Math
posted by Anonymous .
A team has been working to convert dieselpowered cars to run just as efficiently on used cooking oil! They want to compare the mileage and speed of their prototype with that of the dieselpowered car.
The prototype is 100 meters south of an intersection, while the diesel car is 100 meters east of the intersection. Both vehicles start moving at the same time. The prototype moves north, toward the intersection, and the diesel car moves east, away from the intersection. If the prototype is traveling at a velocity of 3 meters per second and the diesel car is traveling at 2 meters per second, what is the rate of change of the distance between the cars after four seconds? Round off your answer to two decimal places.

The Ycoordinate of car 1, relative to the intersection, are
Y = 100 + 3t (Its X value is 0)
The Xcoordinate of car 2, relative to the intersection, is
X = 100 + 2t (Its Y value is zero)
The distance between the two cars is
R = sqrt(X^2 + Y^2) = sqrt(2*10^4 600t + 9t^2 +400t +4t^2) = sqrt(2*10^4 +13t^2 200t)
The rate of change of separation distance is
dR/dt = (1/2)*(26t 200)/sqrt(2*10^4 +13t^2 200t)
Compute the value of dR/dt at t = 4s.
(The denominator is the distance apart, which is 139.13 m at t = 4 s) 
area of parallelogram
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