The intensity​ L(x) of light x feet beneath the surface of the ocean satisfies the differential equation StartFraction dL Over dx EndFraction
equalsminuskL. From​ experience, a diver knows that diving to 16 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?
Question content area bottom
Part 1
He can work approximately 
  
37.1 ft deep without artificial light.
​(Simplify your answer. Round to the nearest tenth as​ needed.)

The given problem involves the differential equation for light intensity \( L(x) \) beneath the surface of the ocean:

\[
\frac{dL}{dx} = -kL
\]

This is a separable differential equation, and solving it will help us find the depth at which the diver can work without artificial light.

### Step 1: Solve the differential equation

We separate variables:

\[
\frac{dL}{L} = -k \, dx
\]

Now integrate both sides:

\[
\int \frac{1}{L} \, dL = -k \int dx
\]

This yields:

\[
\ln|L| = -kx + C
\]

Exponentiate both sides to solve for \( L(x) \):

\[
L(x) = e^{-kx + C} = e^C \cdot e^{-kx}
\]

Let \( L_0 = e^C \), representing the initial light intensity at the surface (when \( x = 0 \)). Thus:

\[
L(x) = L_0 e^{-kx}
\]

### Step 2: Use given conditions to find \( k \)

We are told that diving to 16 feet cuts the intensity in half, meaning:

\[
L(16) = \frac{L_0}{2}
\]

Substitute into the equation \( L(x) = L_0 e^{-kx} \):

\[
\frac{L_0}{2} = L_0 e^{-16k}
\]

Cancel \( L_0 \) from both sides:

\[
\frac{1}{2} = e^{-16k}
\]

Take the natural logarithm of both sides:

\[
\ln \left( \frac{1}{2} \right) = -16k
\]

\[
- \ln 2 = -16k
\]

Thus:

\[
k = \frac{\ln 2}{16}
\]

### Step 3: Determine the maximum depth

The diver cannot work without artificial light when the intensity falls below one-fifth of the surface value. So we want to find the depth \( x \) such that:

\[
L(x) = \frac{L_0}{5}
\]

Substitute this into the equation \( L(x) = L_0 e^{-kx} \):

\[
\frac{L_0}{5} = L_0 e^{-kx}
\]

Cancel \( L_0 \) from both sides:

\[
\frac{1}{5} = e^{-kx}
\]

Take the natural logarithm of both sides:

\[
\ln \left( \frac{1}{5} \right) = -kx
\]

Substitute \( k = \frac{\ln 2}{16} \):

\[
\ln \left( \frac{1}{5} \right) = -\frac{\ln 2}{16} \cdot x
\]

Now solve for \( x \):

\[
x = \frac{16 \ln \left( \frac{1}{5} \right)}{-\ln 2}
\]

Calculate the value of \( x \):

\[
x \approx \frac{16 \times (-1.609)}{-0.693} \approx 37.1 \, \text{feet}
\]

### Conclusion

The diver can work to a depth of approximately **37.1 feet** without artificial light.

---

Would you like more details, or do you have any questions?

### Related Questions:
1. How does changing the initial light intensity affect the depth at which the diver can work without artificial light?
2. What would happen if the light intensity fell to one-fourth instead of one-fifth of the surface value?
3. How is the constant \( k \) related to the properties of the water and light absorption?
4. What assumptions are made in this model about how light decreases with depth?
5. How would the depth change if the diver used a different type of water with less light absorption?

#### Tip:
When solving differential equations involving exponential decay, always carefully interpret the constants by using known boundary or initial conditions.

The intensity​ L(x) of light x feet beneath the surface of the ocean satisfies the differential equation dL/dx = -kL. A diver knows that diving to 16 ft 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?

Discover how to calculate the depth at which a diver can work without artificial light based on the exponential decay of light intensity underwater. This problem uses differential equations and natural logarithms, suitable for university-level calculus students.

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