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.
Hint: The angle of elevation is that angle that the line of sight makes with the horizontal. Construct a triangle representing the situation. Use this to get a equation first involving x and y, then differentiate to involve dx/dt and dy/dt. Think about what it means that the plane is getting closer to the kangaroo.
apparently you didn't bother to follow the hint, so I'll give you a push...
dx/dt = -600 mi/hr
h = 2
tanθ = 2/x
sec^2θ dθ/dt = -2/x^2 dx/dt
Now plug in your numbers and find dθ/dt.
watch the units.
To solve this problem, let's follow the hint and construct a triangle to represent the situation.
Let's start by assuming that the kangaroo is at point K on the ground, the airplane is at point A in the sky directly above the kangaroo, and the line of sight between them forms the hypotenuse of a right triangle. Let the distance between the kangaroo and the airplane be x miles, and let the altitude of the airplane (vertical distance from the ground) be y miles.
We are given that the altitude of the airplane is constant at 2 miles, and the speed of the airplane is constant at 600 miles per hour. We want to find the rate of change of the angle of elevation of the kangaroo's line of sight (θ) with respect to time (t) when x = 3 miles.
Using the triangle, we can relate the variables x, y, and θ as follows:
tan(θ) = y/x
Differentiating both sides of this equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2
Since we are given that the altitude of the airplane is constant at 2 miles, dy/dt = 0. Hence, the equation becomes:
sec^2(θ) * dθ/dt = -y * dx/dt / x^2
We need to find dθ/dt when x = 3 miles. To do this, we need to find the corresponding values for y and θ.
Using the triangle, we can apply the Pythagorean theorem:
x^2 + y^2 = (2 miles)^2
Substituting x = 3 miles into the equation, we can solve for y:
(3 miles)^2 + y^2 = (2 miles)^2
9 + y^2 = 4
y^2 = 4 - 9
y^2 = -5 (which is impossible)
Since we cannot have a negative value for y, we conclude that the airplane is not directly above the kangaroo at x = 3 miles. Therefore, we cannot find the rate of change of the angle of elevation in this specific scenario.
However, if you are given a different value for x where the airplane is directly above the kangaroo, you can substitute that value into the equation and solve for dθ/dt.