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Find a polar equation in the form \(r \cos \left(\theta-\theta_{0}\right)=r_{0}\) for each of the lines in Exercises \(9-12\) . $$ y=-5 $$

Short Answer

Expert verified
Polar equation: \( r \cos(\theta - \frac{\pi}{2}) = -5 \).

Step by step solution

01

Identify the Line Equation

The given line equation is in the Cartesian coordinate system: \( y = -5 \). This represents a horizontal line which is 5 units below the x-axis.
02

Express in General Cartesian Form

Any equation of a line can be expressed in the form \( Ax + By = C \). For \( y = -5 \), this becomes \( 0 \cdot x + 1 \cdot y = -5 \). Hence, \( A = 0, B = 1, C = -5 \).
03

Convert the Line Equation to Polar Form

In polar coordinates, the relation between Cartesian and polar is given by:\( x = r \cos \theta \) and \( y = r \sin \theta \). Substitute these into the Cartesian equation: \( 0 \cdot (r \cos \theta) + 1 \cdot (r \sin \theta) = -5 \). This simplifies to \( r \sin \theta = -5 \).
04

Express in Required Polar Equation Form

The required form is \( r \cos(\theta - \theta_0) = r_0 \). Use the identity: \( \cos(\theta - \theta_0) = \cos\theta \cos\theta_0 + \sin\theta \sin\theta_0 \). In \( r \sin \theta = -5 \), choose \( \theta_0 = \frac{\pi}{2} \) as this makes \( \sin\theta \) equivalent to \( \cos(\theta - \frac{\pi}{2}) \). Thus, the equation becomes \( r \cos(\theta - \frac{\pi}{2}) = -5 \).

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Key Concepts

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

Cartesian Coordinates
Cartesian coordinates are a fundamental part of mathematics and are used to determine a point's position on a plane. The system uses two perpendicular axes, usually labeled as the x-axis and y-axis, to define the coordinates of any given point. Each point is represented by a pair of numerical values:
  • x-coordinate: This is the horizontal value that shows how far away the point is from the y-axis. Positive values are to the right, and negative values are to the left.
  • y-coordinate: This is the vertical value that shows how far away the point is from the x-axis. Positive values are above, and negative values are below the axis.
The equation "y = -5" is in Cartesian form, where y is -5, meaning all points on this line are situated horizontally 5 units below the x-axis. This straight, simple representation is one of the reasons Cartesian coordinates are popularly used in geometry and graphing.
Polar Equations
Polar equations offer a different and sometimes more intuitive way to represent points and curves, especially those involving angles or circular patterns. Unlike Cartesian coordinates that use a grid system, polar coordinates rely on a center point (the pole) and an angle from a reference direction:
  • r: The radius or distance from the pole (origin) to the point.
  • \( \theta \): The angle measured from a fixed direction, usually the positive x-axis in counterclockwise direction.
To convert from the Cartesian to polar system, equations often involve trigonometric functions. For instance, the given line "y = -5" in the polar form becomes \( r \sin \theta = -5 \). Here, \( r \) represents the distance from the origin, while \( \theta \) represents the angle.
The polar equation is eventually expressed in a special form: \( r \cos(\theta - \theta_0) = r_0 \). This form often uses trigonometric identities to switch between angle components, providing a smooth conversion and representation.
Trigonometric Identities
Trigonometric identities are essential in transforming and simplifying equations, particularly when working with polar coordinates. These identities are formulas involving trigonometric functions, like sine and cosine, that always hold true. Here are some key identities:
  • Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
  • Angle addition formula: \( \begin{align*} \cos(a \pm b) &= \cos a \cdot \cos b \mp \sin a \cdot \sin b \ \sin(a \pm b) &= \sin a \cdot \cos b \pm \cos a \cdot \sin b \end{align*}\)
In our exercise, we use the identity \( \cos(\theta - \theta_0) = \cos \theta \cos \theta_0 + \sin \theta \sin \theta_0 \) to aid the conversion of the line equation into polar form. By choosing \( \theta_0 = \frac{\pi}{2} \), the equation \( r \sin \theta = -5 \) is transformed to \( r \cos(\theta - \frac{\pi}{2}) = -5 \), completing the process of expressing a Cartesian line in a polar equation format.

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Most popular questions from this chapter

Sketch the regions in the \(x y\) -plane whose coordinates satisfy the inequalities or pairs of inequalities in Exercises \(69-74 .\) $$ 9 x^{2}+16 y^{2} \leq 144 $$

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ x^{2}=8 y, \quad \text { right } 1, \text { down } 7 $$

The asymptotes of \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) Show that the vertical distance between the line \(y=(b / a) x\) and the upper half of the right-hand branch \(y=(b / a) \sqrt{x^{2}-a^{2}}\) of the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) approaches 0 by showing that $$\lim _{x \rightarrow \infty}\left(\frac{b}{a} x-\frac{b}{a} \sqrt{x^{2}-a^{2}}\right)=\frac{b}{a} \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}-a^{2}}\right)=0$$ Similar results hold for the remaining portions of the hyperbola and the lines \(y=\pm(b / a) x\)

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin, along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$ e=5, \quad y=-6 $$

Exercises \(49-52\) give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an equation for the new ellipse, and find the new foci, vertices, and center. $$ \frac{x^{2}}{16}+\frac{y^{2}}{25}=1, \quad \text { left } 4, \text { down } 5 $$

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