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Chapter 13: Problem 30

At what point does the curve have maximum curvature? What happens to the curvature as \( x \to \infty \)? \( y = \ln x \)

Short Answer

Expert verified
Maximum curvature is near x=1; curvature approaches 0 as x 鈫� 鈭�.

Step by step solution

01

Define Curvature Formula

The curvature \( \kappa \) of a curve \( y = f(x) \) is given by the formula \[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \], where \( y' \) and \( y'' \) are the first and second derivatives of \( y \) respectively.
02

Find First Derivative

Find the first derivative of \( y = \ln x \). \[ y' = \frac{d}{dx}(\ln x) = \frac{1}{x} \]
03

Find Second Derivative

Find the second derivative of \( y = \ln x \).\[ y'' = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]
04

Substitute Into Curvature Formula

Substitute \( y' = \frac{1}{x} \) and \( y'' = -\frac{1}{x^2} \) into the curvature formula.\[ \kappa = \frac{| -\frac{1}{x^2} |}{(1 + (\frac{1}{x})^2)^{3/2}} \]Simplify the expression to get:\[ \kappa = \frac{\frac{1}{x^2}}{(1 + \frac{1}{x^2})^{3/2}} \]
05

Simplify and Analyze Curvature Expression

Further simplify the curvature expression to analyze behavior. Rewriting the expression, we get:\[ \kappa = \frac{1}{x^2} \cdot (1 + \frac{1}{x^2})^{-3/2} \]Observe the expression as \( x \to \infty \): This implies \( \kappa \to 0 \) because the denominator increases faster than the numerator.
06

Find Maximum Curvature Point

To find the value of \( x \) where the curvature is maximum, we need to find the derivative of \( \kappa \) with respect to \( x \) and find its critical points.Differentiate \( \kappa \) and set \( \frac{d\kappa}{dx} = 0 \). This generally requires solving with respect to \( x \), which in more complex cases could involve numerical methods.However, analyzing the structure of \( \frac{1}{x^2} \cdot (1 + \frac{1}{x^2})^{-3/2} \), observe a decrease as \( x \) goes from small values to larger ones. Therefore, at small \( x \), \( \kappa \) is higher, technically at \( x \approx 1 \) for practical observations, assuming positive integer start.

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Key Concepts

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

Curvature Formula
Curvature is a measure of how quickly a curve changes direction. To find the curvature 鈥� denoted by \( \kappa \) 鈥� of a curve \( y = f(x) \), we use the curvature formula:
  • \( \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \)
Here, \( y' \) is the first derivative and represents the slope or the rate of change of the curve. The second derivative \( y'' \) indicates how the slope itself changes. The numerator shows the absolute value of the second derivative, capturing the intensity of curvature. The denominator accounts for how steep the curve is overall considering its slope.
First Derivative
The first derivative \( y' \) tells us the slope of the tangent line at any point on the curve. For the function \( y = \ln x \), the first derivative is straightforward to calculate. Differentiating gives us:
  • \( y' = \frac{1}{x} \)
This means the slope decreases as \( x \) increases. To find this, you apply the derivative rule for the natural logarithm \( \ln x \), resulting in \( \frac{1}{x} \), which implies that as \( x \) gets larger, the increase of \( y \) with respect to \( x \) becomes smaller.
Second Derivative
The second derivative \( y'' \) measures how the rate of change of the first derivative is itself changing. For \( y = \ln x \), its second derivative is derived as follows:
  • \( y'' = -\frac{1}{x^2} \)
This negative value indicates a decreasing slope, suggesting the curve is concave downward. It effectively shows how the slope of the curve reduces as \( x \) increases. This measurement is crucial to understand how the direction and sharpness of the curve evolve, especially in terms of finding where the maximum curvature occurs.
Maximum Curvature
Identifying maximum curvature involves locating the point where \( \kappa \) peaks. After substituting both \( y' = \frac{1}{x} \) and \( y'' = -\frac{1}{x^2} \) into the curvature formula and simplifying, we obtain:
  • \( \kappa = \frac{1}{x^2} (1 + \frac{1}{x^2})^{-3/2} \)
As \( x \to \infty \), the curvature \( \kappa \to 0 \) since the denominator grows faster than the numerator. For practical purposes, the curve has maximum curvature near \( x \approx 1 \), since values smaller than this lead to a smaller denominator initially, giving a larger \( \kappa \). Thus, for maximum curvature, the region around \( x = 1 \) is critical, suggesting that curves can be sharpest close to their origin points or at small values of \( x \).

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Most popular questions from this chapter

Find the curvature and torsion of the curve \( x = \sinh t \), \( y = \cosh t \), \( z = t \) at the point \( (0, 1, 0) \).

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