Problem 26 - Algebra Form and Function

Question

The volume \(V\) of a sphere is a function of its radius \(r\) given by $$ V=f(r)=\frac{4}{3} \pi r^{3} $$ (a) Find \(\frac{f(2 r)}{f(r)}\). $$ \text { (b) Find } \frac{f(r)}{f\left(\frac{1}{2} r\right)} \text { . } $$ (c) What do you notice about your answers to (a) and (b)? Explain this result in terms of sphere volumes.

Step-by-Step Solution

Answer

Answer: The relationship between the radius of a sphere and its volume is that the volume of a sphere is proportional to the cube of its radius. Any change in the radius results in a corresponding change in volume based on the cube of that change factor.

Step 1: (a) Calculate \(\frac{f(2r)}{f(r)}\)

Replace \(r\) in the volume formula with \(2r\). It becomes: $$ f(2r) = \frac{4}{3} \pi(2r)^3 $$ Plug the volume formulas for \(f(2r)\) and \(f(r)\) into the ratio and simplify: $$ \frac{f(2r)}{f(r)} = \frac{\frac{4}{3} \pi(2r)^3}{\frac{4}{3} \pi r^3} = \frac{\cancel{\frac{4}{3}\pi} (2r)^3}{\cancel{\frac{4}{3}\pi} r^3} $$ $$ = \frac{8r^3}{r^3} = 8 $$

Step 2: (b) Calculate \(\frac{f(r)}{f\left(\frac{1}{2} r\right)}\)

Replace \(r\) in the volume formula with \(\frac{1}{2}r\). It becomes: $$ f\left(\frac{1}{2}r\right) = \frac{4}{3} \pi\left(\frac{1}{2}r\right)^3 $$ Plug the volume formulas for \(f(r)\) and \(f(\frac{1}{2}r)\) into the ratio and simplify: $$ \frac{f(r)}{f\left(\frac{1}{2} r\right)} = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3} \pi \left(\frac{1}{2}r\right)^3} = \frac{\cancel{\frac{4}{3}\pi} r^3}{\cancel{\frac{4}{3}\pi} \left(\frac{1}{8}r^3\right)} $$ $$ = \frac{r^3}{\frac{1}{8}r^3} = 8 $$

Step 3: (c) Interpret the Results

Notice how both results give us the same ratio, 8. This result illustrates the fact that the volume of a sphere is proportional to the cube of its radius: - When the radius is doubled (going from \(r\) to \(2r\)), the volume is 8 times larger. - When the radius is halved (going from \(r\) to \(\frac{1}{2}r\)), the volume becomes one-eighth of the original volume. In terms of sphere volumes, the cube of the radius directly correlates to the volume, so any change in the radius results in a corresponding change in volume based on the cube of that change factor.

Key Concepts

Volume Formula

Understanding how to calculate the volume of a sphere is essential. The formula provided for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( V \) represents the volume, \( r \) is the radius, and \( \pi \) is a mathematical constant.

• This formula tells us that the volume of a sphere is dependent on the cube of its radius.
• The factor \( \frac{4}{3} \pi \) remains constant, serving as a multiplier that adjusts for the spherical shape.

This implies that any variation in the sphere's radius will have an exponential impact on the volume, given that the volume scales with \( r^3 \). Understanding the formula is a key step in exploring how geometric and mathematical concepts interplay in determining the space contained within a sphere.

Radius Effect on Volume

The radius of a sphere has a powerful influence on its volume. Due to the \( r^3 \) term in the volume formula, even minor changes in the radius cause significant changes in volume.

• If the radius is doubled, as outlined in the first part of the exercise, the resulting volume becomes \( 2^3 = 8 \) times larger.
• Contrastingly, if the radius is halved, then as seen in the second part of the exercise, the volume shrinks to \( \left( \frac{1}{2} \right)^3 = \frac{1}{8} \) of its original size.

These calculations show us that the relationship between radius and volume is not linear but cubic, meaning small changes to the radius can lead to large changes in volume. This emphasizes the importance of understanding how dimensions scale in three-dimensional space.

Mathematical Proportionality

Proportionality is a fundamental concept in understanding mathematical relationships in geometry. Through the exercises, it's clear that the volume of a sphere is directly proportional to the cube of its radius.

• This means any factor change in the radius results in that factor raised to the third power change in the volume.
• The results where both variations (doubling and halving the radius) gave a factor of 8 reinforce this understanding.

Understanding this concept can be critical when applying these principles in real-world applications. For example, recognizing that a small increase in the radius can immensely increase the volume is important in industries dealing with spherical objects. Thus, mastering mathematical proportionality helps in better predicting how geometric changes influence spatial properties.