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 744: 6

Answer

The velocity vector: ${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$ The speed at $s=2$ is $v\left( 2 \right) = \frac{1}{5}$ The acceleration vector: ${\bf{a}}\left( s \right) = {\bf{v}}'\left( s \right) = \left( {\frac{{6{s^2} - 2}}{{{{\left( {1 + {s^2}} \right)}^3}}},\frac{{2{s^3} - 6s}}{{{{\left( {1 + {s^2}} \right)}^3}}}} \right)$ The acceleration vector at $s=2$ is ${\bf{a}}\left( 2 \right) = \left( {\frac{{22}}{{125}},\frac{4}{{125}}} \right)$

Work Step by Step

We have ${\bf{r}}\left( s \right) = \left( {\frac{1}{{1 + {s^2}}},\frac{s}{{1 + {s^2}}}} \right)$. Write ${\bf{r}}\left( s \right) = \left( {{{\left( {1 + {s^2}} \right)}^{ - 1}},s{{\left( {1 + {s^2}} \right)}^{ - 1}}} \right)$. The velocity vector: ${\bf{v}}\left( s \right) = {\bf{r}}'\left( t \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},{{\left( {1 + {s^2}} \right)}^{ - 1}} - \frac{{2{s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$ ${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 + {s^2} - 2{s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$ ${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$ The velocity vector at $s=2$, ${\bf{v}}\left( 2 \right) = \left( { - \frac{4}{{25}}, - \frac{3}{{25}}} \right)$ So, the speed at $s=2$ is $v\left( 2 \right) = \sqrt {{\bf{v}}\left( 2 \right)\cdot{\bf{v}}\left( 2 \right)} = \sqrt {{{\left( { - \frac{4}{{25}}} \right)}^2} + {{\left( { - \frac{3}{{25}}} \right)}^2}} = \frac{1}{5}$ From the above result we obtain ${\bf{v}}\left( s \right) = \left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}},\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$. Evaluate $\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$ $\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = - \frac{{2{{\left( {1 + {s^2}} \right)}^2} - \left( {2s} \right)4s\left( {1 + {s^2}} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$ $\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = - \frac{{\left( {1 + {s^2}} \right)\left( {2\left( {1 + {s^2}} \right) - 8{s^2}} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$ $\frac{d}{{ds}}\left( { - \frac{{2s}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = - \frac{{ - 6{s^2} + 2}}{{{{\left( {1 + {s^2}} \right)}^3}}} = \frac{{6{s^2} - 2}}{{{{\left( {1 + {s^2}} \right)}^3}}}$ Evaluate $\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right)$ $\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{ - 2s{{\left( {1 + {s^2}} \right)}^2} - \left( {1 - {s^2}} \right)4s\left( {1 + {s^2}} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$ $\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{\left( {1 + {s^2}} \right)\left( { - 2s\left( {1 + {s^2}} \right) - 4s\left( {1 - {s^2}} \right)} \right)}}{{{{\left( {1 + {s^2}} \right)}^4}}}$ $\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{ - 2s - 2{s^3} - 4s + 4{s^3}}}{{{{\left( {1 + {s^2}} \right)}^3}}}$ $\frac{d}{{ds}}\left( {\frac{{1 - {s^2}}}{{{{\left( {1 + {s^2}} \right)}^2}}}} \right) = \frac{{2{s^3} - 6s}}{{{{\left( {1 + {s^2}} \right)}^3}}}$ Thus, the acceleration vector: ${\bf{a}}\left( s \right) = {\bf{v}}'\left( s \right) = \left( {\frac{{6{s^2} - 2}}{{{{\left( {1 + {s^2}} \right)}^3}}},\frac{{2{s^3} - 6s}}{{{{\left( {1 + {s^2}} \right)}^3}}}} \right)$ The acceleration vector at $s=2$ is ${\bf{a}}\left( 2 \right) = \left( {\frac{{22}}{{125}},\frac{4}{{125}}} \right)$
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