A plane flies at a speed 500 km/hr at a constant height of 12 km. How rapidly is the angle of elevation to the plane changing when the plane is directly above a point 110 km away from the observer?
The angle of elevation is changing at
radians/hr (enter a positive value). Round your answer to 3 decimal places.
tanθ = 12/x
sec^2 θ dθ/dt = -12/x^2 dx/dt
so plug in your numbers, and note that sec^2 θ = 1 + tan^2 θ = 1 + 144/x^2
To find the rate at which the angle of elevation is changing, we can use the concept of trigonometry. Let's assume that the angle of elevation is represented by θ.
We know that the plane is flying at a constant height of 12 km, and it is directly above a point 110 km away from the observer.
In this situation, we have a right triangle formed by the observer, the point directly below the plane, and the plane itself. The side opposite to angle θ is the height of the plane, which is 12 km, and the side adjacent to angle θ is the horizontal distance between the observer and the point directly below the plane, which is 110 km.
We can use the tangent function to relate the angle θ with the sides of the triangle:
tan(θ) = opposite / adjacent
tan(θ) = 12 / 110
Now, we need to find the rate at which the angle of elevation is changing, with respect to time. Let's represent this rate as dθ/dt.
To find dθ/dt, we can differentiate the equation tan(θ) = 12 / 110 with respect to time t:
sec^2(θ) * dθ/dt = 0
Since the plane is flying at a constant speed and height, we can assume that the angle of elevation is changing at a constant rate. Therefore, dθ/dt is constant.
Now, we can solve for dθ/dt:
dθ/dt = 0 / sec^2(θ)
dθ/dt = 0
Therefore, the angle of elevation is not changing. The rate at which the angle of elevation is changing is 0 radians/hr.
To solve this problem, we can use trigonometry and related rates. Let's start by drawing a diagram:
/
h /
/
x /
A/
/
/
O/
Note: A is the observer, O is the plane, and x is the horizontal distance between the observer and the plane. The angle of elevation is the angle between the horizontal line AO and the line OA.
We are given:
- Plane's speed: 500 km/hr
- Height: 12 km
- Distance between the observer and the plane: 110 km
To find how rapidly the angle is changing, we need to find the rate of change of the angle with respect to time dθ/dt.
Now, we can use trigonometry to relate the variables:
sin(θ) = h / x
Differentiating both sides with respect to time (t), we get:
cos(θ) * dθ/dt = (dh/dt * x - h * dx/dt) / x^2
We need to find dθ/dt, so let's rearrange the equation:
dθ/dt = (dh/dt * x - h * dx/dt) / (x^2 * cos(θ))
Now, substitute the values into the equation:
dh/dt = 0 (since the plane is at a constant height)
dx/dt = -500 km/hr (negative because the plane is moving away from the observer)
h = 12 km
x = 110 km
θ = ?
To find θ, we can use the right triangle:
sin(θ) = h / x
sin(θ) = 12 km / 110 km
θ = arcsin(12/110)
Substitute these values into the equation and solve for dθ/dt:
dθ/dt = (0 * 110 km - 12 km * -500 km/hr) / (110 km^2 * cos(arcsin(12/110)))
Now we can calculate dθ/dt.