Calculus
posted by Kristy .
At noon, ship A is 180 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 30 km/h. How fast is the distance between the ships changing at 4:00 PM?

draw the right triangle.
start the origin at B at noon.
a is the distance ship A moves in 4 hours.
a= 180+40*4=20km
da/dt= + 40
b is the distance from the B starting point at noon.
b= 30*4=120
db/dt=30\
r=sqrt(a^2+b^2)
dr/dt= 1/2 *1/(a^2+b^2)* (2a da/dt+2b db/dt)
solve for dr/dt 
I tried plugging in the numbers for the variables, but I am not coming up with the right answer. I think the equation might be wrong.