Question

The illumination of a small plane surface by a luminous point is proportional to the cosine of the angle between the rays of light and the normal to the surface, and inversely proportional to the square of the distance of the luminous point from the surface. At what height on the wall should an arc light be placed in order to light most brightly a portion of the floor \(a\) ft. distant from the wall?

Short answer

Answer: The arc light should be placed at a height equal to the distance from the wall in order to provide maximum brightness to the portion of the floor.

Step-by-step solution

Express the illumination in terms of \(\theta\) and \(d\)

Let's denote the illumination as \(I\). Given that the illumination is proportional to the cosine of the angle \(\theta\) and inversely proportional to the square of the distance \(d\), we can write the relationship as: \(I = k\cdot \frac{\cos\theta}{d^2}\), where \(k\) is a constant of proportionality.

Write down the relationship between \(d\), \(h\), and \(a\) using trigonometry

We can use the Pythagorean theorem to write the relationship between \(d\), \(h\), and \(a\): \(d^2 = h^2 + a^2\). Also, using trigonometry, we can express \(\cos\theta\) in terms of \(h\) and \(d\): \(\cos\theta = \frac{h}{d}\).

Use calculus to find the maximum illumination

First, let's substitute the expression for \(d^2\) and \(\cos\theta\) into the illumination equation: \(I = k\cdot \frac{\frac{h}{d}}{h^2 + a^2}\) Now, let's differentiate I with respect to h: \(\frac{dI}{dh} = k\frac{d}{dh}\left(\frac{h}{h^2 + a^2}\right)\) We can use the quotient rule for differentiation: \(\frac{dI}{dh} = k\frac{(h^2 + a^2) - 2h^2}{(h^2 + a^2)^2}\) To find the maximum value of I, we need to find the critical points, i.e., where \(\frac{dI}{dh} = 0\). Therefore, \((h^2 + a^2) - 2h^2 = 0\). Solving for \(h\), we get: \(h^2 = a^2\). Hence, \(h = a\).

Conclusion

The arc light should be placed at a height equal to the distance \(a\) from the wall in order to provide maximum brightness to the portion of the floor.

Key concepts

Calculus Applications

In the world of illumination mathematics, calculus plays a vital role, especially when we aim to optimize certain conditions, such as maximizing brightness. The original problem deals with positioning an arc light optimally to achieve the maximum illumination on a floor area.
To tackle this, calculus, specifically differentiation, is utilized. Given the illumination equation \(I = k \cdot \frac{\cos\theta}{d^2}\), we substitute expressions for \(\cos\theta\) and \(d\) in terms of measurable quantities like height \(h\) and distance \(a\).
The core application of calculus here is to find the derivative of the illumination function \(I\) with respect to \(h\), the height at which the arc light is fixed. By using the quotient rule, the derivative \(\frac{dI}{dh}\) is found to determine the critical points where the illumination is maximized. Setting \(\frac{dI}{dh} = 0\) helps find the optimal height, demonstrating the power of calculus in making informed decisions in real-world scenarios.

Trigonometric Functions

Trigonometric functions are essential tools in solving problems involving angles and distances. In this exercise, the illumination received on a surface is partly dependent on the cosine of the angle \(\theta\) between the light ray and the normal to the surface.
Trigonometry allows us to express angles in terms of distances. Specifically, \(\cos\theta\) is described using the equation \(\cos\theta = \frac{h}{d}\), where \(h\) is the height of the arc light and \(d\) is the slant distance from the light to the specific point on the floor.
Furthermore, the Pythagorean theorem helps relate \(d\), \(h\), and \(a\). By understanding these trigonometric identities, the problem solver is equipped to transform real-world angles and distances into usable mathematical terms. This conversion is crucial in setting up equations that can be manipulated using calculus to find optimal solutions.

Problem-Solving Techniques

Effective problem-solving often involves breaking a complex problem into manageable parts and using well-established techniques to solve each part. In this exercise, the aim is to adjust the positioning of a light fixture for maximum illumination.
The approach begins by understanding the relationships involved, notably how illumination is affected by the cosine of the angle and the distance squared. By clearly identifying these factors, we establish a foundational equation for illumination \(I\).
Next, using trigonometry and calculus, we substitute known values and differentiate to find critical points. This step-by-step breakdown employs techniques such as Pythagorean relationships for distance and quotient rule for differentiation.
Finally, logically solving the derivative equation \((h^2 + a^2) - 2h^2 = 0\) gives us the critical height \(h\) where illumination is optimized. This structured process showcases the importance of following a logical order and using mathematical tools to navigate and solve intricate real-world problems.