Problem 85 - Thomas Calculus

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 are equal as both are \(\frac{1}{a^2}\sinh(ab)\).

Step 1: Understand the Problem

We are asked to demonstrate that the area under the curve \(y = \frac{1}{a} \cosh(ax)\) from \(x=0\) to \(x=b\) in the first quadrant is equal to the area of a rectangle with a known height and computed length. Additionally, we need to draw a figure to illustrate this relationship.

Step 2: Set Up the Area Integral

The area \(A\) under the curve \(y = \frac{1}{a} \cosh(ax)\) from \(x=0\) to \(x=b\) can be calculated using the integral formula:\[A = \int_{0}^{b} \frac{1}{a} \cosh(ax) \, dx\]

Step 3: Evaluate the Integral

Evaluate the integral \( \int \cosh(ax) \, dx \). We know that the integral of \(\cosh(ax)\) is \(\frac{1}{a}\sinh(ax)\). Therefore:\[A = \frac{1}{a} \left[ \frac{1}{a} \sinh(ax) \right]_0^b = \frac{1}{a^2} \left( \sinh(ab) - \sinh(0)\right) = \frac{1}{a^2} \sinh(ab)\]

Step 4: Set Up the Arc Length Integral

The length \(s\) of the curve \(y = \frac{1}{a} \cosh(ax)\) from \(x=0\) to \(x=b\) is given by the arc length formula:\[s = \int_{0}^{b} \sqrt{1 + \left(\frac{d}{dx}\left(\frac{1}{a} \cosh(ax)\right)\right)^2} \, dx\] Compute \( \frac{d}{dx}\left(\frac{1}{a} \cosh(ax)\right) = \sinh(ax)\). Hence:\[s = \int_{0}^{b} \sqrt{1 + \sinh^2(ax)} \, dx = \int_{0}^{b} \cosh(ax) \, dx\]

Step 5: Evaluate the Arc Length Integral

Evaluate the integral \( \int \cosh(ax) \, dx \), which we've already found to be \(\frac{1}{a}\sinh(ax)\). Thus, for the arc length:\[s = \left[ \frac{1}{a} \sinh(ax) \right]_0^b = \frac{1}{a} \sinh(ab)\]

Step 6: Relate the Area and Rectangle

The area of the rectangle with height \(\frac{1}{a}\) and length \(s\) is:\[\text{Area} = \frac{1}{a} \times \frac{1}{a} \sinh(ab) = \frac{1}{a^2} \sinh(ab)\]This matches the area under the curve found in Step 3, confirming that the areas are equal.

Step 7: Draw the Figure

Sketch the graph of the function \(y = \frac{1}{a} \cosh(ax)\) in the first quadrant from \(x=0\) to \(x=b\). Add a rectangle with height \(\frac{1}{a}\) and the same area, showing the relationship between the curve and the rectangle.

Key Concepts

Hyperbolic functions

Hyperbolic functions are mathematical functions that resemble trigonometric functions but are associated with hyperbolas instead of circles. Just like how sine and cosine relate to the unit circle, hyperbolic sine ( \( \sinh \)) and hyperbolic cosine ( \( \cosh \)) relate to the hyperbola. In the context of this exercise, we are dealing with the hyperbolic cosine function: \( y = \frac{1}{a} \cosh(ax) \).
This particular function describes a curve that opens upward and is symmetrical about the y-axis. Hyperbolic functions are utilized in various branches of mathematics and physics, especially in scenarios involving growth processes, wave motion, and relativity theory. They have significant applications due to their unique properties and relationships, such as:

• Identity: \( \cosh^2(x) - \sinh^2(x) = 1 \).
• Derivative: The derivative of \( \cosh(x) \) is \( \sinh(x) \).
• Integral: The integral of \( \cosh(x) \) is \( \sinh(x) \).

In solving this exercise, understanding the properties and behaviors of hyperbolic functions is crucial as they directly relate to both area and arc length calculations.

Integral calculus

Integral calculus is a branch of mathematics that deals with integrals and their properties. It plays a crucial role in calculating areas under curves, solving differential equations, and finding other quantities that accumulate over time or space. In this exercise, we are using integral calculus to find the area under the curve defined by \( y = \frac{1}{a} \cosh(ax) \) from \( x=0 \) to \( x=b \). This involves setting up an integral: \[ A = \int_{0}^{b} \frac{1}{a} \cosh(ax) \, dx \].
This definite integral provides the precise area beneath the curve over a specified interval. Integral calculus is essential for such calculations because it sums infinitely small quantities over a continuous interval, effectively providing an accumulation of space under a curve. When evaluating the integral, knowledge of hyperbolic function integrals simplifies the process significantly, as we know the integral of \( \cosh(ax) \) is \( \frac{1}{a} \sinh(ax) \). By applying these integration techniques, we obtain the area under the curve, demonstrating the power of integral calculus in mathematical problem-solving.

Arc length

The arc length is the distance along a curve from one point to another. In this exercise, we calculate the length \( s \) of the curve given by \( y = \frac{1}{a} \cosh(ax) \) from \( x=0 \) to \( x=b \). For a differentiable curve, the formula for arc length is: \[ s = \int_{0}^{b} \sqrt{1 + \left( \frac{d}{dx} \left( \frac{1}{a} \cosh(ax) \right) \right)^2} \, dx \].
Here, we calculate the derivative \( \sinh(ax) \) and use it in the arc length formula, simplifying: \[ s = \int_{0}^{b} \cosh(ax) \, dx \].
Understanding arc length is critical for accurately measuring the distance along curves, which is useful in numerous fields such as physics, engineering, and computer graphics. The beauty arises when solving our specific integral for arc length, as it results in a value directly related to the physical dimension of the area (as calculated previously) enclosed by the curve. It neatly mirrors the mathematical elegance and interconnected nature of different calculus concepts.

Definite integral

A definite integral is a fundamental concept in calculus that represents the accumulation of quantities, such as area, over a particular interval. It is typically denoted as \( \int_{a}^{b} f(x) \, dx \), indicating the sum of infinitesimally small areas under the curve \( f(x) \) from \( x=a \) to \( x=b \). In this problem, the definite integral helps calculate the area \( A \) beneath the hyperbolic function curve: \[ A = \int_{0}^{b} \frac{1}{a} \cosh(ax) \, dx = \frac{1}{a^2} \sinh(ab) \].
Definite integrals are crucial for many practical applications beyond mathematical theory, such as calculating net change, determining volumes, and analyzing economic data. The beauty of using definite integrals in calculating areas lies in their ability to translate dynamic curves into precise, manageable numbers. This exercise showcases how definite integrals serve not only to compute an area but also play a vital role in revealing geometric relationships between curves and rectangular regions—demonstrating the essential, harmonious role these integrals play in mathematical analysis and application.