Problem 26 - Algebra 2

Question

Industrial Design A storage tank will have a circular base of radius \(r\) and a height of \(r .\) The tank can be either cylindrical or hemispherical (half a sphere). a. Write and simplify an expression for the ratio of the volume of the hemispherical tank to its surface area (including the base). For a sphere, \(V=\frac{4}{3} \pi r^{3}\) and \(S .\) A. \(=4 \pi r^{2} .\) b. Write and simplify an expression for the ratio of the volume of the cylindrical tank to its surface area (including the bases). c. Compare the ratios of volume to surface area for the two tanks. d. Compare the volumes of the two tanks.

Step-by-Step Solution

Answer

The ratio of volume to surface area is greater for a hemisphere than for a cylinder when the radius is greater than zero (\(\frac{r}{3}>\frac{r}{4}\)). Additionally, the volume of a hemisphere is greater than the volume of a cylinder for a given radius (\(\frac{4}{3}\pi r^{3} > \pi r^{3}\)).

Step 1: Ratio of Volume to Surface Area for Hemisphere

Let's find the ratio of volume to surface area for a hemisphere. The volume of the hemisphere is given as \(\frac{4}{3} \pi r^{3}\), and the surface area is \(4 \pi r^{2} .\) This includes the bottom base as well. We simplify this ratio using algebra.\n\nRatio (Hemisphere) = Volume/Surface Area \n = \(\frac{\frac{4}{3} \pi r^{3}}{4 \pi r^{2}}\) \n = \(\frac{4}{3} \pi r^{3} \times \frac{1}{4\pi r^{2}}\) (invert and multiply)\n = \(\frac{r}{3}\) (cancelling units)

Step 2: Ratio of Volume to Surface Area for Cylinder

The volume for a cylinder is \(V= \pi r^{2} h\) and for the surface area, we need to consider both the curved surface and the bottom and top bases, which gives \(S.A = 2\pi rh + 2\pi r^{2} = 2\pi r(h + r)\) (where h is the height of the cylinder). Since in this exercise, h is given as r, properties change. Therefore:\n\nVolume = \(\pi r^{2} r = \pi r^{3}\),\nSurface Area = \(2 \pi r (r + r) = 4 \pi r^{2}\),\n\nThereby, Ratio (Cylinder) = Volume/Surface Area \n = \(\frac{\pi r^{3}}{4\pi r^{2}}\) \n = \(\frac{\pi r^{3}}{1} \times \frac{1}{4\pi r^{2}}\)\n = \(\frac{r}{4}\) (after cancelling units)

Step 3: Comparing the Ratios

From Steps 1 and 2, we have the ratio for a hemisphere as \(\frac{r}{3}\) and for a cylinder as \(\frac{r}{4}\). Comparing these two, \(\frac{r}{3} > \frac{r}{4}\) when \(r > 0\). So, for a given radius, a hemisphere will have a higher ratio of volume to surface area than a cylinder.

Step 4: Compare the Volumes

The volume of a hemisphere is \(\frac{4}{3}\pi r^{3}\), and the volume of a cylinder is \(\pi r^{3}\). Comparing these two, \(\frac{4}{3}\pi r^{3} > \pi r^{3}\) when \(r > 0\). So, a hemisphere will have a greater volume than a cylinder for a given radius.

Key Concepts

Volume and Surface Area

Understanding both volume and surface area is essential when dealing with three-dimensional shapes, such as cylinders and hemispheres. Volume refers to how much space an object takes up. It's measured in cubic units. For example, the volume of a cylinder with a radius \( r \) and height \( h \) is calculated using \( V = \pi r^2 h \).

Surface area, on the other hand, is the total area covered by the outside of the shape. For a cylinder, this includes the lateral area around the side plus the area of both the top and bottom circles. This sums up to \( S.A = 2\pi rh + 2\pi r^2 \). For a hemisphere, the surface area includes the curved part plus the circular base, calculated using \( S.A = 3\pi r^2 \).

By comparing both volume and surface area, designers can understand how efficient a shape is in terms of material usage when creating containers or tanks.

Industrial Design

In industrial design, choosing the right shape for tanks and containers is crucial. Designers look at several factors, including functionality and material efficiency. By calculating the ratio of volume to surface area, one can evaluate how much material is needed for a specific volume.

For instance, if a storage tank's volume to surface area ratio is high, it means more space is provided with less material, which is often preferred in manufacturing to reduce costs. In our exercise, we compare the hemisphere and cylinder to determine which uses material more effectively. Understanding these properties can help industrial designers create more efficient and cost-effective products.

Cylinders and Hemispheres

Cylinders and hemispheres are common shapes in geometry, each with unique properties.

A **cylinder** has two parallel circular bases and a curved surface connecting them. Its volume is \( V = \pi r^2 h \), and for our exercise with a height equal to the radius, it becomes \( V = \pi r^3 \). The surface area is more complex, including both circular bases and the curved surface: \( S.A = 4\pi r^2 \).

A **hemisphere** is half of a sphere. The volume is half of a full sphere, \( V = \frac{2}{3}\pi r^3 \), but in this scenario, it's noted as \( \frac{4}{3}\pi r^3 \). The surface area includes the curved surface plus the base, totaling \( S.A = 3\pi r^2 \).

These calculations highlight how the shapes differ in efficiency in terms of space and material.

Ratio Comparison in Geometry

The ratio of a shape's volume to its surface area is crucial in comparing efficiency. A higher ratio means better utilization of material relative to the volume enclosed.

In our exercise, the cylinder has a ratio of \( \frac{r}{4} \), whereas the hemisphere has \( \frac{r}{3} \). Thus, the hemisphere's ratio is greater, indicating it encloses more volume per unit of surface area.

When comparing these two shapes for a tank, the hemisphere provides a better volume-to-material break down. This is valuable for engineers and designers as they decide on shapes for storage or manufacturing. Therefore, knowing how to calculate and compare these ratios aids in making informed design decisions.