At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 24 knots. How fast (in knots) is the distance between the ships changing at 3 PM?
Make a sketch to get a right angled triangle
let the time passes since noon be t hrs
let the noon position of ship B be the origin O
AO = 10 + 16t miles
OB = 24t miles
AB^2 = AO^2 + OB^2
= (10+16t)^2 + (24t)^2
2 AB d(AB)/dt = 2(10+16t)(16) + 2(24t)((24)
when t = 3 ( at 3:00 pm)
AB^2 = (58)^2 + 72^2 = 8548
AB = √8548
d(AB)/dt = (32(58)+ 48(72) )/(2√8548)
= 28.72
at that moment they are separating at 28.7 knots
To find the speed at which the distance between the ships is changing at 3 PM, we need to use the concept of related rates. We can start by understanding the positions of the ships at that time.
From noon to 3 PM, 3 hours have passed. Since Ship A is sailing west at a constant speed of 16 knots, it has traveled a distance of 16 * 3 = 48 nautical miles due west from the starting point.
On the other hand, Ship B is sailing north at a constant speed of 24 knots. In 3 hours, it has traveled 24 * 3 = 72 nautical miles due north from the starting point.
Now, let's assume that at 3 PM, the distance between the ships is D.
Using the Pythagorean theorem, we can set up a right-angled triangle with sides of 48 and 72 as the legs, and D as the hypotenuse. The equation becomes:
D^2 = 48^2 + 72^2
Simplifying:
D^2 = 2304 + 5184
D^2 = 7488
Now, let's differentiate both sides of the equation with respect to time, t:
2D * dD/dt = 0 + 0
2D * dD/dt = 0
dD/dt = 0
This implies that the distance between the ships, D, is not changing with respect to time. Therefore, the speed at which the distance between the ships is changing at 3 PM is 0 knots.