Problem 24 - Algebra and Trigonometry Real Mathematics, Real People

Question

Identify the type of conic represented by the equation. Use a graphing utility to confirm your result. $$r=\frac{10}{3+9 \sin \theta}$$

Step-by-Step Solution

Answer

The equation represents a circle with its center at point (\(\frac{5}{3}, 5\)) and radius \(\frac{10}{3}\) in cartesian coordinates. This is confirmed both algebraically and graphically.

Step 1: Rewrite the equation

First, let's do some work on the equation to see if it can be simplified or rewritten in a more recognizable form. Multiply both the numerator and the denominator by \( \frac{1}{3} \) to get \( r= \frac{10/3}{1+ 3\sin \theta} \).

Step 2: Convert into Cartesian Form

Now, convert the equation from polar to Cartesian form. This can be done using the conversion formulas \(\sin \theta= \frac{y}{r}\) and \(r^2= x^2 + y^2\), yielding \( r= \frac{10/3}{1+ 3\frac{y}{r}} \). Cross multiplying gives \( r^2 = \frac{10r}{3}(1+ 3\frac{y}{r}) = \frac{10r}{3} + 10y \). Substituting \( r^2 = x^2 + y^2 \) gives \( x^2 + y^2 = \frac{10x}{3} + 10y \).

Step 3: Rearrange to Identify Conic Section

Rearrange the equation to a standard form of a conic section. Bring all terms to one side to get \( x^2 - \frac{10x}{3} + y^2 - 10y = 0 \). To make it easier to identify, complete the square for both x and y: \((x - \frac{5}{3})^2 + (y - 5)^2 = (\frac{5}{3})^2 + 5^2\) . Simplify to get \((x - \frac{5}{3})^2 + (y - 5)^2 = \frac{100}{9}\) . This is the standard form for a circle with a radius square of \(\frac{100}{9}\).

Step 4: Verify with a Graphing Utility

Input the original polar equation \(r=\frac{10}{3+9 \sin \theta}\) into a graphing calculator or online graphing tool to visually confirm. The graph should show a circle with center at (\(\frac{5}{3}, 5\)) and radius of \(\frac{10}{3}\), which confirms the algebraic solution.

Key Concepts

Polar Coordinates

Polar coordinates offer a unique way to represent points on a plane using a distance and an angle. It’s quite different from the more common Cartesian coordinates, which use x and y values. In polar coordinates:

• Each point is represented by \(r, \theta\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle measured counterclockwise from the positive x-axis.
• The radius can be zero or positive, and the angle can be any real number, making polar coordinates quite flexible.

This system is particularly useful in cases involving circular and spiral shapes, and is widely used in fields like physics and engineering.
When dealing with polar equations, it is often essential to convert them to Cartesian coordinates for easier analysis or graphing.

Cartesian Coordinates

Cartesian coordinates are used to pinpoint a location on a plane using two perpendicular values, typically noted as \(x, y\). This system creates a grid where:

• The \(x\) coordinate represents the horizontal position,
• The \(y\) coordinate represents the vertical position.

These coordinate systems are fundamental in algebra and geometry.
To convert polar coordinates to Cartesian, you can use the formulas:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

Knowing both systems allows flexibility in approaching various mathematical problems, especially those involving conic sections.

Graphing Utility

A graphing utility is a digital tool that can plot equations and visualize their graphs. It is incredibly helpful for verifying analytical solutions. These utilities offer:

• Quick graphical representations of complex equations,
• Insights into the shape and nature of equations,
• Ease of use—just input the equation in its respective form.

In this particular exercise, a graphing utility confirms that the polar equation \(r=\frac{10}{3+9 \sin \theta}\) indeed forms a circle.
By inputting equations directly, students can confirm their solutions or gain a better understanding of how changes in equations affect graph shapes.

Circle Equation

A circle equation in a Cartesian coordinate system typically follows the form:\[ (x - h)^2 + (y - k)^2 = r^2 \]where:

• \(h, k\) are the coordinates of the circle's center,
• \(r\) is the radius of the circle.

In standard form, the equation conveys essential information at a glance. This form allows for easy identification of the circle's properties.
In this exercise, converting the polar equation to its Cartesian form reveals the circle's center at \(\left(\frac{5}{3}, 5\right)\) and radius \(\frac{10}{3}\). Recognizing these aspects in equations is crucial for understanding and visualizing conic sections.