Question

(a) Write the volume \(V\) of a cube as a function of the side length \(s\) . (b) Find the (instantaneous) rate of change of the volume V with respect to a side s. (c) Evaluate the rate of change of \(V\) at \(s=1\) and \(s=5\) (d) If \(s\) is measured in inches and \(V\) is measured in cubic inches, what units would be appropriate for \(d V / d s ?\)

Short answer

The volume function of a cube is \(V = s^3\). The rate of change of \(V\) with respect to \(s\) is \(dV/ds = 3s^2\). When \(s=1\), \(dV/ds = 3\) and when \(s=5\), \(dV/ds = 75\). The units for the rate of change are inches squared per inch.

Step-by-step solution

Write Volume Function

The volume \(V\) of a cube can be written as a function of the side length \(s\) . For a cube, the volume is given by \(V = s^3\) .

Find Rate of Change

The rate of change of the volume \(V\) with respect to the side length \(s\) is given by the derivative of the volume function with respect to \(s\). This can be computed as follows: \(dV/ds = 3s^2 \)

Evaluate Rate of Change at Specific Points

The rate of change of \(V\) is evaluated at \(s=1\) and \(s=5\). Substituting these values into the derivative function, we get \(dV/ds (s=1) = 3*1^2 = 3\) and \(dV/ds (s=5) = 3*5^2 = 75\)

Identify Appropriate Units

If \(s\) is measured in inches and \(V\) is measured in cubic inches, then the rate of change of volume with respect to side length, \(dV/ds\), would be measured in cubic inches per inch which simplifies to inches squared per inch.

Key concepts

Derivative of Volume Function

Understanding the derivative of a volume function is fundamental in calculus as it provides the rate at which volume changes with respect to another variable. For the volume of a cube expressed as a function of its side length, written as V = s^3, the derivative with respect to the side length s is mathematically represented as dV/ds. To calculate this derivative, we apply the power rule which states that the derivative of s^n is n*s^(n-1). As such, the derivative of the volume function of the cube is 3s^2. This derivative signifies how quickly the volume of the cube is changing as its sides change.

In practical terms, if you imagine a cube whose sides are expanding or contracting, the rate of this expansion or contraction at any given moment is given by this derivative. This has applications in various fields, including physics and engineering, where understanding how a change in dimension affects volume can be critical.

Instantaneous Rate of Change

The concept of instantaneous rate of change is comparable to measuring the speed of a car at a specific moment. In calculus, it describes how fast a given quantity changes at a particular instant. It is calculated using derivatives, which provide a snapshot of this rate of change at any given point. When we found the derivative of the volume function of a cube, we were finding its instantaneous rate of change with respect to its side length.

To grasp this at a specific point, consider our cube example. The rate of change of the cube's volume when its side length is equal to 1 inch is found by evaluating the derivative at s=1, which gives us a value of 3 cubic inches per inch. This result indicates that at the exact moment when the side length is 1 inch, if the side length increases by a tiny bit, the volume will increase by approximately three times that bit.

Volume of a Cube

The volume of a cube is a simple yet important geometric concept that lays the foundation for more complex volume calculations in mathematics. It is calculated by raising the length of a cube's side, s, to the third power (cubed), yielding the formula V=s^3. This formula assumes all sides of the cube are equal, which is an essential characteristic of a cube.

Understanding the volume of a cube not only helps in mathematical problems but also in real-world applications, such as determining how much space an object will occupy or the capacity of a container. Visualizing a cube growing or shrinking in size is a concrete way to understand how changes in dimensions affect the overall volume of a shape, which in turn fosters a better understanding of the relationship between linear dimensions and volume.

Application of Derivatives

Derivatives are a powerful tool in calculus with numerous applications across various fields such as physics, engineering, economics, and more. They enable us to determine the rate at which one quantity changes relative to another. By finding a function's derivative, we can calculate the instantaneous rate of change, forecast future trends, and optimize processes.

In the context of the cube problem, the application of derivatives helps us understand how a change in the length of the sides of a cube will affect its volume. This has practical implications, for example, in determining the amount of material required for manufacturing or the expansion of materials with temperature changes. From an academic standpoint, mastering this concept is an essential step for students as they tackle real-world problems that involve rates of change and understanding how various quantities are interrelated.