Question

A plane flying horizontally at an altitude of 1 mi and a speed of 450 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 3 mi away from the station. (Round your answer to the nearest whole number.)

Answer 1

Given

The plane horizontal altitude is 1 mi and speed is 450 mi/h.

The plane distance from the station is 3 mi.

Explanation

Draw the figure,

P is the plane's position

R is the radar station's position

V is the point located vertically of the radar station at the plane's height.

h is the plane's height

d is the distance between the plane and the radar station

x is the distance between the plane and the V point

Since the plane flies horizontally, we can conclude that PVR is a right triangle. Therefore, the pythagoreas theorem is used to find x,

$d=\sqrt{h^2+x^2}$

Substitute the values,

$\begin{gathered} x=\sqrt{3^2-1^2} \\ x=\sqrt{9-1} \\ x=\sqrt{8} \\ x=2\sqrt{2} \end{gathered}$

When d=3 mi, and, since the plane flies horizontally, we know that h=1mi regardless of the situation. We are looking for

$\frac{dd}{dt}=d$$\begin{gathered} d^2=h^2+x^2 \\ 2d\frac{dd}{dt}=2h\frac{dh}{dt}+2x\frac{dx}{dt} \\ d\frac{dd}{dt}=x\frac{dx}{dt} \end{gathered}$

Substitute the values.

$\begin{gathered} \frac{dd}{dt}=\frac{x}{d}\frac{dx}{dt} \\ \frac{dd}{dt}=\frac{2\sqrt{2}}{3}\times450 \\ \frac{dd}{dt}=150\times2\times\sqrt{2} \\ \frac{dd}{dt}=300\sqrt{2} \end{gathered}$

Answer

Hence the rate at which the distance from the plane to the station is increasing when it has a total distance of 3 mi away from the station is

$424.6mi\text{ /h}$