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.

Hello!
Denote the elevation angle as alpha(t) (in radians, t is in hours). Denote the known speed as V and the (unknown) horizontal distance between the airplane and the kangaroo that corresponds to t=0 as L. Then the horizontal distance at any time t is equal to L - Vt.
This way the tangent of alpha(t) is  2/(L - Vt), so  alpha(t) = arctan(2/(L - Vt)).
The increase of the angle of elevation is the derivative of alpha(t), and we are interested of its value for t such that the height 2mi, the horizontal distance L - Vt and the distance 3mi form a right triangle. In other words, L - Vt = sqrt(3^2-2^2) = sqrt(5) (mi).
Find the derivative:
d/dt(alpha(t)) = 1/(1+(2/(L - Vt))^2)*(2V)/((L - Vt)^2) =(2V)/((L - Vt)^2+4).
When L - Vt = sqrt(5) it will be (2V)/9 approx 133 (radians per hour). In radians per minute it will be 60 times less, about 2.2. This is the answer.