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: Find the Points on the Line

First, we need to find the points on the line at the specified values of \(x\). For \(x = 1\), substitute into the line equation \(y = 2x + 3\):\[y = 2(1) + 3 = 5.\]Therefore, the point is \((1, 5)\).For \(x = 3\), substitute into the line equation:\[y = 2(3) + 3 = 9.\] Thus, the point is \((3, 9)\).

Step 2: Use the Distance Formula

We can now calculate the length of the line segment between the points \((1,5)\) and \((3,9)\) using the distance formula:The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\]Substituting the points \((1, 5)\) and \((3, 9)\) gives:\[d = \sqrt{(3 - 1)^2 + (9 - 5)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}.\]

Step 3: Verify with Integration

Using integration involves the concept of arc length for a straight line being the same as the straight-line distance.For a curve described as \(y = f(x)\), the arc length \(L\) from \(x = a\) to \(x = b\) is:\[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx.\]For the line \(y = 2x + 3\), the derivative \(\frac{dy}{dx} = 2\). Substituting, we get:\[L = \int_{1}^{3} \sqrt{1 + (2)^2} \, dx = \int_{1}^{3} \sqrt{5} \, dx.\]Solve this integral:\[L = \sqrt{5} \times [x]_{1}^{3} = \sqrt{5} \times (3 - 1) = 2\sqrt{5}.\]Thus, the length obtained using integration is \(2\sqrt{5}\), confirming the distance calculation.

Key Concepts

Distance Formula

The distance formula is a fundamental concept in mathematics for finding the length of a line segment connecting two points in a plane. This formula is derived from the Pythagorean theorem and ensures that students can transition smoothly from understanding simple geometrical concepts to more complex mathematical notions.
Using this formula, you can easily calculate the distance between any two given points in a coordinate plane. The formula is:

• \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Here, \((x_1, y_1)\) and \((x_2, y_2)\) represent the coordinates of the two points. Essentially, this formula measures the direct distance "as the crow flies" between two points.
In the given problem, the distance formula was used to find the segment length of the line \(y=2x+3\) between \(x=1\) and \(x=3\). By simply substituting the coordinates obtained from the line equation, the calculation resulted in a length of \(2\sqrt{5}\). Understanding how to use this formula is crucial for grasping other mathematical concepts like integration and arc length.

Integration

Integration is a logical mathematical process that helps to find the collective whole from the sum of individual parts. It's especially used to calculate areas under curves and to find the arc lengths of curves. For students discovering calculus, integration can first appear complex, but it unfolds the depth of mathematics elegantly.
In the context of finding lengths, integration helps to find the sum of infinitesimal parts that make up a whole. For instance, computed integrally, the arc length formula from \(x=a\) to \(x=b\) is:

• \(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

For a straight line such as our example, where \(y=2x+3\), the derivative \(\frac{dy}{dx}=2\), remains constant. This provides a straightforward solution when calculating length integrally. Despite the complex appearance of integration, it's essentially about understanding how small things add up to form something bigger.

Arc Length

Arc length is a concept that measures the distance along a curved path. However, when the curve is straight, as in our line equation \(y=2x+3\), the arc length becomes equivalent to the segment's straight-line distance.
The formula for arc length can pose a challenge when involving non-linear curves. For a line \(y=f(x)\), the arc length from \(x=a\) to \(x=b\) is:

• \(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

In our scenario, checking the arc length using integration gave a result of \(2\sqrt{5}\), which matched perfectly with the result from the distance formula. Understanding that arc length uses calculus to measure curved lines emphasizes the elegance of why simple linear equations fit snugly into our existing knowledge of distance.

Calculus

Calculus is a versatile component of mathematics that marks the start of understanding change and motion. Despite its intimidating reputation, it's structured around two main ideas: differentiation and integration.
In problems involving line segment lengths, calculus allows for handling complex scenarios involving variable curves. It's via calculus that students can approach the arc length formula to assess not just straight segments but also more convoluted pathways. For the line in this exercise, applying calculus confirmed a simple straight-line result. Despite the complexity calculus can handle, this instance illustrated the discipline’s transparency in validating basic geometric truths.
By having grasped the integration of the line \(y=2x+3\), we evidenced the continuity between algebra and calculus, ensuring that students can build upon known concepts to tackle unfamiliar problems.