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 segment length is \(2\sqrt{2}\).

Step 1: Rearrange the Line Equation

You are given the line equation \(2y - 2x + 3 = 0\). Rearrange it to express \(x\) in terms of \(y\). Start by isolating \(x\): \[2x = 2y + 3\] Then divide by 2 to get \(x\): \[x = y + \frac{3}{2}\]

Step 2: Find Derivative dx/dy

Calculate the derivative of \(x\) with respect to \(y\): \[\frac{dx}{dy} = \frac{d}{dy}(y + \frac{3}{2}) = 1\]

Step 3: Express Arc Length Formula

The arc length \(L\) of the curve between \(y = 1\) and \(y = 3\) is given by: \[L = \int_{y=1}^{y=3} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\] Substituting \(\frac{dx}{dy} = 1\), you get: \[L = \int_{1}^{3} \sqrt{1 + 1^2} \, dy = \int_{1}^{3} \sqrt{2} \, dy\]

Step 4: Integrate to Find Length

Evaluate the integral \(L = \int_{1}^{3} \sqrt{2} \, dy\): \[L = \sqrt{2} \times [y]_{1}^{3} = \sqrt{2} \times (3 - 1)\] \[L = 2\sqrt{2}\]

Step 5: Verify Using Distance Formula

Locate the points where the line intersects \(y = 1\) and \(y = 3\). For \(y = 1\), \[x = 1 + \frac{3}{2} = \frac{5}{2}\] giving point \((\frac{5}{2}, 1)\). For \(y = 3\), \[x = 3 + \frac{3}{2} = \frac{9}{2}\] giving point \((\frac{9}{2}, 3)\). The distance \(d\) between these points is: \[d = \sqrt{\left(\frac{9}{2} - \frac{5}{2}\right)^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = 2\sqrt{2}\].

Step 6: Conclusion: Verify Result

Both methods, integration and distance formula, confirm that the length of the segment is \(2\sqrt{2}\).

Key Concepts

Line Segment

Understanding a line segment is fundamental in geometry. A line segment is simply a part of a line that is bounded by two distinct endpoints. Think of it as a "slice" of a line. Unlike lines, which extend infinitely in both directions, line segments have a fixed length.

To find the length of a line segment in mathematics, we can use various methods. One method is by applying the principles of integration if the line segment is part of a curve. This involves calculating the arc length of the curve between two points. Another way, especially in coordinate geometry, is by using the distance formula when dealing with straight segments between known points. Regardless of the method, the goal is to determine how long this "slice" or section of the line truly is.

Integration

Integration is a key concept often used to find areas under curves or to calculate arc lengths, amongst other things. In essence, integration helps in finding the "total" effect when adding up many tiny parts—whether they're distances, areas, or others.

For finding an arc length like in our exercise, integration provides a way to sum up an infinite number of infinitesimally small line segments along a curve. The arc length of a function between two points can be derived using the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]
This formula inputs the derivatives of the function and the limits of integration to effectively provide the total length of the curve from the start point, \( a \), to the end point, \( b \).

By integrating over a specified interval, we achieve a precise measurement of the desired segment.

Distance Formula

The distance formula comes handy to find the straight-line distance between two points in coordinate geometry. This algebraic method offers an easier way to compute distances using coordinates.

The formula is expressed as:
\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points in question.

In our specific exercise, once we determined the coordinates of the line segment endpoints, we used this formula to verify the length determined by integration. The joyful result? Both methods yielded the same length, affirming the segment's true measure.

Coordinate Geometry

Coordinate Geometry, or analytic geometry, combines algebra with geometry to explore geometric shapes, like lines and curves, using an ordered pair of numbers.

In this system, each point is defined in a plane with an \((x, y)\) address, thanks to the Cartesian coordinate system. This allows for precise calculations of distances, slopes, and angles, and provides a powerful method for analyzing and describing geometric figures.

In our line segment problem, we used coordinate geometry to translate the equation of a line into a format that makes calculations straightforward. The process of rearranging the line equation to isolate \(x\) was an essential step. It enabled us to determine the specific \(x\) values associated with given \(y\) values, thus leading us smoothly to the line segment's endpoints.

This approach underlies many other explorations in mathematics, making it a pivotal tool in the toolkit of aspiring mathematicians and scientists.