At noon, ship a is 50 nautical miles due west of ship b. Ship a is sailing west at 16 knots and ship b is sailing north at 15 knots. How fast in knots is the distance between the ships changing at 4 pm?

Question

anonymous · Answer

Let s be the total distance between them, x horizontal distance and y the vertical distance $$
s^2=x^2+y^2
$$By chain rule, we get:$$
2ss'=2xx'+2yy'
$$

anonymous · Answer

We want to know $s'$.

anonymous · Answer

You got to figure out $x, y, z$ at 4 pm.  Though we already know $x'=16, y'=15$.

anonymous · Answer

I mean $s$ not $z$.

anonymous · Answer

I am not certain I know how to put it together to find x,y, and s...

anonymous · Answer

Well, they tell you starting at noon, so at 4pm, 4 hours have passed.  We have $x = 50+4(16)$ and $y=4(15)$ and $z = \sqrt{x^2+y^2}$

anonymous · Answer

Again, when I say z or s I mean the same thing, sorry.

anonymous · Answer

So i just needed to draw a right triangle and solve for the hypotenuse (spellingsorry)? Take the derivative of that equation plug in t=4 and get my answer... Took me 4 hours to figure it out.. Thank you!!!