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From a car traveling east at 40 mi.per hr., an airplane traveling horizontally
north at 100 mi.per hr, is visible 1 mi. east, 2 mi. south and 2 mi. up.
Find when the two will be nearest together.
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We place the initial position of the car at the origin of the coordinate system (x,y,z) = (0,0,0).
Then the airplane's initial position is (x,y,z) = (1,-2,2).
The trajectory of the car in time is (40t,0,0).
The trajectory of the plane in time is (0,-2+100t,2).
The vector from the car to the airplane in the coordinate form is (-40t,-2+100t,2).
The square of the length of this vector (= the distance between the objects) is
d^2(t) = = =
= .
The distance d(t) is minimum when d^2(t) is minimum.
d^2(t) is minimum at t = " " = - = 0.017241 of an hour = 1.0345 minutes.
At this time moment, the square of the distance is d^2(t) = = 8.0069 mi^2,
hence, the distance itself is d(t) = = 2.830 miles.
Compare it with the initial distance d(0) = = = 3 miles.
Solved.
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On finding minimum of a quadratic function, see the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
in this site.