Problem 93 - Algebra and Trigonometry

Question

The number of gallons of water, \(W,\) used when taking a shower varies directly as the time, \(t\), in minutes, in the shower. A shower lasting 5 minutes uses 30 gallons of water. How much water is used in a shower lasting 13 minutes?

Step-by-Step Solution

Answer

78 gallons is the amount of water used in a 13-minute shower.

Step 1: Find the constant of variation \(k\)

We know that a 5-minute shower uses 30 gallons of water. Thus, we can find the constant of variation \(k\) by dividing the amount of water used by the duration of the shower, i.e., \(k = W/t = 30/5 = 6\).

Step 2: Find the amount of water used in a 13-minute shower

Having computed the value of \(k\) in Step 1, we can find the amount of water used during a 13-minute shower by substituting the values of \(k\) and \(t\) into the equation \(W = kt\), to give \(W = 6 * 13 = 78\).

Key Concepts

Constant of Variation

In direct variation problems, the constant of variation, often denoted as \(k\), is a crucial element. It's the factor that relates two variables that change at the same rate. For example, in our original exercise, the number of gallons of water used, \(W\), is directly proportional to the time spent showering, \(t\), leading to an equation of the form \(W = kt\). The constant \(k\) helps us understand how much water is used per unit of time.

To determine \(k\), you divide the result by the measure that's causing the variation. In this case, 5 minutes of shower uses 30 gallons of water, so \(k = \frac{30}{5} = 6\). This means for every minute in the shower, 6 gallons of water is used. It's worth noting that the constant of variation remains the same regardless of how long the shower takes, as long as the rate of water usage per unit time stays unchanged. This is true across all scenarios of direct variation.

Proportional Relationships

Proportional relationships describe the relationship between two quantities where one is a constant multiple of the other. In simpler terms, if two quantities are proportional, they increase or decrease together at the same rate. Going back to the shower example, the amount of water \(W\) and the time \(t\) spent in the shower have a proportional relationship.

When you have a constant of proportionality, like \(k = 6\), it means that no matter how you change the time \(t\), the water usage \(W\) will change in such a way that \(W = kt\) remains true. This forms a linear relationship, which can be represented graphically as a straight line passing through the origin on a coordinate plane with \(t\) on the x-axis and \(W\) on the y-axis. Such relationships are easy to work with mathematically since they simplify prediction and analysis of how changes in one variable will directly affect the other.

Algebra Problems

Algebra problems involving direct variation often require you to identify relationships between quantities and use this understanding to solve equations. These problems typically require setting up an equation using the concept of direct variation. The given problem about shower duration and water usage is a classic algebra problem.

Here's how you can approach them:

• Identify the two variables that change together, such as \(W\) and \(t\).
• Determine the constant of variation \(k\), by dividing one variable by the other when values are given.
• Use the equation \(W = kt\) to find unknown values by substituting given values.

By mastering these skills in algebra, you can efficiently handle not only similar problems but also more complex situations where direct variation is involved. Techniques learned from solving direct variation problems are fundamental and applicable in many real-life contexts, such as in physics or economics.