Chapter 11, Problem 18PS

a.

To determine

To find: To find the distance between the point and the line as a function of $s$ .

Answer to Problem 18PS

  $D\left(s\right)=\sqrt{\frac{5\left(s^2+2s+22\right)}{21}}$ is the function of $s$ which gives the distance between the point and the line.

Explanation of Solution

Given: The line is given by the parametric equations,

  $x=-t+3$

  $y=\frac{1}{2}t+1$

  $z=2t-1$

and a point $\left(4,3,s\right)$ for any real number $s$ .

To find the two points on the line $P_0$ and $P_2$ when $t=0\text{and }2$ ,

Where $P_2-P_0$ is direction vector of the line $\stackrel{\rightharpoonup }{u}$ .

  $\begin{array}{c}x=-t+3,y=\frac{1}{2}t+1,z=2t-1\\ Q=\left(4,3,s\right)\\ P_0=\left(3,1,-1\right)P_2=\left(1,2,3\right)\\ \stackrel{\rightharpoonup }{u}=P_2-P_0=\left(-2,1,4\right)\end{array}$

Finding the vector $\stackrel{\rightharpoonup }{PQ}$ ,

  $\begin{array}{l}\stackrel{\rightharpoonup }{PQ}=Q-P_0\\ \stackrel{\rightharpoonup }{PQ}=\left(4,3,s\right)-\left(3,1,-1\right)\\ \stackrel{\rightharpoonup }{PQ}=\left(1,2,s+1\right)\end{array}$

By taking the cross product of $\stackrel{\rightharpoonup }{PQ}$ and $\stackrel{\rightharpoonup }{u}$ , we get

  $\begin{array}{c}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1& 2& s+4\\ -2& 1& 4\end{array}\right|=\left(8-s-1\right)\text{i}-\left(4+2s+2\right)\text{j}+\left(1+4\right)\text{k}\\ \stackrel{\rightharpoonup }{PQ}\times \stackrel{\rightharpoonup }{u}=\left(7-s\right)\text{i+}\left(-6-2s\right)\text{j}+5\text{k}\end{array}$

Finding the distance $D$ ,

  $\begin{array}{l}D=\frac{\Vert \stackrel{\rightharpoonup }{PQ}\times \stackrel{\rightharpoonup }{u}\Vert }{\Vert \stackrel{\rightharpoonup }{u}\Vert }\\ D=\sqrt{\frac{{\left(7-s\right)}^2+{\left(-6-s\right)}^2+5^2}{21}}\\ D=\sqrt{\frac{49-14s+s^2+36+24s+4s^2+25}{21}}\\ D=\sqrt{\frac{110+10s+5s^2}{21}}\\ D=\sqrt{\frac{5\left(s^2+2s+22\right)}{21}}\end{array}$

Thus we can find the distance between the point and the line as a function of $s$ .

b.

To determine

To find: To find the value of $s$ when the distance between the point and the line is minimum.

Answer to Problem 18PS

The value of $s$ at the minimum distance at $s=-1$ is $\sqrt{5}$ .

Explanation of Solution

Given: The line is given by the parametric equations,

  $x=-t+3$

  $y=\frac{1}{2}t+1$

  $z=2t-1$

and a point $\left(4,3,s\right)$ for any real number $s$ .

By using the graphing utility,

  

From the graph the curve’s bottom is at $s=-1$ .

The minimum distance between the point and the line is,

  $D=\sqrt{5}$

The distance between the point and the line is the minimum when $s=-1$ .

Therefore, we can conclude that the minimum distance between the point and the line is $D=\sqrt{5}$ at $s=-1$ .

c.

To determine

To find: To find the asymptotes of the graph appear slant.

Answer to Problem 18PS

Yes, the graph appears to have a slant asymptoes $y=1\pm \frac{\sqrt{5}}{\sqrt{21}}\left(s+1\right)$ .

Explanation of Solution

Given: The line is given by the parametric equations,

  $x=-t+3$

  $y=\frac{1}{2}t+1$

  $z=2t-1$

and a point $\left(4,3,s\right)$ for any real number $s$ .

  

We know the $s$ is a $z-$ coordinate and the point is always $x=4,y=3$ .

Because of the coordinates we can understand that the far ranges of $s$ is affected by one variable.

So,

  $\begin{array}{c}D\left(s\right)=\frac{\sqrt{5}}{\sqrt{21}}\sqrt{s^2+2s+22}\\ m=\pm \frac{\sqrt{5}}{\sqrt{21}}\end{array}$

By using the point slope form we can conclude that the graph appears to have a slant asymptoes $y=1\pm \frac{\sqrt{5}}{\sqrt{21}}\left(s+1\right)$ .