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Chapter 10: Problem 54

Give polar coordinates for their centers and identify their radii. $$r=6 \sin \theta$$

Short Answer

Expert verified
Center: (0,3); Radius: 3.

Step by step solution

01

Identify the type of curve

The given equation is in polar form: \( r = 6 \sin \theta \). The form \( r = a \sin \theta \) represents a circle centered on the y-axis in polar coordinates.
02

Determine the circle's center

For the equation \( r = 6 \sin \theta \), the center is located at the point \( (0, a/2) \) in polar coordinates, which translates to \( (0, 3) \) since \( a = 6 \).
03

Calculate the circle's radius

The formula \( r = a \sin \theta \) also indicates that the radius of the circle is half of \( a \). Since \( a = 6 \), the radius is \( 6/2 = 3 \).

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Key Concepts

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

Circle in Polar Coordinates
Polar coordinates offer a unique way to represent curves by using a fixed point and an angle from the point. In polar coordinates, equations like \( r = 6 \sin \theta \) describe circles.

This form \( r = a \sin \theta \) is specific for circles that are centered along the y-axis. Unlike Cartesian coordinates, where you have \( x^2 + y^2 = r^2 \), polar coordinates directly use the radius \( r \) and an angle \( \theta \).
  • The angle \( \theta \) is measured from the positive x-axis.
  • The radius \( r \) depends on \( \theta \), changing as the angle changes.
This method simplifies calculations and representations for circular paths and structures.
Radius Calculation
The radius in polar coordinates can sometimes be hidden in the equation, like in \( r = 6 \sin \theta \). To find the radius from this form, recognize that for \( r = a \sin \theta \), \( a \) determines size characteristics of the circle.

Here's how to calculate the radius:
  • Identify the parameter \( a \) from the given equation.
  • The radius of the circle is half the value of \( a \).
For the specific equation \( r = 6 \sin \theta \), the radius turns out to be \( \frac{6}{2} = 3 \). This gives you the exact size of the circle portrayed by the equation.
Equation of Circles
In polar coordinates, circle equations reveal characteristics at a glance. The equation \( r = 6 \sin \theta \) is expressed in the standard form \( r = a \sin \theta \), signifying special cases:

  • The circle is symmetric about the line \( \theta = \frac{\pi}{2} \).
  • The center is expressed as \( (0, \frac{a}{2}) \), polar form reflecting its vertical position.
This uses \( a/2 \) to determine the vertical offset of the circle from the origin.

From the equation, the steps to derive these characteristics become intuitive, reinforcing understanding of profile and placement in a circular definition. This form favors an analysis that aligns naturally with polar symmetry, offering a clear, succinct method to elaborate on circle dynamics.

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Most popular questions from this chapter

Find a parametrization for the curve. the ray (half line) with initial point (2,3) that passes through the point (-1,-1)

Identify the symmetries of the curves. Then sketch the curves in the \(x y\) -plane. $$r=\cos (\theta / 2)$$

Find the area enclosed by the ellipse $$ x=a \cos t, \quad y=b \sin t, \quad 0 \leq t \leq 2 \pi $$

Replace the Cartesian equations with equivalent polar equations. $$x-y=3$$

Give parametric equations and parameter intervals for the motion of a particle in the \(x y\) -plane. Identify the particle's path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion. $$x=3-3 t, \quad y=2 t, \quad 0 \leq t \leq 1$$
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