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Many physical quantities are connected by inverse square laws, that is, by power functions of the form $ f(x) = kx^{-2} $. In particular, the illumination of an object by a light source is inversely proportional to the square of the distance from the source. Suppose that after dark you are in a room with just one lamp and you are trying to read a book. The light is too dim and so you move halfway to the lamp. How muchbrighter is the light?

If $x$ is the original distance from the source, then the illumination is $f(x)=k x^{-2}=k / x^{2}$. Moving halfway to the lamp gives us an illumination of $f\left(\frac{1}{2} x\right)=k\left(\frac{1}{2} x\right)^{-2}=k(2 / x)^{2}=4\left(k / x^{2}\right),$ so the light is 4 times as bright.

01:32

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 2

Mathematical Models: A Catalog of Essential Functions

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Johns Hopkins University

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University of Michigan - Ann Arbor

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Okay, so we have this function which represents the brightness of light as a function of the distance from the source and f of X is the brightness. The illumination and X is the distance. So what happens if you take half of your distance from the light source? How much brighter will it be? So we're going to find f of 1/2 of X and comparative f of X, so we're going to substitute 1/2 of X in place of X. So 1/2 of exes raised the negative second power. So each of those factors, the 1/2 and the X is going to be raised to the negative second power on 1/2 to the negative second power. That's the reciprocal of 1/2 to the second power. So that would be two to the second power, which is for two to the second powers for so let's look at what we have. And compared to the original brightness, we have four times K to the negative second power. Originally, we had just k to the negative second power K X to the negative second power, so it's four times brighter

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