# 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?

May 18, 2015

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 $\left(0 , 0\right)$.

Aircraft carrier Alpha at noon is 100 km due East from Beta, so its coordinates are $\left(100 , 0\right)$ with unit of measurement to be 1 km.

At time $t$ (unit of time is 1 hour) the coordinates of Alpha that sails due West (negative direction of the X-axis) at speed $12$ km/h will be $\left(100 - 12 t , 0\right)$.
At the same time $t$ the coordinates of Beta that sails due South (negative direction of the Y-axis) at speed $10$ km/h will be $\left(0 , - 10 t\right)$.

The square of a distance between two points with coordinates $A \left({x}_{1} , {y}_{1}\right)$ and $B \left({x}_{2} , {y}_{2}\right)$ is calculated using the Pythagorean formula
${d}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$

In our case the square of a distance between the ships at moment $t$ equals:
${d}^{2} = {\left(100 - 12 t\right)}^{2} + {\left(10 t\right)}^{2} = 244 {t}^{2} - 2400 t + 10000$
Graphically, it is a parabola that tends to positive infinity as $t$ increases to infinity. It intersects the Y-axis at point $Y = 10000$ at time $t = 0$, as it should.

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 $t$ that minimizes the square of a distance expressed by the above formula.

For a general quadratic function $a {x}^{2} + b x + c$ the extreme (minimum or maximum) is reached when $x = - \frac{b}{2 a}$ since the first derivative of this function is $2 a x + b$ and it's equal to zero at $x = - \frac{b}{2 a}$.

The square of a distance is a quadratic polynomial with negative discriminant $D = {2400}^{2} - 4 \cdot 244 \cdot 10000 = - 4000000$. It means, it never equal to zero and remains always positive (as at $t = 0$). It's minimum is reached at point $t = \frac{2400}{2 \cdot 244} \cong 4.918$.

So, the minimum distance will be achieved after approximately $4.918$ (almost $5$) hours into the trip.
The square of a distance at $t = 4.918$ hours equals to
$244 \cdot {4.918}^{2} - 2400 \cdot 4.918 + 10000 \cong 4098.36$
And the minimum distance $d \cong \sqrt{4098.36} \cong 64$ km.