Chapter 4.3, Problem 59PS

Solution

a.

Write the equation and graph the percent of surface light that filters down through clear water as a function depth.

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

Given information:

The exponential function $L\left(x\right)=100e^{kx}$ where $k$ is the measure of murkiness of the water and $x$ is the depth.

  $k=-0.02$

Concept Used:

Substitute the value of a $k$ on the exponential function $L\left(x\right)=100e^{kx}$.

Calculation:

As per the given problem

Note that $L\left(x\right)$ is the percent of surface light at depth $x$ . In water with murkiness of $-0.02$

Therefore the equation is

  $L\left(x\right)=100e^{kx}$

Substitute the value of $k=-0.02$

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

Graph:

  

Conclusion:

Hence, the equation is $L\left(x\right)=100e^{\left(-0.02\right)x}$

b.

Estimate the percent of surface light available at depth of $40$ meters.

  $=44\%$

Given:

The exponential function $L\left(x\right)=100e^{kx}$ where $k$ is the measure of murkiness of the water and $x$ is the depth.

  $x=40m$

Concept Used:

Substitute the value of a $k$ on the exponential function $L\left(x\right)=100e^{kx}$.

Calculation:

As per the given problem

Note that $L\left(x\right)$ is the percent of surface light at depth $x$.

Therefore $L\left(x\right)$ is around $44$ . At a depth of $40$ meters, there is approximately $44\%$ of surface light available,

Graph:

  

Conclusion:

Hence, the percent of surface light is $44\%$.

c.

How deep the submersible can descend in clear water.

  $=35m$

Given:

The exponential function $L\left(x\right)=100e^{kx}$ where $k$ is the measure of murkiness of the water and $x$ is the depth.

  $L\left(x\right)=50$

Concept Used:

Substitute the value of a $k$ on the exponential function $L\left(x\right)=100e^{kx}$.

Calculation:

As per the given problem

Note that $L\left(x\right)$ is the percent of surface light at depth $x$.

When $L\left(x\right)=50$ , $x$ is around $35$.

Note that the submersible can descend about $35$ before only $50\%$ of surface light is available,

Graph:

  

Conclusion:

Hence, the solution is $35m$.