Calculus
posted by Vivi on .
Car A traveling southward towards Point P at 45 mi/hr. Car B is traveling east away from P at 30mi/hr. AT the instant when the distance AP is 60 mi and PB is 80 mi, what is the rate of change of distance AB?

We have a nice 345 triangle at the moment in question, so AB = 100
If we let x and y be the distances of the two cars (B and A) then the distance d=AB is
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
so, when x=80 and y=60, d=100, and we have
100 dd/dt = 80(30) + 60(45)
dd/dt = 3