91Ó°ÊÓ

We are given rate of airplane A at 120 miles per hour with an initial distance of 100 miles and given rate of airplane B at 600 miles per hour with an initial distance of 550 miles.

For airplane A

We have $d=vt$

Since we have given initial distance of 100 miles that starts from positive y-axis and rate of 120mi/hr, the horizontal distance in time $t$

role="math" localid="1649065691653" $y_a=120t+100and{x}_a=0$

For airplane B

Since we have given initial distance of 550 miles that starts from positive x-axis and rate of 600mi/hr, the horizontal distance in time $t$

role="math" localid="1649065725693" $x_b=600t+500and{y}_b=0$

By the formula of distance $d=\sqrt{{\left(x_b-x_a\right)}^2+{\left(y_b-y_a\right)}^2}$

localid="1649067432606" $$\begin{gathered} d=\sqrt{{\left(600t+550-0\right)}^2+{\left(0-(120t+100\right)}^2} \\ d=\sqrt{{(600t+550)}^2+{\left(-120t-100\right)}^2} \\ d=\sqrt{{(600t+550)}^2+{\left(-120t-100\right)}^2} \end{gathered}$$

The graph is shown in the figure

From the graph, we see that the minimum distance between airplane is $205.92$miles and minimum time when the two planes are closed is 0.85 hours.

Enter the parametric equations

$$\begin{gathered} X_{1T}=0,Y_{1T}=120t+100 \\ {X}_{2T}=600t+550,Y_{2T}=0 \end{gathered}$$

Select the viewing window

$$\begin{gathered} T_{min}=-0.915,T_{,max}=0.85,T_{step}=0.1 \\ {X}_{min}=-100,X_{max}=100,X_{step}=1 \\ {Y}_{min}=-100,Y_{,max}=100,Y_{step}=1 \end{gathered}$$

Select the graph window.