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A plane flies horizontally at an altitude of 6 km and passes directly over a tracking telescope on the ground.
When the angle of elevation is π/3, this angle is decreasing at a rate of π/3 rad/min.
How fast is the plane traveling at that time?
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We have horizontal line y = 0 representing the ground surface.
We also have horizontal line y = 6 km representing the trajectory of the plane.
The telescope (the observer) is point T on the ground.
We have point V on the line y = 6 km representing the plane when it is directly over an observer,
so the line TV is perpendicular to the ground surface y = 0.
Let point P on the line y = 6 represents the current position of the plane P = P(t).
The line VP is the trajectory of the plane, and triangle TVP is a right-angled triangle
with angle VPT = = which represents the elevation angle.
The length of the leg |VT| = L(t) is the covered distance, and the derivative
is the speed of the plane along the line y = 6 km at time moment t.
We can write = , or L = L(t) = .
Take the time derivative, considering L(t) as a composite function. You will get
= = ) = * ' . (1)
We are given that at the time moment t = radians and ' = radians per minute.
We substitute these values into formula (1), and we get the speed of the plane
= [ ] * [ ] = [ ] * [ ] = () * = = 8.3776 kilometers per minute,
or 8.3776*60 = 502.65 kilometers per hour.
At this point, the problem is solved completely.