Rate Of Change Of A Sphere's Surface Area A Comprehensive Guide

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In the realm of calculus and related rates, understanding how different properties of a geometric shape change with respect to time is a fundamental concept. This article delves into a specific problem: determining the rate of increase in the surface area of a sphere when its radius is increasing at a constant rate. We'll explore the underlying principles, the mathematical formulas involved, and provide a step-by-step solution. Let's embark on this mathematical journey together.

Problem Statement

Imagine a sphere, like a balloon being inflated, whose radius is expanding over time. The problem states that the radius of this sphere is increasing at a rate of $3 rac{cm}{s}$. Our goal is to find out how quickly the surface area of the sphere is increasing at the precise moment when the radius is 2 cm. This problem elegantly combines geometry and calculus, requiring us to relate the rate of change of the radius to the rate of change of the surface area.

Key Concepts and Formulas

Before we dive into the solution, let's solidify our understanding of the essential concepts and formulas:

  1. Surface Area of a Sphere: The surface area (A) of a sphere is given by the formula $A = 4

  2. Related Rates: Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. The chain rule from calculus is a crucial tool in solving these problems. It allows us to express the derivative of a composite function in terms of the derivatives of its constituent functions.

  3. Differentiation: Differentiation is a fundamental operation in calculus that allows us to find the instantaneous rate of change of a function. In this problem, we will differentiate both sides of the surface area formula with respect to time (t) to relate the rates of change of the surface area and the radius.

Step-by-Step Solution

Now, let's break down the problem and solve it step by step:

Step 1: Write down the given information and what we need to find.

We are given that the rate of increase of the radius, $ rac{dr}{dt}$, is $3 rac{cm}{s}$. We need to find the rate of increase of the surface area, $ rac{dA}{dt}$, when the radius, r, is 2 cm.

Step 2: Write the formula for the surface area of a sphere.

The surface area (A) of a sphere is given by:

A = 4 This formula is the cornerstone of our solution, as it directly relates the surface area to the radius. Understanding this relationship is crucial for solving the problem. ### Step 3: Differentiate both sides of the equation with respect to time (t). To relate the rates of change, we differentiate both sides of the surface area formula with respect to time (t). This is where the power of calculus comes into play. Applying the chain rule, we get: $ rac{dA}{dt} = rac{d}{dt}(4 Using the chain rule, we differentiate the right side with respect to r and then multiply by $ rac{dr}{dt}$: $ rac{dA}{dt} = 4 Simplifying, we get: $ rac{dA}{dt} = 8 This equation is the heart of our solution. It tells us how the rate of change of the surface area is related to the radius and the rate of change of the radius. This is a classic example of a related rates problem, where we use calculus to connect the rates of change of different quantities. ### Step 4: Substitute the given values. We are given that $ rac{dr}{dt} = 3 rac{cm}{s}$ and r = 2 cm. Substitute these values into the equation we derived in the previous step: $ rac{dA}{dt} = 8 $ rac{dA}{dt} = 8 ### Step 5: Calculate the rate of increase of the surface area. Performing the calculation, we get: $ rac{dA}{dt} = 48 Therefore, the rate of increase in the surface area when the radius is 2 cm is $48 ## The Correct Answer Comparing our result with the given options, we see that the correct answer is: * C. $48 ## Expanding on the Concept of Related Rates Related rates problems are a staple in calculus courses because they demonstrate the power of calculus in solving real-world problems. These problems often involve geometric shapes, physical quantities, and their rates of change with respect to time. The key to solving related rates problems lies in identifying the relationships between the quantities involved and using calculus to connect their rates of change. ## Common Mistakes and How to Avoid Them When tackling related rates problems, it's easy to make mistakes. Here are some common pitfalls and how to avoid them: 1. **Forgetting the Chain Rule:** The chain rule is essential for differentiating composite functions. Make sure to apply it correctly when differentiating with respect to time. 2. **Substituting Too Early:** Avoid substituting the given values too early in the problem. It's best to differentiate first and then substitute the values to avoid complications. 3. **Incorrectly Identifying Relationships:** Make sure you correctly identify the relationships between the quantities involved. A clear diagram can often help in visualizing these relationships. 4. **Units:** Pay close attention to units throughout the problem. Make sure your units are consistent and that your final answer has the correct units. ## Real-World Applications of Related Rates Related rates have numerous applications in various fields, including: * **Physics:** Calculating the rate of change of velocity or acceleration. * **Engineering:** Designing structures and systems that can withstand changing conditions. * **Economics:** Modeling the rate of change of economic indicators. * **Computer Graphics:** Creating realistic animations and simulations. ## Practice Problems To master related rates problems, practice is key. Here are a few practice problems to test your understanding: 1. A ladder is sliding down a wall. If the top of the ladder is sliding down at a rate of 2 m/s, how fast is the bottom of the ladder sliding away from the wall when the ladder is 5 m long and the top is 4 m from the ground? 2. A conical tank is being filled with water. If the water is flowing in at a rate of 10 cubic meters per minute, how fast is the water level rising when the water is 3 meters deep? ## Conclusion In this comprehensive guide, we've explored the problem of finding the rate of increase in the surface area of a sphere when its radius is increasing at a constant rate. We've delved into the key concepts, the mathematical formulas, and provided a step-by-step solution. We've also discussed common mistakes, real-world applications, and practice problems. With a solid understanding of related rates, you'll be well-equipped to tackle a wide range of calculus problems. Remember, practice makes perfect, so keep exploring and applying these concepts to deepen your understanding.