Problem 53 - College Algebra with Modeling and Visualization

Question

Find the volume, the surface area of the side, and the total surface area of the cylinder that satisfies the given conditions, where \(r\) is the radius and \(h\) is the height. Approximate values to the nearest tenth. \(\mathbf{r}=12\) millimeters, and \(\mathrm{h}\) is twice \(\mathrm{r}\)

Step-by-Step Solution

Answer

Height is 24 mm, volume is 10859.5 mm³, lateral area is 1809.6 mm², total area is 2711.2 mm².

Step 1: Calculate the height (h)

Given that the height \( h \) is twice the radius \( r \), calculate \( h \) using the formula:\[ h = 2 \times r \]. Substitute \( r = 12 \) mm to find \( h \).\[ h = 2 \times 12 = 24 \text{ mm} \].

Step 2: Calculate the Volume of the Cylinder

The formula for the volume \( V \) of a cylinder is:\[ V = \pi r^2 h \]. Substitute \( r = 12 \) mm and \( h = 24 \) mm into the formula:\[ V = \pi \times 12^2 \times 24 \]. Calculate:\[ V = 3456\pi \]. Approximate to the nearest tenth:\[ V \approx 10859.5 \text{ cubic millimeters} \].

Step 3: Calculate the Lateral Surface Area

The lateral (side) surface area \( A_L \) of a cylinder is given by:\[ A_L = 2\pi rh \]. Substitute \( r = 12 \) mm and \( h = 24 \) mm:\[ A_L = 2\pi \times 12 \times 24 \]. Calculate:\[ A_L = 576\pi \]. Approximate to the nearest tenth:\[ A_L \approx 1809.6 \text{ square millimeters} \].

Step 4: Calculate the Total Surface Area

The total surface area \( A_T \) of a cylinder includes the lateral area plus the areas of the two circular bases:\[ A_T = 2\pi rh + 2\pi r^2 \]. Substitute \( r = 12 \) mm and \( h = 24 \) mm:\[ A_T = 576\pi + 2\pi \times 12^2 \]. Calculate \( 2\pi r^2 = 288\pi \), then:\[ A_T = 576\pi + 288\pi = 864\pi \]. Approximate to the nearest tenth:\[ A_T \approx 2711.2 \text{ square millimeters} \].

Key Concepts

Understanding Cylinder Geometry

Cylinders are a fundamental part of geometry involving shapes with straight lines and a circular base. To imagine a cylinder, think of a common object like a soda can. It consists of two parallel circular bases at the top and bottom, connected by a curved surface.
A few key features define a cylinder:

• The base is a circle, and its size is determined by the radius \( r \).
• Height \( h \) is the distance between the two bases.

The simple configuration of the cylinder makes it easy to calculate both its volume and surface area. These calculations require basic understanding of referencing the circular base and the space between them.

Performing Mathematical Calculations

Mathematical calculations give life to our understanding of shapes like cylinders. When solving for a cylinder's volume and surface areas, we rely on specific formulas. Let’s delve into the essential calculations:
- **Volume Calculation:** The formula \( V = \pi r^2 h \) calculates the space inside the cylinder. It requires:

• \( r \), the radius of the base.
• \( h \), the height of the cylinder.

By plugging in these values, you get the solid's volume. In our exercise, substituting \( r = 12 \) mm and \( h = 24 \) mm results in approximately 10859.5 cubic millimeters.
- **Lateral Surface Area:** Calculated by \( A_L = 2\pi rh \), it represents the area covering the sides, excluding the top and bottom bases. The provided dimensions lead to 1809.6 square millimeters.
- **Total Surface Area:** Given by \( A_T = 2\pi rh + 2\pi r^2 \), this formula includes the sides and the two bases. The exercise shows that the complete surface area is around 2711.2 square millimeters.

Algebraic Problem Solving Approach

Algebraic problem solving refers to breaking down problems using mathematical expressions and equations. With cylinders, this involves:

• Analyzing the relation between variables.
• Applying formulas independently for each required measurement.

In our example, the first algebraic step was determining the cylinder’s height, given that it was twice the radius \( h = 2r \). Setting a concrete value for \( r \), 12 mm, made the height solid at 24 mm.
This approach of methodically solving for each property is crucial.
Once the relationship and key figures are clear, plugging them into respective formulas allows swift calculations. This exercise highlights how understanding the relationship between dimensions helps in establishing exact numeric values.