Answer:
465 mi/h
Step-by-step explanation:
[tex]\frac{da}{dt}[/tex] = Velocity of plane = 480 mi/h
a = Distance plane travels when it is 4 mi away from radar
[tex]\frac{db}{dt}[/tex] = Velocity of plane with respect to height = 0 mi/h (altitude is not changing)
b = Horizontal distance between plane and radar = 1 mi
c = Distance between plane and radar after some time = 4 mi
From Pythagoras theorem
a²+b² = c²
⇒a = √(c²-b²)
⇒a = √(4²-1²)
⇒a = √15 mi
a²+b² = c²
Now, differentiating with respect to time
[tex]2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{\sqrt{15}\times 480+1\times 0}{4}\\\Rightarrow \frac{dc}{dt}=464.75=465\ mi/h[/tex]
∴ Rate at which the distance from the plane to the station is increasing when it is 4 mi away from the station is 465 mi/h
