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...
See
This Answer NowStart your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
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.
Further Reading