An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. one airplane is 150 miles from the point and has a speed of 600 miles per hour. the other is 200 miles from the point and has a speed of 800 miles per hour.

(a)at what rate is the distance between the planes changing?

(b) how much time does the controller have to get one of the airplanes on a different flight path?

They are traveling at right angles to each other so we can say one is traveling north to south and the other west to east.  Then we can say that there positions, y and x are:

y=150-600t  x=200-800t

By using the Pythagorean Theorem we can find the distance between these two planes as a function of time:

d^2=y^2+x^2, using y and x from above

d^2=(150-600t)^2+(200-800t)^2

d^2=22500-180000t+360000t^2+40000-320000t+640000t^2

d^2=1000000t^2-500000t+62500

d=√(1000000t^2-500000t+6250)

So the rate of change is the derivative of d

dd/dt=(1/2)(2000000t-500000)/√(1000000t^2-500000t+6250)

dd/dt=(1000000t-250000)/√(1000000t^2-500000t+6250)

So the rate depends upon t and is not a constant, so for the instantaneous rate you would plug in a specific value of t...

...

To find how much time the controller has to change the airplanes flight path, we only need to solve for when d=0, or even d^2=0...

1000000t^2-500000t+62500=0

6250(16t^2-8t+1)=0

6250(16^2-4t-4t+1)=0

6250(4t(4t-1)-1(4t-1))=0

6250(4t-1)(4t-1)=0

6250(4t-1)^2=0

4t-1=0

4t=1

t=1/4 hr

Well technically, the controller has t<1/4 because at t=1/4 impact will occur :)

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12