Problem 28 - College Algebra

Question

Use the four-step procedure for solving variation problems given on page 356 to solve. \(a\) is directly proportional to \(b\) and inversely proportional to the square of \(c . a=7\) when \(b=9\) and \(c=6 .\) Find \(a\) when \(b=4\) and \(c=8\)

Step-by-Step Solution

Answer

The value of 'a' when \(b=4\) and \(c=8\) is 14.

Step 1: Determine the Constant

Given that \(a = kb/c^2\), and \(a=7\), \(b=9\) and \(c=6\) are provided, substitute these values into the equation resulting in \(7 = k*9/6^2\). After calculation, \(k = (7 * 6^2) / 9\), which in turn equates to \(28\).

Step 2: Calculate 'a' Using New Values for 'b' and 'c' and the Constant

Plugging in the calculated \(k=28\), \(b=4\) and \(c=8\) into the equation results in \(a = 28*4/8^2\). Which simplifies to \(a = 14\).

Key Concepts

Direct Variation

Understanding direct variation is essential for solving problems involving relationships between two variables that increase or decrease together. This concept indicates that two variables change at a consistent rate when compared to one another.

In mathematical terms, if we say that variable 'a' varies directly as variable 'b', we express this relationship as: \[\begin{equation} a = k \times b, \text{where } k \text{ is the constant of proportionality.} \text{This constant } k \text{ remains fixed as } a \text{ and } b \text{ change.} \text{For instance, if } a \text{ is doubled, } b \text{ is also doubled, maintaining the ratio of } a \text{ to } b. \end{equation}\].
In the provided exercise, the value of 'a' increases as the value of 'b' increases, which is a classic example of direct variation.

Inverse Variation

In contrast to direct variation, where two quantities increase or decrease together, inverse variation describes a situation where one quantity increases while the other decreases.

This relationship can be mathematically represented as: \[\begin{equation} a = \frac{k}{c^2}, \text{where } k \text{ is again the constant of proportionality, and } a \text{ inversely varies with the square of } c. \text{As } c \text{ increases, } a \text{ decreases, and vice versa.} \end{equation}\]

In our exercise, 'a' is inversely proportional to the square of 'c', highlighting how 'a' decreases when the square of 'c' increases.

Proportional Relationships

When we talk about proportional relationships, we're referring to the situation where two quantities maintain a constant ratio to one another. This is true for both direct and inverse variations.

It means there is a consistent, predictable relationship between the variables. For example, in direct proportionality, tripling one variable triples the other if the constant of proportionality remains the same. Conversely, with inverse proportionality, doubling one variable results in halving the other, according to the constant of proportionality.

Identifying Proportional Relationships

You can determine whether a relationship is proportional by checking whether the ratio between variables stays consistent no matter the values they take. Graphically, a proportional relationship between two quantities is represented by a straight line through the origin if directly proportional or a hyperbola if inversely proportional.

Constant of Proportionality

The constant of proportionality plays a crucial role in any proportional relationship, as it represents the unchanging ratio between two variables.

In the context of our exercise problem, the constant of proportionality, denoted by 'k', scales the relationship between 'a' and 'b' or 'a' and 'c'. It's the multiplier that relates the dependent variable to the independent variable(s).

Calculating the Constant

To find this constant, one must have known values for the other variables in the equation. The provided steps illustrate this process: By substituting known values into the equation, we isolated 'k' and found its value to be 28. This constant then allows for the calculation of unknown variable values within the same proportional framework.