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Chapter 8: Problem 47

In psychology, the WeberFechner model of stimulus-response asserts that the rate of change \(d R / d S\) of the reaction \(R\) with respect to a stimulus \(S\) is inversely proportional to the stimulus. That is, $$ \frac{d R}{d S}=\frac{k}{S}, $$ where \(k\) is some positive constant. We also assume that \(S>0\). Let \(S_{0}\) be the detection threshold value, so that \(R\left(S_{0}\right)=0 .\) Solve for \(R\) as a function of \(S\). Your answer will involve \(k\) and \(S_{0}\).

Short Answer

Expert verified
R(S) = k \ln \bigg| \frac{S}{S_0} \bigg|.

Step by step solution

01

Write Down the Given Differential Equation

The problem provides the differential equation \(\frac{d R}{d S} = \frac{k}{S}\). This equation relates the rate of change of the reaction \(R\) with respect to the stimulus \(S\).
02

Separate the Variables

To solve the differential equation, rewrite it by separating the variables \(R\) and \(S\): \(\frac{d R}{d S} = \frac{k}{S} \implies dR = \frac{k}{S} dS\).
03

Integrate Both Sides

Integrate both sides of the separated equation: \(\begin{aligned} \int dR &= \int \frac{k}{S} dS \end{aligned}\). This yields \(\begin{aligned} R &= k \ln |S| + C \end{aligned}\), where \(C\) is the constant of integration.
04

Apply the Boundary Condition

Use the boundary condition \(R(S_0) = 0\) to find the constant \(C\). Substitute \(S_0\) for \(S\) and set \(R\) to 0: \(\begin{aligned} 0 &= k \ln |S_0| + C \end{aligned}\). Solving for \(C\) gives \(\begin{aligned} C &= -k \ln |S_0| \end{aligned}\).
05

Substitute Back the Constant

Put the constant \(C\) back into the expression for \(R\): \(\begin{aligned} R &= k \ln |S| - k \ln |S_0| \end{aligned}\).
06

Simplify the Expression

Combine the logarithmic terms: \(\begin{aligned} R &= k \ln \bigg| \frac{S}{S_0} \bigg| \end{aligned}\). Thus, the function \(R(S)\) is given by \(\begin{aligned} R(S) &= k \ln \bigg| \frac{S}{S_0} \bigg| \end{aligned}\).

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Key Concepts

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

Differential Equations
A differential equation is an equation that involves an unknown function and its derivatives. In the exercise, we are given the differential equation: \(\frac{d R}{d S} = \frac{k}{S}\).
It describes how the reaction \(R\) changes with respect to the stimulus \(S\). Differential equations are crucial in modeling real-world situations where rates of change are involved.
To solve this differential equation, we follow systematic steps: separating variables, integrating both sides, and applying boundary conditions. This method helps us find a specific solution based on the given conditions.
Rate of Change
The rate of change tells us how a quantity changes over another quantity. Here, the rate of change of reaction \(R\) to stimulus \(S\) is described by \(\frac{d R}{d S}\).
The proportionality constant \(k\) indicates how strongly the reaction changes in response to the stimulus. The higher the value of \(k\), the bigger the change in \(R\) for a given change in \(S\).
This model, known as the Weber-Fechner law, is widely used in psychology to describe sensory responses to stimuli.
Logarithmic Function
A logarithmic function is one that involves the logarithm of a variable. In our solution, we see that \(\frac{d R}{d S} = \frac{k}{S}\) leads to an integral that involves a logarithm: \(\begin{aligned} R &= k \ln|S| + C \end{aligned}\).
Logarithmic relationships are common in many natural processes, including growth rates and sensory perception. They allow for modeling relationships where changes have a diminishing effect as the input grows.
The term \[ \ln| \frac{S}{S_0} | \] shows how the reaction \(R\) depends logarithmically on the ratio of \(S\) to the threshold \(S_0\).
Integration
Integration is the process of finding the antiderivative, or the original function, from its derivative. In solving our differential equation, we integrated both sides to find \(R\): \[\begin{aligned} \int dR &= \int \frac{k}{S} dS \end{aligned}\].
By performing the integration, we get: \[\begin{aligned} R &= k \ln |S| + C \end{aligned}\]. Integrating helps us to determine the relationship between the quantities involved in the differential equation.
Boundary conditions like \(R(S_0) = 0\) are used to find the constant of integration \(C\), providing a complete, precise solution to the problem at hand.

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Most popular questions from this chapter

Solve the differential equation. Be sure to check for possible constant solutions. If necessary, write your answer implicitly. $$ \frac{d y}{d t}=t^{2} e^{t^{3}} y^{3} $$

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Let \(y(t)\) be the proportion of crystallizable fat in a sample after \(t\) hours. Then \(y\) satisfies the differential equation $$y^{\prime}=k\left(y^{n}-y\right).$$ where \(k\) is a constant and \(n\) is the Avrami exponent for decrystallization reactions. In practice, \(n\) is an integer greater than 1 that is computed from the time dependence of nucleation and the number of dimensions in which crystal growth occurs. \({ }^{15}\) Use this differential equation to solve Exercises \(29-32\). Although this differential equation for \(y\) is nonlinear, we will solve it using Theorem 3 after an initial modification. a) Divide both sides of the differential equation by \(y^{n} .\) Show that $$y^{-n} \frac{d y}{d t}+k y^{1-n}=k.$$ b) Let \(z(t)=[y(t)]^{1-n} .\) Use the Chain Rule to find \(z^{\prime}\) in terms of \(y\) and \(y^{\prime}\). c) Use the result of part (b) to show that $$\frac{d z}{d t}+k(1-n) z=k(1-n).$$ d) Find the general solution for \(z(t)\). e) Find the general solution for \(y(t)\). f) Suppose that the initial condition is \(y(0)=p\), where \(p\) is the initial proportion of crystallizable fat. Show that $$ y(t)=\left[1+\left(p^{1-n}-1\right) e^{(n-1) k t}\right]^{1 /(1-n)}. $$

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