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Math Help - Calculus: The Intensity of Light Change Question?

  1. #1
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    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
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  2. #2
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    Hello Kayla_N
    Quote Originally Posted by Kayla_N View Post
    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,

    Grandad
    Last edited by Grandad; February 19th 2010 at 11:28 PM.
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