Chapter 4, Problem 5P

a.

To determine

What is the light intensity $I_o$ .

Answer to Problem 5P

The solution is,GRAPH.

Explanation of Solution

Given information:

As sunlight passes through the waters of lakes and oceans, the light is absorbed, and the deeper it penetrates, the more its intensity diminishes. The light intensity $I$ at depth $x$ is given by the Beer-Lambert Law:

  $I=I_o{e}^{-kx}$

Where $I_o$ is the light intensity at the surface and $k$ is a constant that depends on the murkiness of the water (see page 336). A biologist uses a photometer to investigate light penetration in a northern lake, obtaining the data in the table.

  

Use a graphing calculator to find an exponential function of the form given by the BeerLambert Law to model these data. What is the light intensity $I_o$ at the surface on this day, and what is the “murkiness” constant $k$ for this lake? [Hint: If your calculator gives you a function of the form $I=a{b}^x$ convert this to the form $I=a{b}^x$ you want using the identities $b^x=e^{\ln \left(b^x\right)}=e^{x\ln b}$

Calculation:

  $\boxed{STAT}\boxed{CALC}$ and enter Zero for exponential regression.

  

Press Enter the exponential model for the data.

  

The exponential model for the data.

  $y=ab\text{'}$

Where $a=1.181\times {10}^{-15}$ and $b=1.020,$

Hence, the solution is,GRAPH.

b.

To determine

What is the light intensity $I_o$ .

Answer to Problem 5P

The solution is,GRAPH.

Explanation of Solution

Given information:

As sunlight passes through the waters of lakes and oceans, the light is absorbed, and the deeper it penetrates, the more its intensity diminishes. The light intensity $I$ at depth $x$ is given by the Beer-Lambert Law:

  $I=I_o{e}^{-kx}$

Where $I_o$ is the light intensity at the surface and $k$ is a constant that depends on the murkiness of the water (see page 336). A biologist uses a photometer to investigate light penetration in a northern lake, obtaining the data in the table.

  

Make a scatter plot of the data, and graph the function that you found in part (a) on your scatter plot.

Calculation:

  $\boxed{STAT}\boxed{CALC}$ and enter Zero for exponential regression.

  

Press Enter the exponential model for the data.

  

The exponential model for the data.

  $y=ab\text{'}$

Where $a=1.181\times {10}^{-15}$ and $b=1.020,$

Hence, the solution is,GRAPH.

c.

To determine

Determine the depth below which there is insufficient light to support these algae.

Answer to Problem 5P

  $\boxed{\ln \left(\frac{x\sqrt{3-x^4}}{{\left(x^2+1\right)}^2}\right)}$

Explanation of Solution

Given information:

As sunlight passes through the waters of lakes and oceans, the light is absorbed, and the deeper it penetrates, the more its intensity diminishes. The light intensity $I$ at depth $x$ is given by the Beer-Lambert Law:

  $I=I_o{e}^{-kx}$

Where $I_o$ is the light intensity at the surface and $k$ is a constant that depends on the murkiness of the water (see page 336). A biologist uses a photometer to investigate light penetration in a northern lake, obtaining the data in the table.

  

If the light intensity drops below $0.15$ lumen (lm), a certain species of algae can’t survive because photosynthesis is impossible. Use your model from part (a) to determine the depth below which there is insuf?cient light to support this algae.

Calculation:

Cosider the logrithm expression,

  $\ln x-2\ln \left(x^2+1\right)+\frac{1}{2}\ln \left(3-x^4\right)$

Combine in to single logrithm,

  $\begin{array}{l}=\ln x-\ln {\left(x^2+1\right)}^2+\ln {\left(3-x^4\right)}^{1/2}\\ =\ln \left(\frac{x}{{\left(x^2+1\right)}^2}\right)+\ln \sqrt{3-x^4}\\ =\ln \left(\frac{x}{{\left(x^2+1\right)}^2}\right)\cdot \sqrt{3-x^4}\\ =\ln \left(\frac{x\sqrt{3-x^4}}{{\left(x^2+1\right)}^2}\right)\end{array}$

Hence, the solution is $\boxed{\ln \left(\frac{x\sqrt{3-x^4}}{{\left(x^2+1\right)}^2}\right)}$