Derivative/Rate QuestionA 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...
Derivative/Rate Question
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
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:
`tan(theta)=y/x`
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.
`dx/dt=(-5*sec^2(pi/3)*(-pi/6))/(tan^2(pi/3))`
`dx/dt=(-5*2^2*(-pi/6))/(sqrt(3)^2)`
`dx/dt=((20pi)/6)/3=((10pi)/3)*(1/3)`
`dx/dt=((10pi)/9)=3.491`
The plane is traveling at 10pi/9 or 3.491 km/min.