At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
After t hours,
A is at (-(30+17t),0)
B is at (0,16t)
So the distance d is found by
d^2 = (30+17t)^2 + (16t)^2
= 545t^2 + 1020t + 900
at t=6, d = 163.22
So, now we can find the rate of change for d:
2d dd/dt = 1090t + 1020
2*163.22 dd/dt = 7560
dd/dt = 23.16 knots
To find the rate at which the distance between the ships is changing at a given time, we can use the concept of the rate of change. In this case, we need to find the derivative of the distance function with respect to time, and then substitute the given values to find the rate of change.
Let's define the distance between the ships as a function of time: D(t). We are asked to find how fast the distance is changing at 6 PM, so we need to find D'(t) at t = 6 PM.
Given information:
- Ship A is sailing west at 17 knots.
- Ship B is sailing north at 16 knots.
- At noon, ship A is 30 nautical miles due west of ship B.
We can break down the velocity of each ship into east-west (horizontal) and north-south (vertical) components.
- Ship A's velocity components are: 17 knots to the west and 0 knots to the north.
- Ship B's velocity components are: 0 knots to the west and 16 knots to the north.
Since the ships are initially 30 nautical miles apart due west, we can set up a right triangle with the distance between the ships as the hypotenuse, and the horizontal and vertical distances as the legs of the triangle.
At noon, the horizontal distance is 30 nautical miles (given), and the vertical distance is 0 nautical miles (since ship B is not moving horizontally). This forms a right triangle with a 30-nautical-mile hypotenuse.
Using the Pythagorean theorem, we can find the distance between the ships as a function of time:
D(t) = sqrt[(30 + 17t)^2 + (16t)^2]
Now, to find the derivative of D(t), we apply the chain rule and simplify:
D'(t) = [2(30 + 17t)(17) + 2(16t)(16)] / 2sqrt[(30 + 17t)^2 + (16t)^2]
D'(t) = (34(30 + 17t) + 16^2t) / sqrt[(30 + 17t)^2 + (16t)^2]
Substituting t = 6 PM (which is 6 hours after noon), we can find the rate at which the distance between the ships is changing:
D'(6) = (34(30 + 17(6)) + 16^2(6)) / sqrt[(30 + 17(6))^2 + (16(6))^2]