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 22 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.)
To solve this problem, we can use the concept of related rates. We'll start by finding the position of each ship at 6 PM, and then calculate the rate at which the distance between them is changing.
Step 1: Calculate the distance traveled by each ship by 6 PM.
Assuming that the coordinates for ship A's position are (0, 0), we can calculate the positions of ship A and ship B at 6 PM using their respective speeds and the time difference.
Ship A: Since ship A is sailing west at a speed of 18 knots for 6 hours, it would have traveled a distance of 18 × 6 = 108 nautical miles due west.
Ship B: Similarly, since ship B is sailing north at a speed of 22 knots for 6 hours, it would have traveled a distance of 22 × 6 = 132 nautical miles due north.
Step 2: Calculate the coordinates of ship B at 6 PM.
Since ship A starts at the origin (0, 0) and ship B is 10 nautical miles due west, ship B's final x-coordinate at 6 PM would be -10, while its y-coordinate would be 0 + 132 = 132.
Step 3: Calculate the distance between the ships at 6 PM.
Using the distance formula, we can find the distance between the two ships at 6 PM.
Distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
Distance = sqrt[(-10 - 0)^2 + (132 - 0)^2]
Distance = sqrt[100 + 17424]
Distance ≈ sqrt(17524)
Distance ≈ 132.369 nautical miles
Step 4: Calculate the rate at which the distance between the ships is changing at 6 PM.
To find the rate at which the distance between the ships is changing, we can differentiate the distance equation with respect to time and evaluate it at 6 PM.
The rate of change of distance between the ships = d(Distance)/dt
Using the chain rule of differentiation,
d(Distance)/dt = [d(Distance)/dx] * [dx/dt] + [d(Distance)/dy] * [dy/dt]
Since dx/dt represents the westward velocity of ship A (18 knots) and dy/dt represents the northward velocity of ship B (22 knots), we can substitute these values.
d(Distance)/dt = [d(Distance)/dx] * 18 + [d(Distance)/dy] * 22
To find the rates of change of distance with respect to x and y, we can differentiate the distance equation:
d(Distance)/dx = (x2 - x1) / Distance
d(Distance)/dy = (y2 - y1) / Distance
Substituting the values obtained earlier:
d(Distance)/dx = (-10 - 0) / 132.369 ≈ -0.075 nautical miles
d(Distance)/dy = (132 - 0) / 132.369 ≈ 1 nautical mile
Substituting the derived values:
d(Distance)/dt = (-0.075) * 18 + (1) * 22
d(Distance)/dt ≈ -1.35 + 22
d(Distance)/dt ≈ 20.65 knots
Therefore, the distance between the ships is changing at a rate of approximately 20.65 knots at 6 PM.
To solve this problem, we can use the concept of related rates. The distance between the two ships is changing as both ships are moving. We need to find the rate at which this distance is changing at 6 PM.
Let's set up a coordinate system. We'll place ship B at the origin (0,0). Since ship A is 10 nautical miles due west of ship B at noon, we can place ship A at (-10,0).
Let's assume that the x-axis points east and the y-axis points north. Ship A is sailing directly west, so its velocity vector can be written as (-18,0) because it is moving only along the x-axis. Ship B is sailing directly north, so its velocity vector can be written as (0,22) because it is moving only along the y-axis.
To find the distance between the two ships at any time t, we can use the distance formula:
d(t) = sqrt((x_A(t) - x_B(t))^2 + (y_A(t) - y_B(t))^2)
where:
(x_A(t), y_A(t)) is the position of ship A at time t
(x_B(t), y_B(t)) is the position of ship B at time t
Now, let's find the positions of both ships at 6 PM.
Since ship A started at noon and is moving west at a constant velocity of 18 knots, its x-coordinate at 6 PM can be found by the equation:
x_A(t) = -10 - 18(t - 12)
where t is the number of hours after noon. Substituting t = 18, we get:
x_A(18) = -10 - 18(18 - 12) = -10 - 18(6) = -10 - 108 = -118
So at 6 PM, ship A is located at (-118, 0).
Since ship B started at the origin and is moving north at a constant velocity of 22 knots, its y-coordinate at 6 PM can be found by the equation:
y_B(t) = 22(t - 12)
Substituting t = 18, we get:
y_B(18) = 22(18 - 12) = 22(6) = 132
So at 6 PM, ship B is located at (0, 132).
Now we can substitute the coordinates of ships A and B into the distance formula to find the distance between them at 6 PM:
d(6PM) = sqrt((-118 - 0)^2 + (0 - 132)^2) = sqrt(118^2 + 132^2) = sqrt(13924 + 17424) = sqrt(31348) ≈ 177.11 nautical miles
To find how fast the distance is changing at 6 PM, we can differentiate the distance formula with respect to time t:
d'(t) = (x_A'(t) - x_B'(t))(x_A(t) - x_B(t)) + (y_A'(t) - y_B'(t))(y_A(t) - y_B(t)) / d(t)
where x'(t) denotes the derivative of x(t) with respect to t.
Let's find x_A'(t) and y_B'(t):
x_A'(t) = -18
y_B'(t) = 22
Substituting the values into the derivative formula at 6 PM, we get:
d'(6PM) = (-18 - 0)(-118 - 0) + (0 - 22)(0 - 132) / 177.11
Simplifying the expression:
d'(6PM) = (18)(118) + (-22)(132) / 177.11
= 2124 - 2904 / 177.11
= -780 / 177.11 ≈ -4.4
Hence, at 6 PM, the distance between the two ships is changing at a rate of approximately -4.4 knots, which means it is decreasing by 4.4 knots per hour.