Math Problem Statement

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 19 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    enter your response here ft deep without artificial light.

Solution

We are given that the intensity L(x)L(x) of light at a depth xx feet beneath the surface satisfies the differential equation:

dLdx=kL\frac{dL}{dx} = -kL

where kk is a constant. The general solution to this differential equation is:

L(x)=L0ekxL(x) = L_0 e^{-kx}

where L0L_0 is the intensity of light at the surface (i.e., x=0x = 0).

Step 1: Use the given information about intensity at 19 feet

We know that at x=19x = 19 feet, the intensity is half of the surface intensity. This gives the equation:

L(19)=12L0L(19) = \frac{1}{2} L_0

Substituting into the general solution:

12L0=L0ek×19\frac{1}{2} L_0 = L_0 e^{-k \times 19}

We can cancel L0L_0 from both sides (since L00L_0 \neq 0):

12=ek×19\frac{1}{2} = e^{-k \times 19}

Taking the natural logarithm of both sides:

ln(12)=k×19\ln \left( \frac{1}{2} \right) = -k \times 19

This simplifies to:

ln(2)=k×19-\ln(2) = -k \times 19

Thus,

k=ln(2)19k = \frac{\ln(2)}{19}

Step 2: Find the depth where intensity falls below one-fifth of the surface value

We want to find the depth xx at which the intensity falls to one-fifth of the surface intensity. That is, we need to solve for xx in:

L(x)=15L0L(x) = \frac{1}{5} L_0

Substitute this into the general solution:

15L0=L0ekx\frac{1}{5} L_0 = L_0 e^{-kx}

Cancel L0L_0 from both sides:

15=ekx\frac{1}{5} = e^{-kx}

Taking the natural logarithm of both sides:

ln(15)=kx\ln \left( \frac{1}{5} \right) = -kx

This simplifies to:

ln(5)=kx-\ln(5) = -k x

Thus,

x=ln(5)kx = \frac{\ln(5)}{k}

Step 3: Substitute the value of kk

We found earlier that k=ln(2)19k = \frac{\ln(2)}{19}, so substitute this into the equation:

x=ln(5)ln(2)19=19×ln(5)ln(2)x = \frac{\ln(5)}{\frac{\ln(2)}{19}} = 19 \times \frac{\ln(5)}{\ln(2)}

Now, calculate this value:

ln(5)1.6094,ln(2)0.6931\ln(5) \approx 1.6094, \quad \ln(2) \approx 0.6931

x19×1.60940.693119×2.32544.2x \approx 19 \times \frac{1.6094}{0.6931} \approx 19 \times 2.325 \approx 44.2

Final Answer:

The diver can work approximately 44.2 feet deep without artificial light.

Would you like further details or have any questions?


Here are some related questions to expand on this topic:

  1. What happens to the intensity of light at greater depths?
  2. How can the constant kk be derived experimentally?
  3. What other factors might influence how deep a diver can go without artificial light?
  4. How does the rate of change of light intensity differ in other mediums, such as air or water with different properties?
  5. How would the solution change if the intensity decreased at a different rate?

Tip: Understanding exponential decay like this one can be useful in many fields, such as physics and biology, where processes like radioactive decay, population growth, or absorption follow similar patterns.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Exponential Decay
Logarithms

Formulas

General solution of first-order linear differential equation: L(x) = L_0 e^(-kx)
Natural logarithm to solve for k: k = ln(2)/19
Depth calculation using logarithmic equation: x = ln(5)/k

Theorems

Exponential Decay Law
Properties of Logarithms

Suitable Grade Level

College-Level Calculus