Problem 23 - Algebra and Trigonometry

Question

Use the four-step procedure for solving variation problems given on page 445 to solve Exercises 21–36. The height that a ball bounces varies directly as the height from which it was dropped. A tennis ball dropped from 12 inches bounces 8.4 inches. From what height was the tennis ball dropped if it bounces 56 inches?

Step-by-Step Solution

Answer

The tennis ball was dropped from a height of 80 inches.

Step 1: List the known quantities

We know the height from which the ball was dropped (12 inches) and the corresponding bounce height (8.4 inches), as well as the new bounce height (56 inches).

Step 2: Identify the constant of variation

In a direct variation, the ratio between the two quantities is constant. Therefore, if we take the ratio of drop height to bounce height for the known quantity pair, we have: \(k = \frac{h}{b} = \frac{12 inches}{8.4 inches}\)

Step 3: Compute the constant of variation

Calculating the above expression gives us \(k = \frac{12 inches}{8.4 inches} = 1.428571429 inches\).

Step 4: Use the constant of variation to find the unknown quantity

We substitute the new bounce height (56 inches) and the constant of variation into our initial relation: \(h = kb = 1.428571429 inches * 56 inches\).

Step 5: Compute the result

Solving the expression gives us \(h = 80 inches\). This is the height from which the tennis ball was dropped.

Key Concepts

Direct Variation

When two quantities are related in such a way that an increase or decrease in one quantity results in a proportional increase or decrease in the other, this relationship is known as direct variation. This pivotal mathematical concept is often described by the equation \( y = kx \), where \( y \) and \( x \) are the two variables that vary directly, and \( k \) is the constant of variation, representing the ratio of \( y \) to \( x \) for all pairs of values.

For example, if a car travels at a constant speed, the distance it covers (\( y \) variable) varies directly with the travel time (\( x \) variable). If we know the car travels 60 miles in 1 hour, the constant of variation (\( k \) in this case) would be \( k = \frac{distance}{time} = \frac{60 \text{ miles}}{1 \text{ hour}} = 60 \text{ miles per hour} \).

Constant of Variation

The constant of variation, often denoted as \( k \) in direct variation problems, acts as a multiplier that remains unchanged. It provides a straightforward method to link two variables that vary directly. When one variable is multiplied by the constant of variation, the result is the second variable. It's the 'constant' part of the term 'direct variation.'

Returning to the previous car example, if the car keeps traveling at 60 miles per hour (the constant of variation), and we want to find out how far it goes in 3 hours, we would directly use the constant of variation: \( \text{Distance} = 60 \text{ miles per hour} \times 3 \text{ hours} = 180 \text{ miles} \). This clearly illustrates how the constant of variation plays a crucial role in solving direct variation problems.

Ratio and Proportion

Proportionality is at the heart of many algebraic problems, especially when dealing with variation. Ratio and proportion are tools that assist us in expressing the relationship between quantities. A ratio is a comparison of two numbers by division, while a proportion states that two ratios are equal.

Let's apply this to an exercise: Suppose we have a situation where the amount of ingredients in a recipe is directly proportional to the number of servings. If the recipe requires 2 cups of sugar for 4 servings, the ratio of sugar to servings is \( \frac{2}{4} \). If you need to find out how much sugar is needed for 10 servings, we set up a proportion: \( \frac{2 \text{ cups}}{4 \text{ servings}} = \frac{x \text{ cups}}{10 \text{ servings}} \), where \( x \) represents the unknown amount of sugar needed for 10 servings.

Algebraic Problem Solving

Algebraic problem solving encompasses a systematic approach involving various steps to arrive at a solution. This often includes defining variables, setting up equations based on the given problem situation, and then solving these equations using algebraic methods.

As exemplified in the textbook exercise solution, after identifying the constant of variation, we set up an equation based on the relationship of direct variation and solve for the unknown. We can follow a structured method such as the four-step procedure offered in the textbook: List the known quantities, identify the constant variation, compute the constant of variation, and finally, use it to find the unknown. This structured approach provides a reliable pathway to solve algebraic problems dealing with variation.