Question

An airplane is flying at an altitude of 5 mi and passes directly over a radar antenna as shown in the figure. When the plane is 10 mi from the antenna (s = 10), the radar detects that the distance s is changing at the rate of 300mph. What is the rate of change of the angle of depression (θ)?

a) 17.5 degrees per hour
b) 19.5 degrees per hour
c) 20 degrees per hour
d) 22 degrees per hour

Answer 1

The rate of change of the angle of depression (θ) is approximately 20 degrees per hour.

Step-by-step explanation:

To find the rate of change of the angle of depression (θ), we can use the relationship between the rate of change of distance (ds/dt) and the rate of change of the angle of depression (dθ/dt). The tangent of the angle of depression is equal to the altitude divided by the distance from the antenna (tan(θ) = 5/s).

We know that ds/dt = 300 mph, so we can differentiate the tangent equation with respect to time to find dθ/dt. Differentiating both sides gives sec^2(θ) * dθ/dt = -5/s^2 * ds/dt.

Plugging in the given values (ds/dt = 300 mph and s = 10 mi), we can solve for dθ/dt, which is approximately 20 degrees per hour (c).

Answer 2

To find the rate of change of the angle of depression (θ) in a related rates problem, one must apply implicit differentiation to the function that relates θ to the distance s, which is determined by the altitude of the airplane and its horizontal distance from the radar antenna.

Step-by-step explanation:

The student is dealing with a related rates problem in trigonometry, a fundamental concept in calculus. We need to find the rate of change of the angle of depression (θ) as the plane moves away from the radar antenna. Given the information and using the concept that θ is related to the distance s via the tangent function (tan(θ) = altitude / s), we can apply implicit differentiation with respect to time (t) to find dθ/dt, which represents the rate of change of the angle of depression.

Using the given data:
Altitude = 5 miles
s (distance from the antenna when the rate is measured) = 10 miles
ds/dt (rate of change of s) = 300 mph

To find the rate of change of θ, we differentiate the equation tan(θ) = 5 / s with respect to time, which gives us sec2(θ) * (dθ/dt) = -5 / s2 * (ds/dt). Given that s = 10 miles and ds/dt = 300 mph, we can use the value of tan(θ) at s = 10 miles to find θ and then calculate sec2(θ). This will allow us to solve for dθ/dt and determine the rate of change of the angle of depression.