Question

Find the area of the hyperbolic quarter-circle enclosed by \(x=2\) and \(y=2\) above \(y=1 / x\). [T] Find the arc length of \(y=1 / x\) from \(x=1\) to \(x=4\).

Short answer

The area is \(2 - \ln 2\). The arc length is typically computed numerically.

Step-by-step solution

Understand the Problem

We need to find two things: the area of the region bounded by the lines \(x=2\), \(y=2\), and the curve \(y=\frac{1}{x}\) above \(y=1/x\). We also want to find the arc length of the curve \(y=\frac{1}{x}\) from \(x=1\) to \(x=4\).

Identify the Bounds for the Area

To find the area, determine the bounds: the left boundary is \(x=1\), as the curve \(y=\frac{1}{x}\) intersects the line \(y=2\) at \(x=\frac{1}{2}\), but within the constraint \(x=2\), so the right boundary is \(x=2\). The top boundary is \(y=2\) and the bottom boundary is \(y=\frac{1}{x}\).

Calculate the Area Using Integration

The area under the curve \(y=\frac{1}{x}\) from \(x=1\) to \(x=2\) is given by: \[ A = \int_{1}^{2} \left(2 - \frac{1}{x}\right) \: dx \]Evaluate the integral as follows:\[ A = \int_{1}^{2} 2\:dx - \int_{1}^{2} \frac{1}{x}\:dx \]First integral: \(\int_{1}^{2} 2\:dx = 2x \bigg|_1^2 = 4 - 2 = 2\).Second integral: \(\int_{1}^{2} \frac{1}{x}\:dx = \ln|x| \,\bigg|_1^2 = \ln 2\).Thus, the area is \(2 - \ln 2\).

Set Up the Arc Length Formula

The arc length \(L\) of a curve \(y=f(x)\) from \(x=a\) to \(x=b\) is given by:\[ L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \]For the curve \(y=\frac{1}{x}\), the derivative \(f'(x) = -\frac{1}{x^2}\).

Calculate the Arc Length

Use the formula from the previous step to calculate the arc length:\[ L = \int_1^4 \sqrt{1 + \left(-\frac{1}{x^2}\right)^2} \, dx = \int_1^4 \sqrt{1 + \frac{1}{x^4}} \, dx \]This integral does not have a simple antiderivative and will typically be evaluated using numerical methods or definite integrals.

Key concepts

Arc length calculation

When dealing with curves, a common task is to calculate their arc length. For a given function, the arc length from one point to another is the total distance along the curve. This calculation involves integration and typically requires finding the derivative of the function.
In general, the arc length \(L\) of a function \(y = f(x)\) over an interval \([a, b]\) can be computed using the formula:

• \[ L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \]

For our specific problem with \(y = \frac{1}{x}\), the derivative \(f'(x)\) is \(-\frac{1}{x^2}\).
When you plug this derivative into the arc length formula, you end up needing to evaluate an integral with \(\sqrt{1 + \frac{1}{x^4}}\), which might require numerical methods for a precise result due to its complexity.

Hyperbolic functions

Hyperbolic functions might sound complex, but they're just as important as the familiar trigonometric functions. They appear frequently in various areas of calculus, such as when working with hyperbolas.
Just like sine and cosine, hyperbolic functions have equivalents known as hyperbolic sine \(\sinh(x)\) and hyperbolic cosine \(\cosh(x)\). They are defined as follows:

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

Although the example given deals with \(y = \frac{1}{x}\), which is not a hyperbolic function, understanding these functions gives you greater insight into other problems involving hyperbolas. They serve as a useful reference and provide symmetrical properties, much like their circular cousins.

Definite integrals

Definite integrals are a fundamental tool in calculus used to find areas, among other applications. For a given function \(f(x)\), the definite integral from \(a\) to \(b\) is represented as \(\int_{a}^{b} f(x) \, dx\), yielding the net area between the curve and the x-axis.
In the exercise, we calculate the area under the curve \(y = \frac{1}{x}\) from \(x=1\) to \(x=2\) by evaluating the definite integral:

• \[ \int_{1}^{2} \left(2 - \frac{1}{x}\right) \, dx \]

This involves splitting it into two integrals for ease:

• \(\int_{1}^{2} 2\,dx = 2x\big|_1^2 = 2\)
• \(\int_{1}^{2} \frac{1}{x} \, dx = \ln|x| \,\big|_1^2 = \ln 2\)

Hence, the total area is \(2 - \ln 2\), highlighting the effectiveness of definite integrals in area calculation.

Area under a curve

Calculating the area under a curve is a common application of integration in calculus. It involves finding the region bounded by the curve and usually the x-axis over a specific interval.
For our problem, the task was to find the area enclosed by certain boundaries, including the hyperbolic curve \(y = \frac{1}{x}\) from \(x = 1\) to \(x = 2\) and the line \(y = 2\).
The area is not just a plain integral of \(y = \frac{1}{x}\), but rather:

• \[ A = \int_{1}^{2} (2 - \frac{1}{x}) \, dx \]

This approach subtracts the area under the curve from the encompassing rectangle's area, providing the area bound by the stated lines and curves. Understanding how to break down the problem into component areas makes integration a powerful tool for area calculation.