Answer:
The diver can expect to work up to a depth of 37.151 ft without artificial light.
Step-by-step explanation:
Given differential equation:
[tex]\dfrac{\text{d}L}{\text{d}x}=-kL[/tex]
where L is the intensity of light x feet beneath the surface of the ocean, and k is a constant.
To determine how deep the diver can expect to work without the need for artificial light, we first need to solve the differential equation.
Separate the variables:
[tex]\dfrac{1}{L}\;\text{d}L=-k\;\text{d}x[/tex]
Integrate both sides:
[tex]\begin{aligned}\displaystyle \int\dfrac{1}{L}\;\text{d}L &=\int-k\;\text{d}x\\\\\ln|L| &=-kx+C\\\\e^{\ln|L|} &=e^{-kx+C}\\\\L &=e^{-kx}\cdot e^{C}\\\\L &=Ae^{-kx}\end{aligned}[/tex]
A is the intensity of light at surface level (when x = 0). Rewrite this as L₀:
[tex]L=L_0e^{-kx}[/tex]
Given that the intensity is halved when x = 16, then:
[tex]\dfrac{L_0}{2}=L_0e^{-16k}[/tex]
Solve for k:
[tex]\begin{aligned}e^{-16k} &=\dfrac{1}{2}\\\\\ln(e^{-16k}) &=\ln\left(\dfrac{1}{2}\right)\\\\-16k\ln(e) &=\ln(1)-\ln(2)\\\\-16k(1) &=0-\ln(2)\\\\-16k &=-\ln(2)\\\\k &=\dfrac{1}{16}\ln(2)\end{aligned}[/tex]
So, the function for the intensity L(x) of light x feet beneath the surface of the ocean is:
[tex]\Large\text{$L(x)=L_0e^{-\left(\frac{1}{16}\ln(2)\right)x}$}[/tex]
If the diver cannot work without artificial light when the intensity falls below 1/5 of the surface value, we need to find the value of x when the intensity is (1/5)L₀:
[tex]\begin{aligned}L_0e^{-\left(\frac{1}{16}\ln(2)\right)x}&=\dfrac{1}{5}L_0\\\\e^{-\left(\frac{1}{16}\ln(2)\right)x}&=\dfrac{1}{5}\\\\\ln\left(e^{-\left(\frac{1}{16}\ln(2)\right)x}\right)&=\ln\left(\dfrac{1}{5}\right)\\\\-\left(\frac{1}{16}\ln(2)\right)x\ln\left(e\right)&=\ln(1)-\ln(5)\\\\-\left(\frac{1}{16}\ln(2)\right)x&=-\ln(5)\\\\x&=\dfrac{\ln(5)}{\frac{1}{16}\ln(2)}\\\\x&=\dfrac{16\ln(5)}{\ln(2)}\\\\x&=37.151\; \sf ft\;(3\;d.p.)\end{aligned}[/tex]
Therefore, the diver can expect to work up to a depth of 37.151 ft without artificial light.
