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An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 225 miles from the point and is moving at 450 miles per hour. The other plane is 300 miles from the point and has a speed of 600 miles per hour." "a. At what rate is the distance between the planes decreasing? b. If the controller does not intervene, how close will the planes come to each other?"

I understand that we can use implicit differentiation to solve part a, which shows that the difference is decreasing at 750 mph. However, I wonder why in part b, the answer is 30 minutes. In other words, the difference between the two planes (the hypotenuse of their right angle) is decreasing at a constant rate. Why is that so?

Haim
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1 Answers1

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Set of the first planes distance from the point as a function of time, do the same with the second plane. Then use the pythagorean thereom to get the distance inbetween them on terms of t. Finally differentiate that distance to get the rate of change of the distance between them.
And then it makes since that that distance is decreasing linearly because it is dependent on two linearly changing rates.
Although implicit differentiation may be quicker I would say that x=225-450t and y=300-600t. Therefore the distance between them is given by w=sqrt(x^2+y^2) which you can easily differentiate without having the derivative reference itself. Then for b, how close they come, it's an optimization problem. Find when dw/dt=0 then determine if it's a minimum (it may turn out there's no minimum and the closest they come is 0 ft apart when the graph crosses the x axis, if it does).

Will Fisher
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  • So we have w^2 = y^2 + x^2 , where w is the distance between them and y and x are the respective differences from the point.

    2w * dw/dt = 2y (dy/dt) + 2x (dx/dt)

    dw/dt = (y(dy/dt) + x(dx/dt)) / w

    But in this derivative, all three variables y,x,and w are changing at different rates, no?

     – Haim Nov 26 '15 at 10:16