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Express the given equations in polar coordinates. $$ \begin{array}{ll}{\text { (a) } x=3} & {\text { (b) } x^{2}+y^{2}=7} \\\ {\text { (c) } x^{2}+y^{2}+6 y=0} & {\text { (d) } 9 x y=4}\end{array} $$

Short Answer

Expert verified
(a) \(r \cos \theta = 3\), (b) \(r^2 = 7\), (c) \(r(r + 6 \sin \theta) = 0\), (d) \(r^2 \sin(2\theta) = \frac{4}{9}\).

Step by step solution

01

Convert Equation (a) to Polar Coordinates

The equation given is \(x = 3\). In polar coordinates, \(x = r \cos \theta\). To express this in polar coordinates, substitute \(x = r \cos \theta\) to get \(r \cos \theta = 3\). This is the polar representation of the line \(x=3\).
02

Convert Equation (b) to Polar Coordinates

The given equation is \(x^2 + y^2 = 7\). In polar coordinates, \(x^2 + y^2 = r^2\). Thus, the equation becomes \(r^2 = 7\), representing a circle with radius \(\sqrt{7}\).
03

Convert Equation (c) to Polar Coordinates

The given equation is \(x^2 + y^2 + 6y = 0\). Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the equation to get \(r^2 + 6(r \sin \theta) = 0\), simplifying to \(r(r + 6 \sin \theta) = 0\). Thus either \(r = 0\) or \(r = -6 \sin \theta\).
04

Convert Equation (d) to Polar Coordinates

The given equation is \(9xy = 4\). Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) to get \(9(r \cos \theta)(r \sin \theta) = 4\), which simplifies to \(9r^2 \cos \theta \sin \theta = 4\). Therefore, \(r^2 \sin(2\theta) = \frac{4}{9}\).

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Key Concepts

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

Equations in Polar Form
In mathematics, equations can be expressed in different coordinate systems, one of which is the polar coordinate system. This system uses radial and angular coordinates, known as \( r \) and \( \theta \), respectively, to describe a point's position. These two values correspond to the distance from the origin and the angle relative to the positive x-axis.

When converting equations into polar form, it's essential to switch from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\).
  • For the x-coordinate: \(x = r\cos\theta\)
  • For the y-coordinate: \(y = r\sin\theta\)
  • For the equation of a circle: \(x^2 + y^2 = r^2\)
By using these relations, you can convert any Cartesian equation into its polar form. This transformation often simplifies problems, especially when the equation involves circular or spiral shapes.
Coordinate Transformation
Coordinate transformation involves changing one set of coordinates into another. In the context of turning Cartesian coordinates into polar coordinates, this transformation helps to describe points in terms of angles and distances.

The relationship between these two systems is essential:
  • \(x = r \cos \theta\)
  • \(y = r \sin \theta\)
  • \(x^2 + y^2 = r^2\)
  • \(\tan \theta = \frac{y}{x}\)
These equations allow for a seamless transition from describing linear positions to understanding circular or radial structures. Transformations like this are crucial in fields such as physics and engineering, where rotational dynamics are considered.

Understanding the transformation helps distinguish how points relate spatially in different coordinate systems and can clarify the underlying geometry of the equations.
Circular Equations
Circles are fundamental shapes in geometry, and understanding their governing equations is crucial. In Cartesian coordinates, a circle's general equation is \(x^2 + y^2 = r^2\), where \(r\) represents the radius.

By expressing this equation in a polar coordinate form as \(r = \text{constant}\), it becomes especially intuitive. Every point along the circle maintains an equal distance from the center, making it simple to visualize or graph when dealing with polar forms.

Through the polar lens:
  • The center of a circle is at the origin \((0, 0)\).
  • The radius is \(r\), which equals a constant value across all theta.
  • Any deviation in the equation, such as additional linear terms, describes shifts or distortions.
Using polar equations to represent circles can simplify complex problems by focusing on symmetry and rotational aspects. This simplification is why circle equations often appear in various mathematical and real-world applications.

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Most popular questions from this chapter

Let an \(x^{\prime} y^{\prime}\) -coordinate system be obtained by rotating an \(x y\) -coordinate system through an angle \(\theta .\) Prove: For every value of \(\theta,\) the equation \(x^{2}+y^{2}=r^{2}\) becomes the equation \(x^{\prime 2}+y^{\prime 2}=r^{2} .\) Give a geometric explanation.

Find the area of the region described. $$ \begin{array}{l}{\text { The region inside the rose } r=2 a \cos 2 \theta \text { and outside the }} \\ {\text { circle } r=a \sqrt{2} \text { . }}\end{array} $$

Use the following values, where needed: radius of the Earth \(=4000 \mathrm{mi}=6440 \mathrm{km}\) 1 year (Earth year) \(=365\) days (Earth days) \(1 \mathrm{AU}=92.9 \times 10^{6} \mathrm{mi}=150 \times 10^{6} \mathrm{km}\) (a) Let \(a\) be the semimajor axis of a planet's orbit around the Sun, and let \(T\) be its period. Show that if \(T\) is measured in days and \(a\) is measured in kilometers, then \(T=\left(365 \times 10^{-9}\right)(a / 150)^{3 / 2}\). (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is \(a=57.95 \times 10^{6} \mathrm{km} .\) (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is \(e=0.206 .\) (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c).

Show that the graph of the given equation is a parabola. Find its vertex, focus, and directrix. $$x^{2}-2 \sqrt{3} x y+3 y^{2}-8 \sqrt{3} x-8 y=0$$

Find an equation of the ellipse traced by a point that moves so that the sum of its distances to \((4,1)\) and \((4,5)\) is 12 .

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