calculus
posted by Mr. goodstuff on .
A police cruiser, approaching a right –angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 mi north of the intersection and the car 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60mph at the instant of measurement, what is the speed of the car?

make a sketch
At a time of t hrs,
let the distance the police car is from the intersection be x miles
let the distance the speeding car is from the intersection be y miles
let the distance between them be d miles
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
d dd/dt = x dx/dt + y dy/dt
given: x = .6, y = .8 , dx/dt = 60 , dd/dt = 20
( I made dx/dt negative because the value of x is decreasing)
d^2 = .6^2 + .8^2 = 1
d = 1
1(20) = .6(60) + .8dy/dt
20 + 36 = .8 dy/dt
dy/dt = 56/.8 = 70
The car was speeding at 70 mph 
70 mph