Question

, find the point of the curve at which the curvature is a maximum. $$ y=\sinh x $$

Short answer

Maximum curvature occurs at the point where \(x = 0\) on the curve \(y = \sinh x\).

Step-by-step solution

Understand the Problem

We need to find the point on the curve defined by the equation \(y = \sinh x\) where its curvature is maximum. Curvature, in general, measures how rapidly a curve changes direction.

Recall the Curvature Formula

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

Find First Derivative \(y'\)

Given \(y = \sinh x\), the first derivative is calculated using the derivative of hyperbolic sine:\[y' = \frac{d}{dx}(\sinh x) = \cosh x\]

Find Second Derivative \(y''\)

The second derivative is found by differentiating \(y' = \cosh x\):\[y'' = \frac{d}{dx}(\cosh x) = \sinh x\]

Substitute Derivatives into Curvature Formula

Substitute \(y'\) and \(y''\) into the curvature formula:\[\kappa = \frac{|\sinh x|}{(1 + (\cosh x)^2)^{3/2}}\]

Simplify the Denominator

Using the hyperbolic identity \((\cosh x)^2 - (\sinh x)^2 = 1\), simplify the denominator:\[1 + (\cosh x)^2 = 2(\cosh x)^2 - 1\]So, \[(1 + (\cosh x)^2)^{3/2} = ((2(\cosh x)^2 - 1)^{3/2}\]

Analyze the Expression for Maximum Curvature

To find the maximum, examine the behavior of: \[\kappa = \frac{|\sinh x|}{((2(\cosh x)^2 - 1)^{3/2})\]By analyzing critical points or through symmetry, deduce that maximum curvature occurs when \(|\sinh x|\) is maximized relative to \(\cosh x\).

Conclusion

In hyperbolic functions, at \(x = 0\), \(\sinh x\) is 0 and has significant relative significance due to symmetry and function characteristics. Thus, the maximum curvature occurs at \(x = 0\).

Key concepts

Curvature

In calculus, curvature refers to how sharply a curve bends. It provides a way to understand the geometry of curves by evaluating the rate at which a tangent line to the curve changes direction. This is crucial when analyzing the shape and behavior of different curves.
To calculate the curvature, you need the first and second derivatives of the curve's function. The general formula for curvature, denoted as \(\kappa\), is:

• \(\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\)

Here:

• \(y'\) is the first derivative, indicating the slope or the rate of change of the function.
• \(y''\) is the second derivative, showing how the rate of change itself changes over the curve.

The larger the curvature, the sharper the bend. Curves with constant curvature are actually circles, where the curvature is simply the reciprocal of the radius.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola. The most commonly used are hyperbolic sine (\(\sinh\)) and hyperbolic cosine (\(\cosh\)). These functions are essential in calculus for expressing relationships in both algebraic and geometric forms.
The hyperbolic sine function (\(\sinh x\)) is defined as:

• \(\sinh x = \frac{e^x - e^{-x}}{2}\)

While the hyperbolic cosine function (\(\cosh x\)) is:

• \(\cosh x = \frac{e^x + e^{-x}}{2}\)

These functions often appear in the solutions to differential equations and in the hyperbolic identity:

• \((\cosh x)^2 - (\sinh x)^2 = 1\)

This identity is valuable for simplifying calculations, including those involving curvature as seen in the exercise.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. They provide critical information about the behavior and properties of functions, such as slopes, tangents, and curvature.
For hyperbolic functions like \(y = \sinh x\):

• The first derivative \(y'\) is \(\cosh x\), showing how \(\sinh x\) grows as \(x\) increases.

Differentiation gives us information about the function's velocity, or rate of change, along the curve.
The second derivative, \(y''\), is \(\sinh x\) again, indicating acceleration or how the slope is changing. In the context of the given problem, it is used to find curvature, highlighting how these mathematical tools interconnect.
Understanding the relationships between a function and its derivatives through calculus gives us deeper insights into the mathematics underlying physical phenomena and geometric structures.