Understanding related rates problems can be a challenge for many students in calculus. These problems involve finding the rate of change of one quantity with respect to time, given the rate of change of another related quantity. One of the key steps in solving these problems is to understand how to calculate the derivative of the distance between two objects. This article will explore the process of solving related rates problems involving distances, with a focus on calculating the derivative of the distance function.
Understanding the Problem Setup
Related rates problems often involve scenarios where objects are moving and their positions are changing with time. The goal is to determine how quickly the distance between these objects is changing. To solve these problems, we need to:
- Identify the variables: These typically include the distances between objects, their rates of change (velocities), and time.
- Establish a relationship between the variables: This usually involves a geometric formula, such as the Pythagorean theorem or the formula for the area of a triangle.
- Differentiate the relationship with respect to time: This step involves using implicit differentiation to find the rates of change of the variables.
- Solve for the desired rate of change: We will use the given information and the results of the differentiation to find the specific rate of change we are looking for.
Calculating the Derivative of the Distance Function
The most common scenario for distance-related rates problems involves two objects moving in different directions. To find the derivative of the distance between these objects, we can use the Pythagorean theorem.
Imagine two objects, A and B, moving along perpendicular paths. At any given time, let:
- x be the distance traveled by object A
- y be the distance traveled by object B
- z be the distance between objects A and B
The Pythagorean theorem states: z² = x² + y².
To find the rate of change of the distance (z) with respect to time (t), we differentiate both sides of the equation with respect to t using implicit differentiation:
2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)
Solving for dz/dt, the rate of change of the distance:
*(dz/dt) = (x(dx/dt) + y(dy/dt)) / z
This formula tells us how the distance between the objects changes with time, given the rates of change of their individual distances (dx/dt and dy/dt) and the current distances (x, y, and z).
Example Problem:
Let's consider a concrete example to illustrate how to use this formula.
Problem: A car is traveling east at 60 mph and a truck is traveling north at 40 mph. Both vehicles are moving toward an intersection. The car is 3 miles from the intersection and the truck is 4 miles from the intersection. How fast is the distance between the vehicles changing?
Solution:
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Identify the variables:
- x = distance traveled by the car (miles)
- y = distance traveled by the truck (miles)
- z = distance between the car and truck (miles)
- dx/dt = 60 mph (speed of the car)
- dy/dt = 40 mph (speed of the truck)
-
Establish a relationship between the variables:
- Using the Pythagorean theorem: z² = x² + y²
-
Differentiate the relationship with respect to time:
- 2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)
-
Solve for the desired rate of change (dz/dt):
- We need to find dz/dt when x = 3, y = 4, dx/dt = 60, and dy/dt = 40.
- First, use the Pythagorean theorem to find z: z² = 3² + 4² = 25, so z = 5.
- Substitute the known values into the differentiated equation: 2(5)(dz/dt) = 2(3)(60) + 2(4)(40)
- Simplify and solve for dz/dt: 10(dz/dt) = 360 + 320 = 680
- Therefore, (dz/dt) = 68 mph.
Conclusion: The distance between the car and truck is increasing at a rate of 68 mph.
Important Notes
- Units: Always pay attention to the units of the variables and rates of change in the problem. The units of the final answer should reflect the units of the variables involved.
- Sign: The sign of the rate of change indicates whether the distance is increasing or decreasing. A positive sign means the distance is increasing, while a negative sign means the distance is decreasing.
- Visualization: It can be helpful to draw a diagram to visualize the situation described in the problem. This will help you identify the variables and their relationships.
Applications of Related Rates Problems
Related rates problems have numerous applications in various fields, including:
- Physics: Calculating the rate of change of the volume of a balloon, the rate of change of the area of a circle, or the rate of change of the speed of a moving object.
- Engineering: Determining the rate of change of the flow of water through a pipe, the rate of change of the pressure in a tank, or the rate of change of the stress in a beam.
- Economics: Analyzing the rate of change of supply and demand, the rate of change of profit, or the rate of change of interest rates.
Conclusion
Solving related rates problems involving distance requires understanding how to calculate the derivative of the distance function. This involves applying the Pythagorean theorem, differentiating the relationship with respect to time, and substituting the known values to solve for the desired rate of change. By mastering these concepts, you will be equipped to handle a wide range of problems involving the rate of change of distance. Remember to pay attention to units, signs, and visualization to ensure accurate and effective problem-solving.
