Answer:
86.6 km/h
Explanation:
[tex]\frac{da}{dt}[/tex] = Velocity of Plane = 600 km/h
a = Distance Plane travels
[tex]\frac{db}{dt}[/tex] = Vertical velocity of Plane = 0
b = Altitude of plane = 10 km
c = Distance between town and plane = 20 km
[tex]a=\sqrt{c^2-b^2}\\\Rightarrow a=\sqrt{20^2-10^2}\\\Rightarrow a=17.32\ km[/tex]
From Pythogoras theorem
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{17.32\times 100+10\times 0}{20}\\\Rightarrow \frac{dc}{dt}=86.6\ km/h[/tex]
The rate at which the distance from the plane to Quinton is increasing 86.6 km/h
