Problem 18 - Precalculus: Functions and Graphs

Question

Dimensions of a human limb A circular cylinder is sometimes used in physiology as a simple representation of a human limb. (a) Express the volume \(V\) of a cylinder in terms of its length \(L\) and the square of its circumference \(C\). The formula obtained in part (a) can be used to approximate the volume of a limb from length and circumference measurements. Suppose the (average) circumference of a human forearm is 22 centimeters and the average length is 27 centimeters. Approximate the volume of the forearm to the nearest \(\mathrm{cm}^{3}\).

Step-by-Step Solution

Answer

The approximate volume of the forearm is 1040 cm³.

Step 1: Understand the Relationship

The volume of a cylinder can be computed using the formula: \( V = \pi r^2 L \), where \( r \) is the radius and \( L \) is the length. The circumference \( C \) is given by \( C = 2 \pi r \). Our goal is to express the volume \( V \) in terms of \( L \) and \( C^2 \).

Step 2: Express the Radius in Terms of Circumference

From the circumference formula \( C = 2 \pi r \), solve for the radius \( r \): \( r = \frac{C}{2 \pi} \).

Step 3: Substitute Radius into Volume Formula

Substitute \( r = \frac{C}{2 \pi} \) into \( V = \pi r^2 L \): \[V = \pi \left( \frac{C}{2\pi} \right)^2 L\]

Step 4: Simplify the Volume Expression

Simplify the expression: \[V = \pi \frac{C^2}{4\pi^2} L = \frac{C^2 L}{4 \pi}\] Thus, the volume \( V \) in terms of circumference \( C \) and length \( L \) is \( \frac{C^2 L}{4 \pi} \).

Step 5: Calculate the Approximate Volume of the Forearm

Given \( C = 22 \) cm and \( L = 27 \) cm, substitute these values into the formula \( V = \frac{C^2 L}{4 \pi} \): \[V = \frac{22^2 \times 27}{4 \pi} \]Calculate \( V \): \[V = \frac{484 \times 27}{4 \pi} \approx \frac{13068}{12.57} \approx 1040 \text{ cm}^3\]

Step 6: Round to Nearest Integer

Round the calculated volume to the nearest \( \text{cm}^3 \): The approximate volume of the forearm is \( 1040 \text{ cm}^3 \).

Key Concepts

Cylinder Surface Area

Understanding the cylinder surface area is a delightful exploration into geometry. The total surface area of a cylinder consists of two parts: the curved surface area and the areas of the two circular ends (bases). Each base has an area of \( \pi r^2 \) where \( r \) is the radius. For the curved surface area, imagine unrolling the cylinder into a rectangle. Its width is the circumference of the base \( (2 \pi r) \) and its height is the cylinder's length \( L \). Thus, the curved surface area is \( 2 \pi r L \).
The total surface area \( A \) therefore combines these:

• Area of both bases: \( 2 \pi r^2 \)
• Curved surface area: \( 2 \pi r L \)

Adding these yields the formula: \[ A = 2\pi r (r + L).\] This equation helps in various problems beyond cylinders, portraying the elegance and utility of precalculus.

Radius and Circumference

The concepts of radius and circumference are foundational in the study of circles and cylinders. The circumference \( C \) of a cylinder's base is an essential part of defining its shape. This is accessible through the simple formula \( C = 2 \pi r \), which relates the circumference directly to the radius \( r \).
For instance, measuring the circumference can often be easier than measuring the radius directly in practical situations. Knowing \( C \), one can solve for \( r \) using \( r = \frac{C}{2\pi} \).
This relationship is instrumental when working on cylinder volume problems, especially when the task asks to derive the volume in terms of circumference as an intermediate step. This technique is a common strategy in precalculus problem-solving, enhancing one's mathematical toolkit.

Volume Approximation

Volume approximation offers powerful tools in estimating the dimensions and capacities of shapes we encounter in the real world. In the case of a cylinder, knowing the average circumference and length provides an effective way to approximate its volume.
The formula \( V = \frac{C^2 L}{4 \pi} \) encapsulates this idea, beautifully derived by expressing the radius \( r \) in terms of circumference \( C \), and substituting it back into the volume formula \( V = \pi r^2 L \).
This step results in an approximation that is easily computed without direct access to the radius.

• Use the formula \( V = \frac{C^2 L}{4 \pi} \)
• Substitute known values of \( C \) and \( L \)
• Solve for \( V \), providing a practical approximation of volume.

This simplistic yet effective approach is prevalent in physiology where precise measurements of body parts such as limbs are necessary.

Precalculus Problem-Solving

Precalculus problem-solving equips students with strategies to tackle complex mathematical challenges. Exercises like the cylinder volume calculation promote analytical thinking by integrating various mathematical concepts. It involves understanding geometry, algebraic manipulation, and formula derivation.
The process often involves:

• Breaking down the problem into smaller, manageable parts
• Identifying relationships between different mathematical elements
• Deriving new formulas to simplify solutions

For instance, expressing cylinder volume in terms of circumference illustrates the skill of transforming equations to meet specific problem requirements. Such exercises underline the importance of flexibility and creativity in mathematical problem-solving, empowering students to handle unexpected scenarios and real-world applications.