Problem 41 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

MICROBIOLOGY A spherical cell of radius \(r\) has volume \(V=\frac{4}{3} \pi r^{3}\) and surface area \(S=4 \pi r^{2}\). Express \(V\) as a function of \(S\). If \(S\) is doubled, what happens to \(V\) ?

Step-by-Step Solution

Answer

When the surface area is doubled, the volume becomes \( \sqrt{2} \) times the original volume.

Step 1 - Write down the given formulas

The volume of a spherical cell is given by \[V = \frac{4}{3} \, \pi \, r^{3}\] and the surface area is given by \[S = 4 \, \pi \, r^{2}\]

Step 2 - Express radius r in terms of S

From the surface area formula, rearrange to solve for the radius, r. \[S = 4 \, \pi \, r^{2} \Rightarrow r^{2} = \frac{S}{4 \, \pi} \Rightarrow r = \sqrt{\frac{S}{4 \, \pi}}\]

Step 3 - Substitute the expression for r into the volume formula

Substitute \( r = \sqrt{\frac{S}{4 \, \pi}}\) into the volume formula \[V = \frac{4}{3} \, \pi \, \left(\sqrt{\frac{S}{4 \, \pi}}\right)^{3} \Rightarrow V = \frac{4}{3} \, \pi \, \left(\frac{S}{4 \, \pi}\right)^{3/2} \Rightarrow V = \frac{4}{3} \, \pi \, \frac{S^{3/2}}{(4 \, \pi)^{3/2}}\]

Step 4 - Simplify the expression for V

Simplify the expression \[V = \frac{4}{3} \, \pi \, \frac{S^{3/2}}{(4 \, \pi)^{3/2}} \Rightarrow V = \frac{4}{3} \, \pi \, \frac{S^{3/2}}{4^{3/2} \, \pi^{3/2}} \Rightarrow V = \frac{4 \, S^{3/2}}{3 \, \pi^{1/2} \, 8} \Rightarrow V = \frac{S^{3/2}}{6 \, \sqrt{\pi}}\]

Step 5 - Analyze the change in V when S is doubled

If the surface area, S, is doubled, let the new surface area be 2S. Substitute 2S into the volume formula \[V_{new} = \frac{(2S)^{3/2}}{6 \, \sqrt{\pi}} \Rightarrow V_{new} = \frac{2^{3/2} \, S^{3/2}}{6 \, \sqrt{\pi}} \Rightarrow V_{new} = \frac{2 \sqrt{2} \, S^{3/2}}{6 \, \sqrt{\pi}} \Rightarrow V_{new} = \sqrt{2} \, V\] So, the new volume is \( \sqrt{2} \) times the original volume.

Key Concepts

sphere volume formula

The volume of a sphere can be found using a specific formula. This formula is derived from integral calculus but is straightforward to use. The volume formula is:

\[ V = \frac{4}{3} \, \pi \, r^{3} \]
Where:

• V is the volume of the sphere
• r is the radius of the sphere
• π (Pi) is a constant approximately equal to 3.14159

This formula tells us that the volume of a sphere depends on the cube of its radius. As the radius increases, the volume increases dramatically because the radius is raised to the third power.

For example, if you have a sphere with a radius of 3 units, the volume would be:
\[ V = \frac{4}{3} \, \pi \, (3^{3}) = 36 \pi \ \text{cubic units} \]

sphere surface area formula

The surface area of a sphere can be calculated using another specific formula. This gives us the total area that covers the outer surface of the sphere. The formula is:

\[ S = 4 \, \pi \ r^{2} \]
Where:

• S is the surface area of the sphere
• r is the radius of the sphere
• π (Pi) is the constant approximately equal to 3.14159

This formula shows that the surface area is proportional to the square of the radius. As the radius increases, the surface area increases, but only in the same way that the radius is squared.

For example, if the sphere has a radius of 3 units, the surface area would be:
\[ S = 4 \, \pi \ (3^{2}) = 36 \pi \ \text{square units} \]

relationship between volume and surface area

Understanding the relationship between the volume and the surface area of a sphere can help us solve many practical problems. Let's express the volume, V, as a function of the surface area, S. First, from the surface area formula, we determine the radius:

\[ S = 4 \pi \ r^{2} \ \rightarrow \ r = \sqrt{\frac{S}{4 \ \pi}} \]
Next, substitute this radius value into the volume formula:
\[ V = \frac{4}{3} \ \pi \ \left(\sqrt{\frac{S}{4 \ \pi}}\right)^{3} = \frac{S^{3/2}}{6 \ \sqrt{\ \pi}} \]
This shows that the volume, V, depends on the surface area, S, raised to the 3/2 power.
For instance, if the surface area, S, is doubled, the new volume is:
\[ V_{new} = \frac{(2S)^{3/2}}{6 \sqrt{\ \pi}} = \sqrt{2} \ V \]
So, the volume becomes \sqrt{2} times the original volume. This tells us that a small increase in surface area leads to a relatively larger increase in volume.