Chapter 11, Problem 19PS

a)

To determine

The distance between the two insects at time $t=0$ .

Answer to Problem 19PS

  $8.37\text{inches}$

Explanation of Solution

Given: Two insects are crawling along different lines.

The line given by first insect at point $\left(x,y,z\right)$ is

  $\begin{array}{l}x=6+t\\ y=8-t\\ z=3+t\end{array}$

Line for Second insect is given by

  $\begin{array}{l}x=1+t\\ y=2+t\\ z=2t\end{array}$

Concept Used: Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

Calculation: Points $P_1=\left(6,8,3\right)$ , $P_2=\left(1,2,0\right)$

Using distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

  $\begin{array}{c}\sqrt{{\left(6-1\right)}^2+{\left(8-2\right)}^2+{\left(3-0\right)}^2}=\sqrt{{\left(5\right)}^2+{\left(6\right)}^2+{\left(3\right)}^2}\\ =\sqrt{25+36+9}\\ =8.37\text{inches}\end{array}$

Conclusion:

b)

To determine

To graph the distance between insects from $t=0$ to $t=10$ .

Answer to Problem 19PS

  

Explanation of Solution

Given: Two insects are crawling along different lines.

The line given by first insect at point $\left(x,y,z\right)$ is

  $\begin{array}{l}x=6+t\\ y=8-t\\ z=3+t\end{array}$

Line for Second insect is given by

  $\begin{array}{l}x=1+t\\ y=2+t\\ z=2t\end{array}$

Concept Used: Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

Calculation: Using Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

  $\begin{array}{c}\sqrt{{\left(5\right)}^2+{\left(6-2t\right)}^2+{\left(3-t\right)}^2}=\sqrt{\left(25\right)+\left(4t^2-24t+36\right)+\left(t^2-6t+9\right)}\\ =\sqrt{t^2+4t^2-6t-24t+9+25+36}\\ =\sqrt{5t^2-30t+70}\end{array}$

For $0\le t\le 10$

The graph is

  

c)

To determine

The distance between two insects from the graph.

Answer to Problem 19PS

Distance can never be zero.

Explanation of Solution

Given: Two insects are crawling along different lines.

The line given by first insect at point $\left(x,y,z\right)$ is

  $\begin{array}{l}x=6+t\\ y=8-t\\ z=3+t\end{array}$

Line for Second insect is given by

  $\begin{array}{l}x=1+t\\ y=2+t\\ z=2t\end{array}$

Concept Used: Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

Calculation: Using Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

  $\begin{array}{c}\sqrt{{\left(5\right)}^2+{\left(6-2t\right)}^2+{\left(3-t\right)}^2}=\sqrt{\left(25\right)+\left(4t^2-24t+36\right)+\left(t^2-6t+9\right)}\\ =\sqrt{t^2+4t^2-6t-24t+9+25+36}\\ =\sqrt{5t^2-30t+70}\end{array}$

Conclusion: Distance can never be zero

d)

To determine

The minimum time between two insects from the graph.

Answer to Problem 19PS

  $\begin{array}{l}Minimum\\ \left(3,5\right)\\ \end{array}$

  $3$ minutes, $5$ inches

Explanation of Solution

Given: Two insects are crawling along different lines.

The line given by first insect at point $\left(x,y,z\right)$ is

  $\begin{array}{l}x=6+t\\ y=8-t\\ z=3+t\end{array}$

Line for Second insect is given by

  $\begin{array}{l}x=1+t\\ y=2+t\\ z=2t\end{array}$

Concept Used: Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

Calculation: Using Distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

  $\begin{array}{c}\sqrt{{\left(5\right)}^2+{\left(6-2t\right)}^2+{\left(3-t\right)}^2}=\sqrt{\left(25\right)+\left(4t^2-24t+36\right)+\left(t^2-6t+9\right)}\\ =\sqrt{t^2+4t^2-6t-24t+9+25+36}\\ =\sqrt{5t^2-30t+70}\end{array}$

For $t=0$

  $\begin{array}{l}Minimum\\ \left(3,5\right)\\ \end{array}$