Hello Kayla_N Originally Posted by

**Kayla_N** 1. The intensity of light at a distance r from a source is given by L = I*r^(-2) where I is the illumination at the source. Start with the values I = 20, r = 60. Suppose we increase the distance by 3 and the illumination by 4.

By how much does the intensity of light change?

dL =?

Here is what i got: Both answers i got are wrong?? Can someone please explain. Thanks

1st attempt:

L = 20(60^(-2)) = 1/180

L = 24(63^(-2)) = 8/1323

(1/180)-(8/1323)=dl= **-4.91307634E-4. **

2nd attempt:

L = 20*60^(-2) = 1/180

L = 23*64^(-2) = 23/4096

(1/180)-(23/4096)=dL= **-5.96788194E-5**

Your first answer is correct except for the sign. The increase in $\displaystyle L$ is $\displaystyle 4.913 \times 10^{-4}$.

However, since you've put this question in the Calculus section, I wonder whether you're expected to use a method involving small changes. If so, then if we take logs of both sides of the equation and then differentiate, replacing the $\displaystyle d$'s by $\displaystyle \delta$'s we get: $\displaystyle L = Ir^{-2}$

$\displaystyle \Rightarrow \ln(L) = \ln(I) -2\ln(r)$

$\displaystyle \Rightarrow \frac{\delta L}{L}\approx \frac{\delta I}{I}-2\frac{\delta r}{r}$

This will give you the approximate fractional increase in $\displaystyle L$, which you can multiply by $\displaystyle 100$ to give the percentage increase; thus:Percentage increase in $\displaystyle L\approx 100\left(\frac{4}{20}-\frac{6}{60}\right)\approx 10$%

This is not very accurate in this case, since the changes are relatively large,

Grandad