At what rate is the distance from the plane to the radar station increasing 3 minutes later?

Question

A plane flying at constant speed of 24 km/min passes over a ground radar station at an altitude 10 km and climbs at an angle of 40 degree?

Answer 1

Constant speed of the plane v = 24 km/min

Climbing angle = 40°

Distance from radar station to plane climbing position = 10 Km.

O is Radar station Position.

A is Position of plane at time t.

B is Position of plane at radar station.

OB = 10 km

Plane speed v = distance/time = x/t

x = vt = 24t

Here x is distance from radar station to plane at t minuites.So BA = x = 24t

By using cosine rule for OAC traingle :

OA² = OB² + BA² - 2(OA)(BA)cos(90+40)

OA² = 10² + (24t)² - 2(10)(24t)cos(130)

OA² = 100 + 576t² - 480tcos(130)                   --------------------- (1)

At time t = 3 min.

OA² = 100 + 576(3)² - 480(3)cos(130) = 100 + 5184 + 925.614 = 6209.614

OA = √6209.614 = 78.801

 

(1) ⇒ OA = √[100 + 576t² - 480tcos(130)]

Apply derivative with respect to time.

d(OA)/dt = { 1/2√[100 + 576t² - 480tcos(130)] } { 576*2t - 480cos(130) }

d(OA)/dt = [ 576*2t - 480cos(130)]/2OA

d(OA)/dt = [576*2*3 - 480cos(130)] / (2*278.801) = 3764.538 / 557.602 = 6.7513

The distance increasing rate from the plane to the radar station is 6.7513 km/min.