SOLUTION: One car leaves an intersection traveling north at 50 mph; another is driving west toward the intersection at 40 mph. At one point, the north-bound car is three-tenths of a mile no

Question 1193574: One car leaves an intersection traveling north at 50 mph; another is driving west toward the intersection
at 40 mph. At one point, the north-bound car is three-tenths of a mile north of the intersection and the
west-bound car is four-tenths of a mile east of it. At this point, how fast is the distance between the
cars changing?
Answer by yurtman(42) (Show Source): You can put this solution on YOUR website!
**1. Visualize and Define Variables:**
* **Draw a diagram:**
* Represent the intersection as the origin (0,0) on a coordinate plane.
* Let the northbound car's position be point A (0, y) where y is the distance north of the intersection.
* Let the westbound car's position be point B (x, 0) where x is the distance east of the intersection.
* Let 's' be the distance between the two cars.
* **Variables:**
* x: Distance of the westbound car from the intersection (decreasing)
* y: Distance of the northbound car from the intersection (increasing)
* s: Distance between the cars
* dx/dt: Rate of change of x (speed of the westbound car) = -40 mph (negative because x is decreasing)
* dy/dt: Rate of change of y (speed of the northbound car) = 50 mph
* ds/dt: Rate of change of the distance between the cars (what we need to find)
**2. Relate the Variables:**
* Use the Pythagorean theorem to relate x, y, and s:
s² = x² + y²
**3. Differentiate Implicitly:**
* Differentiate both sides of the equation with respect to time (t):
2s * ds/dt = 2x * dx/dt + 2y * dy/dt
**4. Simplify:**
* Divide both sides of the equation by 2:
s * ds/dt = x * dx/dt + y * dy/dt
**5. Find the Current Values:**
* x = 0.4 miles
* y = 0.3 miles
* dx/dt = -40 mph
* dy/dt = 50 mph
* Calculate s using the Pythagorean theorem:
s = √(x² + y²) = √(0.4² + 0.3²) = √(0.16 + 0.09) = √0.25 = 0.5 miles
**6. Substitute and Solve:**
* 0.5 * ds/dt = 0.4 * (-40) + 0.3 * 50
* 0.5 * ds/dt = -16 + 15
* 0.5 * ds/dt = -1
* ds/dt = -1 / 0.5
* ds/dt = -2 mph
**7. Interpret the Result:**
* The negative sign for ds/dt indicates that the distance between the cars is decreasing at a rate of 2 mph.
**Therefore, at the given point, the distance between the cars is changing at a rate of 2 mph.**

RELATED QUESTIONS
Two cars are driving towards each other. The first car is driving at 30 mph north and is... (answered by ankor@dixie-net.com)

A car leaves an intersection traveling west. Its position 4 sec later is 24 ft from the... (answered by pinkslips1717)

A car leaves an intersection traveling west. Its position 4 sec later is 18 ft from the... (answered by Alan3354)

Two cars leave an intersection at the same time. One travels north, and the other heads... (answered by Fombitz)

a car traveling north at 60 mph and a truck traveling east at 45 mph leave an... (answered by Alan3354)

Two cars are approaching an intersection. One is 3 miles South of the intersection and is (answered by mananth,jrfrunner)

a car left an intersection and travel led east at 32 mph. Another car left the same... (answered by math33)

A car leaves Oak Corners at 11:30 AM traveling south at 60 mph. At the same time, another (answered by Alan3354,josmiceli)

A car traveling goes through an intersection. The car is traveling north at a constant... (answered by richard1234)