Problem 78 - A Graphical Approach to Precalculus with Limits

Question

The illumination \(I\) in foot-candles produced by a light source is related to the distance \(d\) in feet from the light source by the equation $$d=\sqrt{\frac{k}{I}},$$ where \(k\) is a constant. If \(k=400,\) how far from the source will the illumination be 14 foot-candles? Round to the nearest hundredth of a foot.

Step-by-Step Solution

Answer

The distance is approximately 5.34 feet.

Step 1: Identify Variables and Constants

We are given the equation for illumination as a function of distance: \(d = \sqrt{\frac{k}{I}}\). Here, \(k = 400\) is the constant, and \(I = 14\) is the illumination in foot-candles.

Step 2: Substitute Values into the Equation

Substitute \(k = 400\) and \(I = 14\) into the equation. This gives us: \(d = \sqrt{\frac{400}{14}}\).

Step 3: Perform Division

Calculate \(\frac{400}{14}\). This simplifies to approximately 28.57 since 400 divided by 14 is 28.5714.

Step 4: Take the Square Root

Calculate the square root of 28.57. \(\sqrt{28.57} \approx 5.34\).

Step 5: Round to the Nearest Hundredth

Round the result of 5.34 to two decimal places, which remains 5.34.

Key Concepts

Inverse Square Law

The inverse square law helps us understand how light diminishes over distance. As you move away from a light source, the intensity of the light decreases significantly. This law states that the illumination, or brightness, is inversely proportional to the square of the distance from the source. In simpler terms, when you double the distance from a light, its brightness becomes a quarter of its original value. Here's a quick breakdown:

• The formula is represented as: \( I \propto \frac{1}{d^2} \), meaning illumination \( I \) is inversely related to the square of the distance \( d \).
• Thus, as the distance \( d \) increases, the illumination \( I \) decreases at a faster rate.
• This relationship is crucial for calculations involving lights and helps engineers and designers in determining proper lighting conditions for spaces.

Understanding this concept provides the basis for our problem, which revolves around calculating the distance given a specific level of illumination from a light source.

Solving Equations

When tackling equations like \( d = \sqrt{\frac{k}{I}} \), we need to focus on isolating \( d \) and simplifying the equation step-by-step. Solving equations often involves substitution and simplification, which are essential skills in mathematics.
Let's dissect how to solve the equation in our exercise:

• The equation given is \( d = \sqrt{\frac{k}{I}} \), representing how distance \(d\) is calculated based on illumination \(I\).
• Firstly, substitute the known values: \( k = 400 \) and \( I = 14 \).
• This substitution simplifies our equation to \( d = \sqrt{\frac{400}{14}} \).
• Breaking it down further by calculating \( \frac{400}{14} \) results in approximately \( 28.57 \).
• Lastly, solving for \( d \) involves taking the square root of \( 28.57 \).

Through each step, logical processes are applied to maneuver through mathematical problems efficiently, promoting a better understanding of how to derive a result from a given equation.

Square Roots

Understanding square roots is fundamental when dealing with equations involving roots, like our exercise. The square root of a number \( x \) is a value that, when multiplied by itself, gives \( x \). It’s often represented by the symbol \( \sqrt{} \).

Key points about square roots include:

• Square roots are the opposite of squaring a number. If \( 5^2 = 25 \), then \( \sqrt{25} = 5 \).
• In the exercise, finding \( \sqrt{28.57} \) is crucial to determine the distance.
• To calculate the root, you need either a calculator or knowledge of approximation techniques.
• It is integral to round decimal figures correctly; here, \( \sqrt{28.57} \approx 5.34 \), which rounds to 5.34.

Square roots form the cornerstone of many mathematical calculations, especially those leading to practical real-world applications like measuring distances in our illumination problem.