Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the point of the curve at which the curvature is a maximum. \(y=\cosh x\)

Step-by-Step Solution

Answer

The point of maximum curvature is (0, 1).

Step 1: Understand the Curvature Formula

The curvature \( \kappa \) of a curve given by a function \( y(x) \) is defined as \( \kappa = \frac{|y''(x)|}{(1 + (y'(x))^2)^{3/2}} \). We need to compute the first and second derivatives of \( y = \cosh x \) to use this formula.

Step 2: Compute the First Derivative

The derivative of \( y = \cosh x \) is \( y'(x) = \sinh x \).

Step 3: Compute the Second Derivative

The derivative of \( y'(x) = \sinh x \) is \( y''(x) = \cosh x \).

Step 4: Write the Expression for Curvature

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

Step 5: Simplify the Expression

Recognize that \( 1 + (\sinh x)^2 = \cosh^2 x \) and simplify the expression for curvature to \( \kappa = \frac{\cosh x}{(\cosh^3 x)} = \frac{1}{\cosh^2 x} \).

Step 6: Find the Maximum Curvature

Since \( \cosh^2 x \) is a positive increasing function, \( \frac{1}{\cosh^2 x} \) is a decreasing function. The maximum curvature occurs at \( x = 0 \), where \( \cosh(0) = 1\), giving maximum curvature \( \kappa = 1 \).

Step 7: Identify the Point of Maximum Curvature

The point on the curve with maximum curvature is at \( x = 0 \). Substituting back into the original equation \( y = \cosh x \), we find \( y(0) = 1 \). Thus, the point is \( (0, 1) \).

Key Concepts

Curvature of a Curve

Curvature is a measure of how sharply a curve bends at any given point. For a curve described by the equation \( y = f(x) \), the curvature \( \kappa \) is determined by the formula:\[\kappa = \frac{|y''(x)|}{(1 + (y'(x))^2)^{3/2}}\]In this context, the first derivative \( y'(x) \) tells us how the slope of the curve changes, and the second derivative \( y''(x) \) tells us how the slope itself is changing, which is related to the acceleration of the curve. The function \( 1 + (y'(x))^2 \) forms the basis for the normalization term, ensuring that curvature is dimensionless and suitable for geometric interpretation.

• Importance of Curvature: In practical terms, roads, roller coasters, and natural phenomena like rivers need such analysis for structural and safety planning.
• Calculation Focus: For any given point \( x \) on a curve, finding the curvature involves evaluating these derivatives and using the curvature formula.

Hyperbolic Functions

Hyperbolic functions, like \( \cosh(x) \) and \( \sinh(x) \), are analogs of trigonometric functions. They arise naturally in the study of calculus and have unique properties that relate to exponential functions:

• For \( y = \cosh(x) \), the function can also be expressed as \( \frac{e^x + e^{-x}}{2} \).
• The derivative \( \sinh(x) \) is given by \( \frac{e^x - e^{-x}}{2} \), similar but distinct from its cosine counterpart.

Hyperbolic functions are essential in defining curves in hyperbolic geometry and are significant in various fields such as engineering and physics.
Their shape is related to a hyperbola, hence these names, and their graphs show symmetric, smooth lines reflecting this geometric trait. The calculation of curvature focuses mainly on identifying and utilizing these derivatives properly to understand how sharply the graph of \( \cosh(x) \) bends.

Derivatives

Derivatives represent the rate of change of a function and are foundational to calculus. For a function \( y = f(x) \), the first derivative \( y'(x) \) indicates how the function is changing at a specific point. For hyperbolic functions like \( y = \cosh(x) \):

• The first derivative \( y'(x) = \sinh(x) \) shows the slope of the tangent to the curve at any point.
• The second derivative \( y''(x) = \cosh(x) \) tells us how this slope is changing, leading to insights about the curvature of the curve.

Calculating derivatives of hyperbolic functions follows the same rules as derivatives of simpler functions, with a focus on the exponential forms underlying \( \cosh \) and \( \sinh \). These insights help solve real-world problems involving rates of change, such as predicting the trajectory of objects or optimizing architectural structures.