when doing this problem make sure to use linearization The volume V of a solid right circular cylinder is given by 
V = 𝜋r2h
 where r is the radius of the cylinder and h is its height. A soda can has inner radius 
r = 1.5 inches,
 height 
h = 9 inches,
 wall thickness 0.02 inches, and top and bottom thickness 0.05 inches. Use linearization to compute the volume, in in3, of metal in the walls and top and bottom of the can. Give your answer to 2 decimal places.

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**.

---

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.

When doing this problem make sure to use linearization. The volume V of a solid right circular cylinder is given by V = 𝜋r²h, where r is the radius of the cylinder and h is its height. A soda can has inner radius r = 1.5 inches, height h = 9 inches, wall thickness 0.02 inches, and top and bottom thickness 0.05 inches. Use linearization to compute the volume, in in³, of metal in the walls and top and bottom of the can. Give your answer to 2 decimal places.

Learn how to use linearization to approximate the volume of metal in the walls and top/bottom of a soda can. This problem involves geometry and calculus concepts, including the volume formula for a cylinder and circular disks. The can has an inner radius of 1.5 inches, height of 9 inches, and wall thickness of 0.02 inches. Solution provided step-by-step.

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