vector is ${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {2t,4} \right)}}{{\sqrt {\left( {2t,4} \right)\cdot\left( {2t,4} \right)} }} = \frac{1}{{\sqrt {4{t^2} + 16} }}\left( {2t,4} \right)$ ${\bf{T}} = \left( {\frac{t}{{\sqrt {{t^2} + 4} }},\frac{2}{{\sqrt {{t^2} + 4} }}} \right)$ By Eq. (2) of Theorem 1 we have ${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( {2,0} \right)\cdot\left( {\frac{t}{{\sqrt {{t^2} + 4} }},\frac{2}{{\sqrt {{t^2} + 4} }}} \right)$ ${a_{\bf{T}}} = \frac{{2t}}{{\sqrt {{t^2} + 4} }}$ Next, we use Eq. (3) of Theorem 1 to find ${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$ ${a_{\bf{N}}}{\bf{N}} = \left( {2,0} \right) - \frac{{2t}}{{\sqrt {{t^2} + 4} }}\left( {\frac{t}{{\sqrt {{t^2} + 4} }},\frac{2}{{\sqrt {{t^2} + 4} }}} \right)$ ${a_{\bf{N}}}{\bf{N}} = \left( {2,0} \right) - \left( {\frac{{2{t^2}}}{{{t^2} + 4}},\frac{{4t}}{{{t^2} + 4}}} \right)$ ${a_{\bf{N}}}{\bf{N}} = \left( {\frac{8}{{{t^2} + 4}}, - \frac{{4t}}{{{t^2} + 4}}} \right)$ Since ${\bf{N}}$ is an unit vector, so ${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = \sqrt {{{\left( {\frac{8}{{{t^2} + 4}}} \right)}^2} + {{\left( { - \frac{{4t}}{{{t^2} + 4}}} \right)}^2}} $ ${a_{\bf{N}}} = \sqrt {\frac{{64 + 16{t^2}}}{{{{\left( {{t^2} + 4} \right)}^2}}}} = \frac{4}{{\sqrt {{t^2} + 4} }}$ We find ${\bf{N}}$ by using the equation ${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \frac{{\sqrt {{t^2} + 4} }}{4}\left( {\frac{8}{{{t^2} + 4}}, - \frac{{4t}}{{{t^2} + 4}}} \right)$ ${\bf{N}} = \left( {\frac{2}{{\sqrt {{t^2} + 4} }}, - \frac{t}{{\sqrt {{t^2} + 4} }}} \right)$ Thus, we obtain the decomposition of ${\bf{a}}$: ${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}}\left( t \right) + {a_{\bf{N}}}{\bf{N}}\left( t \right)$ ${\bf{a}}\left( t \right) = \left( {\frac{{2t}}{{\sqrt {{t^2} + 4} }}} \right){\bf{T}}\left( t \right) + \left( {\frac{4}{{\sqrt {{t^2} + 4} }}} \right){\bf{N}}\left( t \right)$, where ${\bf{T}}\left( t \right) = \left( {\frac{t}{{\sqrt {{t^2} + 4} }},\frac{2}{{\sqrt {{t^2} + 4} }}} \right)$ and ${\bf{N}}\left( t \right) = \left( {\frac{2}{{\sqrt {{t^2} + 4} }}, - \frac{t}{{\sqrt {{t^2} + 4} }}} \right)$.
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