Problem 12 - University Calculus: Early Transcendentals

Question

A cube's surface area increases at the rate of 72 in \(^{2} /\) sec. At what rate is the cube's volume changing when the edge length is \(x=3\) in?

Step-by-Step Solution

Answer

The volume increases at a rate of 54 in³/sec.

Step 1: Understand the Given Problem

We know that the surface area of a cube is given by \( S = 6x^2 \), where \( x \) is the length of the edge of the cube. We are told that the rate at which the surface area increases, \( \frac{dS}{dt} \), is 72 in²/sec. We need to find the rate at which the volume of the cube is changing, \( \frac{dV}{dt} \), when the edge length \( x = 3 \) in.

Step 2: Write the Formula for Volume

The formula for the volume \( V \) of a cube is \( V = x^3 \). This will be used later to find the derivative with respect to time.

Step 3: Differentiate the Surface Area Formula

Differentiate the surface area formula with respect to time \( t \): \[ \frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt} \]We know \( \frac{dS}{dt} = 72 \) in²/sec, so substitute this value and solve for \( \frac{dx}{dt} \).

Step 4: Solve for \( \frac{dx}{dt} \)

Using the equation \( 72 = 12x \frac{dx}{dt} \) and substituting \( x = 3 \): \[ 72 = 12 \times 3 \times \frac{dx}{dt} \] This simplifies to \( 72 = 36 \frac{dx}{dt} \), giving \( \frac{dx}{dt} = 2 \) in/sec.

Step 5: Differentiate the Volume Formula

Differentiate the volume formula \( V = x^3 \) with respect to time \( t \):\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt} \]

Step 6: Substitute and Calculate \( \frac{dV}{dt} \)

Substitute \( x = 3 \) and \( \frac{dx}{dt} = 2 \) into the differentiated volume formula:\[ \frac{dV}{dt} = 3 \times (3)^2 \times 2 = 3 \times 9 \times 2 = 54 \] So, \( \frac{dV}{dt} = 54 \) in³/sec.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function changes as its input changes. In the context of the problem, derivatives are used to determine how the cube's volume changes over time, based on changes in its surface area and edge length. When you find a derivative, you're essentially finding the slope of the tangent line at any point on a function.
This slope indicates how sharp or flat the function is at that point. For the cube's edge, we're exploring how efficiently the volume changes, given a particular edge growth rate. Differentiation allows us to convert these static formulas for surface area and volume into dynamic tools that track changes over time.

Related Rates

Related rates problems involve finding the rate at which one quantity changes in relation to another. This often involves several interconnected parts of a problem, like in our cube example. We can express the change of one measurement in terms of another varying measurement.
In the cube problem, we know how fast the surface area is increasing and use this to find how fast the volume is increasing. We achieve this by finding relationships between surface area, volume, and edge length and then applying calculus techniques to discover how these rates interact. By combining given rates and unknowns using derivatives, we can solve for the desired rate of change.

Differentiation

Differentiation is the process of finding a derivative. It allows us to analyze how one quantity changes in response to changes in another. By applying the rules of differentiation to the formulas for surface area and volume, we can interpret these changes over time.
In this problem, differentiation converts the static equations for surface area (\(6x^2\)) and volume (\(x^3\)) into dynamic forms that express their rates of change (\(\frac{dS}{dt}\) and \(\frac{dV}{dt}\)). By taking the derivative with respect to time, we glean insights into how a cube's geometric properties evolve as it grows.

Geometry

Geometry explores the properties and relations of points, lines, surfaces, and solids. The cube is a classic geometric shape, and understanding its properties is key to solving the problem.
The surface area of a cube is determined by its six square faces (\(S = 6x^2\)), and its volume is found by cubing the edge length (\(V = x^3\)). These geometric formulas form the foundation upon which we build our calculations. By understanding the geometric implications of these formulas, we can effectively apply calculus techniques to practically solve related rates problems involving shapes and volumes.

Mathematics Problem Solving

Mathematics problem solving involves deciphering and resolving complex problems using logical reasoning and mathematical techniques. Our cube problem illustrates this by requiring us to apply derivatives to find the rate of volume change given a change in surface area.
By following a structured approach — understanding the problem, employing differentiation, and systematically solving for unknown rates — we develop a clear path to the solution. Consistent practice with such problems enhances analytical thinking and strengthens our ability to tackle varied mathematical challenges. Clear problematics handling demonstrates how mathematical theories translate into practical solutions for real-world scenarios.