Problem 35 - Calculus Early Transcendentals

Question

Use differentials to estimate the amount of tin in a closed tin can with diameter 8 \(\mathrm{cm}\) and height 12 \(\mathrm{cm}\) if the tin 0.04 \(\mathrm{cm}\) thick.

Step-by-Step Solution

Answer

The volume of tin is approximately \(5.124\\pi \, \text{cm}^3\).

Step 1: Understand the Problem

We need to estimate the volume of tin required to make a closed tin can. The can is a cylinder, with an outer diameter of 8 cm and a height of 12 cm. The tin thickness is 0.04 cm, meaning both the diameter and height will reduce by this thickness from all sides when considering the inner volume.

Step 2: Set Up the Formula for Volume of a Cylinder

The formula for the volume of a cylinder is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. The outer dimensions are diameter 8 cm and height 12 cm, so the outer radius \( R = 4 \) cm and height \( H = 12 \) cm.

Step 3: Calculate Outer Volume of the Can

Using the outer dimensions, the outer volume of the can is: \[ V_{outer} = \pi R^2 H = \pi (4)^2 (12) = 192\pi \, \text{cm}^3 \]

Step 4: Define Inner Dimensions of the Can

The inner dimensions will be reduced by the thickness of the tin from all sides: \[ r = R - t = 4 - 0.04 = 3.96 \, \text{cm} \] \[ h = H - 2t = 12 - 2(0.04) = 11.92 \, \text{cm} \] We subtract \( 2t \) from the height because the thickness affects both the top and the bottom.

Step 5: Calculate Inner Volume of the Can

Using the reduced dimensions, the inner volume of the can is: \[ V_{inner} = \pi r^2 h = \pi (3.96)^2 (11.92) \] \[ V_{inner} \approx \pi (15.6816)(11.92) \] \[ V_{inner} \approx 186.876 \pi \, \text{cm}^3 \]

Step 6: Calculate the Volume of Tin

The volume of tin used for forming the can is the difference between the outer volume and the inner volume: \[ V_{tin} = V_{outer} - V_{inner} \] \[ V_{tin} = 192\pi - 186.876\pi \] \[ V_{tin} = 5.124\pi \, \text{cm}^3 \] This is the estimated volume of tin required to form the can.

Key Concepts

Understanding the Volume of a Cylinder

When estimating the volume of tin required for a closed tin can, it's crucial to understand how to calculate the volume of a cylinder, whether for the entire can (outer volume) or just the space inside it (inner volume). The formula for the volume of a cylinder is given by \( V = \pi r^2 h \), where:

• \( r \) is the radius of the cylinder’s base.
• \( h \) is the height of the cylinder.
• \( \pi \) is approximately 3.14159, a constant in mathematics associated with circles.

For a tin can with an outer diameter of 8 cm, the outer radius \( R = 4 \) cm. The height \( H \) is 12 cm.
This makes the outer volume \( V_{outer} = \pi (4)^2 (12) = 192\pi \text{ cm}^3 \). Remember that the tin thickness alters these outer measurements for computing the inner volume.
Understanding these basics can help in visualizing the entire cylinder and analyzing how changes, like thickness, impact the overall measures.

Surface Area and Its Influence

Knowing how to calculate the surface area of a cylinder is just as important as understanding its volume, especially when considering materials like tin. The surface area gives insight into how much surface is exposed or covered by the tin.
The formula for the surface area of a cylinder involves both its lateral area and its top/bottom circles:

• Lateral area: \( 2\pi rh \)
• Top and bottom areas (two circles): \( 2\pi r^2 \)

Thus, the total surface area of a cylinder is \( 2\pi rh + 2\pi r^2 \).
For our can with the adjusted inner radius \( r = 3.96 \) cm and height \( h = 11.92 \) cm, calculating the surface area helps you understand the tin needed to cover the entire structure, both outside and inside.
Remember how surface area changes with varying thickness, impacting how materials are managed and utilized in real-world applications.

Estimation Using Derivatives

Derivatives are powerful tools in mathematics for estimating small changes in a shape’s dimensions and understanding their impact on volume and surface area.
When dealing with a change in dimensions such as an increase or decrease in radius and height (due to tin thickness), the derivative helps approximate the difference in volumes without recalculating everything from scratch.

Why Use Derivatives?

• Easy approximation of small changes.
• Understanding sensitivity of the volume to changes in dimensions.

Applying differentials in this context, we consider the slight reduction in the can's dimensions due to the tin thickness. We use this concept to estimate the volume lost within the cylinder without direct physical measurement. This is especially useful when precision is critical, and slight variations in dimensions can significantly affect material volume like in \( V_{tin} = V_{outer} - V_{inner} \).
Understanding differentials and how they connect to derivatives equips students with the analytic skills necessary for precise calculations in engineering and material science fields.