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Chapter 27: Problem 41

Solve the given problems by finding the appropriate derivative. The speed \(s\) of signaling by use of a certain communications cable is directly proportional to \(x^{2} \ln x^{-1},\) where \(x\) is the ratio of the radius of the core of the cable to the thickness of the surrounding insulation. For what value of \(x\) is \(s\) a maximum?

Short Answer

Expert verified
The value of \( x \) that maximizes \( s \) is \( \frac{1}{\sqrt{e}} \).

Step by step solution

01

Understand the Problem

We need to find the value of \( x \) that maximizes the speed \( s \). Since \( s \) is directly proportional to \( x^2 \ln{x^{-1}} \), we can express \( s \) as \( s = kx^2 \ln{x^{-1}} \), where \( k \) is a proportionality constant.
02

Set up the Function

To maximize \( s \), we start by setting up the function \( f(x) = x^2 \ln{x^{-1}} \), which simplifies to \( f(x) = x^2(-\ln{x}) = -x^2 \ln{x} \). We will first differentiate this function with respect to \( x \).
03

Differentiate the Function

Find the derivative \( f'(x) \) using the product rule. Let \( u = x^2 \) and \( v = \ln{x^{-1}} \), giving \( v = -\ln{x} \). The derivative is \( f'(x) = u'v + uv' \). Compute \( u' = 2x \) and \( v' = -\frac{1}{x} \). Thus, \( f'(x) = 2x(-\ln{x}) + x^2\left(-\frac{1}{x}\right) \).
04

Simplify the Derivative

Simplify the expression found in Step 3: \( f'(x) = -2x \ln{x} - x \). This results from combining like terms \( 2x(-\ln{x}) + (-x) \).
05

Find Critical Points

To find the critical points, set \( f'(x) = 0 \), which gives us \(-2x \ln{x} - x = 0 \). Factor out \( x \), resulting in \( x(-2 \ln{x} - 1) = 0 \). Since \( x = 0 \) is not meaningful in the context, solve \(-2 \ln{x} - 1 = 0 \).
06

Solve for \( x \)

From \(-2 \ln{x} - 1 = 0\), we have \(-2 \ln{x} = 1 \). Solving for \( \ln{x} \), we get \( \ln{x} = -\frac{1}{2} \). Raise \( e \) to both sides to solve for \( x \): \( x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \).
07

Verify Maximum

Confirm that this critical point results in a maximum value by using the second derivative test. Compute \( f''(x) \) and evaluate it at \( x = \frac{1}{\sqrt{e}} \). For simplicity, examine the behavior of \( f'(x) \) around this point to ensure it goes from positive to negative.

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Key Concepts

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

Critical Points
Critical points are important in calculus because they help us find where a function has a maximum or minimum value. A critical point occurs where the derivative of a function is zero or undefined. This often indicates a peak (maximum) or trough (minimum) in a curve. In the context of the provided problem, we found a critical point by first setting up the derivative function
  • We derived \(-2x \ln{x} - x = 0\) from the prime function.
  • We factor the equation, revealing \(x(-2 \ln{x} - 1) = 0\).
  • This process is essential to determine potential values of \(x\) where the function's behavior changes.
Once we have this result, we discard \(x = 0\) as it doesn't make sense for the problem, leaving us to explore further solutions to \(-2 \ln{x} - 1 = 0\) for locating the critical point.
Maximum Value
Finding the maximum value of a function is often the ultimate goal in optimization problems. Once we locate critical points, we need to determine whether they correspond to maximum or minimum values. In this problem:
  • We solve \(-2 \ln{x} - 1 = 0\) finding the solution \(\ln{x} = -\frac{1}{2}\).
  • By exponentiating, we find \(x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}\).
  • To check if this corresponds to a maximum value, we observe changes in the slope around the critical point, noting if the derivative \(f'(x)\) transitions from positive to negative.
This check confirms whether \(x = \frac{1}{\sqrt{e}}\) gives a maximum speed \(s\), simply by verifying the behavior of the function near this value.
Product Rule
The product rule is a vital differentiation technique used when finding the derivative of a function that's a product of two or more functions. In simplistic terms, for two functions \(u\) and \(v\), the derivative of their product is given by \(u'v + uv'\).
  • In our problem, \(u = x^2\) and \(v = \ln{x^{-1}}\), simplifies to \(-\ln{x}\).
  • Then, we calculated derivatives: \(u' = 2x\) and \(v' = -\frac{1}{x}\).
  • The product rule is applied to formulate \(f'(x) = u'v + uv' = 2x(-\ln{x}) + x^2(-\frac{1}{x})\).
This method helps us assemble the derivative necessary to locate critical points.
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is a fundamental function in mathematics that uses the base \(e\). It has unique properties that make it useful for various calculus applications. In this problem:
  • The function involves \(\ln{x^{-1}}\), simplifying to \(-\ln{x}\).
  • Natural logarithms transform multiplicative processes into additive ones, useful for differentiation.
  • Solving \(-2 \ln{x} - 1 = 0\), we use properties of logarithms to find \(x\): write \(\ln{x} = -\frac{1}{2}\) and solve \(x = e^{-\frac{1}{2}}\).
Understanding natural logs is crucial in unraveling equations for smooth transformations and optimizations.

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Most popular questions from this chapter

Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{x \rightarrow 0} \frac{\sqrt{1-x}-\sqrt{1+x}}{x}$$

Solve the given problems. Show that \(y=e^{-x} \sin x\) satisfies the equation \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0\).

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