# Thread: related rates of change

1. ## related rates of change

a red car travelling east away from an interesection at a speed of 40km/hr, while a blue car is simmultaneously travelling north towards the intersection at a speed of 60km/hr. If the red car is 8km from the intersection and the blue car is 6km from the intersection, at what rate is the distance between the cars changing?

dont even know where to start.. i drew a diagram.. didnt help

2. Your diagram should be a right triangle, with these:
---top leg, horizontal leg = say, r
---left leg, vertical leg = say, b
---right side, hypotenuse = say, s

where
r = distance travelled by red car---getting longer as time increases.
b = distance to be travelled by blue car----getting shorter as time increases.
s = distance between red and blue cars

So, by Pythagorean Theorem,
s^2 = r^2 +b^2 --------------------------------------(i)
Differentiate both sides of (i) with respect to time t,
2s(ds/dt) = 2r(dr/dt) +2b(db/dt)
s(ds/dt) = r(dr/dt) +b(db/dt) -------------------(ii)

At the said instant in the Problem, in (ii) we know:
r = 8 km
b = 6 km
And it is given that:
dr/dt = 40 km/hr
db/dt = -60 km/hr -----negative because b is decreasing with time.

We don't know:
ds/dt ----------we are spolving for it.
s -----we can get via (i)
s^2 = r^2 +b^2 --------------------------------------(i)
s^2 = 8^2 +6^2
s^2 = 64 +36 = 100
s = 10 km

So, substitute all those into (ii),
10(ds/dt) = 8(40) +6(-60)
ds/dt = 8(4) -6(6)
ds/dt = 32 -36 = -4 km/hr

Therefore, at that instant, the distance between the two cars is decreasing at the rate of 4 km/hr. --------------answer.