Cube Surface Area Increase Problem A Calculus Exploration

Jul 11, 2025 by Admin 58 views

The surface area of a cube is increasing at a rate of 10 cm^2/s. Find the rate of increase of the edge when the edge is 12 cm

Understanding the Problem

In this problem, we are dealing with a cube whose surface area is changing over time. Surface area is increasing at a rate of 10 square centimeters per second (cm²/s). Our goal is to determine how quickly the edge length of the cube is increasing at the instant when the edge length is exactly 12 centimeters. This is a classic related rates problem in calculus, where we use derivatives to relate the rates of change of different variables.

To solve this, we need to establish a relationship between the surface area of a cube and its edge length. Let's denote the edge length of the cube as 'x' and the surface area as 'S'. A cube has six faces, each of which is a square. The area of one face is x², so the total surface area S is given by:

S = 6x²

We are given that dS/dt = 10 cm²/s, which represents the rate of change of the surface area with respect to time. We want to find dx/dt, the rate of change of the edge length with respect to time, when x = 12 cm. To do this, we'll use the chain rule of differentiation. We will differentiate the surface area equation with respect to time, which will allow us to relate dS/dt and dx/dt.

This problem highlights the practical applications of calculus in understanding how different quantities are related and how their rates of change influence each other. In many real-world scenarios, understanding these relationships is crucial for making predictions and informed decisions. For instance, in engineering, understanding how the dimensions of an object change with temperature or pressure is essential for designing stable and reliable structures. Similarly, in economics, understanding how different economic indicators are related can help in forecasting market trends.

Setting up the Equation

As mentioned before, the surface area (S) of a cube with edge length (x) is given by:

S = 6x²

This equation forms the basis for our problem. Now, we need to relate the rates of change of the surface area and the edge length. To do this, we differentiate both sides of the equation with respect to time (t). Remember that both S and x are functions of time, so we'll need to apply the chain rule when differentiating x² with respect to t.

Differentiating both sides of S = 6x² with respect to t, we get:

dS/dt = d/dt (6x²)

Using the chain rule, we have:

dS/dt = 6 * (2x) * (dx/dt)

Simplifying, we get:

dS/dt = 12x (dx/dt)

This equation is the key to solving the problem. It relates the rate of change of the surface area (dS/dt) to the rate of change of the edge length (dx/dt) and the current edge length (x). We are given dS/dt and x, and we want to find dx/dt. So, we can rearrange this equation to solve for dx/dt:

dx/dt = (dS/dt) / (12x)

This equation tells us that the rate at which the edge length is changing is directly proportional to the rate at which the surface area is changing, and inversely proportional to the current edge length. This makes intuitive sense: if the surface area is increasing rapidly, the edge length will also increase rapidly, but if the cube is already large, a given increase in surface area will result in a smaller increase in edge length.

This setup phase is crucial in related rates problems. It involves identifying the relevant variables, establishing the relationship between them, and differentiating with respect to time to relate their rates of change. A clear understanding of the problem and careful application of differentiation rules are essential for success.

Solving for the Rate of Increase of the Edge

Now that we have the equation relating the rates of change:

dx/dt = (dS/dt) / (12x)

We can plug in the given values to find the rate of increase of the edge when the edge is 12 cm. We are given:

dS/dt = 10 cm²/s (the surface area is increasing at this rate)
x = 12 cm (the edge length at the instant we are interested in)

Substituting these values into the equation, we get:

dx/dt = (10 cm²/s) / (12 * 12 cm)

dx/dt = 10 / (12 * 12) cm/s

dx/dt = 10 / 144 cm/s

dx/dt = 5 / 72 cm/s

Therefore, the rate of increase of the edge when the edge is 12 cm is 5/72 centimeters per second. This is approximately 0.0694 cm/s. This result tells us that at the instant when the cube's edge is 12 cm, the edge is growing at a rate of about 0.0694 cm for every second that passes.

It's important to include units in our answer to ensure that it is physically meaningful. The units of dx/dt are centimeters per second (cm/s), which makes sense since we are measuring the rate of change of a length (in centimeters) with respect to time (in seconds).

This final calculation step demonstrates how we use the equation derived from the relationship between surface area and edge length to answer the specific question posed in the problem. By plugging in the given values, we can find the desired rate of change.

Conclusion

In conclusion, the rate of increase of the edge of the cube when the edge is 12 cm is 5/72 cm/s, or approximately 0.0694 cm/s. This problem illustrates the power of related rates in calculus. By understanding how different quantities are related and applying the principles of differentiation, we can solve problems that involve changing quantities.

The key steps in solving this problem were:

Establishing the relationship between the surface area and the edge length of a cube (S = 6x²).
Differentiating both sides of the equation with respect to time to relate the rates of change (dS/dt = 12x (dx/dt)).
Solving for the rate of change of the edge length (dx/dt = (dS/dt) / (12x)).
Plugging in the given values to find the specific rate of increase when the edge is 12 cm.

This type of problem is common in calculus courses and highlights the importance of understanding the chain rule and its applications. It also reinforces the idea that calculus can be used to model and solve real-world problems involving changing quantities. Understanding related rates problems not only enhances one's calculus skills but also provides a foundation for more advanced mathematical modeling and analysis in various fields such as physics, engineering, and economics.

By carefully analyzing the relationships between variables and their rates of change, we can gain valuable insights into dynamic systems and make accurate predictions about their behavior. This problem, although seemingly simple, encapsulates the core principles of related rates and their significance in understanding change over time.