Question

Suppose a blue car is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph (miles per hour). At the same time, a red car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. The the straight-line distance between the blue and red car is increasing at a rate of 15 mph. What is the speed of the red car?

Answer 1

dx/dt = - 18,79 mph

Explanation:

The two cars with the intersection point and the straight-line distance between the cars make up a right triangle. In that right triangle, the legs are the distance between each car and the intersection point, and the distance between cars is the hypothenuse

If we call x and y distances between blue car and red car respectively and L the hypothenuse by Pythagoras theorem we have:

L² = x² + y² (1)

Tacking derivatives on both sides of the equation

2*L*dL/dt = 2*x*dx/dt + 2*y*dy/dt

And from equation (1)

L² = (0,5)² + (0,5)² ⇒ L = √(0,5)² + (0,5)² ⇒ L = 0,5*√2

By subtitution in equation (2)

2*(0,5*√2)*15 = 2*0,5*dx/dt + 2*0,5*40

(15*√2 - 40 ) / 1 = dx/dt [mph]

dx/dt = - 18,79 mph

Note the( - ) sign is equivalent to say that the car is driving away from the intersection point