The given problem involves the use of linearization to compute the volume of metal in the walls, top, and bottom of a soda can. Linearization is a technique used to approximate the value of a function near a given point. Let's break the problem into steps and solve it carefully using this method.

Step 1: Understanding the Problem

The formula for the volume $V$ of a right circular cylinder is: $V = \pi r^2 h$ where:

• $r$ is the radius of the cylinder,
• $h$ is the height of the cylinder.

We are given:

• The inner radius $r = 1.5$ inches,
• The height $h = 9$ inches,
• Wall thickness $t_{\text{wall}} = 0.02$ inches,
• Top and bottom thickness $t_{\text{top}} = 0.05$ inches.

Step 2: Decomposing the Can into Different Parts

We need to compute the volume of the metal in two parts:

1. The walls of the can (cylindrical surface).
2. The top and bottom of the can (circular disks).

Step 3: Using Linearization for Volume Approximation

We will use linearization to approximate the change in volume when thickness is added to the inner radius $r$ and height $h$.

The total volume of the can including the thickness can be calculated by considering a small change $dr$ in the radius and $dh$ in the height.

(A) Volume of the Walls:

For the walls of the cylinder, the thickness $t_{\text{wall}}$ modifies the radius from $r$ to $r + t_{\text{wall}}$. The change in the volume of the cylindrical part due to the thickness can be approximated using linearization: $dV_{\text{wall}} = \frac{\partial V}{\partial r} dr$ where $dr = t_{\text{wall}}$.

The volume $V$ is given by: $V = \pi r^2 h$ Thus, the derivative with respect to $r$ is: $\frac{\partial V}{\partial r} = 2\pi r h$ Now, using $dr = t_{\text{wall}} = 0.02$: $dV_{\text{wall}} = 2\pi r h \cdot dr = 2\pi (1.5)(9)(0.02)$ Calculating this: $dV_{\text{wall}} = 2\pi (1.5)(9)(0.02) = 5.6529 \, \text{in}^3$ This is the volume of metal in the walls of the can.

(B) Volume of the Top and Bottom:

The top and bottom are two circular disks, each with a thickness of $t_{\text{top}} = 0.05$. The volume of each disk is: $V_{\text{disk}} = \pi r^2 \cdot t_{\text{top}}$ For both top and bottom disks: $V_{\text{top\_bottom}} = 2 \cdot \pi r^2 \cdot t_{\text{top}} = 2 \cdot \pi (1.5)^2 (0.05)$ Calculating this: $V_{\text{top\_bottom}} = 2 \cdot \pi (1.5)^2 (0.05) = 0.70686 \, \text{in}^3$

Step 4: Total Volume of Metal

The total volume of metal in the can is the sum of the volume of the walls and the volume of the top and bottom: $V_{\text{total}} = dV_{\text{wall}} + V_{\text{top\_bottom}}$ $V_{\text{total}} = 5.6529 + 0.70686 = 6.36 \, \text{in}^3$

Thus, the volume of metal in the walls, top, and bottom of the can is approximately 6.36 in³.

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Would you like any further details on this? Here are some related questions:

1. How does linearization work more generally for volume-related problems?
2. Can you explain how the derivative $\frac{\partial V}{\partial r}$ was computed?
3. What happens if the wall thickness is increased?
4. How would you apply linearization to a different shape, like a sphere?
5. What other approximations can be made using linearization for similar geometry?

Tip: Linearization is especially useful when dealing with small changes or perturbations in parameters because it simplifies complex functions into a manageable linear form for quick approximations.