Problem 43 - Algebra A Combined Function

Question

The intensity of light (in foot-candles) varies inversely as the square of \(x\), the distance in feet from the light source. The intensity of light 2 feet from the source is 80 foot-candles. How far away is the source if the intensity of light is 5 foot-candles?

Step-by-Step Solution

Answer

The source is 8 feet away when the intensity is 5 foot-candles.

Step 1: Identifying the Relationship Type

The problem states that the intensity of light varies inversely as the square of the distance. This means that if the intensity is \( I \) and the distance is \( x \), then \( I = \frac{k}{x^2} \), where \( k \) is a constant of proportionality.

Step 2: Determine the Constant of Proportionality

We know that the intensity is 80 foot-candles when the distance is 2 feet. Plug these values into the relationship:\[ 80 = \frac{k}{2^2} \]Solving for \( k \), we get:\[ 80 = \frac{k}{4} \]\[ k = 80 \times 4 = 320 \]

Step 3: Set up the Equation for New Intensity

Now we need to find the distance when the intensity is 5 foot-candles. Use the same relationship \( I = \frac{k}{x^2} \) with the known \( k \):\[ 5 = \frac{320}{x^2} \]

Step 4: Solve for the New Distance

To find \( x^2 \), rearrange the equation:\[ 5x^2 = 320 \]\[ x^2 = \frac{320}{5} = 64 \]Finally, take the square root of both sides:\[ x = \sqrt{64} = 8 \]

Step 5: Verify the Solution

Substitute \( x = 8 \) back into the intensity equation \( I = \frac{320}{x^2} \) to verify:\[ I = \frac{320}{8^2} = \frac{320}{64} = 5 \]The calculation confirms that at 8 feet, the intensity is 5 foot-candles.

Key Concepts

Intensity of Light

The intensity of light refers to the brightness observed at a particular distance from a light source. It is measured in units called foot-candles. The intensity indicates how much light reaches a surface, and it changes with distance. In the context of inverse variation, where the intensity of light varies inversely as the square of the distance, the farther you are from the light source, the less intense the light becomes. This is because the intensity decreases as the square of the distance increases.

To better understand, imagine standing 2 feet away from a light source with an intensity of 80 foot-candles. If you move farther away, the intensity drops significantly. Calculating the intensity at a specific distance involves the formula: \[ I = \frac{k}{x^2} \]where:

• \(I\): Intensity of light in foot-candles
• \(x\): Distance in feet
• \(k\): Constant of proportionality

Constant of Proportionality

The constant of proportionality, denoted as \(k\), is a crucial part of understanding inverse variation relationships. In our specific example, \(k\) is used in the formula \( I = \frac{k}{x^2} \), representing the consistent factor that relates the intensity of light to the square of the distance.

To determine \(k\), you substitute known values of intensity and distance into the formula. From the exercise, when the distance is 2 feet and the intensity is 80 foot-candles, the formula becomes:\[ 80 = \frac{k}{2^2} \]Solving this equation helps you find that \(k = 320\). The constant remains the same for any subsequent calculations involving different distances where the intensity needs to be found, or vice versa. Inverse variations require this constant to ensure the relationship holds true for all situations.

Square of the Distance

The square of the distance plays a pivotal role when dealing with inverse variation problems in light intensity scenarios. When we say that the intensity of light varies inversely with the square of the distance, it means that as the distance from the light source increases, the brightness decreases rapidly, because it is not just the distance that matters, but its square.

In our formula \( I = \frac{k}{x^2} \), the \(x^2\) term ensures that changes in distance have a more pronounced effect on the intensity than if it were just \(x\). For instance, if the distance doubles, the intensity doesn't just halve; it reduces by a factor of four, as the distance squared is now four times larger than before.

• When distance \(x\) is small, \(x^2\) is also small, causing higher intensity.
• As \(x\) increases, \(x^2\) grows much faster, reducing intensity drastically.

Using the concept of squared distance emphasizes how sensitive light intensity is to changes in distance.