   Chapter 4.7, Problem 82E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Two light sources of identical strength are placed 10 m apart. An object is to be placed at a point P on a line ℓ, parallel to the line joining the light sources and at a distance d meters from it (see the figure). We want to locate P on ℓ. so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source.(a) Find an expression for the intensity I(x) at the point P.(b) If d = 5 m, use graphs of I(x) and I′(x) to show that the intensity is minimized when x = 5 m, that is, when P is at the midpoint of ℓ.(c) If d = 10 m, show that the intensity (perhaps surprisingly) is not minimized at the midpoint.(d) Somewhere between d = 5 m and d = 10 m there is a transitional value of d at which the point of minimal illumination abruptly changes. Estimate this value of d by graphical methods. Then find the exact value of d. (a)

To determine

To find: The expression for the intensity I(x) at the point P.

Explanation

Given:

Distance between two light source = 10m.

Distance between the point P and the line joining light source = d.

Formula used:

Pythagorean Theorem:

If c denotes the length of the hypotenuse and a and b denote the length of the other two sides of a right angle triangle, then the Pythagorean theorem can be expressed as the Pythagorean equation: a2+b2=c2

Calculation:

Let the strength of the source be k.

Let the intensity on the point P be I.

In Figure 1,

The light sources are at the points A and B.

I(x)strength of source(distance from source)2.

By Pythagorean Theorem on the triangle ΔPAC,

PA2=x2+d2.

Hence the intensity due to the light source at the points A is kx2+d2

By Pythagorean Theorem on the triangle ΔPBC,

PB2=(10x)2+d2

(b)

To determine

To show: The intensity is minimized when x=5, that is when P is at the midpoint l.

(c)

To determine

To show: The intensity is not minimized when d=10 at the midpoint l.

(d)

To determine

To estimate: The value of d.

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 