An airplane in Australia is flying 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.
To find the rate at which the angle of elevation of the kangaroo's line of sight is changing, we can use trigonometry and related rates.
Let's define the following variables:
- Let "θ" be the angle of elevation of the kangaroo's line of sight.
- Let "x" be the horizontal distance between the kangaroo and the plane.
- Let "y" be the vertical distance between the kangaroo and the plane.
We are given the following information:
- The plane is flying at a constant altitude of 2 miles.
- The plane is flying at a constant speed of 600 miles per hour.
- We need to find the rate at which the angle of elevation (θ) is changing when the distance "x" is 3 miles.
First, let's draw a diagram to illustrate the situation:
/|
/ |
/ | y
/ |
/ θ |
/____|
x
From the diagram, we can see that:
- tan(θ) = y/x
- y = 2 miles (constant altitude of the plane)
Differentiating both sides of the equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (1/x) * dx/dt
Now, we need to find dx/dt. Since the plane is traveling at a constant speed of 600 miles per hour, dx/dt = 600 mph.
Substituting the given values into the equation, we have:
sec^2(θ) * dθ/dt = (1/3) * (600/60) [converting mph to miles per minute]
Simplifying the equation, we get:
sec^2(θ) * dθ/dt = 10
To find dθ/dt, we need to find sec^2(θ):
From the right triangle, we have:
sec(θ) = hypotenuse (h) / adjacent side (x)
Using the Pythagorean theorem, we have:
h^2 = x^2 + y^2
h^2 = x^2 + (2 miles)^2
h^2 = x^2 + 4 miles^2
Taking the square root of both sides, we get:
h = sqrt(x^2 + 4 miles^2)
Substituting the values into the equation, we have:
sec(θ) = sqrt(x^2 + 4 miles^2) / x
Taking the reciprocal of both sides, we have:
cos(θ) = x / sqrt(x^2 + 4 miles^2)
Finally, substituting cos(θ) into the previous equation, we have:
(1 + tan^2(θ)) * dθ/dt = 10
Simplifying further, we get:
1 + tan^2(θ) = 10 / dθ/dt
tan^2(θ) = 10 / dθ/dt - 1
Now, we can substitute the value of θ when x = 3 miles into the equation and solve for dθ/dt.
Once we have dθ/dt, we will have the rate at which the angle of elevation of the kangaroo's line of sight is changing, given in radians per minute.
Draw a diagram. If x is the distance from the kangaroo, then
tanθ = 2/x
sec^2θ dθ/dt = -2/x^2 dx/dt
so, when x=3, using dx/dt = -10 mi/min
13/3 dθ/dt = -2/9 * -10 = 20/9
dθ/dt = 2/3 rad/min