Problem 85 - Thomas Calculus in SI Units

Question

Area Show that the area of the region in the first quadrant enclosed by the curve \(y=(1 / a) \cosh a x,\) the coordinate axes, and the line \(x=b\) is the same as the area of a rectangle of height \(1 / a\) and length \(s,\) where \(s\) is the length of the curve from \(x=0\) to \(x=b .\) Draw a figure illustrating this result.

Step-by-Step Solution

Answer

The areas of the region under the curve and the rectangle are both \(\frac{1}{a^2} \sinh(ab)\).

Step 1: Identify the Area of the Region

To find the area of the region enclosed by the curve, the x-axis, and the line \(x=b\), we need to integrate the given function \(y=\frac{1}{a} \cosh(ax)\) from \(x=0\) to \(x=b\). The formula for the area under a curve from \(x=0\) to \(x=b\) is \(\int_0^b y\, dx\), which becomes \(\int_0^b \frac{1}{a} \cosh(ax)\, dx\).

Step 2: Evaluate the Integral for the Area

Evaluate the integral \(\int_0^b \frac{1}{a} \cosh(ax)\, dx\). The antiderivative of \(\cosh(ax)\) is \(\frac{1}{a} \sinh(ax)\). Thus, the integral becomes:\(\frac{1}{a} \int_0^b \cosh(ax)\, dx = \frac{1}{a^2} \sinh(ax) \bigg|_0^b = \frac{1}{a^2} \left[ \sinh(ab) - \sinh(0) \right] = \frac{1}{a^2} \sinh(ab).\)So, the area is \(\frac{1}{a^2} \sinh(ab)\).

Step 3: Calculate the Length of the Curve (s)

The length of the curve from \(x=0\) to \(x=b\) is denoted as \(s\) and can be found using the arc length formula for a curve \(y=f(x)\): \(s = \int_0^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\). Given \(y=\frac{1}{a}\cosh(ax)\), we find \(\frac{dy}{dx} = \sinh(ax)\). Thus, the length becomes:\(\int_0^b \sqrt{1 + \sinh^2(ax)} \, dx = \int_0^b \cosh(ax) \, dx = \frac{1}{a} \sinh(ax) \bigg|_0^b = \frac{1}{a} \sinh(ab).\)Hence, \(s = \frac{1}{a} \sinh(ab)\).

Step 4: Relate Area to Rectangle Dimensions

The problem states that the area under the curve is equal to the area of a rectangle with height \(\frac{1}{a}\) and length \(s\). The area of this rectangle would be \(\frac{1}{a} \times s = \frac{1}{a} \times \frac{1}{a} \sinh(ab) = \frac{1}{a^2} \sinh(ab)\).We've already calculated that the area under the curve is also \(\frac{1}{a^2} \sinh(ab)\).

Step 5: Illustrate the Areas

Draw a graph showing the region enclosed by the curve \(y=\frac{1}{a} \cosh(ax)\), the x-axis, and the vertical line at \(x=b\). Next to it, draw a rectangle with height \(\frac{1}{a}\) and length \(s\). Label both areas as \(\frac{1}{a^2} \sinh(ab)\) to illustrate they are equal.

Key Concepts

Integration

Integration is a fundamental concept in calculus used to find areas under curves, among other applications. It involves summing an infinite number of infinitesimally small quantities. When we talk about the area under a curve, integration allows us to calculate this by considering the function that describes the curve.

For the given problem, we are dealing with a hyperbolic function, specifically, the curve given by the equation \(y = \frac{1}{a} \cosh(ax)\). To find the area under this curve from \(x = 0\) to \(x = b\), we use the integral formula:

• The integral from \(0\) to \(b\) of \(y\,dx\) equals \(\int_0^b \frac{1}{a} \cosh(ax)\, dx\).

This integral represents the sum of all the little vertical slices under the curve from \(0\) to \(b\) along the \(x\)-axis. Completing this integration process gives us the total area enclosed by the curve, the x-axis, and the vertical line at \(x = b\). This method is crucial for calculating geometric properties of curves.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. They include the hyperbolic sine \(\sinh\), hyperbolic cosine \(\cosh\), and others. These functions frequently appear in calculations involving exponential growth and certain curve equations in mathematics and physics.

In the exercise, we encounter the function \(\cosh(ax)\), where \(\cosh(x)\) is defined as \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). It describes a smooth curve that is always positive and symmetric about the y-axis. This function is used to define the curve whose area we calculate.

• The antiderivative of \(\cosh(ax)\) is \(\frac{1}{a} \sinh(ax)\).

This property allows us to solve the integration and find the area under the curve. Understanding hyperbolic functions aids in solving various problems in calculus involving growth or curve areas.

Arc Length

In geometry, arc length is the distance along a curve between two points. It's a bit different from linear distance because it accounts for the curvature of the path. To find the arc length of a given curve, we use the arc length formula, which integrates the curve function's rate of change over its interval.

For the curve \(y = \frac{1}{a}\cosh(ax)\), the derivative \(\frac{dy}{dx}\) is given by \(\sinh(ax)\). To find the arc length \(s\) from \(x = 0\) to \(x = b\), we use this formula:

• \(s = \int_0^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)
• This simplifies for hyperbolic functions because \(\cosh^2(x) - \sinh^2(x) = 1\).
• The calculation becomes \(\int_0^b \cosh(ax) \, dx\), resulting in \(\frac{1}{a} \sinh(ab)\) for the curve's arc length.

This process of finding an arc length is critical in many areas of mathematics and engineering, where understanding actual path length is required.

First Quadrant

The first quadrant is the section of a Cartesian coordinate system where both \(x\) and \(y\) values are positive. It is of great importance in graphing because many problems, especially in applied mathematics, only consider positive values as they relate to real-world scenarios, such as distance, area, or quantity.

In this problem, the area of interest is specifically in the first quadrant because we are working with positive \(x\) and \(y\) values bounded by the axes and the line \(x = b\). The concept simplifies the problem to considering only these positive values, making the calculations concentrative solely on the meaningful section of the curve.

Function graphs that are open upward, like \(\cosh(x)\), naturally extend into the first quadrant. This is often where real-world data is graphed, focusing on non-negative results. Working within the first quadrant is a common approach as it deals with actual measurable and observable phenomena.