Chapter 4.5, Problem 84E

Transparency of a Lake Environmental scientists measure the intensity of light at various depths in a lake to find the “transparency” of the water. Certain levels of transparency are required for the biodiversity of the submerged macrophyte population. In a certain lake the intensity of light at depth x is given by

$I=10e^{-0.008x}$

where I is measured in lumens and x in feet.

1. (a) Find the intensity I at a depth of 30 ft.
2. (b) At what depth has the light intensity dropped to I = 5?

To determine

To find: The intensity $I$ at a depth of $30\text{ft}$.

Answer to Problem 84E

The intensity $I$ at a depth of $30\text{ft}$ is $I=7.87\text{lumens}$.

Explanation of Solution

Given:

The equation for intensity of light at depth $x$ is, $I=10e^{-0.008x}$.

Depth $x=30\text{ft}$.

Calculation:

The given equation for intensity of light at depth $x$ is,

$I=10e^{-0.008x}$

Substitute $30$ for $x$ in given equation,

$\begin{array}{c}I=10e^{-0.008\left(30\right)}\\ =10\left(e^{-0.24}\right)\\ =10\left(0.7866\right)\left(\because e^{-0.24}=0.7866\right)\\ =7.866\approx 7.87\end{array}$

Thus, the intensity $I$ at a depth of $30\text{ft}$ is $I=7.87\text{lumens}$.

To determine

To find: The depth that has the light intensity dropped to $I=5$.

Answer to Problem 84E

The intensity dropped to $I=5$ at a depth of $x=86.64\text{ft}$.

Explanation of Solution

Given:

The function for intensity of light at depth $x$ is, $I=10e^{-0.008x}$.

Intensity $I=5$.

Calculation:

The given function for the intensity of light at depth $x$ is,

$I=10e^{-0.008x}$

Substitute $5$ for $I$ in the given function,

$\begin{array}{l}5=10e^{-0.008x}\\ \frac{1}{2}=e^{-0.008x}\end{array}$

Simplify the above expression taking $\ln$ to each side,

$\begin{array}{c}\ln \left(0.5\right)=\ln \left(e^{-0.008x}\right)\\ -0.69315=-0.008x\left(\because \ln e^x=x\right)\\ x=86.6434\\ x\approx 86.64\end{array}$

Thus, the intensity dropped to $I=5$ at a depth of $x=86.64\text{ft}$.