Rate of Change

Question

At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 20 knots and ship B is sailing north at 18 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

Draw a diagram of where the ships are at noon and 4 PM.  At noon, let B be at (0,0), then A is at (-30,0).  At 4 PM B is at (0,72) and A is at (-110,0).  A, B and the origin form a right triangle that is getting bigger with time.  By the Pythagorean Theorem, the distance between A and B is given by 

 

Z^2 = X^2 + Y^2  

 

In particular, at 4 PM

 

Z^2 = (-110)^2 + 72^2 = 12100 + 5184 = 17284 so 

 

Z = 131.47 nm at 4 PM

 

Note that X is negative number since A is sailing west.  

 

Implicitly differentiate the distance equation with respect to time to get 

 

2Z(dZ/dt) = 2X(dX/dt) + 2Y(dY/dt)

 

Now plug in the values that you know 

 

2(131.47)(dZ/dt) = 2(-110)(-30) + 2(72)(18)

262.94(dZ/dt) = 6600 + 162 = 6762

 

dZ/dt = 25.72 knots