An airplane in australia is flying at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. how fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? give your answer in radians per minute.

We model the problem as a rectangle triangle.

We have on the one hand:

hypotenuse = z = 3

On the other hand:

vertical side = y = 2

The speed of the plane is:

dz / dt = 600

The flight height is constant, then:

dy / dt = 0

The angle with the horizontal will be:

sin (θ) = 2/3

θ = arcsin (2/3) = 0.729728 radians

On the other hand:

sin (θ) = y / z

z sin (θ) = y

We derive both sides with respect to time:

dz / dt sin (θ) + z cos (θ) dθ / dt = dy / dt

Substituting values:

(600) (2/3) + (3) cos (0.729728) dθ / dt = 0

Rewriting:

400 + [3 / cos (0.729728) ] dθ / dt = 0

400 + 4.0249 dθ / dt = 0

We cleared dθ / dt:

dθ / dt = - 98.38 radians / hour

= - 99.38 / 60 radians / min

= - 1.66 radians / min = 2PI - 1.66 radians / min

= 4.62 radians / min

Answer:

The angle of elevation of the Kangaroo's line of sight is increasing at:

= 4.62 radians / min