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Chapter 2: Problem 73

If the polar coordinates of a point are \((r, \varphi)\) and its rectangular coordinates are \((x, y),\) determine the polar coordinates of the following points: (a) \((-x, y),\) (b) \((-2 x,\) \(-2 y\) ), and \((c)(3 x,-3 y)\)

Short Answer

Expert verified
The polar coordinates for the points are: (a) (-x, y): \(r_1 = \sqrt{(-x)^2 + (y)^2}\), \(\varphi_1 = \tan^{-1}\left(\frac{y}{-x}\right)\) (b) (-2x, -2y): \(r_2 = \sqrt{(-2x)^2 + (-2y)^2}\), \(\varphi_2 = \tan^{-1}\left(\frac{-2y}{-2x}\right)\) (c) (3x, -3y): \(r_3 = \sqrt{(3x)^2 + (-3y)^2}\), \(\varphi_3 = \tan^{-1}\left(\frac{-3y}{3x}\right)\)

Step by step solution

01

Write down the relationship between polar and rectangular coordinates

The relationship between polar and rectangular coordinates is given by the following formulas: \(x = r \cos(\varphi)\) (1) \(y = r \sin(\varphi)\) (2) Note that r is the distance from the origin to the point, and 蠁 is the angle measured counterclockwise from the positive x-axis. #------------------------------------------------------------------------------
02

Determine the polar coordinates for the point (-x, y)

For point (-x, y), we can substitute the values of x and y into the formulas given in step 1: \(-x = r_1 \cos(\varphi_1)\) (3) \(y = r_1 \sin(\varphi_1)\) (4) Now, starting with the two square equations: \((-x)^2 = (r_1 \cos(\varphi_1))^2\) (5) and \((y)^2 = (r_1 \sin(\varphi_1))^2\) (6) Sum up these equations: \((-x)^2 + (y)^2 = (r_1 \cos(\varphi_1))^2 + (r_1 \sin(\varphi_1))^2\) Apply Pythagorean identity: \((-x)^2 + (y)^2 = r_1^2 (\sin^2(\varphi_1) + \cos^2(\varphi_1))\) \((-x)^2 + (y)^2 = r_1^2\) Now, solve for r1: \(r_1 = \sqrt{(-x)^2 + (y)^2}\) (7) Now, use (3) and (4) to obtain 蠁1: \(\tan(\varphi_1) = \frac{y}{-x}\) (8) The polar coordinates for the point (-x, y) are (r1, 蠁1). #------------------------------------------------------------------------------
03

Determine the polar coordinates for the point (-2x, -2y)

For point (-2x, -2y), we can substitute the values of x and y into the formulas given in step 1: \(-2x = r_2 \cos(\varphi_2)\) (9) \(-2y = r_2 \sin(\varphi_2)\) (10) Now, square these equations: \((-2x)^2 = (r_2 \cos(\varphi_2))^2\) (11) and \((-2y)^2 = (r_2 \sin(\varphi_2))^2\) (12) Sum up these equations: \((-2x)^2 + (-2y)^2 = (r_2 \cos(\varphi_2))^2 + (r_2 \sin(\varphi_2))^2\) Apply Pythagorean identity: \((-2x)^2 + (-2y)^2 = r_2^2 (\sin^2(\varphi_2) + \cos^2(\varphi_2))\) \((-2x)^2 + (-2y)^2 = r_2^2\) Now, solve for r2: \(r_2 = \sqrt{(-2x)^2 + (-2y)^2}\) (13) Now, use (9) and (10) to obtain 蠁2: \(\tan(\varphi_2) = \frac{-2y}{-2x}\) (14) The polar coordinates for the point (-2x, -2y) are (r2, 蠁2). #------------------------------------------------------------------------------
04

Determine the polar coordinates for the point (3x, -3y)

For point (3x, -3y), we can substitute the values of x and y into the formulas given in step 1: \(3x = r_3 \cos(\varphi_3)\) (15) \(-3y = r_3 \sin(\varphi_3)\) (16) Now, square these equations: \((3x)^2 = (r_3 \cos(\varphi_3))^2\) (17) and \((-3y)^2 = (r_3 \sin(\varphi_3))^2\) (18) Sum up these equations: \((3x)^2 + (-3y)^2 = (r_3 \cos(\varphi_3))^2 + (r_3 \sin(\varphi_3))^2\) Apply Pythagorean identity: \((3x)^2 + (-3y)^2 = r_3^2 (\sin^2(\varphi_3) + \cos^2(\varphi_3))\) \((3x)^2 + (-3y)^2 = r_3^2\) Now, solve for r3: \(r_3 = \sqrt{(3x)^2 + (-3y)^2}\) (19) Now, use (15) and (16) to obtain 蠁3: \(\tan(\varphi_3) = \frac{-3y}{3x}\) (20) The polar coordinates for the point (3x, -3y) are (r3, 蠁3).

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Key Concepts

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

Rectangular Coordinates
Rectangular coordinates, commonly known as Cartesian coordinates, provide a method of rendering a point on a plane through two values: the x-coordinate and the y-coordinate. These coordinates can be visually represented on a graph as the intersection of a vertical and horizontal line originating from an axis labeled as the x-axis and y-axis, respectively.

In practical terms, the x-coordinate indicates the position of the point along the horizontal axis, while the y-coordinate signifies its location along the vertical axis. When combined, these two values uniquely identify the location of a point in two-dimensional space. The coordinates are often written as an ordered pair \( (x, y) \) .

This system is especially useful because it allows for straightforward algebraic manipulation and visual representation of geometric figures, lines, curves, and regions within the plane.
Coordinate Conversion
Coordinate conversion is a crucial concept in mathematics that enables us to transition between different coordinate systems, such as from polar to rectangular coordinates (or vice versa). This flexibility is important when tackling problems where a particular coordinate system may offer a more simplified approach or insight that is not readily apparent in the other.

To convert from polar coordinates \( (r, \varphi) \) to rectangular coordinates \( (x, y) \) , one uses the following relationships:
  • \[ x = r \cos(\varphi) \]
  • \[ y = r \sin(\varphi) \]
Conversely, to convert from rectangular coordinates back to polar, the following formulae are applied:
  • \[ r = \sqrt{x^2 + y^2} \]
  • The angle \( \varphi \) is determined by \( \text{tan}^{-1}(\frac{y}{x}) \) provided \( x \) is not zero.
Understanding these relationships is vital when dealing with complex problems involving shapes, curves, and angles in different contexts.
Pythagorean Identity
The Pythagorean identity is a fundamental element of trigonometry that emerges from the Pythagorean theorem's relationship to the unit circle. It states that for any angle \(\varphi\), the sum of the squares of sine and cosine of the angle is always equal to 1:

\[\sin^2(\varphi) + \cos^2(\varphi) = 1\]

This identity is particularly helpful when working with polar and rectangular coordinates as it allows the determination of a point's radial distance \(r\) from the origin. When given a point's coordinates \( (x, y) \) , one can apply the Pythagorean identity to derive the corresponding polar coordinate \( r \) using the sum of the squares of \(x\) and \(y\). By incorporating the Pythagorean identity, calculations become manageable, and the fundamental connection between trigonometric functions and geometry becomes apparent.

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Most popular questions from this chapter

Does the odometer in an automobile indicate a scalar or a vector quantity?

Four force vectors each have the same magnitude \(f\) What is the largest magnitude the resultant force vector may have when these forces are added? What is the smallest magnitude of the resultant? Make a graph of both situations.

You fly \(32.0 \mathrm{km}\) in a straight line in still air in the direction \(35.0^{\circ}\) south of west. (a) Find the distances you would have to fly due south and then due west to arrive at the same point. (b) Find the distances you would have to fly first in a direction \(45.0^{\circ}\) south of west and then in a direction \(45.0^{\circ}\) west of north. Note these are the components of the displacement along a different set of axes-namely, the one rotated by \(45^{\circ}\) with respect to the axes in (a).

Which of the following is a vector: a person's height, the altitude on Mt. Everest, the velocity of a fly, the age of Earth, the boiling point of water, the cost of a book, Earth's population, or the acceleration of gravity?

Distances between points in a plane do not change when a coordinate system is rotated. In other words, the magnitude of a vector is invariant under rotations of the coordinate system. Suppose a coordinate system S is rotated about its origin by angle \(\varphi\) to become a new coordinate system \(\mathrm{S}^{\prime},\) as shown in the following figure. \(\mathrm{A}\) point in a plane has coordinates \((x, y)\) in \(S\) and coordinates \(\left(x^{\prime}, y^{\prime}\right)\) in \(\mathrm{S}^{\prime}\) (a) Show that, during the transformation of rotation, the coordinates in \(S^{\prime}\) are expressed in terms of the coordinates in S by the following relations: \(\left\\{\begin{array}{l}x^{\prime}=x \cos \varphi+y \sin \varphi \\\ y^{\prime}=-x \sin \varphi+y \cos \varphi\end{array}\right.\) (b) Show that the distance of point \(P\) to the origin is invariant under rotations of the coordinate system. Here, you have to show that \(\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+y^{2}}\) (c) Show that the distance between points \(P\) and \(Q\) is invariant under rotations of the coordinate system. Here, you have to show that $$\sqrt{\left(x_{P}-x_{Q}\right)^{2}+\left(y_{P}-y_{Q}\right)^{2}}=\sqrt{\left(x_{P}^{\prime}-x^{\prime} \rho\right)^{2}+\left(y_{P}^{\prime}-y^{\prime}_{Q}\right)^{2}}$$
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