Speed Trap (Applications of Derivatives)?

Question

A highway patrol airplane flies 3 mi above a level, straight road at a constant rate of 120 mph. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi the line-of-sight distance is decreasing at the rate of 160 mph. Find the car's speed along the highway.

The answer in the back of the book is 80 mph. I have no idea where to start.

Answer 1

R = line-of-sight distance = sqrt(3^2 + x^2)
dR/dt = (dx/dt)*(dR/dx)
= (dx/dt)* (2x/(2*sqrt(3^2 + x^2))
= (dx/dt)* (x/R)

At that moment:
dR/dt = -160 mph
R = 5
x = sqrt(R^2 - 3^2)
= sqrt(25 - 9)
= sqrt(16)
= 4
So:
dx/dt = (R/x)*dR/dt
= -(5/4) * 160
= - 200

Note that x is the difference between the x-coordinate of the car and of the plane. Since the plane is headed towards the car at 120 mph, the velocity of the car is
-200 - (-120) = -200 + 120 = -80 mph
or 80 mph towards the plane.

Answer 2

First, develop a function that expresses the line-of-sight distance, s, between the plane and the car as a function of the x-coordinate of the plane (call it p) and the x-coordinate of the car (call it c). If we assume c > p with the plane moving towards the right (increasing p) and the car moving left (decreasing c), the Pythagorean theorem will let you show that s = sqrt((c - p)^2 + 3^2), since the plane is 3 miles above the car and a horizontal distance of c - p from it.

Using what you know about derivatives, calculate ds/dt. Remember to use the chain rule. You should end up with an expression for ds/dt that is a function of dc/dt and dp/dt. As an example, d/dt of (c - p)^3 would be (3*(c - p)^2)*(dc/dt - dp/dt)

Now, consider the significance of the data you were given initially. The plane's horizontal velocity of 120 mph is actually dp/dt. The rate of change of the line-of-sight distance, -160 mph, is ds/dt. The car's unknown speed is -dc/dt, because we defined the car to be moving in the negative x-direction. So in your expression with ds/dt, dp/dt, and dc/dt, the only unknown is dc/dt, which is exactly what you are trying to solve for. You should get a negative value for dc/dt, which makes sense because the car has a speed of -dc/dt due to its negative direction of travel.

Answer 3

Let A be the vector denoting the positon of the aeroplane and C the patral car
A= [ x(0) - 120t , 3] , C= [vt , 0]
Here we have chosen co-ordinates such that at time t = 0 the patrol car is at the origin and the plane at point [x(0),3]
We denote r the line of sight distance as
r = || C - A || , where || . || is euc norm ( sqrt(x^2 + y^2...))
at t = 0 we have r = 5 = sqrt(x(0)^2 + 9)
so x(0) = +/-4 and for simplicity we choose x(0) = 4
Now
r(t) = sqrt(( t(v+120) - 4)^2 + 9)
dr/dt = ((v+120)*( t(v+120)-4)) / r(t)
We have when t = 0 dr/dt = -160
so (4*(v+120))/5=160

v=80mph