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Show that the shortest path between two given points in a plane is a straight line, using plane polar coordinates.

Short Answer

Expert verified
The shortest path is a straight line, as shown by minimizing the polar coordinate distance formula with \( \theta_2 - \theta_1 = \pi \) or \( -\pi \).

Step by step solution

01

Define the Points in Polar Coordinates

Consider two points in polar coordinates. Let the first point, \( P_1 \), have coordinates \( (r_1, \theta_1) \) and the second point, \( P_2 \), have coordinates \( (r_2, \theta_2) \). In polar coordinates, a point is described by its radial distance \( r \) from the origin and angle \( \theta \) from the positive x-axis.
02

Express Distance in Terms of Polar Coordinates

The distance \( d \) between two points \( (r_1, \theta_1) \) and \( (r_2, \theta_2) \) can be expressed using the polar form of the Euclidean distance formula. This formula is:\[ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)} \]This step involves using the law of cosines where \( \theta_2 - \theta_1 \) is the angle between the line connecting the origin to \( P_1 \) and the line connecting the origin to \( P_2 \).
03

Simplify the Distance Expression

To simplify, we know that the distance formula represents a straight path when the expression inside the square root is minimized. For a fixed \( r_1 \) and \( r_2 \), the expression \[ -2r_1r_2\cos(\theta_2 - \theta_1) \]is minimized (most negative) when \( \cos(\theta_2 - \theta_1) = -1 \), indicating that \( \theta_2 - \theta_1 = \pi \) or \( -\pi \). These correspond to the points lying on a straight line through the origin.
04

Conclude the Shortest Path as a Straight Line

Once it is shown that the expression is minimized under these conditions, the path minimizing the distance \( d \) between \( P_1 \) and \( P_2 \) aligns with the simplest linear path on the polar plane. Thus, the shortest path between the points is indeed a straight line.

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Key Concepts

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

Polar Coordinates
In mathematics, polar coordinates offer a fascinating method to describe the position of a point in a plane using a pair of numbers. These numbers are represented as \, \((r, \theta)\), where \(r\) is the radial distance from the origin (the center of the coordinate system) and \(\theta\) is the angle at which the point is located with respect to the positive x-axis.

Key features of polar coordinates include:
  • The radial coordinate \, \(r\) tells you how far away the point is from the origin.
  • The angular coordinate \, \(\theta\) specifies the direction of the point, measured in radians or degrees.

These coordinates are especially beneficial in dealing with problems involving circular and rotational symmetry, as they simplify the equations when dealing with curves and circular paths. By applying polar coordinates, you can easily convert between Cartesian and polar systems when necessary.
Euclidean Distance
Euclidean distance is a measure of the straight line distance between two points in a plane or space. In a two-dimensional Cartesian plane, this distance is calculated using the Pythagorean theorem:
  • For two points, \(A(x_1,y_1)\) and \(B(x_2,y_2)\), the Euclidean distance \, \(d\) is calculated as: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

However, for polar coordinates, the Euclidean distance \, \(d\) formula is adapted to involve trigonometric functions. The distance between two polar coordinates \, \((r_1, \theta_1)\) and \, \((r_2, \theta_2)\) can be expressed as:\[ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)} \]This formula effectively uses the distance formula by integrating the direction (angles) and magnitudes (radii) of the coordinates. The transformation of coordinates and use of trigonometric identities make computation straightforward and useful in various applications, including physics and engineering.
Law of Cosines
The Law of Cosines is akin to the Pythagorean theorem but extends to cases where a triangle is not necessarily right-angled. It relates the lengths of the sides of a triangle to the cosine of one of its angles. In the context of the shortest path problem, it's applied to connect polar coordinates.

Given a triangle with sides \, \(a\), \, \(b\), and \, \(c\) and the angle \, \(C\) opposite the side \, \(c\), the Law of Cosines states:\[ c^2 = a^2 + b^2 - 2ab\cos(C) \]
  • This relation helps solve for an unknown side when two sides and the included angle are known.
  • It simplifies problems involving polar coordinates by reducing them to algebraic equations.

In polar coordinate geometry, the Law of Cosines provides a bridge to calculating the straight-line distance between two points by transforming angular differences into linear distances. This forms the base for calculating Euclidean distance in the polar coordinate system.
Angle Minimization
Angle minimization is crucial in determining the shortest path between two points using polar coordinates. When finding such paths, we aim to minimize expressions involving angles to ensure the calculated distance is the smallest possible.

One critical part of angle minimization is understanding how to use the cosine function's properties. For the cosine function, minimizing involves making \, \(\cos(\theta_2 - \theta_1)\) as negative as possible. Since the most negative cosine value happens at \, \(-1\), the angle difference \, \(\theta_2 - \theta_1\) must be \, \(\pi\) or \, \(-\pi\) for minimal distance. This implies:
  • The straight line condition: When two points align so that their angles with respect to the origin differ by exactly \, \(\pi\), they lie on a straight line.

Angle minimization therefore aids in proving that the shortest path between two points in a polar coordinate system, essentially, follows a direct, linear course.

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Most popular questions from this chapter

Consider a right circular cylinder of radius \(R\) centered on the \(z\) axis. Find the equation giving \(\phi\) as a function of \(z\) for the geodesic (shortest path) on the cylinder between two points with cylindrical polar coordinates \(\left(R, \phi_{1}, z_{1}\right)\) and \(\left(R, \phi_{2}, z_{2}\right) .\) Describe the geodesic. Is it unique? By imagining the surface of the cylinder unwrapped and laid out flat, explain why the geodesic has the form it does.

Find and describe the path \(y=y(x)\) for which the integral \(\int_{x_{1}}^{x_{2}} \sqrt{x} \sqrt{1+y^{\prime 2}} d x\) is stationary.

Consider a ray of light traveling in a vacuum from point \(P_{1}\) to \(P_{2}\) by way of the point \(Q\) on a plane mirror, as in Figure \(6.8 .\) Show that Fermat's principle implies that, on the actual path followed, \(Q\) lies in the same vertical plane as \(P_{1}\) and \(P_{2}\) and obeys the law of reflection, that \(\theta_{1}=\theta_{2} .[\) Hints: Let the mirror lie in the \(x z\) plane, and let \(P_{1}\) lie on the \(y\) axis at \(\left(0, y_{1}, 0\right)\) and \(P_{2}\) in the \(x y\) plane at \(\left(x_{2}, y_{2}, 0\right)\) Finally let \(Q=(x, 0, z)\). Calculate the time for the light to traverse the path \(P_{1} Q P_{2}\) and show that it is minimum when \(Q\) has \(z=0\) and satisfies the law of reflection.]

A ray of light travels from point \(P_{1}\) in a medium of refractive index \(n_{1}\) to \(P_{2}\) in a medium of index \(n_{2},\) by way of the point \(Q\) on the plane interface between the two media, as in Figure \(6.9 .\) Show that Fermat's principle implies that, on the actual path followed, \(Q\) lies in the same vertical plane as \(P_{1}\) and \(P_{2}\) and obeys Snell's law, that \(n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2} .\) [Hints: Let the interface be the \(x z\) plane, and let \(P_{1}\) lie on the \(y\) axis at \(\left(0, h_{1}, 0\right)\) and \(P_{2}\) in the \(x, y\) plane at \(\left(x_{2},-h_{2}, 0\right) .\) Finally let \(Q=(x, 0, z)\) Calculate the time for the light to traverse the path \(P_{1} Q P_{2}\) and show that it is minimum when \(Q\) has \(z=0\) and satisfies Snell's law.]

An aircraft whose airspeed is \(v_{\mathrm{o}}\) has to fly from town \(O\) (at the origin) to town \(P\), which is a distance \(D\) due east. There is a steady gentle wind shear, such that \(\mathbf{v}_{\text {wind }}=V y \hat{\mathbf{x}},\) where \(x\) and \(y\) are measured east and north respectively. Find the path, \(y=y(x),\) which the plane should follow to minimize its flight time, as follows: (a) Find the plane's ground speed in terms of \(v_{o}, V, \phi\) (the angle by which the plane heads to the north of east), and the plane's position. (b) Write down the time of flight as an integral of the form \(\int_{0}^{D} f d x .\) Show that if we assume that \(y^{\prime}\) and \(\phi\) both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand \(f\) takes the approximate form \(f=\left(1+\frac{1}{2} y^{2}\right) /(1+k y)\) (times an uninteresting constant) where \(k=V / v_{\mathrm{o}}\). (c) Write down the Euler-Lagrange equation that determines the best path. To solve it, make the intelligent guess that \(y(x)=\lambda x(D-x),\) which clearly passes through the two towns. Show that it satisfies the EulerLagrange equation, provided \(\lambda=(\sqrt{4+2 k^{2} D^{2}}-2) /\left(k D^{2}\right) .\) How far north does this path take the plane, if \(D=2000\) miles, \(v_{\mathrm{o}}=500 \mathrm{mph},\) and the wind shear is \(V=0.5 \mathrm{mph} / \mathrm{mi} ?\) How much time does the plane save by following this path? [You'll probably want to use a computer to do this integral.]

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