Question

Another word problem. See attached please

Answer 1

So what I have are 3 sides, 110, 40, and 100. I have the speeds of both ships 22 and 20 knots/hour, and using the law of cos, I deduced the angle to be 21.27996647 degrees. I then plugged this in to my derived formula and get 8 knots. Any help? @dpaInc

Answer 2

dA/dt= 22, dB/dt=20, a=110, b=100, c=40

Answer 3

nop, three sides will be 70, 100 and 122.0655
So, the distance between two ships at 5pm is 122.0655 knot

Answer 4

can you explain how you got that?

Answer 5

at 5 hour, a travels 110 knot, b travels 100 knot
but a is 40 knot behind b at start
draw a figure for that

Answer 6

I think that you've misunderstood the question. The question gives the rate at which the ships are travelling. the 40 is the difference from A to B.

Answer 7

see the pic. may be it can help you

Answer 8

I don't think this is that type of problem. This is calculus and the drawing that you've attached is trig. Thanks anyway though.

Answer 9

since ship A is heading due west, and ship B is going due north, it makes a right angle thus to find distance you can use pythagorean thm
\[d = \sqrt{ (40+22t)^{2}+(20t)^{2}}\]
differentiate to get rate distance is changing
use chain rule,
\[u = (40+22t)^{2} +(20t)^{2}\]
\[du = (2)(22)(40+22t) + (2)(20)(20t) = 1768t+1760\]
\[\frac{d}{du} \sqrt{u} = u^{1/2} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}\]
\[\frac{dd}{dt} = \frac{1768t+1760}{2\sqrt{(40+22t)^{2}+(20t)^{2}}}\]
to find rate at 5pm, plug in t=5
hope this helps