Problem 80 - 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 \(x=b\) . (See accompanying figure.) Graph cannot copy

Step-by-Step Solution

Answer

The areas are equal: \( \frac{1}{a^2} \sinh(ab) \).

Step 1: Area Under the Curve

To calculate the area under the curve, we need to integrate the function given by the curve. The function is given as \( y = \frac{1}{a} \cosh(ax) \). The area \( A \) under the curve from \( x = 0 \) to \( x = b \) is given by the integral:\[A = \int_{0}^{b} \frac{1}{a} \cosh(ax) \, dx.\]

Step 2: Evaluate the Integral

To solve the integral \( \int \frac{1}{a} \cosh(ax) \, dx \), we first note that the derivative of \( \sinh(ax) \) is \( \cosh(ax) \times a \). Therefore, the integral is:\[\frac{1}{a} \int \cosh(ax) \, dx = \frac{1}{a^2} \sinh(ax) + C.\]For the definite integral from 0 to \( b \), it becomes:\[A = \frac{1}{a^2} \left[ \sinh(ab) - \sinh(0) \right] = \frac{1}{a^2} \sinh(ab).\]

Step 3: Calculate the Arc Length

The arc length \( s \) of the curve from \( x=0 \) to \( x=b \) can be found by using the formula:\[s = \int_{0}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx.\]First, find \( \frac{dy}{dx} = \sinh(ax) \), so:\[s = \int_{0}^{b} \sqrt{1 + \sinh^2(ax)} \, dx = \int_{0}^{b} \cosh(ax) \, dx.\]

Step 4: Evaluate the Arc Length Integral

Solve the integral \( \int \cosh(ax) \, dx \):\[\int \cosh(ax) \, dx = \frac{1}{a} \sinh(ax).\]Thus, the arc length from \( x=0 \) to \( x=b \) is:\[s = \frac{1}{a} \left[ \sinh(ab) - \sinh(0) \right] = \frac{1}{a} \sinh(ab).\]

Step 5: Compare Areas

The area of the rectangle is given by its height \( \frac{1}{a} \) and length \( s \). Therefore, the area of the rectangle \( A_r \) is:\[A_r = \frac{1}{a} \times s = \frac{1}{a} \times \frac{1}{a} \sinh(ab) = \frac{1}{a^2} \sinh(ab).\]Since this is equal to the area under the curve \( A \), the areas are the same.

Key Concepts

Area Under a Curve

Understanding the area under a curve is important in calculus, as it involves finding the integral of a function over a certain interval. In this exercise, we are asked to find the area under the curve given by the function
\( y = \frac{1}{a} \cosh(ax) \).
To determine this area from \( x = 0 \) to \( x = b \), we perform the integral:

• \[A = \int_{0}^{b} \frac{1}{a} \cosh(ax) \, dx.\]

The integral calculates the accumulated value beneath the curve, considering both height and width over the defined interval.

By solving this integral, we determine the total region enclosed, a fundamental process in determining areas in calculus. It is crucial for comparing this area to other geometric shapes, such as rectangles in this context.

Hyperbolic Functions

Hyperbolic functions play a similar role to the trigonometric functions, but they relate to hyperbolas rather than circles. In this exercise, the hyperbolic cosine function
\( \, \cosh(x) \, \)
comes into play, where \( y = \frac{1}{a} \cosh(ax) \).

The hyperbolic cosine function is defined as:

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

This function is important as its derivatives and integrals feature prominently in many mathematical models.

In this exercise, knowing that the derivative of \( \sinh(ax) \) is \( a \cosh(ax) \)helps in integrating and solving for area and arc length. Hyperbolic functions' definitions make calculations involving the unit hyperbola straightforward and illuminate their various applications in calculus.

Arc Length

The arc length of a curve gives the actual 'distance' along the curve, from one point to another. Calculating arc length requires evaluating an integral that includes the square root of one plus the derivative squared. For our curve defined by
\( y = \frac{1}{a} \cosh(ax) \),
we find the derivative:

• \( \frac{dy}{dx} = \sinh(ax) \)

The formula for arc length \( s \) between points \( x=0 \) and \( x=b \) is:

• \[s = \int_{0}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx.\]

Simplifying, we have:

• \[s = \int_{0}^{b} \cosh(ax) \, dx,\]

which leads us to the arc length calculation as shown in the steps. The arc length is crucial when comparing it to the linear dimensions of a rectangle, showing the depth of integral calculus in real-world measurements.

Integrals

Integral calculus is a central topic in mathematics, concerned with the accumulation of quantities, such as areas and lengths. In this exercise, integrals allow us to find the area under the curve and the arc length of the hyperbolic function.

For the function \( y = \frac{1}{a} \cosh(ax) \),
we performed the integral:

• \[ \int_{0}^{b} \frac{1}{a} \cosh(ax) \, dx = \frac{1}{a^2} \sinh(ab). \]

This computation determines the area under the curve, which we showed is equal to the area of a rectangle with dimensions derived from the arc length.

Further, integrals help in computing arc length, emphasizing their versatility:

• \[ s = \frac{1}{a} \sinh(ab), \]

where the same process of calculus tools reveals natural connections between different measures and shapes in mathematics.