The intensity L(x) of light x feet beneath the surface of the ocean decreases at

Question

The intensity L(x) of light x feet beneath the surface of the ocean decreases at a rate proportional to its value at that location. That is. L(x) satisfies the differential equation dL/dx = -kL, for some k > 0 (the constant of proportionality). An experienced diver has determined that the weather conditions on the day of her dive will be such that the light intensity will be cut in half upon diving 13 ft under the surface of the water. She also knows that, once the intensity of the light falls below 1/9 of the surface value, she will have to make use of artificial light. How deep can the diver go without having to resort to the use of artificial light?

Explanation / Answer

dL/dx = -kL

==> dL /L = -k dx

integrating on both sides

==> dL /L = -k dx

==> ln (L) = -kx +c

==> L = e-kx +c

==> L = e-kx ec        since am+n = am * an

==> L = Ce-kx       let ec = C

at x = 0 intensity is Lo (surface value)

==> Lo = Ce-k(0)

==> Lo = C(1) ==> C = Lo

==> L = Loe-kx  

upon diving 13 feet light intensity becomes half

==> at x = 13 , L = Lo/2

==> Lo/2 = Loe-k(13)

==> e-13k = 1/2

apply natural logarithms on both sides

==> ln e-13k = ln(1/2)

==> -13k (ln e) = ln(1/2)            since ln ab = b ln a ; ln e = 1

==> -13k = ln(1/2)

==> k = (-1/13)ln(2)

==> k = 0.05332

==> L = Loe-0.05332x  

L = Lo/9 at certain x

==> Lo/9 = Loe-0.05332x  

==> 1/9 = e-0.05332x  

==> e-0.05332x   = 1/9

apply natural logarithm on both sides

==> ln e-0.05332x   = ln (1/9)

==> -0.05332x lne = ln(1/9)

==> -0.05332x = ln(1/9)

==> x = (-1/0.05332) ln(1/9)

==> x = 41.21

Hence at 41.21 feet light intensity falls below 1/9 of surface value

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