Related Rates Problems: Understanding and Solving Them Subject: math
In mathematics, related rates problems are a type of calculus problem that deals with the relationships between different quantities that change at the same time or at a constant rate. These problems involve finding the instantaneous rates of change for two or more variables in a given equation and then using these rates to solve for the desired variable.
Related rates problems often connect to the concept of the chain rule in calculus, which states that the derivative of a composite function is found by differentiating each term and applying the rules of differentiation in the order in which they are applied. In other words, the chain rule allows us to find the rate at which one variable changes with respect to another when both variables change simultaneously.
Let's explore how related rates problems can be solved using a step-by-step strategy and by understanding their connection to the chain rule. We will also look at two examples: a ladder sliding down a wall and a balloon being inflated. Throughout this discussion, it is crucial to track units and signs accurately.
Step-by-Step Strategy for Solving Related Rates Problems
-
Draw a diagram and label variables: Before solving the problem, draw a diagram that visualizes the situation, and label all relevant variables. This will help you understand the problem's context and how different quantities are related to each other.
-
Write an equation: After understanding the given variables and their relationship, write an equation that connects them. This step is crucial because it sets up the foundation for finding the instantaneous rates of change in the next steps.
-
Differentiate with respect to time: The key to solving related rates problems lies in applying the chain rule to find the rate at which one variable changes with respect to another as both variables change simultaneously. Differentiate each term in the equation you wrote in step 2 with respect to time (usually represented by t).
-
Solve for the desired rate: Finally, solve the differentiated equation obtained in step 3 to find the instantaneous rate at which the desired variable changes. Make sure to track units and signs accurately, as they are crucial when solving related rates problems.
Example 1: A Ladder Sliding Down a Wall
Consider a ladder leaning against a wall. The top of the ladder is attached to a point on the wall, while the bottom slides along the ground. The length of the ladder remains constant at 20 feet. If the bottom of the ladder moves 5 feet in one minute, how fast is the top of the ladder moving down the wall?
-
Draw a diagram and label variables: Draw a right triangle with two sides measuring 20 feet (the length of the ladder) and an angle between them. Label the base of the ladder as 'b', the height as 'h', the time as 't', and the bottom of the ladder's motion as 'd'.
-
Write an equation: The Pythagorean theorem states that the square of the hypotenuse (the height 'h') is equal to the sum of the squares of the other two sides:
[ b^2 + h^2 = 20^2 ]
- Differentiate with respect to time: Differentiating the equation from step 2 with respect to time, we get:
[ \frac{db}{dt} + 2h\frac{dh}{dt} = 0 ]
- Solve for the desired rate: Given that the bottom of the ladder moves at a constant rate of 5 feet per minute (\(\frac{db}{dt} = 5\)), we can substitute this value into the equation from step 3:
[ 5 + 2h \frac{dh}{dt} = 0 ]
Since the length of the ladder remains constant at 20 feet, we know that \(b^2 + h^2 = 20^2\), which implies that \(h^2 = 20^2 - b^2\). Substituting this into the equation from step 4 gives:
[ 5 + 2(20 - d)(-\frac{dh}{dt}) = 0 ]
Solving for \(\frac{dh}{dt}\), we find that the top of the ladder is moving down the wall at a rate of approximately 13.86 feet per minute.
Example 2: A Balloon Being Inflated
Suppose a balloon has an initial radius of 3 inches and is being inflated such that its surface area increases by 1 square inch every second. How fast is the volume of the balloon increasing when its radius is 5 inches?
-
Draw a diagram and label variables: Draw a circle with radius 'r' to represent the balloon, and let the time be represented by 't'.
-
Write an equation: The surface area of a sphere is given by \(A = 4\pi r^2\), while its volume is given by \(V = \frac{4}{3}\pi r^3\). For the balloon, we have:
[ A(t) = 4\pi (r(t))^2 ] [ V(t) = \frac{4}{3}\pi (r(t))^3 ]
- Differentiate with respect to time: Differentiating each equation with respect to time, we get:
[ \frac{dA}{dt} = 8\pi r \frac{dr}{dt} ] [ \frac{dV}{dt} = \frac{4}{3}\pi 3r^2 \frac{dr}{dt} ]
- Solve for the desired rate: Given that the surface area of the balloon is increasing at a rate of 1 square inch per second (\(\frac{dA}{dt} = 1\)), we can substitute this value into the equation from step 3:
[ 1 = 8\pi r \frac{dr}{dt} ]
Solving for \(\frac{dr}{dt}\), we find that the radius of the balloon is increasing at a rate of approximately 0.075 inches per second. To find the volume's instantaneous rate of change, we use the equation from step 4:
[ \frac{dV}{dt} = \frac{4}{3}\pi 3(r)^2 \frac{dr}{dt} ]
Substituting \(r = 5\) and \(\frac{dr}{dt} = 0.075\), we find that the volume of the balloon is increasing at a rate of approximately 48.56 cubic inches per second when its radius is 5 inches.
Conclusion
Related rates problems are an essential tool in calculus, as they enable us to understand how different variables change simultaneously and find the instantaneous rate of change for these variables. By following a step-by-step strategy and connecting related rates problems to the chain rule, we can solve such problems effectively. The two examples discussed demonstrate how to use this strategy to find the desired rates in different scenarios.
Remember always to track units and signs accurately when solving related rates problems, as they play a crucial role in obtaining the correct answer.