An airplane flies at an altitude of y = 5 miles toward a point directly over an observer. The speed of the plane is 500 miles per hour. Find the rates at which the angle of elevation θ is changing when the angle is θ = 45°, θ = 60°, and θ = 80°.
To find the rates at which the angle of elevation θ is changing, we can use trigonometry and related rates.
Let's consider a right triangle formed by the airplane, the observer, and the point on the ground directly below the airplane. Here, the altitude of the airplane (y) is the opposite side, and the distance between the observer and the point on the ground is the adjacent side.
We are given that the altitude of the airplane is y = 5 miles, and the speed of the plane is 500 miles per hour. Let's denote the distance between the observer and the point on the ground as x.
Now, we can use trigonometry to establish the relationship between x, y, and the angle of elevation θ:
tan(θ) = y / x
Taking the derivative of both sides with respect to time t, we get:
sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2
Since dx/dt represents the rate at which the distance x is changing, and since the plane is flying directly over the observer, we can assume that dx/dt is equal to the speed of the plane (500 miles per hour).
We are asked to find the rates at θ = 45°, θ = 60°, and θ = 80°.
For θ = 45°:
Since tan(45°) = 1, we have:
1 = 5 / x
x = 5 miles
Plugging this into our equation, we have:
sec^2(45°) * dθ/dt = (5 * 5 - 5 * 500) / (5^2)
sec^2(45°) * dθ/dt = -495 / 25
dθ/dt = -495 / (25 * cos^2(45°))
For θ = 60°:
Since tan(60°) = sqrt(3), we have:
sqrt(3) = 5 / x
x = 5 / sqrt(3) miles
Plugging this into our equation, we have:
sec^2(60°) * dθ/dt = (5 * 5 / sqrt(3) - 5 * 500) / (25 / 3)
sec^2(60°) * dθ/dt = (-500sqrt(3) - 2500) / 25
dθ/dt = (-500sqrt(3) - 2500) / (25 * cos^2(60°))
For θ = 80°:
Since tan(80°) is approximately 5.671, we have:
5.671 = 5 / x
x = 5 / 5.671 miles
Plugging this into our equation, we have:
sec^2(80°) * dθ/dt = (5 * 5 / 5.671 - 5 * 500) / (25 / (5.671)^2)
sec^2(80°) * dθ/dt = (-5 * 500 + 5 * 5.671) / (25 / (5.671)^2)
dθ/dt = (-5 * 500 + 5 * 5.671) / (25 * cos^2(80°))
These are the rates at which the angle of elevation θ is changing when θ = 45°, θ = 60°, and θ = 80°, respectively.
To find the rates at which the angle of elevation is changing, we can use the definition of the tangent function. The tangent of an angle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In this case, the angle of elevation is the angle between the observer, the point on the ground directly below the airplane, and the line connecting the observer to the airplane.
Let's denote the distance between the observer and the point directly below the airplane as x miles.
Given that the altitude of the airplane is y = 5 miles, we can form a right triangle with the observer, the point directly below the airplane, and the airplane itself. The side opposite the angle of elevation is the altitude of the airplane, y, and the side adjacent to the angle is the distance between the observer and the point directly below the airplane, x.
Using the tangent function, we have:
tan(θ) = opposite/adjacent = y/x
Solving for x, we get:
x = y/tan(θ)
Now, let's find the rates at which the angle of elevation θ is changing when the angle is θ = 45°, θ = 60°, and θ = 80°.
To do this, we need to find the rates at which x (the distance between the observer and the point directly below the airplane) is changing with respect to time.
Given the speed of the plane is 500 miles per hour, the rate of change of x with respect to time (dx/dt) is equal to the speed of the plane, which is 500 miles per hour.
So, let's calculate the rates at which the angle of elevation θ is changing when the angle is θ = 45°, θ = 60°, and θ = 80°.
When θ = 45°:
x = 5/tan(45°) = 5/1 = 5 miles
dx/dt = 500 miles/hour
When θ = 60°:
x = 5/tan(60°) = 5/sqrt(3) miles
dx/dt = 500 miles/hour
When θ = 80°:
x = 5/tan(80°) = 5/4.70463 miles
dx/dt = 500 miles/hour
So, the rates at which the angle of elevation θ is changing when the angle is θ = 45°, θ = 60°, and θ = 80° are all equal to 500 miles per hour.
MathMate had a very nice solution to this question : )
Let
angle of elevation=θ
height of plane = H
horizontal distance from observer = x
tan(θ)=x/H
Use implicit differentiation
d(tan(θ))/dt = d(x/H)/dt
sec²(θ)dθ/dt = (dx/dt)/H
dθ/dt=(1/(Hsec²(θ))(dx/dt)
dθ/dt=(cos²(θ)/H)*(dx/dt)