A gas station stands at the intersection of a north-south road and an east-west road. A police car is traveling toward the gas station from the east, chasing a stolen truck which is traveling south away from the gas station. The speed of the police car is 130 mph at the moment it is 5 miles from the gas station. At the same time, the truck is 12 miles from the gas station going 100 mph. At this time,
(a) Is the distance between the car and truck increasing or decreasing? How fast? (Distance is measured along a straight line joining the car and the truck.)
(b) How does your answer change if the truck is going 70 mph instead of 80 mph?

Let´s call O the place for the gas station, A the location point of police car ( at 5 miles east from the gas station ), and B the location point of the truck ( 12 miles south from a gas station.

The three points shape a right triangle with L (distance between police car and truck) then

L² = OA² + OB² at the point police car is 5 miles east and truck 12 miles south ( both from gas station)

L² = ( 5)² + (12)²

L = √ 25 + 144

L = √169

L = 13 miles

Pitagoras theorem establishes in a right triangle hypotenuse L is:

L² = a² + b² a and b are the legs then

In general

L² = x² + y ² x and y the legs over distance police-car/gas station and truck/gas station

Differentiation on both sides of the equation with respect to time give us:

2*L*DL/dt = 2*x*Dx/dt + 2*y*Dy/dt

Where Dx/dt = 130 m/h West Dy/dt = 100 m/h south

L = 13 m x = 5 m y = 12

Then by substitution

2*13*DL/dt = 2*5*130 + 2*12*100

DL/dt = (1300+ 2400) / 26

DL/dt = 142 m/h

DL/dt is positive then L s increasing

b) If truck speed is 70 m/h

DL/dt still positive but with smaller module, then L still increasing but at smaller speed

DL/dt = 1300 + 2*12*70

DL/dt = (1300 + 1680)/ 26

DL/dt = 114,6 m