An airplane flys at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will tak
e 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.
Let us say that y = vertical height and z =
hypotenuse <span>
dz/dt = 600
dy/dt = 0 since y (altitude) is constant
sin(θ) = 2/3
Calculating for the angle: θ = arcsin(2/3) = 0.729728 radians
sin(θ) = y/z
z sin(θ) = y <span>We then differentiate the two sides with respect to time t:
dz/dt sin(θ)+ z cos(θ) dθ/dt = dy/dt
(600) (2/3) + (3) cos (0.729728) dθ/dt = 0
400 + [3 / cos (0.729728) ] dθ/dt = 0
400+ 4.0249 dθ/dt = 0
dθ/dt = -98.38 radians/hour
= -99.38/60 radians / min
= -1.66 radians / min = 2PI - 1.66 radians / min <span>= 4.62 radians / min (ANSWER)</span></span></span>
