Chapter 6, Problem 14CT

a.

To determine

A function that gives the percent of surface light that filters down through clear water as a function of depth.

Answer to Problem 14CT

  $L\left(x\right)=100e^{-0.02x}$

Explanation of Solution

Given information: $L\left(x\right)=100e^{kx}$

Concept used: We put the value of $k$ on the equation $L\left(x\right)$

Calculation:

  $L\left(x\right)=100e^{kx}$

Now, $k=-0.02$

So, $L\left(x\right)=100e^{-0.02x}$

b.

To determine

Whether the function in part (a) represents exponential growth or exponential decay.

Answer to Problem 14CT

The function in part (a) represents exponential decay.

Explanation of Solution

Given information: $L\left(x\right)=100e^{-0.02x}$

Concept Used:

An exponential function has the form $y=a{b}^x$ where $a\ne 0$ and $b$ is a positive real number other than $1$ .

For exponential growth function : $a>0$ and $b>1$

For exponential decay function: $a>0$ and $0<b<1$

Calculation:

  $\begin{array}{l}L\left(x\right)=100e^{-0.02x}\\ \Rightarrow L\left(x\right)=100{\left(e^{-0.02}\right)}^x\mathrm{.........}\left(i\right)\end{array}$

If we compare equation (i) with $y=a{b}^x$ , we can write ,

  $a=100$ and $b=e^{-0.02}$

Now,

  $\begin{array}{l}a=100>1\\ b=\frac{1}{e^{0.02}}\approx 0.98\end{array}$

So, $0<b<1$

Therefore, the function in part (a) represents exponential decay function.

c.

To determine

To estimate: The percentage of surface light available at depth of $40$ meters.

Answer to Problem 14CT

The percentage of surface light available at depth of $40$ meters is $44.93\%$ .

Explanation of Solution

Given information:

  $L\left(x\right)=100e^{-0.02x}$

  $x=40$

Concept used: We put the value of $x$ on the equation.

Calculation:

  $L\left(x\right)=100e^{-0.02x}$

So, $L\left(40\right)=100e^{-0.02\times 40}$

  $\Rightarrow L\left(40\right)=100e^{-0.8}=44.93$

So, the percentage of surface light available at a depth of $40$ meters is $44.93\%$ .