Problem 30 - College Algebra

Question

Use the four-step procedure for solving variation problems given on page 424 to solve. The illumination provided by a car's headlight varies inversely as the square of the distance from the headlight. A car's headlight produces an illumination of 3.75 footcandles at a distance of 40 feet. What is the illumination when the distance is 50 feet?

Step-by-Step Solution

Answer

The illumination at a distance of 50 feet is 2.4 footcandles.

Step 1: Understand the Principle of Inverse Square Variation

The principle of inverse square variation states that two variables x and y are inversely proportional to the square of each other, which can be written mathematically as y = k/x^2, where k is the constant of variation.

Step 2: Find the Constant of Variation

From the problem, we know that the illumination (y) is 3.75 footcandles when the distance (x) is 40 feet. We can substitute these values into the formula to find the constant of variation: \(3.75 = k/40^2\). Solving this equation for k we find that \(k = 3.75 * 40^2 = 6000\).

Step 3: Calculate the Illumination at a New Distance

We're asked to find the illumination when the distance is 50 feet. We substitute the constant of variation and the new distance into the formula: \(y = 6000/50^2\). Solving this, we get that the illumination is 2.4 footcandles.

Key Concepts

Variation Problems

When studying variation problems, it is essential to identify how two quantities are related to each other. In many real-world scenarios, one variable will change as a result of another. For instance, the illumination of a car's headlight and the distance from it. In the context of inverse square variation, we observe that as one variable increases, the other variable decreases and vice versa, but with a twist – the change occurs at the square of the rate.

Understanding this relationship helps in setting up an equation that accurately represents the problem. The initial step is to express one variable explicitly in terms of the other. Once this is accomplished, the next phase involves identifying a constant that links these variables, which we call the 'constant of variation'. With this fundamental understanding, solving various practical problems, such as calculating the intensity of light at different distances, becomes manageable.

Constant of Variation

In problems involving inverse square variation, the constant of variation plays a crucial role. It is the fixed number that relates to the variables that are inversely proportional to the square of each other. Mathematically, if 'y' varies inversely as the square of 'x', the equation is written as \( y = \frac{k}{x^2} \), where 'k' is the constant of variation.

To find this constant, you need a pair of corresponding values for 'x' and 'y'. Plugging these values into the equation allows us to solve for 'k'. This constant remains unchanged regardless of the values of 'x' and 'y', which enables us to solve for unknowns. It becomes the linchpin for predicting how 'y' will behave when 'x' changes. For instance, knowing the constant of variation for a car's headlight can help us calculate the level of illumination at any distance.

Solving Algebraic Equations

Effective problem solving in algebra often hinges on dealing with algebraic equations. Such equations depict relationships between variables and constants. To solve them, we typically isolate the variable of interest on one side of the equation using algebraic operations, such as addition, subtraction, multiplication, division, and taking roots.

In the case of inverse square variation, we solve for the constant of variation first by substituting known values and then apply that constant to find other variable values. For example, when calculating a headlight's illumination at a different distance, we rearrange the equation to make 'y', the illumination, the subject. Then we substitute the given 'x' value and the constant of variation 'k' into the formula to find the new 'y' values. This systematic approach to algebraic equations is indispensable across all variation problems.