To find how fast the distance between the ships is changing, we can use the concept of the rate of change, specifically the derivatives from calculus. First, let's break down the problem and establish some variables.
Let's assume the position of ship A is given by the x-coordinate and the position of ship B is given by the y-coordinate. At noon, ship A is 180 km west of ship B, so we can set the initial position of ship A and ship B as (-180, 0) and (0, 0), respectively.
Now, let's find the equations that represent the positions of ship A and ship B at any given time t.
The position of ship A can be represented by the equation: x(t) = -180 + 40t, where t is the time in hours since noon.
The position of ship B can be represented by the equation: y(t) = 30t
Now, let's find the distance between the two ships at time t. Using the distance formula, which is the square root of the sum of the squares of the differences in the coordinates, we have:
D(t) = sqrt((x(t) - y(t))^2 + (0 - 0)^2)
Simplifying this equation gives us:
D(t) = sqrt((x(t) - y(t))^2)
D(t) = sqrt((-180 + 40t - 30t)^2)
D(t) = sqrt((-180 + 10t)^2)
Now, let's find the derivative of D(t) with respect to t in order to find how fast the distance is changing:
dD(t)/dt = d(sqrt((-180 + 10t)^2))/dt
Using the chain rule, the derivative simplifies to:
dD(t)/dt = (1/2) * (2 * (-180 + 10t)) * (10)
dD(t)/dt = 10 * (-180 + 10t)
dD(t)/dt = -1800 + 100t
So, the rate at which the distance between the ships is changing is given by the equation -1800 + 100t.
Now, let's find how fast the distance between the ships is changing at 4:00 PM, which is 4 hours after noon. Substituting t = 4 into the equation, we get:
dD(t)/dt = -1800 + 100 * 4
dD(t)/dt = -1800 + 400
dD(t)/dt = 1600 km/h
Therefore, the distance between the ships is changing at a rate of 1600 km/h at 4:00 PM.