SOLUTION: I'm really struggling with this. Please help me out! (Not actually a test problem: This is Calc.) A plane flies horizontally at an altitude of 6 km and passes directly over a

Algebra ->  Test -> SOLUTION: I'm really struggling with this. Please help me out! (Not actually a test problem: This is Calc.) A plane flies horizontally at an altitude of 6 km and passes directly over a       Log On


   

Question 1167520: I'm really struggling with this. Please help me out! (Not actually a test problem: This is Calc.)
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?
Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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 = alpha = alpha%28t%29  which represents the elevation angle.


The length of the leg  |VT| = L(t)  is the covered distance, and the derivative  %28dL%28t%29%29%2F%28dt%29 
is the speed of the plane along the line y = 6 km at time moment t.


We can write  tan%28alpha%29 = 6%2FL,  or  L = L(t) = 6%2Ftan%28alpha%28t%29%29.


Take the time derivative, considering L(t) as a composite function. You will get

    %28dL%28t%29%29%2F%28dt%29 = d%2F%28dt%29  %286%2Ftan%28alpha%28t%29%29%29 = d%2F%28dt%29 %28%286%2Acos%28alpha%28t%29%29%29%2Fsin%28alpha%28t%29%29%29%29) = %28-6%29%2Fsin%5E2%28alpha%28t%29%29 * alpha' .    (1)


We are given that  at the time moment t  alpha%28t%29 = pi%2F3 radians  and  alpha' = -pi%2F3 radians per minute.


We substitute these values into formula (1),  and we get the speed of the plane

    %28dL%28t%29%29%2F%28dt%29 = [ %28-6%29%2Fsin%5E2%28pi%2F3%29 ] * [ -pi%2F3 ] = [ %28-6%29%2F%28%28sqrt%283%29%2F2%29%29%5E2 ] * [ -pi%2F3 ] = (%286%2A4%29%2F3) * %28pi%2F3%29 = 8%2A%283.14159%2F3%29 = 8.3776 kilometers per minute,

    or  8.3776*60 = 502.65 kilometers per hour.

At this point, the problem is solved completely.