Calculus Problem

Question

One ship is sailing east at 60km an hour and another is sailing south at 50km an hour. The slower ship crosses the path of the faster ship at noon when the faster ship was there one hour earlier. Find the time at which the two ships were closest to each other.

Answer 1

Think of the triangle formed by the two ships and the point where their paths cross.  It's a right triangle where the legs are the line segments from the intersection point to the two ships, and the hypotenuse is the line segment between the ships.  Obviously the length of the hypotenuse is the distance between the ships, and that's what we want to minimize.

 

Let's call t the number of hours since noon, when the slower ship crossed the intersection point.  So the length of that ship's "leg" is 50t.  The length of the other ship's leg is 60 (t + 1) = 60t + 60.  So by the pythagorean theorem, the distance between the ships is √[(50t)² + (60t + 60)²] = √(2500t² + 3600t² + 7200t + 3600) = √(6100t² + 7200t + 3600)

 

Let's make an additional simplifying deduction.  Normally you would differentiate the expression for the distance to dind its minumum.  But the nature of this problem is that the square of the distance is going to be at a minimum at the same time that the distance itself is at a minimum.  That makes the differentiation a little easier.  You can't always simplify min/max problems like this, but the geometry of this problem lets you do it.

 

So you want to minimize 6100t² + 7200t + 3600:

 

Minimize 6100t² + 7200t + 3600

12200 t + 7200 = 0

12200 t = -7200

t = -7200 / 12200 = -36 / 61

 

Since t is measured in hours after noon, -36/61 = approximately 35.41 minutes before noon, or 11:24.59