A plane flies horizontally at an altitude of 5 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?
I assume you made a diagram.
I let the angle of elevation be Ø, and the horizontal distance of the plane as x km
In mine,
tanØ = x/5
xtanØ = 5
dx/dt tanØ + x sec^2 Ø dØ/dt = 0
when Ø = π/3 , (I recognize the 30-60-90° triangle)
so
cosØ = 1/2
secØ = 2
sec^2 Ø = 4
dx/dt (√3) + (5/√3)(4)(-π/3) = 0
√3 dx/dt = -(5/√3)(-π/3) = 5π/(3√3)
dx/dt = 5π/9 km/min
or
(5π/9)(60) km/h which is appr 104.7 km/h
better check my arithmetic, I did not write it on paper first.
From triangle,cotu=x/5 differentiate both side cosec^2udu/dt=dx/dt/5subtitute u and du/dt and solve for dx/dt
To solve this problem, we can use trigonometry and related rates.
Let's consider the situation after a short time interval Δt. The plane will move horizontally a distance Δx, while the angle of elevation decreases by Δθ.
We are given that the angle of elevation is π/3, and it is decreasing at a rate of π/3 radians per minute:
dθ/dt = -π/3 rad/min
We need to find the rate at which the plane is traveling, dx/dt.
We can use the tangent function to relate the angle of elevation to the horizontal distance:
tan(θ) = altitude / horizontal distance
Plugging in the given values:
tan(π/3) = 5 km / x
Now, let's take the derivative of both sides with respect to time t:
sec^2(θ) * dθ/dt = 0 - (5 km / x^2) * dx/dt
Since sec^2(π/3) = 4/3, we have:
(4/3) * (-π/3 rad/min) = -(5 km / x^2) * dx/dt
Now we can solve for dx/dt:
dx/dt = -(4/3) * (π/3 rad/min) * (x^2 / 5 km)
Now, we need to find the value of x when the angle of elevation is π/3.
tan(π/3) = 5 km / x
√3 = 5 km / x
x = 5 km / √3
Substitute this value back into the equation to find dx/dt:
dx/dt = -(4/3) * (π/3 rad/min) * ((5 km / √3)^2 / 5 km)
Simplifying,
dx/dt = - (4/3) * (π/3 rad/min) * (25/3 km^2 / 5 km)
dx/dt = - (4/3) * (π/3 rad/min) * (5/3 km)
dx/dt = - (20/9) * π km/min
Therefore, the plane is traveling at a rate of (20/9) * π km/min.
To find the speed of the plane, we need to use trigonometry and relate the angle of elevation with the rate of change of the angle.
Let's consider the situation as shown below:
A
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Plane
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Tracking Telescope B
We will assume that point A represents the position of the plane at a given time, and point B represents the position of the tracking telescope.
We are given that the altitude of the plane is 5 km. Let's call the distance between A and B as x (see the diagram). The angle of elevation is the angle θ between the line AB and the horizontal ground. According to the problem, at a specific time, the angle θ is π/3 (60 degrees), and it is decreasing at a rate of π/3 rad/min (60 degrees per minute).
To find the speed of the plane, we can use trigonometry. We know that the tangent of the angle of elevation (θ) is equal to the opposite side (5 km) divided by the adjacent side (x).
tan(θ) = 5 km / x
Now, we need to differentiate both sides of the equation with respect to time (t) to find the rate of change of θ (dθ/dt) and x (dx/dt).
d(tan(θ))/dt = d(5/x)/dt
By applying the chain rule, we get:
sec^2(θ) * dθ/dt = -5/x^2 * dx/dt
Given that dθ/dt = -π/3 rad/min (as it is decreasing at that rate), we can substitute the values into the equation:
sec^2(π/3) * (-π/3) = -5/x^2 * dx/dt
sec^2(π/3) is equal to 4/3, so the equation simplifies to:
(4/3) * (-π/3) = -5/x^2 * dx/dt
Now, we can solve for dx/dt, which represents the speed of the plane:
(4/3) * (-π/3) = (-5/x^2) * dx/dt
To find dx/dt, we rearrange the equation:
dx/dt = [(4/3) * (-π/3) * x^2] / -5
The x^2 term represents the distance between the plane and the tracking telescope, which is equal to the altitude of the plane (5 km) squared. Substituting this value, we have:
dx/dt = [(4/3) * (-π/3) * (5^2)] / -5
Simplifying further yields:
dx/dt = -20π/9 km/min
Therefore, the speed of the plane at that time is -20π/9 km/min, where the negative sign indicates that the plane is traveling in the opposite direction of the tracking telescope.