Two cars are traveling along perpendicular roads, car A at 40 mph, car B at 60 mph. At noon, when car A reaches the intersection, car B is 90 miles away and moving toward it. Find the rate at which the distance between the cars is changing at 1 P. M ...

-4 mph

Explanation:

Car A is traveling at 40mph.

Car B is traveling at 60mph

At noon, Car A reaches intersection while car B is 90 miles away and moving towards it.

Distance between car A and B is 90 miles.

da/dt = 40 mph

db/dt = 60mph

At 12 noon, car A is at (0,0) while car B is at (0,-90)

A is traveling along the x axis and B is traveling along the y axis.

At 1 pm, car A and B will be at (40,0) and (0,-30)

Car A has moved to the right along the x axis. Car B has moved up along the y axis by 60 because 1 hour passed since 12 pm

The rate at which the distance is changing is dd/dt

d = √ (ax - bx) ^2 + (ay - by) ^2

d = √ (ax - 0) ^2 + (0 - by) ^ 2

d = √ax^2 + by^2

d^2 = ax^2 + by^2

d^2 = a^2 + b^2

Differentiate implicitly

2d (dd/dt) = 2a (da/dt) + 2b (db/dt)

dd/dt = [a (da/dt) + b (db/dt) ] / d

a = 40, b = - 30

da/dt = 40

db/dt = 60

d^2 = 30^2 + 40^2

= 900 + 1600

= 2500

d = √2500

d = 50

= [40 (40) + - 30 (60) ] / 50

dd/dt = (1600 - 1800) / 50

= - 200/-50

dd/dt = - 4 mph