# The intensity L(x) of light x feet beneath the surface of the ocean satisfies the differential equation dL/dx = – kL, where k is a constant. As a diver you know from experience that diving to 18 ft in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below a tenth of the surface value. About how deep can you expect to work without artificial light?

The aim of this question is to learn how to solve simple ordinary differential equations and then use them to solve different word problems.

A differential equation is an equation that involves derivatives and requires integration during their solution.

While solving such equations, we may encounter integration constants that are calculated using the initial conditions given in the question.

## Expert Anwer

Given:

$\dfrac{ dL }{ dx } \ = \ -kL$

Rearranging:

$\dfrac{ 1}{ L } \ dL \ = \ -k \ dx$

Integrating both sides:

$\int \ \dfrac{ 1}{ L } \ dL \ = \ -k \ \int \ dx$

Using integration tables:

$\int \ \dfrac{ 1}{ L } \ dL \ = \ ln| \ L \ | \ \text{ and } \ \int \ dx \ = \ x$

Substituting these values in the above equation:

$ln| \ L \ | \ = \ -k \ x \ … \ … \ … \ (1)$

Exponentiating both sides:

$e^{ ln| \ L \ | } \ = \ e^{ -k \ x }$

Since:

$e^{ ln| \ L \ | } \ = \ L$

So, the above equation becomes:

$L \ = \ e^{ -k \ x } \ … \ … \ … \ (2)$

Given the following initial condition:

$L \ = \ 0.5 \ at \ x \ = \ 18 \ ft$

Equation (1) becomes:

$ln| \ 0.5 \ | \ = \ -k \ ( \ 18 \ )$

$\Rightarrow k = \dfrac{ ln| \ 0.5 \ | }{ -18 }$

$\Rightarrow k = 0.0385$

Substitute this value in the equation (1) and (2):

$ln| \ L \ | \ = \ -0.0385 \ x \ … \ … \ … \ (3)$

And:

$L \ = \ e^{ -0.0385 \ x } \ … \ … \ … \ (4)$

To find the depth $x$ at which intensity $L$ falls to one-tenth, we put the following values in the equation (3):

$ln| \ 0.1 \ | \ = \ -0.0385 \ x$

$\Rightarrow x \ = \ \dfrac{ ln| \ 0.1 \ | }{ -0.0385 }$

$\Rightarrow x \ = \ 59.8 \ ft$

## Numerical Result

$x \ = \ 59.8 \ ft$

## Example

In the above question, with the same differential equation and initial condition, find the depth at which intensity reduces to 25% and 75%.

Part(a): Substitute $L = 0.25$ in equation no. (3):

$ln| \ 0.25 \ | \ = \ -0.0385 \ x$

$\Rightarrow x \ = \ \dfrac{ ln| \ 0.25 \ | }{ -0.0385 }$

$\Rightarrow x \ = \ 36 \ ft$

Part(b): Substitute $L = 0.75$ in equation no. (3):

$ln| \ 0.75 \ | \ = \ -0.0385 \ x$

$\Rightarrow x \ = \ \dfrac{ ln| \ 0.75 \ | }{ -0.0385 }$

$\Rightarrow x \ = \ 7.47 \ ft$