Answer :
Answer:
[tex]\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}[/tex]
Step-by-step explanation:
Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:
First airplane
[tex]r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t[/tex]
Where t is the time measured in hours.
Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:
[tex]s=\sqrt{r_{A}^{2}+r_{B}^{2}}[/tex]
Rate of change of such distance can be found by the deriving the expression in terms of time:
[tex]\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }[/tex]
Where [tex]\frac{dr_{A}}{dt} = 500\,\frac{mi}{h}[/tex] and [tex]\frac{dr_{B}}{dt} = 550\,\frac{mi}{h}[/tex], respectively. Distances of each airliner at 2:30 PM are:
[tex]r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi[/tex]
[tex]r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi[/tex]
The rate of change is:
[tex]\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }[/tex]
[tex]\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}[/tex]