The illumination from a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance s from the source. At a distance of 2 ft. from a glowing fire, a photographer's light meter registers 120 units. If the photographer backs away from the fire, find the rate of change of the meter reading with respect to s when the photographer is 20 ft. from the fire. Thanks!!

Question

anonymous · Answer

This is the most important part: "inversely proportional to the square of the distance, s"
Do you know how to express that in equation form?

anonymous · Answer

Inversely proportional I've written as 1/s^2.

anonymous · Answer

But as the whole equation

anonymous · Answer

I've been writing the equation like a related rates problem. If I let f = units, s = distance, and i = illumination, then I'd have di/ds = di/df * df/ds. But that is where I am stuck.

anonymous · Answer

I see. Yeah, there doesn't seem to be another rate to relate it to.

anonymous · Answer

Here's what I'm thinking:

"The illumination from a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance s from the source. "

$\large I=kLs^{-2}$

"At a distance of 2 ft. from a glowing fire, a photographer's light meter registers 120 units. "

$\large 120units=kL(2ft)^{-2} \rightarrow kL=480ft^2units.$

anonymous · Answer

$\large I=480s^{-2}$

$\large \frac{dI}{ds}=-960s^{-3}$

anonymous · Answer

Does that help?
(Or am I way off-base here?)

anonymous · Answer

Yes! Thank you very much!

anonymous · Answer

Once I plug in 20 to the equation, I get the correct answer of -.12 units.I guess what may confused me was you didn't take the derivative until AFTER you plugged in the given into the original equation. But in retrospect that makes sense because you are using the Chain Rule.

anonymous · Answer

I guess it can be done either way, but it seemed simpler and more natural to develop the equation directly from the given information.

anonymous · Answer

My question, then, is how did you know to multiply s^-2 by the units you obtained in the first step? It seems like 480 ft^2 units was a constant?

anonymous · Answer

Nevermind, I just realized it was because you plug in the illumination into the original equation per the first line: "The illumination from a light source is directly proportional to the strength of the source". Thanks again!

anonymous · Answer

Yup, I was just solving for the constant during that step.