Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The length of the segment is \(2\sqrt{5}\).

Step 1: Write the Line Equation in Slope-Intercept Form

The line is already in slope-intercept form: \(y = 2x + 3\). This indicates a line with slope \(m=2\) and y-intercept \(b=3\).

Step 2: Determine the Points on the Line

The points on the line for \(x=1\) and \(x=3\) are obtained by plugging these values into the line equation.- When \(x = 1\), \(y = 2(1) + 3 = 5\). Hence, the point is \((1, 5)\).- When \(x = 3\), \(y = 2(3) + 3 = 9\). Thus, the point is \((3, 9)\).

Step 3: Apply the Distance Formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substituting our values: \((x_1, y_1) = (1, 5)\) and \((x_2, y_2) = (3, 9)\):\[\sqrt{(3 - 1)^2 + (9 - 5)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\]Hence, the length of the line segment is \(2\sqrt{5}\).

Step 4: Set Up the Integration to Find Arc Length

The length of a curve between two points can also be found using the formula:\[\int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]Given \(y = 2x + 3\), the derivative \(\frac{dy}{dx} = 2\). Substituting this into the arc length formula, we integrate:\[\int_{1}^{3} \sqrt{1 + 2^2} \, dx = \int_{1}^{3} \sqrt{5} \, dx\]

Step 5: Evaluate the Integral

Now, evaluate the integral:\[\sqrt{5} \times \left[ x \right]_{1}^{3} = \sqrt{5} \times \left(3 - 1\right) = 2\sqrt{5}\]The result from integration matches the distance formula method, confirming the length of the segment is \(2\sqrt{5}\).

Key Concepts

Line Equation

The line equation is an essential component in algebra that defines a straight line. One of the most common ways to express a line equation is by using the slope-intercept form. This form is written as \( y = mx + b \), where:

• \( m \) represents the slope of the line, which indicates how steep the line is.
• \( b \) stands for the y-intercept, which is where the line crosses the y-axis.

In the exercise we have, the line equation is given by \( y = 2x + 3 \). This tells us that the line has a slope \( m = 2 \) and a y-intercept at \( b = 3 \). Therefore, the line rises by 2 units for every 1 unit it moves to the right.

Distance Formula

The distance formula is a mathematical tool used to find the distance between two points on a coordinate plane. It is derived from the Pythagorean theorem and is expressed as: \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. The formula helps in calculating the straight-line distance, also known as the Euclidean distance, between these points. In our exercise, to find the distance between the points \((1, 5)\) and \((3, 9)\), we substitute into the formula:\[ \sqrt{(3 - 1)^2 + (9 - 5)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \] The result \( 2\sqrt{5} \) is the exact length of the line segment between those points.

Integration

Integration is a powerful calculus tool used to compute the area under a curve, among other applications like finding the length of a curve or arc. To find the length of a curve, the arc length formula is often used:\[ \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]This formula requires calculating the derivative of \( y \) with respect to \( x \), represented as \( \frac{dy}{dx} \). In the example, the derivative for the line \( y = 2x + 3 \) is constant, \( \frac{dy}{dx} = 2 \).After substituting \( 2 \) back into the arc length formula, we evaluate:\[ \int_{1}^{3} \sqrt{1 + 2^2} \, dx = \int_{1}^{3} \sqrt{5} \, dx \]This integration calculates to \( \sqrt{5} \times (3 - 1) = 2\sqrt{5} \), which matches the distance found using the distance formula.

Slope-Intercept Form

The slope-intercept form is a user-friendly way of writing the equation of a straight line. It clearly outlines both the slope and the y-intercept. The formula \( y = mx + b \) allows us to quickly identify how a line behaves on a graph.

• The slope \( m \) tells us the direction and angle of the line. A positive slope, like the 2 in the equation \( y = 2x + 3 \), indicates the line moves upwards as it goes from left to right.
• The y-intercept \( b \), here being 3, shows where the line crosses the y-axis. This helps in understanding where the line begins if you start graphing from the y-axis.

In the exercise, the slope-intercept form made it easy to derive the points \((1, 5)\) and \((3, 9)\) by simply plugging various \( x \) values into the equation. Having these points then allowed us to use other methods, such as the distance formula and integration, to find the line segment's length efficiently.