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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.5.47
Chapter 4, Problem 4.5.47

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least?

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Understand the problem: We have two light sources, one stronger than the other, and we need to find the point where the total illumination is least. The intensity of illumination is inversely proportional to the square of the distance from the light source.
Define variables: Let the distance from the stronger light be x meters. The distance from the weaker light will then be (6 - x) meters, since the lights are 6 meters apart.
Express the illumination: The intensity of the stronger light is 8 times that of the weaker light. If the intensity of the weaker light at a distance d is I, then the intensity of the stronger light at the same distance is 8I. The illumination from the stronger light at distance x is \( \frac{8I}{x^2} \) and from the weaker light at distance (6-x) is \( \frac{I}{(6-x)^2} \).
Formulate the total illumination: The total illumination at a point x is the sum of the illuminations from both lights, given by \( \frac{8I}{x^2} + \frac{I}{(6-x)^2} \).
Find the minimum illumination: To find the distance x where the total illumination is least, take the derivative of the total illumination function with respect to x, set it equal to zero, and solve for x. This will give the critical points. Check these points to determine which gives the minimum illumination.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Square Law

The inverse square law states that the intensity of illumination from a point light source is inversely proportional to the square of the distance from the source. This means if you double the distance from the light source, the intensity becomes one-fourth. This principle is crucial for calculating the illumination at a point from multiple light sources.
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Optimization

Optimization involves finding the maximum or minimum value of a function within a given domain. In this problem, we need to determine the point where the total illumination from two light sources is minimized. This requires setting up a function for total illumination and using calculus techniques, such as finding the derivative and critical points, to identify the minimum value.
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Superposition of Light Intensities

The superposition principle in this context means that the total illumination at a point is the sum of the illuminations from each light source. Since one light is eight times stronger than the other, the total illumination function will combine the effects of both lights, taking into account their respective distances and intensities, to find the point of least illumination.
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Example 4: Norman Window
Related Practice
Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

39. y = 8 / (x² + 4) (Witch of Agnesi)

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(3t² + t/2) dt

Textbook Question

Identifying Extrema


In Exercises 19–40:


a. Find the open intervals on which the function is increasing and those on which it is decreasing.


b. Identify the function’s local extreme values, if any, saying where they occur.


f(x) = x¹ᐟ³(x + 8)

Textbook Question

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

__________

y = √ 3 + 2𝓍 ―𝓍²

Textbook Question

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.


y = (𝓍 + 1) / (𝓍² + 2𝓍 + 2)

Textbook Question

Checking Antiderivative Formulas


Right, or wrong? Give a brief reason why.


∫−15(x + 3)² / (x − 2)⁴ dx = ((x + 3)/(x − 2))³ + C

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