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Chapter 0: Problem 16

In two bodies of water, \(L_{1}\) and \(L_{2}\), the light intensities \(I_{1}(x)\) and \(I_{2}(x)\) as functions of depth \(x\) are measured simultaneously and found to be $$ I_{1}(x)=800 e^{-0.04 x} \quad \text { and } \quad I_{2}(x)=700 e^{-0.05 x} $$ Explain the differences in the two formulas in terms of the properties of water in the two bodies.

Short Answer

Expert verified
L1 is clearer with slower light reduction; L2 is more turbid, causing faster attenuation.

Step by step solution

01

Identify the General Form of Light Intensity Functions

Both functions, \( I_1(x) = 800 e^{-0.04x} \) and \( I_2(x) = 700 e^{-0.05x} \), are exponential decay functions.This indicates that as depth \( x \) increases, the light intensity decreases in both bodies of water. The exponential term \( e^{-kx} \) is common in describing attenuation (or reduction) of light as it penetrates through a medium.
02

Compare Initial Intensities

The initial light intensity values (when \( x = 0 \)) can be found by setting \( x = 0 \) in the expressions for \( I_1(x) \) and \( I_2(x) \).- For \( I_1(x) = 800 e^{-0.04 imes 0} = 800 \).- For \( I_2(x) = 700 e^{-0.05 imes 0} = 700 \).This shows that the initial light intensity at the surface is greater in body of water \( L_1 \) compared to \( L_2 \).
03

Analyze Exponential Decay Factors

The exponentials \( e^{-0.04x} \) and \( e^{-0.05x} \) represent decay rates in light intensity:- The decay rate constant for \( L_1 \) is \( 0.04 \).- The decay rate constant for \( L_2 \) is \( 0.05 \).A larger decay constant indicates a faster rate of attenuation. Thus, light intensity decreases more rapidly with depth in \( L_2 \) compared to \( L_1 \).
04

Interpret in Terms of Water Properties

The differences in the exponential terms suggest that \( L_2 \) might be more turbid or denser, causing faster light attenuation, whereas \( L_1 \) could be clearer or less dense allowing light to penetrate deeper with less reduction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Decay
Exponential decay describes how a quantity decreases at a rate proportional to its current value. In the context of light attenuation in water, the intensity of light decreases as it penetrates deeper. Mathematically, this is expressed with functions of the form \( I(x) = I_0 e^{-kx} \), where \( I_0 \) is the initial light intensity, \( e \) is Euler's number, and \( k \) is the decay constant. The negative sign in the exponent indicates decay, meaning the intensity diminishes with increasing depth.In our exercise, we see two exponential decay functions: \( I_1(x) = 800 e^{-0.04x} \) and \( I_2(x) = 700 e^{-0.05x} \). Each represents the light intensity at different depths in two bodies of water. The differences in the equation parameters tell us about the rate at which light decreases with depth in each location.
Light Intensity
Light intensity refers to the amount of light energy that reaches a given area. In the context of underwater environments, this intensity decreases with depth due to light absorption and scattering by the water and its contents. At the surface (when \( x = 0 \)), light intensity is at its maximum, defined by the initial value of the functions, \( I_0 \).For the given water bodies:
  • \( I_1 \) starts with an intensity of 800 units.
  • \( I_2 \) begins with 700 units.
This indicates that initially, light travels more intensely at the surface of the first water body than the second. As light travels deeper, factors inherent to each water body cause the light to become less intense, described by the exponential decay terms.
Water Properties
The physical and chemical properties of water greatly influence how light attenuates. Turbidity, which refers to the presence of particles suspended in water, and water clarity, both affect light penetration. In our exercise:
  • Water body \( L_1 \) with the decay factor \( 0.04 \) is likely clearer, allowing more light to reach deeper regions.
  • Water body \( L_2 \) with a higher decay rate of \( 0.05 \) suggests it is more turbid or contains more suspending particles, causing faster light attenuation.
These properties not only tell us about the environmental conditions but also affect the ecological systems within these bodies of water.
Depth Measurement
Measuring depth in bodies of water allows us to understand light attenuation and the factors affecting it. It is a vital step in evaluating how different water properties influence light intensity and penetration. By examining functions like \( I_1(x) = 800 e^{-0.04x} \) and \( I_2(x) = 700 e^{-0.05x} \), we can determine the change in light intensity at any depth, \( x \).This insight helps scientists and ecologists:
  • Predict how far light can penetrate in different aquatic environments.
  • Analyze how depth affects photosynthesis in submerged plants.
  • Inspect how organisms adapt to varying light conditions underwater.
Understanding depth measurement in light attenuation studies is crucial for marine biology and environmental monitoring projects.

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Most popular questions from this chapter

Consider the following osmosis experiment in biology laboratory. Material; A thistle tube, a 1 liter flask, some 'salt water', and some pure water, a membrane that is impermeable to the salt and is permeable to the water. The bulb of the thistle tube is filled with salt water, the membrane is place across the open part of the bulb, and the bulb is inverted in a beaker of pure water so that the top of the pure water is at the juncture of the bulb with the stem. See the diagram. Because of osmotic pressure the pure water will cross the membrane pushing water up the stem of the thistle tube until the increase in pressure inside the bulb due to the water in the stem matches the osmotic pressure across the membrane.

Find a number \(\omega\) so that the proposed solution satisfies the derivative equation. Solution \(\quad\) Derivative equation a. \(y(t)=3 \cos 5 t\) \(y^{\prime \prime}+\omega^{2} y=0\) b. \(y(t)=2 \sin 3 t+5 \cos 3 t\) \(y^{\prime \prime}+\omega^{2} y=0\) c. \(y(t)=-4 \cos \pi t\) \(y^{\prime \prime}+\omega^{2} y=0\) d. \(y(t)=3 e^{-t} \cos 5 t \quad y^{\prime \prime}+2 y^{\prime}+\omega^{2} y=0\) e. \(y(t)=-4 e^{-2 t} \sin 3 t \quad y^{\prime \prime}+4 y^{\prime}+\omega^{2} y=0\)

Show that the proposed solutions satisfy the equations and initial conditions. Solution \(\quad\) Derivative Equation \(\quad\) Initial conditions $$ \begin{array}{lrr} \text { a. } y(t)=2 \sin t+\cos t & y^{\prime \prime}+y=0 & y(0) & =1 \\ & & & y^{\prime}(0) & =2 \\ \text { b. } y(t)=4 \cos 2 t & y^{\prime \prime}+4 y=0 & y(0) & =4 \\ & & y^{\prime}(0) & =0 \\ \text { c. } y(t)=\cos 3 t-\sin 3 t & y^{\prime \prime}+9 y=0 & y(0) & =1 \\ & & y^{\prime}(0) & =-3 \\ \text { d. } y(t)=-20 \sin 5 t+15 \cos 5 t & y^{\prime \prime}+25 y=0 & y(0) & =15 \\ & & y^{\prime}(0) & =-100 \\ \text { e. } y(t)=4 \cos (3 t+\pi / 3) & y^{\prime \prime}+9 y=2 & y(0) & =2 \\\ & & y^{\prime}(0) & =-6 \sqrt{3} \\ \text { f. } y(t)=\sin 2 t-2 \cos 2 t & y^{\prime \prime}+4 y=0 & y(0) & =-2 \\\ & & y^{\prime}(0) & =2 \\ \text { g. } y(t)=2 \sin 3 t+3 \cos 3 t & y^{\prime \prime}+9 y=0 & y(0) & =3 \\\ & & y^{\prime}(0) & =6 \\ & & & \\ \text { h. } y(t)=3 \sin \pi t+4 \cos \pi t & y^{\prime \prime}+\pi^{2} y=0 & y(0) & =4 \\ & & & y^{\prime}(0) & =3 \pi \end{array} $$

Suppose that \(a=b=1\) and \(u_{0}=3\) and \(v_{0}=4\) in the prey equation D. 27 so that $$ u(t)=-4 \sin (t)+3 \cos (t) $$ \(\left(u_{0}\right.\) and \(v_{0}\) are 'small' disturbances. We might suppose, for example, that the equilibrium populations are \(U_{e}=300, V_{e}=200,\) with 3 and 4 'small' with respect to 300 and \(\left.200\right)\) a. Sketch the graph of \(u(t)=-4 \sin (t)+3 \cos (t)\). b. Let \(\phi\) (the Greek letter phi) denote the angle between 0 and \(2 \pi\) whose sine is \(\frac{3}{5}\) and whose cosine is \(\frac{-4}{5}\). Show that $$ \begin{aligned} u(t) &=5(\cos \phi \sin t+\sin \phi \cos t) \\ &=5 \sin (t+\phi) \end{aligned} $$ c. Plot the graph of \(5 \sin (t+\phi)\) and compare it with the graph of \(-4 \sin t+3 \cos t .\)

Show that if \(B\) and \(\omega\) are constants and \(y(t)=B \cos (\omega t),\) then $$ y(0)=B \quad y^{\prime}(0)=0 \quad \text { and } \quad y^{\prime \prime}(t)+\omega^{2} y(t)=0 $$
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