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 21,120 ft. find the rate (in miles per hour) at which its distance from p is changing one minute later.

The point P is a point below the path of travel of the plane such as the

location of a house close to an airport.

Rate of change of the distance of the airplane from P is

Reason:

The given parameters are;

The speed of the airplane = 360 mi/hr

The degree angle of the airplane = 30°

The height from which the airplane passes the point, P = 21,120 ft.

The distance the airplane travels per minute from the point in the x-direction is given as follows;

The distance the airplane travels per minute from the point in the y-direction is given as follows;

The magnitude of the distance, D, is given, by Pythagoras's theorem, as follows;

Therefore;

D² = (6·t·cos(30°))² + (4 + 6·t·sin(30°))²

D² = 36·t²·cos²(30°) + 16 + 24·t·sin(30°) + 36·t²·sin²(30°)

D² = 36·t²·cos²(30°) + 36·t²·sin²(30°) + 16 + 24·t·sin(30°)

D² = 36·t²·(cos²(30°) + sin²(30°)) + 16 + 24·t·sin(30°)

The distance covered in a minute, is given as follows;

D = √(6 × 1 × cos(30°))² + (4 + 6 × 1 × sin(30°))²) = 2·√19

The rate of change of the distance in a minute is therefore;

t = 1

The rate of change of distance in miles per hour is found as follows;

The rate at which its distance from P is changing one minute later is therefore;

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Step-by-step explanation:

Given  

velocity of airplane is 360 mi/hr

inclination of climbing

Plane velocity can be divided in to x & y component

Velocity in x component is

Velocity in y component is

Thus distance traveled in x axis is

=60cos30=51.96 miles

Distance travel in upward direction i.e. in Y axis

so total altitude gain by plane is 30+4=34 miles

Horizontal distance traveled is 51.96

inclination w.r.t to point p

Initial inclination is

therefore rate at which its distance is changing is

the answer for this question is 38