Draw a figure, showing a right triangle with vertical leg 15 and horizontal leg = x + y, where x is the distance of the man from the pole and y is the length of the shadow.

(x+y)/15 = x/6 => x = (3/2)y

dx/dt = (3/2)dy/dt = (3/2) * 4 = 6

This shows that the shadow lengthens at 3/2 the rate of the walker and is independent of his position.

And d(x+y)/dt = dx/dt + dy/dt --this shows that the tip of the shadow moves at a constant rate = 6 + 4 = 10 ft/sec