Question

You stand 20 ft from a bottle rocket on the ground and watch it as it takes off vertically into the air at a rate of 18 ft/sec. Find the rate at which the angle of elevation from the point on the ground at you feet and the rocket changes when the rocket is 40 ft in the air. At the moment the rocket is 40 ft in the air, the angle of elevation is changing at a rate of rad/se.

Answer 1

Answer:

9/50 rad/sec

Step-by-step explanation:

let the angle be represented as ∅

distance form bottle rocket on the ground = 20 ft

height of rocket = 40 ft

solution

first we differentiate w.r.t.time

= sec²(∅)*(d(∅)/dt))=(1/20)*dh/dt   --------- ( 1 )

=  d(∅)/dt = (1/20)cos²(∅)* (dh/dt)   ------- ( 2 )

note :  1/sec(∅) =cos(∅)

therefore ; dh/dt = 18 ft/sec

applying Pythagoras theorem

Tan ∅ =  40 / 20

 h^2( hypotenuse )  = a^2( height )  + b^2 ( distance from bottle rocket )

                     h = √ 40^2 + 20^2 =  √ 2000

also cos ( ∅ ) = b / h  =  20 / √ 2000

hence cos²∅ = 1/5

Back to equation 2

d(∅)/dt=(1/20)×(1/5)×(18)

= 18/100  = 9/50 ( changing rate of angle of elevation )