Question

A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 13 km and climbs at an angle of 20 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 5 minutes later?

Answer 1

`approx 1.753 (km)/min`

Explanation:

the plane's velocity is

`vec v = ((14 cos alpha),(14 sin alpha))` where `alpha = 20`

its position at time t minutes is
`vec r = ((14 cos alpha * t),(13 + 14 sin alpha * t))` where `alpha = 20`

where the base of the station is `vec r_o = ((0),(0))`

the distance `|vec r|` from (0,0) at time t is

`|vec r | = sqrt ( (14 cos alpha * t)^2 + (13 + 14 sin alpha * t )^2 )`

`|vec r | = sqrt ( (14 cos alpha * t)^2 + (13^2 + 14^2 sin^2 alpha * t^2 + 2(13)14 sin alpha * t ) )`

`|vec r | = sqrt ( (14 t)^2 + 13^2 + 2(13)14 sin alpha * t )`

`d/dt |vec r | = (14t + (13)14 sin alpha ) /sqrt ( (14 t)^2 + 13^2 + 2(13)14 sin alpha * t )`

`approx 1.753 (km)/min` at t = 5

i have no idea why you would wish to use implicit differentiation in this problem, maybe it is in the wrong section? in Cartesian, the flight path is simply `y = 13 + x tan 20^o` but you still need to factor in the speed of the plane and so move away from Cartesian. and in Cartesian, the distance formula makes it a problem in single variable calculus.