Math Problem Statement

Use differentials to estimate the amount of metal (in cm3) in a closed cylindrical can that is 22 cm high and 8 cm in diameter if the metal in the top and the bottom is 0.4 cm thick and the metal in the sides is 0.05 cm thick. (Round your answer to two decimal places.)

Solution

We are tasked with estimating the amount of metal in a closed cylindrical can using differentials. The problem involves a cylinder with a height of 22 cm and a diameter of 8 cm, where the metal in the top and bottom is 0.4 cm thick, and the metal in the sides is 0.05 cm thick.

Step 1: Basic formula for volume of a cylinder

The volume VV of a cylinder is given by the formula:

V=πr2hV = \pi r^2 h

where rr is the radius and hh is the height.

Given:

  • Height of the can h=22cmh = 22 \, \text{cm},
  • Diameter of the can =8cm= 8 \, \text{cm}, so the radius r=82=4cmr = \frac{8}{2} = 4 \, \text{cm}.

Step 2: Calculate the volume of the entire can (outer volume)

The volume of the entire can (assuming it is a solid) can be calculated as:

Vouter=πr2h=π(4)2(22)=π(16)(22)=352πcm3V_{\text{outer}} = \pi r^2 h = \pi (4)^2 (22) = \pi (16) (22) = 352\pi \, \text{cm}^3

So,

Vouter=1106.64cm3V_{\text{outer}} = 1106.64 \, \text{cm}^3

Step 3: Estimate the thickness using differentials

We can break this down into two parts: the volume of the metal in the top and bottom, and the volume of the metal in the sides.

(a) Volume of metal in the top and bottom

Each top and bottom is a disk with a thickness of 0.4cm0.4 \, \text{cm}. The volume of one disk is:

Vdisk=πr2tV_{\text{disk}} = \pi r^2 t

where tt is the thickness (0.4 cm). So for each disk:

Vdisk=π(4)2(0.4)=π(16)(0.4)=6.4πcm3V_{\text{disk}} = \pi (4)^2 (0.4) = \pi (16)(0.4) = 6.4\pi \, \text{cm}^3

For two disks (top and bottom), the total volume of metal is:

Vtop+bottom=2×6.4π=12.8πcm3V_{\text{top+bottom}} = 2 \times 6.4\pi = 12.8\pi \, \text{cm}^3

So,

Vtop+bottom=40.21cm3V_{\text{top+bottom}} = 40.21 \, \text{cm}^3

(b) Volume of metal in the sides

The thickness of the metal in the sides is 0.05cm0.05 \, \text{cm}. The side of the cylinder is essentially a rectangular sheet of metal, which when wrapped forms the side of the cylinder. The surface area of the side (before considering thickness) is:

Aside=2πrhA_{\text{side}} = 2\pi r h

Now, we multiply this area by the thickness tt to get the volume of the metal in the sides:

Vside=2πrht=2π(4)(22)(0.05)=8.8πcm3V_{\text{side}} = 2\pi r h \cdot t = 2\pi (4)(22)(0.05) = 8.8\pi \, \text{cm}^3

So,

Vside=27.64cm3V_{\text{side}} = 27.64 \, \text{cm}^3

Step 4: Total volume of metal

Now, we can add up the volumes of the metal in the top, bottom, and sides to get the total amount of metal used in the can:

Vtotal=Vtop+bottom+Vside=40.21cm3+27.64cm3=67.85cm3V_{\text{total}} = V_{\text{top+bottom}} + V_{\text{side}} = 40.21 \, \text{cm}^3 + 27.64 \, \text{cm}^3 = 67.85 \, \text{cm}^3

Thus, the total amount of metal in the can is approximately:

67.85cm3\boxed{67.85 \, \text{cm}^3}


Do you want more details or clarifications on any part of this solution?

Here are 5 related questions:

  1. How would the answer change if the metal thickness in the sides increased to 0.1 cm?
  2. How would you calculate the volume of metal if the can had no bottom?
  3. How does the volume of the can change if the height increases by 10%?
  4. What is the surface area of the entire can?
  5. How would the volume calculation change if the can's top were conical?

Tip: Understanding how differentials can estimate changes in volumes or areas is a powerful tool in physics and engineering!

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Math Problem Analysis

Mathematical Concepts

Geometry
Cylinders
Differentials
Volume Estimation

Formulas

Volume of a cylinder: V = πr^2h
Volume of a disk: V_disk = πr^2t
Surface area of a cylinder: A_side = 2πrh

Theorems

Differentials in volume estimation

Suitable Grade Level

Grades 10-12