Problem 38 - Thomas Calculus

Question

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the volume \(V=x^{3}\) of a cube when the edge lengths change from \(x_{0}\) to \(x_{0}+d x\)

Step-by-Step Solution

Answer

The differential formula for the change in volume is \(dV = 3x^2 \, dx\).

Step 1: Identify Variables

First, recognize the given function as the volume of a cube: \(V = x^3\). The variable \(x\) represents the edge length of the cube.

Step 2: Recall the Concept of Differentials

The differential \(dV\) represents an infinitesimal change in the function \(V = x^3\), which approximates the change in the volume when \(x\) changes by an infinitesimal amount \(dx\).

Step 3: Find the Derivative

Compute the derivative of \(V\) with respect to \(x\). Since \(V = x^3\), the derivative \(\frac{dV}{dx} = 3x^2\).

Step 4: Calculate the Differential

Apply the differential formula \(dV = \frac{dV}{dx} \, dx\). Substituting the derivative, we have \(dV = 3x^2 \, dx\).

Step 5: Interpret the Differential Formula

The formula \(dV = 3x^2 \, dx\) provides an estimate for the change in volume \(dV\) for a small change in edge length \(dx\) from \(x_0\) to \(x_0 + dx\).

Key Concepts

Differentials

Differentials are like tiny tools in calculus that help us understand small changes in functions. Imagine you're tracking how the volume of a cube changes if you slightly increase its edge length. The differential gives a way to estimate that change.
To use differentials, follow these steps:

• Understand the function you're dealing with. Here, it's the volume of a cube, defined by the formula: \(V = x^3\).
• Identify the key variables. In this case, the volume \(V\) and the edge length \(x\).
• Recognize \(dV\) as the infinitesimally small change in volume when \(x\) is changed by a tiny amount \(dx\).

Differentials become especially handy when the changes in \(x\) are very small, providing an approximation for how the volume alters. This is crucial when precise measurements are difficult or when we want to quickly estimate outcomes.

Derivatives

Derivatives are the backbone of calculus. They tell you the rate at which one quantity changes with respect to another. In simpler terms, a derivative helps us see how a small change in one variable impacts another.
For the volume of a cube, where \(V = x^3\), the derivative with respect to \(x\) tells us how fast the volume changes as the edge length changes. This derivative is calculated as:

• Differentiate \(V = x^3\) with respect to \(x\), giving \(\frac{dV}{dx} = 3x^2\).

This result, \(3x^2\), is crucial because it helps create the differential formula used in estimating changes in volume. Understanding derivatives gives essential insight into how mathematical functions behave in dynamic scenarios. They are used not just in geometry, but in physics, engineering, and economics to model real-world phenomena.

Volume of a Cube

The volume of a cube is a classic geometric concept, calculated by multiplying the cube's edge length by itself three times. This is expressed by the formula \(V = x^3\), where \(x\) is the edge length. Cubes have equal edge lengths, making calculations straightforward.
Here are some key points to remember:

• Changing the edge length alters the volume. Since all sides are equal, a small change in one edge influences the overall size significantly.
• The formula \(V = x^3\) serves as the base for exploring more complex ideas like differentials and derivatives.

Understanding the volume's dependence on edge length helps in both theoretical studies and practical applications, such as packaging or materials engineering. The cube's volume is a foundational concept that links simple geometry with more advanced calculus principles.