/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 In Problems 17-22, find the Cart... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problems 17-22, find the Cartesian equations of the graphs of the given polar equations. \(r-5 \cos \theta=0\)

Short Answer

Expert verified
The Cartesian equation of the graph is \((x - \frac{5}{2})^2 + y^2 = (\frac{5}{2})^2\) (a circle).

Step by step solution

01

Rearrange the Polar Equation

The given polar equation is \( r - 5 \cos \theta = 0 \). Start by solving for \( r \): \( r = 5 \cos \theta \).
02

Convert Using Polar to Cartesian Relationships

Recall the relationships between polar and Cartesian coordinates: \( x = r \cos \theta \) and \( y = r \sin \theta \). Also, \( r = \sqrt{x^2 + y^2} \). Using these, express \( r \) in terms of \( x \): substitute into the equation \( r = 5 \frac{x}{r} \), which simplifies to \( r^2 = 5x \).
03

Substitute for \(r^2\)

From step 2, substitute \( r^2 \) with \( x^2 + y^2 \) in the equation: \( x^2 + y^2 = 5x \).
04

Rearrange to Find the Cartesian Equation

Rearrange \( x^2 + y^2 = 5x \) to get the standard form of the circle's equation: \( x^2 - 5x + y^2 = 0 \) and complete the square for the \( x \) terms: \( (x - \frac{5}{2})^2 - (\frac{5}{2})^2 + y^2 = 0 \).
05

Simplify the Cartesian Equation

Simplify the completed square equation: \( (x - \frac{5}{2})^2 + y^2 = (\frac{5}{2})^2 \). This represents a circle with center \( (\frac{5}{2}, 0) \) and radius \( \frac{5}{2} \).

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Key Concepts

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

Circle Equation
A circle is a simple shape in geometry, defined by all points that are a specific distance, known as the radius, from a central point. In math, the general equation for a circle centered at (h, k) with radius \( r \) is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \]
In this original exercise, we derived the equation of a circle in Cartesian coordinates. After converting the polar equation to its Cartesian form, we found the circle's equation was \( (x - \frac{5}{2})^2 + y^2 = (\frac{5}{2})^2 \).
Here's what this tells us:
  • The center of the circle is at the point \( (\frac{5}{2}, 0) \).
  • The radius of the circle is \( \frac{5}{2} \).
Understanding how to find and interpret the circle equation helps visualize how points relate on a plane.
Completing the Square
Completing the square is a technique used in algebra to transform a quadratic equation into a perfect square trinomial. This is particularly useful to easily graph quadratic expressions, especially when working with circles.
In converting the polar equation to Cartesian form, we were tasked with dealing with the term \( x^2 - 5x \). Through completing the square, this expression was rewritten as \( (x - \frac{5}{2})^2 - (\frac{5}{2})^2 \).
Let's break it down:
  • Take the \( x \) term from the equation and calculate half of its coefficient. For \( -5x \), half is \( -\frac{5}{2} \).
  • Square this number: \( (\frac{5}{2})^2 = \frac{25}{4} \).
  • Rewrite the quadratic part as \( (x - \frac{5}{2})^2 \), which adds \( \frac{25}{4} \) to balance the equation.
Completing the square provides a clear view into the graph's characteristics, such as the center and radius of a circle in this example.
Polar Coordinates
Polar coordinates are an alternative to Cartesian coordinates. They describe a point in terms of its distance from a fixed point (the pole, usually the origin) and the angle from a fixed direction (usually the positive x-axis). This system is especially useful for problems involving circles and angles.
In polar coordinates, each point is represented as \( (r, \theta) \), where \( r \) is the radius, and \( \theta \) is the angle. The equation \( r - 5 \cos \theta = 0 \) defines a circle when we recognize patterns between polar and Cartesian systems.
To convert polar equations into a standard form, understanding key relationships is vital, such as:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( r = \sqrt{x^2 + y^2} \)
By transforming polar coordinates to Cartesian coordinates, circles and other geometric figures become easier to graph on traditional planes.
Cartesian Coordinates
Cartesian coordinates, named after mathematician René Descartes, are used to describe the position of a point in the plane using two numbers. These numbers are distances measured along perpendicular axes (commonly denoted \( x \) and \( y \)).
The transformation of equations from polar to Cartesian form allows for easier interpretation and visualization of shapes on a standard coordinate system.
Key relationships for this conversion are:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( r = \sqrt{x^2 + y^2} \)
Breaking down the example in the exercise: starting with \( r = 5 \cos \theta \), we found \( x \) using \( r \cos \theta \). Then through substitutions and simplifications, the equation \( x^2 + y^2 = 5x \) emerged. This manipulation led us to recognize the circle's characteristics.

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Most popular questions from this chapter

In each of Problems 1-20, a parametric representation of a curve is given. (a) Graph the curve. (b) Is the curve closed? Is it simple? (c) Obtain the Cartesian equation of the curve by eliminating the parameter (see Examples 1-4). $$ x=\sin \theta, y=2 \cos ^{2} 2 \theta ;-\infty<\theta<\infty $$

The path of a projectile fired from level ground with a speed of \(v_{0}\) feet per second at an angle \(\alpha\) with the ground is given by the parametric equations $$ x=\left(v_{0} \cos \alpha\right) t, \quad y=-16 t^{2}+\left(v_{0} \sin \alpha\right) t $$ (a) Show that the path is a parabola. (b) Find the time of flight. (c) Show that the range (horizontal distance traveled) is \(\left(v_{0}^{2} / 32\right) \sin 2 \alpha\). (d) For a given \(v_{0}\), what value of \(\alpha\) gives the largest possible range?

You can generate surfaces by revolving smooth curves, given parametrically, about a coordinate axis. As \(t\) increases from a to b, a smooth curve \(x=F(t)\) and \(y=G(t)\) is traced out exactly once. Revolving this curve about the \(x\)-axis for \(y \geq 0\) gives the surface of revolution with surface area $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ See Section 6.4. Problems 48-54 relate to such surfaces. Derive a formula for the surface area generated by the rotation of the curve \(x=F(t), y=G(t)\) for \(a \leq t \leq b\) about the \(y\)-axis for \(x \geq 0\), and show that the result is given by $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$

In Problems 23-36, name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. \(r=-4 \cos \theta\)

In Problems 23-36, name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. \(r=\frac{6}{4-\cos \theta}\)

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