Introduction to Related Rates
Related rates problems involve quantities that are changing over time and are typically solved using the chain rule. This rule states that if two variables are related, then the rate of change of one variable is related to the rate of change of the other. Understanding the basics of derivatives is crucial before diving into related rates problems.
Key Concepts
- Rate of Change: This refers to the speed at which a variable changes, often represented as dy/dt or dx/dt, where t represents time.
- Related Rates: This concept comes into play when the rates of change of two variables are related by an equation.
- Chain Rule: This is a fundamental principle in calculus used to find the derivative of the composition of two functions. To fully grasp this concept, one should be familiar with differentiation rules.
Example 1: Ladder Sliding Down a Wall
Consider a 10-metre ladder leaning against a wall. If the bottom of the ladder slides away from the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 metres from the wall?
Solution:
- Identify Known Rates:
- dx/dt = 1 m/s (rate at which the bottom of the ladder is moving away from the wall)
- Identify Unknown Rates:
- dy/dt (rate at which the top of the ladder is moving down the wall)
- Relate the Variables:
- Use the Pythagorean Theorem: x2 + y2 = L2, where L = 10 m (length of the ladder). A good understanding of surface area and volume is beneficial here.
- Differentiate with Respect to Time (t):
- Differentiating implicitly gives: 2x(dx/dt) + 2y(dy/dt) = 0.
- Substitute Known Values and Solve for the Unknown:
- When x = 6 m, use the Pythagorean Theorem to find y = 8 m.
- Substitute x, dx/dt, and y into the differentiated equation and solve for dy/dt.
- dy/dt = -(x/y)(dx/dt)
- dy/dt = -(6/8)(1)
- dy/dt = -0.75 m/s
Thus, when the bottom of the ladder is 6 m from the wall, the top of the ladder is sliding down the wall at a rate of 0.75 m/s.
Real-World Applications of Related Rates
Application 1: Population Growth
The rate at which a population grows can be related to the current population. If P represents population and t represents time, the rate of change of the population dP/dt might be proportional to the current population P. Such applications often require a solid understanding of introduction to integrals for solving differential equations.
Example:
If a bacteria colony doubles every hour, its population growth rate can be expressed as dP/dt = kP, where k is a constant. Solving such a differential equation would involve integration, providing a formula to predict future population sizes based on the current population.
Application 2: Economics
In economics, related rates can be used to analyse how different economic variables change with respect to one another. For instance, the rate of change of cost might be related to the rate of change of production.
Example:
If a company produces x items at a cost of C(x), the rate of change of cost with respect to the number of items produced dC/dx can provide insights into how costs evolve as production changes, which is vital for economic planning and analysis.
Application 3: Physics
In physics, related rates can describe how different physical quantities such as position, velocity, and acceleration are related. This concept is closely tied to understanding the volumes of revolution, which can be applied to find the volume of objects with symmetrical rotation.
Example:
If a car’s velocity v is related to time t by a function v(t), the acceleration a of the car is the rate of change of its velocity, expressed as dv/dt. Understanding these relationships is crucial for predicting the motion of objects.
Diving Deeper into Related Rates
The Chain Rule and Related Rates
The chain rule is pivotal in solving related rates problems. It allows us to find the derivative of a composite function. If we have two functions y(u) and u(x), then the derivative of y with respect to x is given by:
dy/dx = dy/du * du/dx
In related rates problems, we often have an equation relating two or more variables, and we know the rate of change of at least one of them. We want to find the rate of change of another variable, and we do this by implicitly differentiating the equation with respect to time t and using the chain rule.
Example 2: Expanding Sphere
Consider a sphere whose radius is increasing at a rate of 2 cm/s. How fast is the volume of the sphere increasing when the radius is 5 cm?
Solution:
- Identify Known Rates:
- dr/dt = 2 cm/s
- Identify Unknown Rates:
- dV/dt (rate at which the volume is increasing)
- Relate the Variables:
- Volume of a sphere: V = (4/3)pi r3
- Differentiate with Respect to Time (t):
- Differentiating implicitly gives: dV/dt = 4pi r2 dr/dt
- Substitute Known Values and Solve for the Unknown:
- When r = 5 cm, substitute r and dr/dt into the differentiated equation and solve for dV/dt.
- dV/dt = 4pi (5)2 (2)
- dV/dt = 200pi cm3/s
Therefore, when the radius of the sphere is 5 cm, the volume of the sphere is increasing at a rate of 200pi cm3/s.
FAQ
Yes, there are several common pitfalls in related rates problems. One common mistake is not correctly setting up the relationship between the variables before differentiating. It's crucial to ensure that the equation relating the variables is accurate and represents the given scenario. Another common error is forgetting to apply the chain rule when differentiating or applying it incorrectly. Additionally, students sometimes confuse the given rates or misinterpret the problem, leading to incorrect solutions. It's essential to read the problem carefully, identify known and unknown rates, and methodically work through the solution.
Absolutely! While many standard related rates problems involve two variables, more complex problems can involve three or more related variables. In such cases, the approach remains the same: establish relationships between the variables using appropriate equations, differentiate with respect to time, and then use the given information to solve for the desired rate. The presence of more variables might require the use of additional equations or more complex differentiation, but the fundamental principles of related rates remain consistent.
Direct rate problems involve finding the rate of change of a single variable, often using straightforward differentiation. For example, given a function f(x), a direct rate problem might ask for the rate of change of f at a specific value of x. In contrast, related rates problems involve two or more variables whose rates of change are interconnected. The challenge is to determine the rate of change of one variable based on the known rate of change of another variable. This requires setting up an equation relating the variables, differentiating implicitly, and then solving for the desired rate using the chain rule.
In related rates problems, we typically differentiate with respect to time, denoted as 't'. This is because most related rates problems involve quantities that are changing over time. By differentiating with respect to time, we can determine how fast one quantity is changing in relation to another as time progresses. It's essential to identify which rates are known and which rates need to be found, and then set up an equation relating the variables. Once this equation is established, differentiating with respect to time allows us to find the desired rate of change.
The chain rule is fundamental in related rates problems because it allows us to relate the rate of change of one variable to the rate of change of another variable. In many real-world scenarios, we often know the rate at which one quantity is changing and want to determine how that affects the rate of change of a related quantity. The chain rule provides the mathematical framework to establish this relationship. By differentiating an equation that relates the two variables and then applying the chain rule, we can express the rate of change of one variable in terms of the rate of change of the other.
Practice Questions
The volume V of a sphere is given by the formula V = (4/3)πr3. To find the rate at which the volume is increasing, we need to find dV/dt. Using the chain rule, we differentiate V with respect to r and then multiply by dr/dt (the rate of change of the radius). So, dV/dt = dV/dr * dr/dt = (4πr2) * (3 cm/min). Substituting r = 5 cm, dV/dt = 4π(5)2(3) = 300π cm3/min. Therefore, when the radius of the balloon is 5 cm, the volume of the balloon is increasing at a rate of 300π cm3/min.
The volume V of a cone is given by V = (1/3)πr2h. Since the tank has a fixed shape, the radius and height of the water in the tank are related by the similarity ratio. Specifically, r/h = 5/10 = 1/2. So, r = h/2. Now, substituting this into the volume formula, we get V = (1/3)π(h/2)2h = (1/12)πh3. Differentiating V with respect to t to get dV/dt, and using the chain rule, we have dV/dt = dV/dh * dh/dt = (1/4)πh2 * dh/dt. We are given dV/dt = 12 cm3/min, and we want to find dh/dt when h = 4 cm. Substituting these values in, 12 = (1/4)π(4)2 * dh/dt. Solving for dh/dt, dh/dt = 12/(π(4)2/4) = 12/(4π) = 3/π cm/min. Therefore, when the water is 4 cm deep, the water level is rising at a rate of 3/π cm/min.
