Drone Path Warning Intervals and Closest Approach
A drone follows the parabolic path y=12−x^2/8, with x measured in kilometres east of a radar station and y in kilometres altitude; the usable path is −8≤x≤8. The radar dish is at (0,4). The drone's horizontal coordinate changes at a constant rate dx/dt=3 km/min. Mission rules require a warning whenever the straight-line distance from the radar to the drone is less than 7 km. The flight controller computes the closest approach by differentiating the distance (not the squared distance) and obtains one critical point, then stops. A second controller notes that the endpoints and the symmetry x↔−x must be considered, and that the distance function is not differentiable in a simple way if written with a square root at the closest point. You are asked to determine the warning interval in time, decide whether the closest approach triggers the warning, and check endpoint behaviour. Take t=0 when x=0.
1. Find the closest approach of the drone to the radar and justify that it is the global minimum, including endpoint checks.
Read the worked answer
Let D(x) be the squared distance: D=x^2+(y−4)^2=x^2+(8−x^2/8)^2. Expanding, D=x^2+64−2x^2+x^4/64=64−x^2+x^4/64. Differentiate: D'(x)=−2x+x^3/16=x(x^2/16−2). Critical points are x=0 and x^2=32, so x=±4sqrt2≈±5.657. Values: D(0)=64, D(±4sqrt2)=64−32+1024/64=48, D(±8)=64−64+4096/64=64. Thus the minimum squared distance is 48, so the closest distance is sqrt48=4sqrt3≈6.928 km. This is less than 7 km, so the closest approach does trigger the warning.
2. Determine the time intervals during which the warning is active. Express the answer exactly and explain how the quartic inequality is reduced.
Read the worked answer
The warning is active when the distance is less than 7, i.e. when D<49: 64−x^2+x^4/64<49, so x^4/64−x^2+15<0. Let u=x^2≥0. Then u^2/64−u+15<0, or u^2−64u+960<0. The roots are u=[64±sqrt(4096−3840)]/2=[64±16]/2, giving u=24 and u=40. Since the quadratic opens upward, the inequality holds for 24<u<40. Hence 2sqrt6<|x|<2sqrt10. With x=3t and t=0 at x=0, 2sqrt6<3|t|<2sqrt10, so the warning intervals are −2sqrt10/3<t<−2sqrt6/3 and 2sqrt6/3<t<2sqrt10/3. Numerically these are about (−2.11,−1.63) and (1.63,2.11) minutes.
3. At x=4, compute the rate of change of the straight-line distance from the radar with respect to time. State whether the drone is approaching or receding and whether the warning is active at that instant.
Read the worked answer
At x=4, D=64−16+256/64=52, so the distance is s=sqrt52=2sqrt13≈7.211 km, which is greater than 7, so the warning is not active at that instant. Differentiate D with respect to x: D'(x)=−2x+x^3/16. At x=4, D'(4)=−8+64/16=−8+4=−4. By the chain rule, dD/dt=D'(x)(dx/dt)=(−4)(3)=−12 km^2/min. Since s=sqrtD, ds/dt=(1/(2s))dD/dt=(1/(2sqrt52))(−12)=−6/sqrt52≈−0.832 km/min. The negative sign means the distance is decreasing, so the drone is approaching the radar at about 0.83 km/min. It has not yet entered the warning zone but is heading toward it.