Question

The illumination (I) from a light source varies inversely as the square of its distance (d) from the light source. The illumination 5 feet from a light source is 0.04 lux.
a. Find the constant of variation, k.
b. Write the formula for finding the illumination (I) given the distance from the light source (d).
c. If you double the distance from the light source, what is the illumination (in lux)?

Answer 1

So the " varies inversely as the square of its distance" part means that you can set up an equation like so\[l={k \over d^2}\]Where \(l\) is the illumination, \(d\) is the distance, and \(k\) is you constant of variation.

Answer 2

So for part a, you have \[.04={k \over 5^2}\]an you solve for \(k\) from there?

Answer 3

How?

Answer 4

You have to multiply both sides by \(5^2\)\[0.04(5^2)={k \over 5^2}\cdot (5^2)\]On the right side, the \(5^2\)'s cancel, and you get\[1=k\]since \(0.04(5^2)=1\)

Answer 5

o ok thanks

Answer 6

so is l=l/d^2 the formula i would use for part B

Answer 7

\[l={1 \over d^2}\]That is correct.

Answer 8

and part c plug in values?

Answer 9

Exactly. Plug in \(2d\) for \(d\), and find the ratio between the original equation, and what you got from plugging in that value.