Find the rate at which its distance from P is changing one minute later.

Question

An airplane, flying at a constant speed of 360 mi/hr and climbing at a 30-degree angle, passes over a point P on the ground at an altitude of 15,840 ft. Find the rate (in miles per hour) at which its distance from P is changing one minute later.

Answer 1

Let x be the horizontal distance, y be the vertical distance and s be the distance from P to the airplane.

dx/dt = 360 cos(30) = 180√3

dy/dt = 360 sin(30) = 180

After 1 minute or 1/60 hours, the horizontal distance is (dx/dt)t = (180√3)/60 = 3√3 miles.

After 1 minute, the vertical distance is (dy/dt)t + 15840 = 180/60 + 15840 = 15843 miles.

The line of sight distance between P and the airplane is: s = √( (3√3)²+15843²) = 15843.0009

To find ds/dt, use

s² = x²+y²

Doing implicit derivation

2sds/dt = 2xdx/dt + 2ydy/dt

ds/dt = (xdx/dt + ydy/dt)/s = (3√3(180√3) + 15843(180))/15843.0009) = (1620+2851740)/15843.009

= 180.1 mph