An airplane flys at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. How fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? Give your answer in radians per minute.
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To find the rate of change of angle of elevation, we can use the concept of related rates. Let's assume that the ground is horizontal, the kangaroo is on the ground, and the angle of elevation is measured from the horizontal plane.
We are given the following information:
- Airplane altitude: 2 miles
- Airplane speed: 600 miles per hour
Let's introduce some variables:
- Let "x" represent the horizontal distance from the kangaroo to the airplane.
- Let "θ" represent the angle of elevation of the kangaroo's line of sight.
Since the altitude of the airplane is constant at 2 miles, the vertical height of the triangle formed by the airplane, kangaroo, and the ground remains constant.
As the airplane flies directly over the kangaroo, the horizontal distance "x" and the altitude of the airplane form sides of a right triangle. The hypotenuse of this triangle is the distance between the kangaroo and the airplane, which is given as 3 miles.
Using the Pythagorean theorem, we can write an equation relating "x" and "θ":
x^2 + 2^2 = 3^2
x^2 + 4 = 9
x^2 = 5
Differentiating both sides of the equation with respect to time (t), we get:
2x * (dx/dt) = 0
Since "x" represents the horizontal distance from the kangaroo to the airplane, the rate of change of "x" with respect to time is the speed of the airplane, which is given as 600 miles per hour.
Therefore, substituting dx/dt = 600 into the equation, we have:
2x * 600 = 0
Simplifying the equation, we find:
x = 0
This implies that the horizontal distance "x" is not increasing or decreasing (i.e., it remains constant) when the distance between the kangaroo and the airplane is 3 miles.
Now, to find the rate of change of the angle of elevation, we need to differentiate the equation relating "x" and "θ" implicitly with respect to time:
2x * (dx/dt) + 0 = 0
2x * (dx/dt) = 0
Since x is constant in this scenario, the rate of change of x with respect to time is zero, resulting in:
0 = 0
This means that the rate of change of the angle of elevation is zero radians per minute when the distance from the kangaroo to the plane is 3 miles.