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hellion
28.07.2019 •
Mathematics
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.
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Ответ:
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
The angle of elevation of the Kangaroo's line of sight is increasing at:
= 4.62 radians / min
Ответ:
.25
Step-by-step explanation: