# At noon, aircraft carrier Alpha is 100 kilometers due east of destroyer Beta. Alpha is sailing due west at 12 kilometers per hour. Beta is sailing south at 10 kilometers per hour. In how many hours will the distance between the ships be at a minimum?

##### 1 Answer

Let's define the coordinates in this space, assuming the Earth is large enough for this problem to follow Euclidean geometry (to take into account spherical shape of our planet would significantly complicate the issue, but, in theory, might be considered).

We will assign East as the positive direction of the X-axis and North as the positive direction of the Y-axis.

The origin of coordinates we put at the initial location at noon of destroyer Beta. So, at noon Beta is at position

Aircraft carrier Alpha at noon is 100 km due East from Beta, so its coordinates are

At time

At the same time

The square of a distance between two points with coordinates

In our case the square of a distance between the ships at moment

Graphically, it is a parabola that tends to positive infinity as

Obviously, the minimum of a square of a distance is reached at the same point as the minimum of a distance.

Therefore, to minimize the distance, we will find a moment in time

For a general quadratic function *first derivative* of this function is

The square of a distance is a quadratic polynomial with negative *discriminant*

So, the minimum distance will be achieved after approximately *hours* into the trip.

The square of a distance at *hours* equals to

And the minimum distance *km*.