1. ## word prob

The illumination of an object (in lumens) created by a certain light source is given by I = 80/D^2 where D is the distance (in meters) of the object from the light source. Use differentials to approximated the decrease in illumination as the distance from the light source changes from 2 meters to 2.5 meters.

How to do this?

2. Originally Posted by kwivo
The illumination of an object (in lumens) created by a certain light source is given by I = 80/D^2 where D is the distance (in meters) of the object from the light source. Use differentials to approximated the decrease in illumination as the distance from the light source changes from 2 meters to 2.5 meters.

How to do this?
Hello,

you have: $\displaystyle I(D)=\frac{80}{D^2}$

you get: $\displaystyle \frac{d(I(D))}{dD}=-\frac{160}{D^3}$

To estimate the decrease in illumination it is sufficient to use the following values: $\displaystyle D = 2, \ dD = \frac12$

$\displaystyle d(I(D))=-\frac{160}{D^3} \cdot dD~\implies~ d(I(2))=-20 \cdot \frac12 = -10$