To solve this problem, we can use the Pythagorean theorem to relate the distance between the two cyclists to the distance each has traveled. Let's call the distance between the two cyclists "D", and let's call the distance Cyclist B has traveled "x" (in miles). Then, the distance Cyclist A has traveled is "x - 15" (since A leaves one hour later than B).
According to the Pythagorean theorem, we know that:
D^2 = (x - 15)^2 + x^2
To find how fast D is changing with respect to time (in this case, three hours after Cyclist B departs), we can take the derivative of both sides of the equation with respect to time:
2D(dD/dt) = 2(x - 15)(dx/dt) + 2x(dx/dt)
Simplifying this equation, we get:
dD/dt = [(x - 15)(dx/dt) + x(dx/dt)] / D
Now, we need to find x and dx/dt three hours after Cyclist B departs. Since Cyclist B travels at a constant speed of 12mph, we know that x = 12t = 36 (since three hours have elapsed). We also know that dx/dt = 12, since B is traveling at a constant speed.
Plugging these values into the equation above, we get:
dD/dt = [(36 - 15)(12) + 36(12)] / D
dD/dt = 21(12) / D
dD/dt = 252 / D
We still need to find D to get our final answer. Using the Pythagorean theorem again, we know that:
D^2 = (36 - 15)^2 + 36^2
D^2 = 441
D = 21
Plugging this value of D into our equation for dD/dt, we get:
dD/dt = 252 / 21
dD/dt = 12 mph
Therefore, the rate at which the distance between the two cyclists is changing three hours after Cyclist B departs is 12 mph.