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.**
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