Related Rates Problem:

Question

Two straight roads intersect at right angles in Newtonville. Roadrunner (Beep Beep) is on one
road lazily moving west toward the intersection at 442 kph. Wyle E. Coyote is on the other
road attached to rocket traveling north toward the intersection at 481 kph. At what rate is the
distance between them changing when Roadrunner is 4 km from the intersection, if we assume
they will collide?

Answer 1

The distance between Roadrunner and Wyle E. is represented by the hypotenuse of a right triangle: c = sqrt(a² + b²). a represents Roadrunner's distance away from the intersection, and b represents Wyle E.'s distance from the intersection. The question asks for the change in the distance between then at a specific point, so it is looking for dc/dt. Therefore, we need to differentiate with respect to time:

 

dc/dt = (1/2)*sqrt(a² + b²)^-1/2*(2a(da/dt) + 2b(db/dt))

 

Now, values for a, b, da/dt, and db/dt need to be determined.

 

da/dt is the rate of change in distance to the intersection for Roadrunner: da/dt = -442 (he is traveling west, which is in the negative direction)

 

db/dt is the rate of change in distance to the intersection for Wyle E.: db/dt = 481

 

a is the distance from Roadrunner to the intersection: a = 4.

 

We need to calculate for b. Since they collide at the intersection, it should take Wyle E. the same amount of time to travel to the intersection as it does Roadrunner. So we solve how long it takes Roadrunner to travel 4km:

4km / 442km/hr = 2/221 hr

 

We then multiply this by the velocity of Wyle E. to determine how much distance he covers in this time:

481km/hr * 2/221hr = 74/17 km = 4.3529 km

 

As such, Wyle E. must be 4.3529 km away from the intersection when Roadrunner is 4 km away: b = -4.3529 (since he is still below the intersection)

 

The values are all placed in the equation:

 

dc/dt = (1/2)*((4)² + (-4.3529)²)^-1/2*(2(4)(-442)+2(-4.3529)(482))

dc/dt = (1/2)*(34.9477)^-1/2*(-7732.24) = -653.98 km/hr