PDA

View Full Version : Solving inverse variation


camerono
Sep 7, 2008, 05:20 AM
I am studying variations (direct, inverse, joint)

Given: the intensity of light, measured in lux, is inversely proportional to the square of the distance between the light source and the object illuminated - solve:

A light meter 7.5 m from a light source register 24 lux. What intensity would it register 15 m from the light source?

I tired to use the inverse variation equation y=k/x.

I thought that y would be 24. And x would be the square of the distance 7.5
24=k/56.25
so 1350=k.

substituting:
24=1350/225 (15x15)
24=6

obviously this is incorrect. Can someone guide me as to what formula I should be using? I have spent a long time rereading the section in my text but I just don't get it. Thanks.

ISneezeFunny
Sep 7, 2008, 05:35 AM
The best way to do inversely proportional, is to think of it this way:

x * y = some constant.

this way, if x goes up, then y must go down.

so

since the intensity of light is inversely proportional to the square of the distance

(intensity of light) (distance)^2 = some constant

(24)(7.5)^2 = some constant (this constant does not change)

= 1350

now, new distance, and unknown intensity of light

(unknown intensity) (15)^2 = 1350

unknown intensity = 6

so... you had it right, you just didn't take the earlier intensity out. And this makes sense, the farther you are away from the light source, the less intense the light will be.