Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

. Use a \(y\)-integration to find the length of the segment of the line \(2 y-2 x+3=0\) between \(y=1\) and \(y=3\). Check by using the distance formula.

Step-by-Step Solution

Answer

The length of the segment is \(2\sqrt{2}\).

Step 1: Rewrite the Line Equation

First, we need to rewrite the given line equation in the form of \( x = f(y) \). The given equation is \( 2y - 2x + 3 = 0 \). Solving for \( x \), we get \( 2y + 3 = 2x \) or \( x = y + \frac{3}{2} \).

Step 2: Determine the Derivative of x with Respect to y

Calculate \( \frac{dx}{dy} \) from the equation \( x = y + \frac{3}{2} \). This results in \( \frac{dx}{dy} = 1 \).

Step 3: Use the Arc Length Formula

The arc length \( L \) can be found using the formula \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \ dy \). Substitute \( \frac{dx}{dy} = 1 \) into this expression. Hence, \( L = \int_{1}^{3} \sqrt{1 + 1^2} \ dy = \int_{1}^{3} \sqrt{2} \ dy \).

Step 4: Evaluate the Integral

Calculate the integral \( \int_{1}^{3} \sqrt{2} \ dy \). This simplifies to \( \sqrt{2} \int_{1}^{3} dy \), which is \( \sqrt{2} \times [y]_1^3 = \sqrt{2} \times (3 - 1) = 2\sqrt{2} \).

Step 5: Use the Distance Formula as Verification

Find the endpoints of the segment on the line. When \( y = 1, x = 1 + \frac{3}{2} = \frac{5}{2} \). When \( y = 3, x = 3 + \frac{3}{2} = \frac{9}{2} \). The endpoints are \( \left( \frac{5}{2}, 1 \right) \) and \( \left( \frac{9}{2}, 3 \right) \). The distance formula for these points is \( \sqrt{\left( \frac{9}{2} - \frac{5}{2} \right)^2 + (3 - 1)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \), confirming the result.

Key Concepts

Integration

Integration is a powerful tool in calculus used to find areas, volumes, central points, and many useful things, one of which is the arc length of curves. In the context of the exercise, integration helps us calculate the length of a segment of a line. When finding arc length using integration, we make use of the formula:

• For a line defined parametrically by functions of some variable, say \( y \), the arc length \( L \) is computed using \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \).

In this exercise, integrating from \( y = 1 \) to \( y = 3 \) allows us to determine the arc length over this interval. Integration simplifies the computation by incorporating changes in both directions (here, \( x \) and \( y \)).
It involves finding the antiderivative of the function, which in this case is quite straightforward because we are dealing with a linear function.

Line Equation

A line equation can be represented in multiple forms. One common way is the slope-intercept form: \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. However, lines can also be expressed in other equivalent forms as needed by different contexts, like when setting equations for integration.

• The given line equation in the exercise is \( 2y - 2x + 3 = 0 \), which is re-organized to express \( x \) in terms of \( y \).
• This conversion makes the equation \( x = y + \frac{3}{2} \), simplifying the use of calculus tools like integration and derivatives.

Converting line equations is a crucial skill that aids in evaluating geometric properties such as slopes and intersections, and when checking features such as parallelism or perpendicularity.

Distance Formula

The distance formula is a key tool in geometry which provides a way to measure the distance between two points in a coordinate plane. It is derived from the Pythagorean Theorem:

• The formula is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

• In the exercise, you determine the distance between the points \(\left( \frac{5}{2}, 1 \right)\) and \(\left( \frac{9}{2}, 3 \right)\), which are endpoints derived from intercepting the line formula with \( y = 1 \) and \( y = 3 \).

This approach provides a robust verification for the result obtained by integration. Using the distance formula reinforces learning by giving an immediate and direct method to confirm calculated arc lengths.

Derivative

The derivative of a function measures how a function changes as its input changes, which is the essence of calculus. Specifically, it represents the rate of change and can indicate the slope of a line for linear functions, or how curves react at any given point.

• In the context of the exercise, we find the derivative of \( x \) with respect to \( y \) from the line equation \( x = y + \frac{3}{2} \).
• The derivative \( \frac{dx}{dy} = 1 \) implies that for each unit increase in \( y \), \( x \) also increases by 1 unit, indicating a constant slope.

Calculating derivatives is vital as it is also used when integrating to find arc lengths, as seen in the exercise. Recognizing \( \frac{dx}{dy} \) is crucial for handling arc length calculations and interpreting geometric relationships.