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/4 rad/min. How fast is the plane traveling at that time? (Round your answer to two decimal places.)
Please stop posting your question on other people's work, If I did the same to you, Im sure you wouldnt like it.
Draw the triangle. h=5km, let x be the the horizontal distance.
Then angle of elevation=arctan h/x
tan angle=h/x
x= h ctn angle
dx/dt=h d ctnTheta/dt * dTheta/dt
so solve for dx/dt, you know dTheta/dt, and you can take the derivative of ctn
a jet plane begins a steady climb at 17 degrees and flies 5 km measured along the ground. Wht is its change of altitude
To solve this problem, we can use the concept of related rates. We need to find the rate at which the distance between the plane and the telescope is changing, which can be determined by finding the derivative of this distance with respect to time.
Let's consider the triangle formed by the plane, the telescope, and the vertical line from the telescope to the plane. The distance between the plane and the telescope represents the opposite side of this triangle, and the altitude of 5 km represents the adjacent side. The angle of elevation is π/3 radians.
Using trigonometric functions, we can establish the relationship between the opposite side, the adjacent side, and the angle π/3:
tan(π/3) = opposite/adjacent
√3 = opposite/5000 (since the adjacent side is 5000 meters or 5 km)
Now, let's differentiate both sides of the equation implicitly with respect to time:
d/dt (√3) = d/dt (opposite/5000)
The derivative of a constant is 0, so the left side simplifies to:
0 = (d/dt opposite)/5000
Now we need to find the rate of change of the opposite side (d/dt opposite). We are given that the angle of elevation is decreasing at a rate of π/4 radians per minute (dθ/dt = -π/4 rad/min).
To find d/dt opposite, we can differentiate with respect to time:
0 = (d/dt opposite)/5000
0 = (dθ/dt) * (d opposite/dθ)
Substituting the given values, we have:
0 = (-π/4) * (d opposite/dθ)
Now we can solve for d opposite/dθ:
d opposite/dθ = 0 / (-π/4)
d opposite/dθ = 0
This means that the opposite side is not changing with respect to the angle of elevation. Therefore, the distance between the plane and the telescope is not changing. This implies that the plane is flying directly towards the telescope.
Given that the distance is constant, the rate at which the plane is traveling is also constant. However, since the question asks us to find the speed of the plane at that specific time, we can conclude that the plane is not moving at that moment (since the distance is not changing).
Therefore, the speed of the plane at that time is 0 km/h.