shortly after taking off, a plane is climbing at an angle of 30° and traveling at a constant speed of 640 ft/sec as it passes over a ground radar tracking station. At that instant of time, the altitude of the plane is 1000 ft. How fast is the distance between the plane and the radar station increasing at that instant of time?
As usual, draw a diagram:
You have a triangle, ABC, with angle A = 30 deg.
The plane is at C
The station is at B
When the plane is overhead, BC = 1000, so AB = 1000sqrt(3)
Angle B subtends side b, and is sweeping through 90 deg at the moment in question
Let a be the distance from the station to the plane.
Now, at any time, b/sinB = a/sin30
b/2 = asinB
b'/2 = a' sinB + acosB B'
Now, when C is directly overhead, B = 90 deg.
b' is the speed of the plane, 640
320 = a' sin 90 + aB' cos 90
320 = a'
Makes sense, since at the moment in question, the hypotenuse of our triangle is twice the altitude of the plane. So, it will be increasing twice as fast as the distance (height) at that time.
To find the rate at which the distance between the plane and the radar station is changing, we can use trigonometry and calculus.
Let's first draw a diagram representing the situation:
Plane
/
/
/ 1000 ft
/
/θ
Radar Station
In this diagram, θ represents the angle of climb, which is 30° in this case, and the side adjacent to the angle (1000 ft) represents the altitude of the plane. We want to find the rate at which the distance between the plane and the radar station is changing, which is the rate of change of the hypotenuse.
To solve this problem, we need to express the relationship between the changing quantities in the problem. Let's define the variables:
- x = distance between the plane and the radar station (in feet)
- h = altitude of the plane (in feet)
Given that the plane is climbing at an angle of 30°, we can determine that:
h = 1000 ft.
Now, we need to find an expression that relates x and h. We can use trigonometry to do that. The tangent function is defined as the ratio of the opposite side (h) to the adjacent side (x). So, we have:
tan(θ) = h / x
Substituting the known values:
tan(30°) = 1000 ft / x
Now, we can solve for x:
x = 1000 ft / tan(30°)
Using a calculator, we find that:
x ≈ 1732.05 ft
Now, we need to differentiate the equation to find the rate at which x is changing with respect to time. Taking the derivative of both sides with respect to time (t) gives us:
d(x) / dt = d(1000 ft / tan(30°)) / dt
To find this derivative, we can use the chain rule. The derivative of 1000 ft with respect to time is zero because it is a constant. So, we only need to differentiate the term tan(30°) with respect to time:
d(x) / dt = -1000 ft / (tan(30°)^2) * d(tan(30°)) / dt
The derivative of tan(30°) with respect to time can be found using calculus or by recognizing that it is a constant. Since the plane is traveling at a constant speed, the angle of climb does not change, so d(tan(30°)) / dt = 0.
Therefore, we have:
d(x) / dt = -1000 ft / (tan(30°)^2) * 0
As the derivative of the constant term is zero, we find that the rate at which the distance between the plane and the radar station is changing is zero.
So, at that instant of time, the distance between the plane and the radar station is not changing.