Problem 27 - Thomas Calculus

Question

Find polar equations for the circles in Exercises \(21-28 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$ x^{2}+y^{2}+y=0 $$

Step-by-Step Solution

Answer

The circle is centered at \((0, -\frac{1}{2})\) with radius \(\frac{1}{2}\). Its polar equation is \(r = -\frac{1}{2}\sin\theta\).

Step 1: Write the Cartesian Equation in Standard Form

The given equation is \(x^2 + y^2 + y = 0\). To transform this into a standard form equation of a circle, we need to complete the square for the \(y\) terms. First, rearrange the equation: \(x^2 + y^2 + y = 0\) into \(x^2 + (y^2 + y) = 0\).

Step 2: Complete the Square for the y Terms

Focus on the \(y\) part: \(y^2 + y\). Take half of the coefficient of \(y\), which is \(\frac{1}{2}\), square it to get \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\), and add and subtract this inside the equation. Thus, rewrite the equation as: \(x^2 + (y^2 + y + \frac{1}{4} - \frac{1}{4}) = 0\), which simplifies to \(x^2 + ((y + \frac{1}{2})^2 - \frac{1}{4}) = 0\).

Step 3: Rearrange into Standard Form of a Circle

Now, simplify the equation obtained in Step 2: \(x^2 + (y + \frac{1}{2})^2 = \frac{1}{4}\). This represents a circle centered at \((0, -\frac{1}{2})\) with radius \(\frac{1}{2}\). The Cartesian equation of the circle is now in its standard form.

Step 4: Convert Cartesian Equation to Polar Equation

Recall that in polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\). Substitute these into the standard circle form from Step 3: \(x^2 + (y + \frac{1}{2})^2 = r^2\). Rewrite as \((r\cos\theta)^2 + (r\sin\theta + \frac{1}{2})^2 = (\frac{1}{2})^2\), which simplifies to \(r^2 \cos^2\theta + (r\sin\theta + \frac{1}{2})^2 = \frac{1}{4}\).

Step 5: Simplify the Polar Equation

Further simplification gives us the polar equation of the circle. First, expand the equation: \(r^2\cos^2\theta + (r^2\sin^2\theta + r\sin\theta + \frac{1}{4}) = \frac{1}{4}\). This simplifies to \(r^2 = -r\sin\theta\) as \(r(\cos^2\theta + \sin^2\theta) + r\sin\theta - \frac{1}{4} = 0\) folds the identity \(\cos^2\theta + \sin^2\theta = 1\).

Step 6: Sketch the Circle

The Cartesian equation \((x - 0)^2 + (y + \frac{1}{2})^2 = (\frac{1}{2})^2\) represents a circle with center at \((0, -\frac{1}{2})\) and radius \(\frac{1}{2}\). In the coordinate plane, the circle is situated with the center below the x-axis along the y-axis, reaching up to the x-axis.

Key Concepts

Completing the Square

Completing the square is a method used to rewrite quadratic expressions in a form that makes them easier to analyze, particularly for identifying characteristics like the center and radius of a circle. Let's say we have the expression \(y^2 + y\). To complete the square, we take the coefficient of the linear term \(y\), which is \(1\), divide it by 2 to get \(\frac{1}{2}\), and then square it, resulting in \(\frac{1}{4}\).
Add \(\frac{1}{4}\) and also subtract \(\frac{1}{4}\) to the expression to maintain its balance:

• Original: \(y^2 + y\)
• Modified: \(y^2 + y + \frac{1}{4} - \frac{1}{4}\)

The expression \(y^2 + y + \frac{1}{4}\) is a perfect square trinomial, which can be written as the square of a binomial: \((y + \frac{1}{2})^2\). This technique is essential in converting a quadratic equation into an easy-to-interpret standard form, especially useful in the context of conic sections, such as circles.

Cartesian Coordinates

Cartesian coordinates are a system in which a point in a plane is determined by an ordered pair of numbers, typically written as \((x, y)\). This system allows us to visualize and solve geometric problems on a flat surface.
In this exercise, the original equation given is in Cartesian form:

• \(x^2 + y^2 + y = 0\)

Completing the square transforms it into the circle's standard equation, which reveals more about its geometric properties in the Cartesian plane. The circle's center and radius become apparent, making graphing straightforward on a coordinate plane.

Circle Equation

The standard form of a circle's equation in Cartesian coordinates is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. By manipulating the initial equation given through completing the square, we derive this form.
For the equation \(x^2 + (y + \frac{1}{2})^2 = \frac{1}{4}\), we identify the following:

• Center: \((0, -\frac{1}{2})\)
• Radius: \(\frac{1}{2}\)

This equation represents a circle centered below the x-axis, with a small radius of \(\frac{1}{2}\). Observing this form helps us understand the circle’s placement and size when plotted.

Coordinate Transformation

Coordinate transformation involves converting points from one coordinate system to another, such as from Cartesian to polar coordinates. This transformation is useful in simplifying equations or visualizing them in different contexts.
In polar coordinates, any point is represented by \((r, \theta)\), where \(r\) is the radius from the pole (origin) and \(\theta\) is the angle from the positive x-axis. Using the relations \(x = r \cos\theta\) and \(y = r \sin\theta\), we convert the circle's Cartesian equation into polar form.

• Transform: \(x^2 + (y + \frac{1}{2})^2 = \frac{1}{4}\)
• Substitute: \(x = r\cos\theta\) and \(y = r\sin\theta\)

This leads to a manipulation that eventually simplifies the equation in polar form as \(r^2 = -r\sin\theta\). This form helps understand the circle's geometric relationships in a radial context. It illustrates the powerful flexibility of switching coordinates to match the needs of specific calculations or visualizations.