Harold is in an airplane that is flying at a constant height of 4505 feet away from a fixed observation point. Maude, whose eyes are 5 feet from the ground, is standing at this point and watching the plane; the angle between her line of sight (the line line between her eyes and the plane) and the horizontal is called the angle of elevation. If, at a particular moment in time, the angle of elevation is 60◦ and the plane is flying at a rate of 22,600 feet per minute, determine the rate at which the angle of elevation is changing. Give you answer in degrees per minute.

Question

Harold is in an airplane that is flying at a constant height of 4505 feet away from a fixed observation point. Maude, whose eyes are 5 feet from the ground, is standing at this point and watching the plane; the angle between her line of sight (the line line between her eyes and the plane) and the horizontal is called the angle of elevation. If, at a particular moment in time, the angle of elevation is 60◦ and the plane is flying at a rate of 22,600 feet per minute, determine the rate at which the angle of elevation is changing. Give you answer in degrees per minute.

Answer 1

Always draw a diagram for these problems. You will see that if the plane is x feet away from Maude's position, then

tanθ = 4500/x
so,
sec^2θ dθ/dt = -4500/x^2 dx/dt
When θ=60°, that gives x = 2250, so
4 dθ/dt = -4500/2250^2 * 22600
dθ/dt = -226/45 ≈ -5.02

Of course, that is in radians/min, so just convert that to °/min

Answer 2

To find the rate at which the angle of elevation is changing, we need to take the derivative of the angle with respect to time.

Let's denote the angle of elevation as θ and the distance between Maude's eyes and the airplane as x.

From the given information, we know that the height of the airplane is 4505 feet and Maude's eyes are 5 feet from the ground. Therefore, the distance x is equal to the height of the airplane minus the height of Maude's eyes:

x = 4505 - 5 = 4500 feet.

We also know that the plane is flying at a rate of 22,600 feet per minute. This means that the distance x is changing with respect to time:

dx/dt = 22,600 feet per minute.

Now, we need to find the relationship between x and θ. From trigonometry, we have:

tan(θ) = x / 4505.

To isolate x, we can rearrange this equation to:

x = 4505 * tan(θ).

Now, we can differentiate both sides of the equation with respect to time t:

dx/dt = d(4505tan(θ))/dt.

Using the chain rule, we can compute the derivative:

dx/dt = 4505 * sec^2(θ) * dθ/dt.

Since we are looking for the rate at which the angle of elevation is changing (dθ/dt), we can rearrange the equation:

dθ/dt = (dx/dt) / (4505 * sec^2(θ)).

Substituting the given values, we have:

dθ/dt = (22600) / (4505 * sec^2(60◦)).

To convert degrees to radians, we can use the conversion factor π/180:

dθ/dt = (22600) / (4505 * sec^2((π/3) radians)).

Finally, we can simplify and calculate the value:

dθ/dt = (22600) / (4505 * (2/3)^2) = 2 radians per minute.

Therefore, the rate at which the angle of elevation is changing is 2 degrees per minute.