Calculus Related Rates-Police Car Example

Question

A police car is located 30 feet to the side of a straight road.

A red car is driving along the road in the direction of the police car and is 190 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 100 feet per second. How fast is the red car actually traveling along the road?

The actual speed (along the road) of the red car is feet per second

Answer 1

Let y be how far the car is up the road. Let x be how far the cop is off the road. Let h be the distance between the cop and the car.

 

This forms a right triangle so we can use the Pythagorean Theorem.

x² + y² = h²

h² = (30ft)² + ( 190ft)²

h² = 900(ft²) + 36,100(ft²)

h² = 37,000(ft²)

h = Sqrt(37,000)ft

 

Let y' be how fast the car is traveling down the road. Let x' be how fast the cop is traveling toward the road. Let h' be how fast the distance between the cop and the car is changing.

 

Now we take the derivative of our equation x² + y² = h² with respect to time.

2x(x') + 2y(y') = 2h(h')

 

Now divide both sides by 2 then plug in what we know.

x(x') + y(y') = h(h')

 

From the problem we know x, y, h' we calculated h and since the cop is not moving we know x' = 0

30ft(0)ft/s + 190ft(y') = Sqrt(37,000)ft(100)ft/s

 

Solve for y'

y' = Sqrt(37,000)(100)/(190) = about 101.23886 ft/s