Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.6, Problem 37E

To determine

To calculate: The least illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three as strong as the other, are placed $10 ft$ apart, where should an object be placed on the line between the sources.

Expert Solution & Answer

Answer to Problem 37E

the object should be placed about $5.9 ft$ from the larger light source

Explanation of Solution

Given information:

The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three as strong as the other, are placed $10 ft$ apart, where should an object be placed on the line between the sources

Formula used:

Let $f$ be a differentiable function defined on an interval $I$ and let $a∈I$ .

Then

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f′(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f′(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changessign from negative to positive as $x$ increases through $a$ , i.e. if $f′(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f′(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

$f′(a)=0$ and If $f′(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

If $f″(a)>0$ then $f$ has a local minimum at $x=a$
If $f″(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of the illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three as strong as the other, are placed $10 ft$ apart, where should an object be placed on the line between the sources

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.6, Problem 37E

Let the object is placed $x ft$ from the bigger source then the object $10−x ft$ from the smaller source

Let the strength of the source be $k$

Hence, the strength of the bigger source is $3k$

Since illumination is directly proportional to the square of the strength and inversely proportional to the distance from the source,

The illumination from the bigger source is $=3kx2$

The illumination from the smaller source is $=k(10−x)2$

Hence, the total illumination is

$I′(x)=−6kx3+2k(10−x)3$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a∈I$ .

Then

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$ and
$f′(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f′(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f′(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima

$x=a$ is a point of local maximum value of $f$ , if

$f′(a)=0$

$f′(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f′(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.

$f′(a)=0$ and If $f′(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

If $f″(a)>0$ then $f$ has a local minimum at $x=a$
If $f″(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

$I′(x)=−6kx3+2k(10−x)3 ......(1)$

Solve for $I′(x)=0$ , and simplified

$−6kx3+2k(10−x)3=0⇒ 2k(10−x)3=6kx3⇒ 1(10−x)3=3x3⇒ 3(10−x)3=x3$

Take cube root on both sides,

$33(10−x)=x⇒33×10−33x=x⇒x+33x=33×10⇒x(1+33)=33×10⇒ x=33×10(1+33) ≈5.9 ft$

Differentiate equation $(1)$ with respect to $x$

$I″(x)=18kx4+2k(10−x)4$

Therefore,

$I″(x)=18kx4+2k(10−x)4≥0$ provided $k>0$

For $x=5.9 ft$ is minimum for $I$

Conclusion:

Thus the object should be placed about $5.9 ft$ from the larger light source.

Chapter 4.6, Problem 36E

Chapter 4.6, Problem 38E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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