# Calculus: The Intensity of Light Change Question?

• February 19th 2010, 07:43 PM
Kayla_N
Calculus: The Intensity of Light Change Question?
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
• February 19th 2010, 10:40 PM
Hello Kayla_N
Quote:

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 $L$ is $4.913 \times 10^{-4}$.

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 $d$'s by $\delta$'s we get:
$L = Ir^{-2}$

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

$\Rightarrow \frac{\delta L}{L}\approx \frac{\delta I}{I}-2\frac{\delta r}{r}$
This will give you the approximate fractional increase in $L$, which you can multiply by $100$ to give the percentage increase; thus:
Percentage increase in $L\approx 100\left(\frac{4}{20}-\frac{6}{60}\right)\approx 10$%
This is not very accurate in this case, since the changes are relatively large,