Question

An airplane is flying at an elevation of 6 miles on a flight path that will take it directly over a radar tracking station. Let S represent the distance between the radar station and the plane. If S is decreasing at a rate of 400 mph when S is 10 miles, what is the velocity of the plane

Answer 1

Answer:

$-500\ \text{mph}$

Explanation:

h = Height at which the plane is flying = 6 miles

S = Distance between plane and radar = 10 miles

$\dfrac{dS}{dt}$ = Rate at which S is decreasing = -400 mph

Distance between S and and the elevation of the plane

$b=\sqrt{S^2-h^2}=\sqrt{10^2-6^2}\\\Rightarrow b=8$

From Pythagoras theorem we get

$S^2=b^2+h^2$

Differentiating with respect to time we get

$2S\dfrac{dS}{dt}=2b\dfrac{db}{dt}+2h\dfrac{dh}{dt}\\\Rightarrow S\dfrac{dS}{dt}=b\dfrac{db}{dt}+h\dfrac{dh}{dt}\\\Rightarrow 10\times -400=8\times \dfrac{db}{dt}+0\\\Rightarrow \dfrac{db}{dt}=\dfrac{10\times -400}{8}\\\Rightarrow \dfrac{db}{dt}=-500\ \text{mph}$

Velocity of the plane is $-500\ \text{mph}$.