Question

the intensity ( l(x) ) of light ( x ) feet beneath the surface of the ocean satisfies the differential equation ( \frac{dl}{dx}=-kl ). from experience, a diver knows that diving to 17 ft in a sea cuts the intensity in half. he cannot work without artificial light when the intensity falls below one - fifth of the surface value. about how deep can he expect to work without artificial light? he can work approximately ( square ) ft deep without artificial light. (simplify your answer. round to the nearest tenth as needed.)

Explanation:

Step1: Solve the differential equation

The differential equation \(\frac{dL}{dx}=-kL\) is a first - order separable differential equation. We can rewrite it as \(\frac{dL}{L}=-kdx\). Integrating both sides:
\(\int\frac{dL}{L}=-\int kdx\)
Using the integral formula \(\int\frac{1}{u}du=\ln|u| + C\), we get \(\ln L=-kx + C\). Exponentiating both sides gives \(L = L_0e^{-kx}\) (where \(L_0 = e^{C}\) is the initial intensity when \(x = 0\)).

Step2: Find the value of \(k\)

We know that when \(x = 17\), \(L=\frac{L_0}{2}\). Substitute into \(L = L_0e^{-kx}\):
\(\frac{L_0}{2}=L_0e^{-17k}\)
Divide both sides by \(L_0\) (since \(L_0
eq0\)): \(\frac{1}{2}=e^{-17k}\)
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=-17k\)
Since \(\ln(\frac{1}{2})=-\ln 2\), then \(k=\frac{\ln 2}{17}\approx\frac{0.693}{17}\approx0.0408\)

Step3: Find the depth \(x\) when \(L=\frac{L_0}{5}\)

Substitute \(L=\frac{L_0}{5}\) and \(k = \frac{\ln 2}{17}\) into \(L = L_0e^{-kx}\)
\(\frac{L_0}{5}=L_0e^{-\frac{\ln 2}{17}x}\)
Divide both sides by \(L_0\) (since \(L_0
eq0\)): \(\frac{1}{5}=e^{-\frac{\ln 2}{17}x}\)
Take the natural logarithm of both sides: \(\ln(\frac{1}{5})=-\frac{\ln 2}{17}x\)
Since \(\ln(\frac{1}{5})=-\ln 5\), then \(x=\frac{17\ln 5}{\ln 2}\)
Using the property of logarithms \(\log_a b=\frac{\ln b}{\ln a}\) and calculating \(\ln 5\approx1.609\), \(\ln 2\approx0.693\)
\(x=\frac{17\times1.609}{0.693}=\frac{27.353}{0.693}\approx39.5\)

Answer:

\(39.5\)