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 January 17th, 2015, 01:11 PM #1 Newbie   Joined: Jan 2015 From: Sacramento Posts: 3 Thanks: 0 I cant figure this out! please help In some murky harbors, the intensity of light d feet below the surface is given approximately by: I = I0e^(-0.23d) What percentage of the surface light will reach a depth of: a. 10 feet? b. 20 feet? Last edited by skipjack; January 17th, 2015 at 07:15 PM. Reason: to provide the exponentiation presumably intended
 January 17th, 2015, 01:29 PM #2 Newbie   Joined: Jan 2015 From: france Posts: 7 Thanks: 1 I=Io*exp(-0.23*d) we suppose that the unity of the term (0.23 are inverse of a foot) for d=10 ft I(10 ft)=Io * 0.10032 Percentage is 10% for d=20 ft I(20 ft)=Io * 0.01005 percentage is 1% Thanks from rdaloian
 January 17th, 2015, 05:13 PM #3 Newbie   Joined: Jan 2015 From: Sacramento Posts: 3 Thanks: 0 Thank you so much, but I still don't understand how you get the answer.
January 17th, 2015, 06:10 PM   #4
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Quote:
 Originally Posted by alistrawberry I=Io*exp(-0.23*d) we suppose that the unity of the term (0.23 are inverse of a foot) for d=10 ft I(10 ft)=Io * 0.10032 Percentage is 10% for d=20 ft I(20 ft)=Io * 0.01005 percentage is 1%
I'm not sure exactly what alistrawberry did, either... I don't know much about light penetration in water, but I didn't get about 0.1 for exp(-.23*10). Not sure what line 2 was supposed to mean (unity?).

 January 17th, 2015, 07:10 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond At d = 0, I = 10. We want a percent so we want "intensity at a given depth"/"intensity at d = 0". With d= 10: $\displaystyle \frac{10e^{-0.23\cdot10}}{10}=e^{-0.23\cdot10}\approx0.1002$ That's pretty close to 10%. With d= 20: $\displaystyle \frac{10e^{-0.23\cdot20}}{10}=e^{-0.23\cdot20}\approx0.0101$ That's pretty close to 1%.
January 17th, 2015, 07:18 PM   #6
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Quote:
 Originally Posted by alistrawberry . . . for d=10 ft I(10 ft)=Io * 0.10032
$e^{-2.3}$ = 0.10026 approximately, not 0.10032.

The question defines $d$ to be dimensionless, so inverse foot units are not required for the constant 0.23.

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