Problem 4 - Advanced Engineering Mathematics

Question

Note that \(x=k\) is the only critical point since \(\ln (x / k)\) is not defined at \(x=0 .\) since \(g^{\prime}(x)=-k-k \ln (x / k)\) \(g^{\prime}(k)=-k<0 .\) Therefore \(x=k\) is an asymptotically stable critical point by Theorem 11.2

Step-by-Step Solution

Answer

The critical point \(x = k\) is asymptotically stable since \(g'(k) = -k < 0\).

Step 1: Understanding the Critical Point

The problem states that the only critical point of the function is at \(x = k\), since \(\ln(x/k)\) is undefined at \(x = 0\). This is because the natural logarithm \(\ln\) is not defined for zero or negative values.

Step 2: Examining the Derivative

The derivative of the function is given as \(g'(x) = -k - k \ln(x/k)\). We need to evaluate this derivative at the critical point \(x = k\). Substituting \(x = k\) into the expression, we have \(g'(k) = -k - k \ln(1) = -k\), since \(\ln(1) = 0\).

Step 3: Checking Stability

Given that \(g'(k) = -k < 0\), the critical point \(x = k\) is asymptotically stable. According to Theorem 11.2, if the derivative at a critical point is negative, the critical point is stable because any small perturbation from this point will decay back to \(k\).

Key Concepts

Asymptotic Stability

Asymptotic stability refers to a condition where, if slightly disturbed from an equilibrium point, a system will return to that point over time. In mathematical terms, a critical point is considered asymptotically stable if small deviations from this point result in the system gradually returning to it.
In the given example, we identify the critical point at \( x = k \). The derivative at this point is negative, which indicates asymptotic stability. This conclusion is supported by Theorem 11.2: if the derivative of a function at its critical point is negative, the function will naturally tend to return to this critical point when perturbed. This concept is fundamental in fields such as physics and engineering, where stability ensures system reliability and predictability.

Natural Logarithm Function

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. The function is defined for positive values of \( x \) and is a crucial element in calculus and analysis.
In this exercise, we have the expression \( \ln(x/k) \). Here, it's important to note that the natural logarithm is undefined for zero or negative values. Hence, at \( x = 0 \), \( \ln(x/k) \) becomes undefined, making \( x = k \) the only valid critical point.
Key properties of the natural logarithm include:

• \( \ln(e) = 1 \)
• \( \ln(1) = 0 \)
• The function increases slowly and without bound as \( x \) increases.

The natural logarithm helps us explore real-world phenomena like growth processes and exponential decay.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to a variable. In simple terms, the derivative provides the slope of the function at any given point and is denoted by \( f'(x) \) or \( \frac{df}{dx} \).
In our problem, the derivative \( g'(x) = -k - k \ln(x/k) \) gives us insight into the behavior of the function around its critical point \( x = k \). When we evaluated this expression at \( x = k \), the result was \( g'(k) = -k \). This negative value indicates that as \( x \) approaches \( k \), the function is decreasing, hence the system is stable.
The calculation of derivatives allows us to understand how variables interact, and is widely used in optimization, dynamics, and several practical applications.

Stability Analysis

Stability analysis is the process of determining the stability of system behavior over time. It often involves examining critical points and understanding the derivative's role in those areas.
In our example, analyzing the function around the critical point \( x = k \) involves checking if perturbations lead to a natural return to this point, which is assessed via the derivative. We found \( g'(x) = -k - k \ln(x/k) \). Checking at \( x = k \) yields \( g'(k) = -k < 0 \), indicating stability.
This technique is invaluable in science and engineering, where ensuring a system remains stable under various conditions is crucial. A thorough stability analysis helps in designing systems that are robust and less susceptible to failures.