At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 18 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Question

radar · Answer

I may be wrong, and if so, I hope someone will post to correct me.  But it appears to me that ship A and B have a speed or rate of 18 knots, but as a velocity they are 90 degrees apart, so the distance change would be the square root of their sum:$$d` = \sqrt{18 + 18}$$

radar · Answer

That is all the question is asking for...

phi · Answer

radar has a nice insight, but the answer is ever so slightly different.

let the distance between the two boats be D, and let x be the distance from the "origin" to boat A,  and similarly , y the distance from the origin to boat B.
|dw:1417049799166:dw|
$$ D^2= x^2 + y^2 $$
take the (implicit) derivative with respect to time t:
$$ 2 D \dot{D} = 2 x \dot{x} + 2y \dot{y}$$
$ \dot{x} $ is the speed of boat A = 18 nm/hr.  and is equal to the speed of boat B. Thus
$$ D \dot{D}= (x+y) 18 \ \dot{D} = \frac{ 18(x+y) }{D} $$
at 7 pm x is 10+7*18= 136  and y = 7*18= 126
use those values to find the velocity of D.  We will get 25.437 which is very close to radar's
$$ \sqrt{18^2 +18^2} = 25.456 $$

radar · Answer

Thanks @phi I had my doubts, somehow I figured those distances had to be brought in, and I did try to do the related rate derivative but I went astray and got an answer which was over twice their individual rate.....so I knew that was wrong, so I took the easy way out.  Thanks for refreshing my memory on the proper steps.