A plane, traveling at a constant height of 5km above the ground, passes directly over a radar station. When the angle from the ground to the plane is (pi/3) radians, the angle is decreasing at a rate of (pi/6) radians/minute. How fast is the plane traveling?
1 Answer | Add Yours
The math model that represents this situation is a triangle. The plane has already flown beyond the radar station which is on the ground. A line segment from the radar station to the plane is the hypotenuse of the triangle. The vertical distance from the plane to ground is constant at 5 km. The horizontal distance (x) changes along the ground.
A relationship between the angle and the sides of the triangle is:
Since the vertical distance is constant, `tan(theta)=5/x`
Solving for x, `x=5/(tan(theta))`
Take the derivative of both sideswith respect to t:
`dx/dt=(-5*sec^2(theta)*(d theta)/dt)/(tan^2 (theta))`
Substiting in values when theta = pi/3 and d(theta)/dt = -pi/6.
The plane is traveling at 10pi/9 or 3.491 km/min.
We’ve answered 317,511 questions. We can answer yours, too.Ask a question