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Chapter 10: Problem 19

Find the Cartesian equations of the graphs of the given polar equations. $$ r \cos \theta+3=0 $$

Short Answer

Expert verified
The Cartesian equation is \( x = -3 \).

Step by step solution

01

Understand the Polar Equation

The given polar equation is \( r \cos \theta + 3 = 0 \). This is a polar equation where \( r \) is the radius (distance from the origin) and \( \theta \) is the angle. Our goal is to convert this into a Cartesian equation using the variables \( x \) and \( y \).
02

Use Polar to Cartesian Conversion

Recall the conversion formulas: \( x = r \cos \theta \) and \( y = r \sin \theta \). Substitute \( x = r \cos \theta \) into the polar equation \( r \cos \theta + 3 = 0 \).
03

Substitute and Simplify

Substitute for \( r \cos \theta \): \( x + 3 = 0 \). This simplifies to \( x = -3 \).
04

Identify the Cartesian Equation

The equation \( x = -3 \) is the Cartesian equation of the line. It represents a vertical line in the coordinate plane, since the value of \( y \) can be any real number.

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Key Concepts

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

polar_coordinates
Polar coordinates are a way to represent a point in the plane using a distance and an angle. It's a system that's particularly handy when dealing with problems involving circular or rotational symmetry.
  • The key components are the radius \( r \), which stretches from the origin to the point, and \( \theta \), the angle formed with the positive x-axis.
  • These values tell you exactly where the point is, much like how Cartesian coordinates use \( x \) and \( y \) axes.
In our given exercise, the equation \( r \cos \theta + 3 = 0 \) tells us about a certain location in polar form. The task often involves converting it into a form we might be more familiar with鈥攕uch as a Cartesian equation. Understanding how polar coordinates map positions is essential before moving them into the framework of \( x \) and \( y \).
polar_to_Cartesian_conversion
Converting from polar coordinates to Cartesian coordinates involves using simple trigonometric relationships between the systems. The relations are derived from the components of a right triangle formed by \( r \), about the origin, and lines parallel to the axes.
  • The equations we use are \( x = r \cos \theta \) and \( y = r \sin \theta \).
  • These expressions decompose the radius \( r \) into how far 'up' and 'across' the point lies from the origin.
In our case, substituting \( x = r \cos \theta \) into the polar equation is key. The step-by-step conversion translates the problem into an intuitive Cartesian format. After substituting, the polar equation \( r \cos \theta + 3 = 0 \) simplifies using the above relation to \( x + 3 = 0 \). Thus, we reach the Cartesian equation \( x = -3 \), representing the same locus in a different coordinate system.
graphing_equations
Graphing equations in Cartesian coordinates offers a visual representation of mathematical relationships. Once we transform polar equations, like \( r \cos \theta + 3 = 0 \), into Cartesian form, they become easier to visualize, especially if you're familiar with the standard coordinate axes.
  • The equation \( x = -3 \) indicates a vertical line, a straightforward Cartesian plot where every point has an \( x \)-value of -3, regardless of the \( y \)-value.
  • This technique helps in easily assessing intercepts, slopes, and other properties visible directly on the graph.
By converting and graphing these polar-derived Cartesian equations, we can better understand geometric and algebraic properties, making complex systems simpler to handle. Through such conversions, seemingly intricate polar forms are expressed in familiar ways that more clearly depict their graphical implications.

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Most popular questions from this chapter

Find the standard equation of each parabola from the given information. Assume that the vertex is at the origin. Directrix is \(x=3\)

Show that the equations of the parabola and hyperbola with vertex \((a, 0)\) and focus \((c, 0), c>a>0,\) can be written as \(y^{2}=4(c-a)(x-a)\) and \(y^{2}=\left(b^{2} / a^{2}\right)\left(x^{2}-a^{2}\right),\) respectively. Then use these expressions for \(y^{2}\) to show that the parabola is always "inside" the right branch of the hyperbola.

Sketch the graph of the given equation, indicating vertices, foci, and asymptotes (if it is a hyperbola). \(16 x^{2}+4 y^{2}=32\)

Name the conic (horizontal ellipse, vertical hyperbola, and so on ) corresponding to the given equation. \(\frac{-x^{2}}{9}+\frac{y}{4}=0\)

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. \(y^{2}=20 x,(2,-2 \sqrt{10})\)
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