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Chapter 6: Problem 2

Show that the shortest distance between two points on a plane is a straight line.

Short Answer

Expert verified
To prove that the shortest distance between two points on a plane is a straight line, we can use the Euclidean distance formula, \(d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\), as a reference. Let's consider two points P(x1, y1) and Q(x2, y2) and an alternate curved path C between them. Breaking C into small straight-line segments, we can calculate the total length (L) of these segments along C. Since the sum of these segments' lengths will always be greater than the straight-line distance (Euclidean distance) between P and Q, it proves that the straight line between P and Q is the shortest distance.

Step by step solution

01

Understanding the Euclidean Distance

The Euclidean distance is the straight-line distance between two points in a plane. If we have two points A(x1, y1) and B(x2, y2), the Euclidean distance between A and B can be calculated using the formula: \(d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\).
02

Proving the Straight Line is the Shortest Distance

Let鈥檚 consider two points P(x1, y1) and Q(x2, y2) on the plane. We need to prove that the distance along a straight line connecting the points P and Q is the shortest. Let's assume another path between P and Q which is not a straight line; we'll call this path the curve C. To find the total distance along curve C, we can break it down into short, straight line segments. The length of any segment (螖L) can be represented by: \(\Delta L = \sqrt{(\Delta x)^2 + (\Delta y)^2}\), where 螖x and 螖y represent the changes in the x and y coordinates between the segment's starting and ending points. Now let's consider the sum of all these segments' lengths along curve C: Total Length of C (L) = \( \sum_{i=1}^n \Delta L_i = \sum_{i=1}^n \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} \) Since each \(\sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}\) term represents the length of a straight line segment, the sum of their lengths will always be greater than the straight-line distance (Euclidean distance) between points P and Q. So, \( L > \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \), which means the straight line between P and Q will always have the shortest distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Euclidean Distance
The concept of Euclidean distance is foundational in understanding how to determine the shortest route between two points in a plane. Imagine you've dropped two pins on a map, labeled A and B, located at coordinates \(A(x1, y1)\) and \(B(x2, y2)\) respectively. To find the shortest path you would draw a straight line from A to B. This line represents the Euclidean distance. Mathematically, we calculate this using the formula \(d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\). This distance calculation is a vital tool used not only in geometry but in various fields such as physics, computer science, and even everyday navigation.

Understanding this distance is critical because it represents the most efficient path between two points. Whenever efficiency is a priority, such as plotting flight routes or optimizing delivery paths, the Euclidean distance is a key player. It's fascinating to see how this concept, rooted in ancient geometry, plays an integral role in our modern, fast-paced world.
Geometry
Geometry, the branch of mathematics concerned with the properties and relations of points, lines, shapes, and solids, becomes a vibrant playground when exploring the shortest distance between two points. At the heart of this exploration is the line鈥攁 seemingly simple yet profoundly important shape. In real-life contexts, understanding lines and their properties can be highly practical. For instance, architects use the principles of straight lines to design buildings, and artists use geometric principles to create perspective in their work.

The shortest distance problem also challenges us to refine our spatial reasoning. By visualizing various paths connecting two points and comparing their lengths, students engage in a practical exercise that helps develop their geometric intuition. This deepens their appreciation for the abstract elements of mathematics as they witness its applications in concrete problems.
Mathematical Proof
A mathematical proof is a logical argument that establishes the truth of a theorem using previously accepted statements, such as axioms and known theorems. In proving the shortest distance between two points is a straight line, we invoke the power of proof to validate a fact that might seem intuitively obvious, but still needs rigorous confirmation.

To illustrate, consider the attempt to prove that any alternative path besides a straight line is longer. Breaking down a curved path into infinitesimally small straight segments allows us to apply the concept of Euclidean distance to each segment. Summing these distances shows that they exceed the singular Euclidean path, thus reinforcing the idea that a straight line represents the minimum distance. This approach not only fortifies our geometrical understanding but also emphasizes the essence of proof in mathematics: to remove all doubt through methodical reasoning and evidence.

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Most popular questions from this chapter

Consider light passing from one medium with index of refraction \(n_{1}\) into another medium with index of refraction \(n_{2}\) (Figure 6 -A). Use Fermat's principle to minimize time, and derive the law of refraction: \(n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}\)

Find the dimensions of the parallelepiped of maximum volume circumscribed by (a) a sphere of radius \(R ;\) (b) an ellipsoid with semiaxes \(a, b, c\)

Find the shortest path between the \((x, y, z)\) points (0,-1,0) and (0,1,0) on the conical surface \(z=1-\sqrt{x^{2}+y^{2}} .\) What is the length of the path? Note: this is the shortest mountain path around a volcano.

Find the ratio of the radius \(R\) to the height \(H\) of a right-circular cylinder of fixed yolume \(V\) that minimizes the surface area \(A\)

Consider the line connecting \(\left(x_{1}, y_{1}\right)=(0,0)\) and \(\left(x_{2}, y_{2}\right)=(1,1) .\) Show explicitly that the function \(y(x)=x\) produces a minimum path length by using the varied function \(y(\alpha, x)=x+\alpha \sin \pi(1-x) .\) Use the first few terms in the expansion of the resulting elliptic integral to show the equivalent of Equation 6.4
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