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 - 14.1 Vector-Valued Functions - Exercises - Page 711: 33

Answer

The two curves collide and intersect at $\left( {12,4,2} \right)$. The two curves intersect at another point, which is $\left( {4,0,6} \right)$.

Work Step by Step

1. Check if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide. Suppose there is some ${t_0}$ such that ${{\bf{r}}_1}\left( {{t_0}} \right) = {{\bf{r}}_2}\left( {{t_0}} \right)$. So, ${{\bf{r}}_1}\left( {{t_0}} \right) = \left( {{t_0}^2 + 3,{t_0} + 1,6{t_0}^{ - 1}} \right)$ ${{\bf{r}}_2}\left( {{t_0}} \right) = \left( {4{t_0},2{t_0} - 2,{t_0}^2 - 7} \right)$ $\left( {{t_0}^2 + 3,{t_0} + 1,6{t_0}^{ - 1}} \right) = \left( {4{t_0},2{t_0} - 2,{t_0}^2 - 7} \right)$ We obtain $x = {t_0}^2 + 3$, ${\ \ }$ $y = {t_0} + 1 = 2{t_0} - 2$, $z = 6{t_0}^{ - 1} = {t_0}^2 - 7$ From the second equation we obtain ${t_0} = 3$, So, $y=4$. Substituting in the first equation gives $x=12$. Substituting in the third equation gives $z=2$. Since ${{\bf{r}}_1}\left( 3 \right) = {{\bf{r}}_2}\left( 3 \right)$, therefore, by definition ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide at ${t_0} = 3$. From Exercise 32, we know that if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ collide, then they intersect. Thus, the point of intersection is ${{\bf{r}}_1}\left( 3 \right) = {{\bf{r}}_2}\left( 3 \right) = \left( {12,4,2} \right)$. 2. Check if ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ intersect. We still need to check if they intersect at some other point. Two curves intersect if there exist parameter values $t$ and $s$ such that ${{\bf{r}}_1}\left( t \right) = {{\bf{r}}_2}\left( s \right)$. So, ${{\bf{r}}_1}\left( t \right) = \left( {{t^2} + 3,t + 1,6{t^{ - 1}}} \right)$ ${{\bf{r}}_2}\left( s \right) = \left( {4s,2s - 2,{s^2} - 7} \right)$ We obtain $x = {t^2} + 3 = 4s$, ${\ \ }$ $y = t + 1 = 2s - 2$, $z = 6{t^{ - 1}} = {s^2} - 7$ From the second equation we obtain $t=2s-3$. Substituting it in the first equation gives ${\left( {2s - 3} \right)^2} + 3 = 4s$ $4{s^2} - 12s + 9 + 3 = 4s$ $4{s^2} - 16s + 12 = 0$ ${s^2} - 4s + 3 = 0$ $\left( {s - 1} \right)\left( {s - 3} \right) = 0$ The solutions are ${s_1} = 1$ and ${s_2} = 3$. The corresponding $t$ values are ${t_1} = - 1$ and ${t_2} = 3$. Substituting these values in the first, the second and the third equation give $\begin{array}{*{20}{c}} {Parameters}&x&y&z\\ {{s_1} = 1,{t_1} = - 1}&4&0&6\\ {{s_2} = 3,{t_2} = 3}&{12}&4&2 \end{array}$ So, there are two points of intersection $\left( {4,0,6} \right)$ and $\left( {12,4,2} \right)$. In part (a) we have already obtained the point $\left( {12,4,2} \right)$, which is also the point of collision.
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