Problem 16 - Calculus Volume 1

Question

Draw and label diagrams to help solve the related-rates problems. The side of a cube increases at a rate of \(\frac{1}{2} \mathrm{~m} / \mathrm{sec}\). Find the rate at which the volume of the cube increases when the side of the cube is \(4 \mathrm{~m}\).

Step-by-Step Solution

Answer

The volume increases at a rate of 24 \(\mathrm{m^3/s}\).

Step 1: Understand the Problem

We need to find the rate at which the volume of a cube changes. Given that the side of the cube increases at a rate of \(\frac{1}{2} \mathrm{~m/s}\). The cube has a side length \(s = 4 \mathrm{~m}\) at a certain instant.

Step 2: Write the Volume Formula

The volume \(V\) of a cube with side length \(s\) is given by \(V = s^3\). We will differentiate this formula with respect to time \(t\).

Step 3: Differentiate with Respect to Time

Using the chain rule, differentiate both sides of the volume equation with respect to \(t\): \[ \frac{dV}{dt} = 3s^2 \frac{ds}{dt} \].

Step 4: Substitute Known Values

We know \(\frac{ds}{dt} = \frac{1}{2} \mathrm{~m/s}\) and \(s = 4 \mathrm{~m}\). Substituting these values into the differentiated equation:\[ \frac{dV}{dt} = 3(4)^2 \left(\frac{1}{2}\right) \].

Step 5: Calculate the Rate of Change of Volume

Calculate the expression: \[ \frac{dV}{dt} = 3 \times 16 \times \frac{1}{2} = 24 \].Thus, the volume of the cube is increasing at a rate of \(24 \mathrm{~m^3/s}\) when \(s = 4 \mathrm{~m}\).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with change. In this specific problem, we're interested in how the volume of a cube changes over time. Calculus helps us to study this by providing tools to compute rates of change. These rates of change are vital for understanding phenomena in physics, engineering, and many other fields. By finding the rate at which the cube's volume increases, we're essentially applying calculus to solve a real-world problem. In calculus, we often use differentiation as a method to find these rates of change. Understanding the role of calculus will help you approach related-rates problems, which involve changes happening at the same time, like in this cube problem, with increased confidence.

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which tells us how a quantity changes when another quantity changes.
For our cube problem, we start by expressing the volume of the cube as a function of its side length:

• The volume of a cube: \(V = s^3\)

The differentiation of this volume function with respect to time is crucial. It helps us find the rate at which the volume changes as the side length of the cube increases. By applying the differentiation rules to the formula \(V = s^3\), we find the equation for the rate of change of volume \(\frac{dV}{dt}\). Differentiation simplifies the equation into a form where we can easily input the given values to find our solution.

Chain Rule

The chain rule is a crucial concept when solving related-rates problems, like the one with the cube. It is used to differentiate composite functions. In this cube example, the rate of change of the volume with respect to time involves differentiating a function of a function.
The volume of the cube is a function of its side length \(s\), which in turn is a function of time \(t\). Therefore, to find \(\frac{dV}{dt}\), we need to apply the chain rule on the function \(V = s^3\).

• Start by differentiating \(V\) with respect to \(s\): \(\frac{dV}{ds} = 3s^2\).
• Next, multiply by the derivative of \(s\) with respect to \(t\): \(\frac{ds}{dt}\).

By doing this, we effectively find \(\frac{dV}{dt} = 3s^2\frac{ds}{dt}\). The chain rule is a powerful tool in calculus that helps us tackle problems involving multiple rates of change.

Geometry

Geometry plays a key role in understanding the shapes and relationships within related-rates problems. In this specific exercise, we're dealing with a cube, a three-dimensional geometric shape. A cube has equal sides, and its volume is calculated by cubing the side length: \(V = s^3\).
Geometry helps us visualize and set up the problem. By drawing a cube, we can label its sides and clearly see how a change in the side length affects the entire shape. Visualizing geometry not only makes it easier to set up equations but also provides insight into the physical changes happening in the problem. Understanding the geometric principles underlying the problems helps in correctly applying the calculus techniques to find accurate solutions.