Cube Surface Area Increase Rate Calculation
This article delves into a classic calculus problem involving related rates. Related rates problems explore how the rates of change of different variables are related when those variables are connected by an equation. Specifically, we will tackle the question: The volume of a cube is increasing at the rate of 6 cm³/min. How fast is the surface area increasing when the length of each edge is 12 cm? This problem requires us to understand the relationships between a cube's volume, surface area, and edge length, and to apply the principles of differential calculus to find the rate of change of the surface area.
Our core task is to determine the rate at which the surface area of a cube is increasing, given that its volume is increasing at a rate of 6 cubic centimeters per minute. The critical moment for this calculation is when the cube's edge length is exactly 12 centimeters. This problem elegantly demonstrates the application of calculus in understanding dynamic geometric relationships. We will methodically dissect the problem, establishing the relevant formulas, applying differentiation, and finally, solving for the desired rate of change. The solution will not only provide a numerical answer but also illuminate the underlying principles of related rates and their practical implications in geometry and beyond.
Before diving into the solution, let's solidify our understanding of the fundamental geometric relationships of a cube. A cube, a three-dimensional solid, is characterized by its six congruent square faces. The volume (V) of a cube is determined by cubing the length of its edge (s), expressed as V = s³. The surface area (A), on the other hand, is calculated by summing the areas of all six faces, giving us A = 6s². These two equations form the bedrock of our problem-solving approach. We are given the rate of change of volume (dV/dt) and are tasked with finding the rate of change of surface area (dA/dt). The link between these rates lies in their shared variable, the edge length (s), which changes over time. By employing the chain rule in calculus, we can connect dV/dt and dA/dt through their mutual dependence on s. This process will involve differentiating both the volume and surface area equations with respect to time (t), a crucial step in solving related rates problems.
Defining Variables and Rates
To begin, let's clearly define the variables and rates involved in this problem. We denote the length of each edge of the cube as s, which is a function of time (t). The volume of the cube, V, is also a function of time and is given by the formula V = s³. Similarly, the surface area of the cube, A, varies with time and is expressed as A = 6s². The problem states that the volume is increasing at a rate of 6 cm³/min, which we represent as dV/dt = 6 cm³/min. Our goal is to find how fast the surface area is increasing, which means we need to determine dA/dt when the edge length s is 12 cm. This careful setup of variables and rates is a crucial first step in solving any related rates problem. It helps to organize the given information and clarifies what we need to find. By establishing these definitions, we lay a solid foundation for the subsequent steps of differentiation and substitution.
Establishing the Formulas
The foundation of solving this related rates problem lies in the correct application of geometric formulas for a cube. The volume (V) of a cube is given by the formula V = s³, where 's' represents the length of an edge. This formula tells us how much space the cube occupies and is crucial because we know the rate at which the volume is changing. The surface area (A) of a cube, which is the total area of all its faces, is given by A = 6s². This is because a cube has six identical square faces, each with an area of s². The surface area is what we're interested in finding the rate of change for. These two formulas, V = s³ and A = 6s², are our primary tools. They link the cube's dimensions (edge length) to its volume and surface area, allowing us to connect the rates of change through calculus. In the next steps, we'll use these formulas to differentiate with respect to time, bringing in the rates of change dV/dt and dA/dt.
Differentiating the Volume Formula
The first crucial step in solving this related rates problem is to differentiate the volume formula with respect to time. We start with the formula for the volume of a cube, V = s³, where V is the volume and s is the length of an edge. Both V and s are functions of time t. To find the relationship between their rates of change, we differentiate both sides of the equation with respect to t. Applying the chain rule, we get dV/dt = d(s³)/dt = 3s² * ds/dt. This equation is fundamental because it connects the rate of change of the volume (dV/dt) to the rate of change of the edge length (ds/dt). We know dV/dt = 6 cm³/min, so we can substitute this value into the equation: 6 = 3s² * ds/dt. This equation now allows us to solve for ds/dt, the rate at which the edge length is changing, at any given value of s. This is a critical piece of information that we will use later to find the rate of change of the surface area.
Differentiating the Surface Area Formula
Now that we've differentiated the volume formula, let's turn our attention to the surface area. We have the formula A = 6s², where A is the surface area and s is the edge length of the cube. Both A and s are functions of time, so we need to differentiate both sides of the equation with respect to t to find the relationship between their rates of change. Differentiating A = 6s² with respect to t, we get dA/dt = d(6s²)/dt. Applying the chain rule, this becomes dA/dt = 12s * ds/dt. This equation is key to solving our problem because it directly relates the rate of change of the surface area (dA/dt) to the rate of change of the edge length (ds/dt) and the edge length itself (s). We already found an expression for ds/dt from the volume equation, so we can substitute that into this equation. This substitution will allow us to find dA/dt when s = 12 cm, which is what the problem asks for.
Finding ds/dt when s = 12 cm
Before we can calculate the rate at which the surface area is increasing (dA/dt), we need to determine the rate at which the edge length is changing (ds/dt) when the edge length (s) is 12 cm. We derived the equation dV/dt = 3s² * ds/dt from the volume formula. We know that dV/dt = 6 cm³/min, so we can substitute this value and s = 12 cm into the equation to solve for ds/dt. Plugging in the values, we get 6 = 3(12)² * ds/dt. This simplifies to 6 = 3 * 144 * ds/dt, or 6 = 432 * ds/dt. Now, we solve for ds/dt by dividing both sides of the equation by 432: ds/dt = 6 / 432 = 1 / 72 cm/min. This tells us that when the edge length is 12 cm, it is increasing at a rate of 1/72 cm per minute. This value is crucial because we will use it to find the rate at which the surface area is increasing.
Calculating dA/dt
Now that we know ds/dt when s = 12 cm, we can calculate dA/dt, the rate at which the surface area is increasing. We derived the equation dA/dt = 12s * ds/dt from the surface area formula. We have s = 12 cm and ds/dt = 1/72 cm/min, so we can substitute these values into the equation. Plugging in the values, we get dA/dt = 12 * 12 * (1/72). This simplifies to dA/dt = 144 / 72, which further simplifies to dA/dt = 2 cm²/min. Therefore, when the length of each edge is 12 cm, the surface area of the cube is increasing at a rate of 2 square centimeters per minute. This is our final answer, and it matches option (c) in the multiple-choice options provided. This result demonstrates how the principles of related rates can be applied to solve practical problems involving geometric shapes and their changing dimensions.
In conclusion, we have successfully solved the problem of finding the rate at which the surface area of a cube is increasing, given that its volume is increasing at a rate of 6 cm³/min, and the edge length is 12 cm. By understanding the relationships between volume, surface area, and edge length, and by applying the principles of differential calculus, we found that the surface area is increasing at a rate of 2 cm²/min. This problem underscores the power of calculus in analyzing dynamic situations and provides a clear example of how related rates problems can be approached and solved. The key steps included establishing the relevant formulas, differentiating with respect to time, and carefully substituting known values to solve for the unknown rate. This approach can be applied to a wide range of similar problems in physics, engineering, and other fields.
