A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/4, this angle is decreasing at a rate of π/4 rad/min. How fast is the plane traveling at that time? (Round your answer to two decimal places.)
That looks hard
Draw a diagram. If the plane's horizontal distance is x,miles, then
tanθ = 5/x
sec^2θ dθ/dt = -5/x^2 dx/dt
So plug in your numbers and find dx/dt
To find the speed of the plane, we need to use the concept of trigonometry and related rates.
Let's consider the given situation:
- The altitude of the plane: 5 km
- The angle of elevation: π/4
- The rate at which the angle of elevation is decreasing: π/4 rad/min
Let's denote the distance from the telescope to the plane as D and the speed of the plane as v.
We can draw a right triangle to represent this situation, where:
- The adjacent side of the triangle represents the horizontal distance from the telescope to the plane (D).
- The opposite side of the triangle represents the altitude of the plane (5 km).
Since the tangent of an angle is equal to the ratio of the opposite side to the adjacent side (tan θ = opposite/adjacent), we can write:
tan (π/4) = 5/D
To find the value of D, we can solve this equation for D:
D = 5/tan (π/4)
Using the fact that tan (π/4) = 1, we have:
D = 5/1 = 5 km
Now that we have the value of D, we can differentiate both sides of the equation with respect to time (t) using the chain rule:
dD/dt = d(5/tan (π/4))/dt
The left side represents the rate at which the distance is changing, which is the speed of the plane (v). The right side represents the rate at which D is changing with time (dD/dt):
v = d(5/tan (π/4))/dt
Taking the derivative of 5/tan (π/4) with respect to t:
v = d(5/tan (π/4))/dt = -5sec^2 (π/4) d(tan (π/4))/dt
Since d(tan (π/4))/dt is given as -π/4 rad/min, we can substitute this value:
v = -5sec^2 (π/4) * (-π/4) rad/min
Using the fact that sec^2 (π/4) = 2, we have:
v = -5 * 2 * (-π/4) rad/min
Simplifying, we get:
v = 5π/4 rad/min
So, the plane is traveling at a speed of 5π/4 rad/min at that time.
Note: The speed is given in rad/min, as the units cancel out during the calculation.
To solve this problem, we can use the concept of related rates.
Let's denote the distance of the plane from the telescope as "x" and the height of the telescope as "h."
From the given information, we have the following:
Altitude of the plane, h = 5 km
The angle of elevation, θ = π/4
Rate of change of angle of elevation, dθ/dt = -π/4 rad/min (negative sign because the angle is decreasing)
We need to find the rate at which the distance from the telescope is changing, dx/dt.
We can use trigonometry to relate the angle of elevation and the distance from the telescope:
tan(θ) = h / x
Differentiating both sides of the equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (1 / x) * dx/dt
Now, substitute the given values:
sec^2(π/4) * (-π/4) = (1 / x) * dx/dt
Simplifying this equation, we have:
(2 / √2) * (-π/4) = (1 / x) * dx/dt
Simplify further:
-π / 2√2 = (1 / x) * dx/dt
Now, rearrange the equation to solve for dx/dt:
dx/dt = -(π / 2√2) * x
Finally, substitute the given altitude of the plane, h = 5 km, into the equation:
x = √(h^2 + x^2)
x = √(5^2 + x^2)
x^2 = 25 + x^2
0 = 25
x = 5 km
Now, substitute x = 5 km into the equation for dx/dt:
dx/dt = -(π / 2√2) * 5
Calculating this expression, we find:
dx/dt = -5π / 2√2 ≈ -5.56 km/min
Therefore, the plane is traveling at a speed of approximately 5.56 km/min at that time.