Find the Speed of a Car Approaching at 190mph: Optimization Problem

[quatrarot]

A police is flying at 150 mph at a constant altitude of .5 miles above a straight rode. The pilot uses radar to determine that an oncoming car is at a distance of exactly one mile from the helicopter, and that this distance is decreasing at 190 mph. Determine the speed of the car.

This is a problem I have tried numerous times but keep coming up with different solutions, any help would be greatly appreciated.

The Attempt at a Solution

let y = the helicopter's altitude
let x = car's distance to directly below the helicopter.
let h = the distance in miles between the helicopter and the car
let dh/dt = 190 miles/hour
150 miles/hour + dx/dt = horizontal component of car and helicopter.
then dx/dt is what we're looking for - the car's speed with respect to the ground.

y² + x² = h²

d(y²)/dt + d(x²)/dt = d(h²)/dt

dy/dt d(.5)²/dt + 150 + 2x dx/dt = d(1²)/dt 190 miles/hour = 0

2x dx/dt = 0 - 150 = -150miles/hour

x = √(1² - (.5)²) = √3/2

2(√3/2) dx/dt = -150 m/h

dx/dt = -150/√3 miles/hour

dx/dt = -86.60254 miles/hour

 

[haruspex]

d(y²)/dt + d(x²)/dt = d(h²)/dt

dy/dt d(.5)²/dt + 150 + 2x dx/dt = d(1²)/dt 190 miles/hour = 0

The first equation is fine, but the second makes no sense. The 2x dx/dt term is fine... why aren't the other terms similar?
Btw, you don't actually need calculus for this problem. Just consider the components of the two speeds along the line connecting the vehicles.

 

[quatrarot]

Okay: Then I get something like

2x*dx/dt = 190 mph

where does 150mph fit into this and is dx/dt the speed of the car?

 

[haruspex]

Okay: Then I get something like
2x*dx/dt = 190 mph

Not quite. You have 2y dy/dt + 2x dx/dt + 2 h dh/dt, right? So the 2's should all cancel.
Remember that x is the distance from the car to immediately below the helicopter at any given time t. This is affected by the helicopter's movement.