Problem 7 - Calculus with Concepts in Calculus

Question

In a transatlantic cable let \(x\) be the ratio of the radius of the core to the thickness of the covering. It can be shown that the speed \(s\) of a signal passing through the cable is given by $$ s=-k x^{2} \ln x \quad \text { for } 0

Step-by-Step Solution

Answer

The value of \(x\) that maximizes \(s\) is \(e^{-\frac{1}{2}}\).

Step 1: Identify the Given Function

We are given the function for speed, \(s = -k x^2 \ln x\), where \(x\) is the ratio of the core's radius to the thickness of the covering and \(k\) is a positive constant. Our task is to find the value of \(x\) that maximizes \(s\).

Step 2: Determine the Derivative of the Function

To find the maximum value, we need to calculate the derivative \(\frac{ds}{dx}\) of the function \(s(x) = -k x^2 \ln x\). We'll apply the product rule and the chain rule for differentiation.

Step 3: Apply Product Rule and Differentiate

Using the product rule \((uv)' = u'v + uv'\), we calculate the derivative: set \(u = x^2\) and \(v = \ln x\). Then \(u' = 2x\) and \(v' = \frac{1}{x}\). The derivative is \(\frac{ds}{dx} = -k\left(2x \ln x + x\right)\).

Step 4: Solve for Critical Points

Set \(\frac{ds}{dx} = 0\) to find critical points: \(-k(2x \ln x + x) = 0\). Simplify the expression and solve for \(x\) to get \(2x \ln x + x = 0\) or \(x(2 \ln x + 1) = 0\). The critical point occurs at \(2 \ln x + 1 = 0\).

Step 5: Solve for \(x\)

From \(2 \ln x + 1 = 0\), solve for \(x\): \(2 \ln x = -1\), which gives \(\ln x = -\frac{1}{2}\). Exponentiate both sides: \(x = e^{-\frac{1}{2}}\).

Step 6: Check Concavity for Maximization

To confirm that \(x = e^{-\frac{1}{2}}\) is a maximum, check the second derivative \(\frac{d^2s}{dx^2}\). If \(\frac{d^2s}{dx^2} < 0\) at \(x = e^{-\frac{1}{2}}\), then \(s\) has a local maximum at this point.

Key Concepts

Critical Points

In the realm of calculus, critical points are pivotal in analyzing the behavior of functions. Critical points occur when the derivative of a function is either zero or undefined. These points help to identify where a function might have a local maximum or minimum. This is crucial, especially in optimization problems, where our goal is to find points that maximize or minimize a particular function.

To find critical points in our exercise, we set the derivative of the speed function \(s = -k x^2 \ln x\) to zero. This involves differentiation and solving the resulting equation. Through this process, we identify potential values of \( x\) that could maximize the speed. The equation detailing where the critical points exist is formed as: \(2x \ln x + x = 0\). Thus, critical points are not just mathematical curiosities; they serve as indicators of important changes in the function's behavior.

Differentiation Techniques

Differentiation is a powerful calculus technique used to find the rate at which one quantity changes with respect to another. It's essential for identifying critical points. In our problem, we start by finding the derivative of the given function \(s = -k x^2 \ln x\). For functions composed of multiple parts, like ours, we often employ both the product rule and chain rule.

• Product Rule: This is used when differentiating the product of two functions. For \(s\), consider \(u = x^2\) and \(v = \ln x\). The product rule states that \((uv)' = u'v + uv'\).
• Chain Rule: This assists in differentiating composite functions. It will often work in tandem with the product rule, especially for the \(\ln x\) component of the function.

Applying these techniques yields the derivative, \(\frac{ds}{dx} = -k(2x \ln x + x)\), which is the linchpin in finding our critical points. Understanding and practicing these differentiation techniques can significantly simplify solving complex calculus problems.

Second Derivative Test

Once critical points are identified, the second derivative test steps in to determine the nature of these points. A critical point could be a local maximum, local minimum, or saddle point. The second derivative test evaluates the concavity of the function at these points.

The test itself is pretty straightforward:

• Compute the second derivative, \(\frac{d^2s}{dx^2}\).
• Evaluate this second derivative at the critical point.
• If \(\frac{d^2s}{dx^2} < 0\), the function is concave down, indicating a local maximum.
• If \(\frac{d^2s}{dx^2} > 0\), the function is concave up, suggesting a local minimum.
• If \(\frac{d^2s}{dx^2} = 0\), the test is inconclusive, and further investigation is needed.

For the given cable signal speed function, after finding the critical point \(x = e^{-\frac{1}{2}}\), we then take the second derivative. Checking this derivative shows if our critical point is indeed a point of maximum speed for signal transmission. Using the second derivative test helps confirm whether optimizations made were successful, providing more mathematical certainty to our conclusions.