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: 10

Answer

The two parametrizations trace the same curve, however as is shown in the figure attached, only the velocity vectors point in the same direction. Since the $x$- and $y$-component of ${{\bf{r}}_1}\left( t \right)$ increase faster than ${\bf{r}}\left( t \right)$, we expect the velocity vector and acceleration vector of ${{\bf{r}}_1}\left( t \right)$ have larger magnitude. $\begin{array}{*{20}{c}} t&{{\bf{v}}\left( t \right)}&{{{\bf{v}}_1}\left( t \right)}&{{\bf{a}}\left( t \right)}&{{{\bf{a}}_1}\left( t \right)}\\ {}&{\left( {2t,3{t^2}} \right)}&{\left( {4{t^3},6{t^5}} \right)}&{\left( {2,6t} \right)}&{\left( {12{t^2},30{t^4}} \right)}\\ 1&{\left( {2,3} \right)}&{\left( {4,6} \right)}&{\left( {2,6} \right)}&{\left( {12,30} \right)} \end{array}$

Work Step by Step

We have ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$ and ${{\bf{r}}_1}\left( t \right) = \left( {{t^4},{t^6}} \right)$. The two parametrizations trace the same curve, however as is shown in the figure attached, only the velocity vectors point in the same direction. Since the $x$- and $y$-component of ${{\bf{r}}_1}\left( t \right)$ increase faster than ${\bf{r}}\left( t \right)$, we expect the velocity vector and acceleration vector of ${{\bf{r}}_1}\left( t \right)$ have larger magnitude. The velocity vectors: ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {2t,3{t^2}} \right)$, ${\ \ }$ ${{\bf{v}}_1}\left( t \right) = {{\bf{r}}_1}'\left( t \right) = \left( {4{t^3},6{t^5}} \right)$ So, ${\bf{v}}\left( 1 \right) = \left( {2,3} \right)$ and ${{\bf{v}}_1}\left( 1 \right) = \left( {4,6} \right)$. The acceleration vectors: ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {2,6t} \right)$, ${\ \ }$ ${{\bf{a}}_1}\left( t \right) = {{\bf{r}}_1}{\rm{''}}\left( t \right) = \left( {12{t^2},30{t^4}} \right)$ So, ${\bf{a}}\left( 1 \right) = \left( {2,6} \right)$ and ${{\bf{a}}_1}\left( 1 \right) = \left( {12,30} \right)$ In summary: $\begin{array}{*{20}{c}} t&{{\bf{v}}\left( t \right)}&{{{\bf{v}}_1}\left( t \right)}&{{\bf{a}}\left( t \right)}&{{{\bf{a}}_1}\left( t \right)}\\ {}&{\left( {2t,3{t^2}} \right)}&{\left( {4{t^3},6{t^5}} \right)}&{\left( {2,6t} \right)}&{\left( {12{t^2},30{t^4}} \right)}\\ 1&{\left( {2,3} \right)}&{\left( {4,6} \right)}&{\left( {2,6} \right)}&{\left( {12,30} \right)} \end{array}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.