AP CALC
posted by Nel .
Two roads intersect at right angles. At a certain moment, one bicyclist is 8 miles due NORTH of the intersection traveling TOWARDS the intersection at a rate of 16 miles/hour. A second bicyclist is 10 miles due WEST of the intersection traveling AWAY from the intersection at 12 miles/hour.
(a) How fast is distance between the two bicyclists changing at this moment?

make a diagram, let the distance the north biker is from the intersection be y miles, let the distance the west biker is from the intersection be x miles
Let the distance between them be D
D^2 = x^2 + y^2
2D dD/dt = 2x dx/dt + 2y dy/dt
given:
dx/dt = 12 mph
dy/dt = 16 mph (y is decreasing)
find dD/dt when x=10 and y=8
then D^2 = 100+64
D = √164
dD/dt =(2(10)(12) + 2(8)(16))/(2√164) = .....
you finish the arithmetic 
My final answer was 8/square root of 164. Is that correct?

I believe the correct answer is 0.635 but i got 0.625.