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Chapter 5: Problem 11

In Section 1.3 we found from the discrete model of light extinction, \(I_{d+1}=I_{d}-0.18 I_{d}\) that the solution \(\quad I_{d+1}=0.82 I_{d}\) Light decrease in water is continuous, however. Find a value of \(k\) for which the solution to the continuous model, \(I^{\prime}(x)=-k I(x),\) matches the data.

Short Answer

Expert verified
The value of \(k\) for the continuous model is approximately 0.1987.

Step by step solution

01

Understand the Problem

We have a discrete model for light extinction where the light intensity decreases by 18% each step, expressed as \(I_{d+1} = 0.82 I_{d}\). We need to find a continuous rate of decrease \(k\) for the continuous model \(I'(x) = -k I(x)\) that aligns with the given discrete model data.
02

Express Discrete Model

The discrete model indicates that the intensity at the next depth, \(I_{d+1}\), is 82% of the current intensity, \(I_{d}\). This can be mathematically expressed as \(I_{d+1} = 0.82 I_{d}\).
03

Formulate the Continuous Model

The continuous model for the rate of change in light intensity with respect to depth \(x\) is given by \(I'(x) = -k I(x)\). This is a differential equation where \(k\) represents the continuous rate of extinction.
04

Solve the Differential Equation

The general solution of the differential equation \(I'(x) = -k I(x)\) is \(I(x) = I(0)e^{-kx}\), where \(I(0)\) is the initial intensity. This indicates exponential decay of intensity.
05

Relate Continuous Model to Discrete Model

To find the value of \(k\) that matches the discrete model, we set one depth step \(x\) in the continuous model to decrease intensity to 82%. Thus, we have \(I(x) = 0.82 I(0)\). This implies \(e^{-kx} = 0.82\).
06

Solve for k

To solve for \(k\), take the natural logarithm of both sides of \(e^{-kx} = 0.82\). This yields \(-kx = \ln(0.82)\). Assuming one depth interval corresponds to \(x = 1\), solve for \(k\): \[k = -\ln(0.82)\]
07

Calculate k Value

Calculate \(-\ln(0.82)\) to find the continuous extinction rate \(k\). Use a calculator: \[k \approx 0.1987\]

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Key Concepts

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

Discrete Model
A discrete model helps us understand processes that occur in separate, distinct steps. In this context, it describes how light intensity reduces as it passes through layers of water. For example, our model, expressed as \(I_{d+1} = 0.82 I_{d}\), tells us that each layer causes an 18% reduction in light. This simple representation makes it easier to predict the intensity change at each specific step.
  • Discrete models are step-based.
  • Suitable for processes occurring in fixed intervals.
  • In our case, they offer a straightforward understanding of light decrement by layers.
This stepwise approach is invaluable in showing light's behavior in segmented depths, highlighting its limitations in continuous environments.
Continuous Model
While the discrete model looks at light reduction in specific steps, the continuous model examines it as an ongoing process. Specifically, it represents how light diminishes smoothly over a continuous depth in water. The mathematical expression for this is given by the differential equation \(I'(x) = -k I(x)\).
  • Continuous models address uninterrupted processes.
  • They use calculus to express changes in a smooth fashion.
  • In this case, continuous models better match water environments.
This model gives a holistic picture of light change, offering insights where step functions can't, benefiting especially in natural settings where variables change continuously.
Differential Equation
Differential equations are key in modeling systems where variables change continuously. In light extinction, our differential equation plays a crucial role: \(I'(x) = -k I(x)\). This equation tells us the instantaneous rate of change of light intensity with depth.
  • Differential equations explain dynamic changes.
  • They involve functions and their derivatives.
  • Essential in expressing how quantities evolve over time or space.
Solutions to such equations describe how a system evolves, as seen by deriving a formula for light intensity over depth鈥擻(I(x) = I(0)e^{-kx}\). This communicates the continuous decay of light through water.
Exponential Decay
Exponential decay is a concept where a quantity reduces at a rate proportional to its current value. In our continuous model of light extinction, this manifests as \(I(x) = I(0)e^{-kx}\). Here, light intensity decreases exponentially with depth in water.
  • Exponential decay describes rapid diminution.
  • Characterized by the equation \(e^{-kx}\).
  • Essential in understanding phenomena like light and radioactivity.
Applying this to our data, a precise calculation of \(k\) helps align the continuous decay prediction with observed discrete data. Thus, \(k = -\ln(0.82)\), calculated as approximately 0.1987, perfectly integrates the discrete findings into a continuous context, offering a comprehensive view of light decay in aquatic settings.

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Most popular questions from this chapter

2 kilos of a fish poison, rotenone, are mixed into a lake which has a volume of \(100 \times 20 \times 2=4000\) cubic meters. A stream of clean water flows into the lake at a rate of 1000 cubic meters per day. Assume that it mixes immediately throughout the whole lake. Another stream flows out of the lake at a rate of 1000 cubic meters per day. What is the amount \(\left(p_{t}\right.\) for discrete time or \(P(t)\) for continuous time) of poison in the lake at time \(t\) days after the poison is applied? a. Treat the problem as a discrete time problem with one-day time intervals. Solve the difference equation $$ p_{0}=2 \quad p_{t+1}-p_{t}=-\frac{1000}{4000} p_{t} $$ b. Let \(t\) denote continuous time and \(P(t)\) the amount of poison in the lake at time \(t\). Let \([t, t+\Delta t]\) denote a short time interval (measured in units of days). An equation for the mathematical model is $$ P(t+\Delta t)-P(t)=-\frac{P(t)}{4000} \Delta t 1000 $$ Show that the units on the terms of this equation balance. c. Argue that $$ P(0)=2, \quad P^{\prime}(t)=-0.25 P(t) $$ d. Compute the solution to this equation. e. Compare the solution to the discrete time problem, \(p_{t},\) with the solution to the continuous time problem, \(P(t)\). f. For what value of \(k\) will the solution, \(Q(t),\) to $$ Q(0)=2, \quad Q^{\prime}(t)=k Q(t) \quad \text { satisfy } $$ \(Q(t)=p_{t}, \quad\) for \(\quad t=0,1,2, \cdots ?\) g. Which of \(P(t)\) and \(Q(t)\) most accurately estimates rotenone levels? h. On what day, \(\bar{t}\) will \(P(\bar{t})=4 \mathrm{~g}\) ?

Use the logarithmic differentiation to compute \(y^{\prime}(t)\) for a. \(y(t)=t^{\pi}\) b. \(y(t)=t^{e}\) c. \(y(t)=\left(1+t^{2}\right)^{\pi}\) d. \(y(t)=t^{3} e^{t}\) e. \(y(t)=e^{\sin t}\) f. \(y(t)=t^{t}\)

Suppose a bacterium Vibrio natriegens is growing in a beaker and cell concentration \(C\) at time \(t\) in minutes is given by $$C(t)=0.87 \times 1.02^{t} \quad \text { million cells per ml }$$ a. Approximate \(C(t)\) and \(C^{\prime}(t)\) for \(t=0,10,20,30,\) and 40 minutes. b. Plot a graph of \(C^{\prime}(t)\) vs \(C(t)\) using the five pairs of values you just computed.

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E. O. Wilson, a pioneer in study of area-species relations on islands, states in Diversity of Life, p 221,: "In more exact language, the number of species increases by the area-species equation, \(S=C A^{z},\) where \(A\) is the area and \(\mathrm{S}\) is the number of species. \(C\) is a constant and \(z\) is a second, biologically interesting constant that depends on the group of organisms (birds, reptiles, grasses). The value of \(z\) also depends on whether the archipelago is close to source ares, as in the case of the Indonesian islands, or very remote, as with Hawaii \(\cdots\) It ranges among faunas and floras around the world from about 0.15 to 0.35 ." Discuss this statement as a potential Mathematical Model.
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