Problem 34 - Precalculus: Functions and Graphs

Question

Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$x^{2}+(y-1)^{2}=1$$

Step-by-Step Solution

Answer

The polar equation is \(r = 2\sin \theta\).

Step 1: Identify the given equation type

The given equation \(x^2 + (y-1)^2 = 1\) is in the form of a circle with center at \((0,1)\) and radius 1.

Step 2: Rewrite the center coordinates in polar form

In polar coordinates, every point is given in terms of \(r\) (radius) and \(\theta\) (angle). The center of the circle \((0, 1)\) has \(r = 1\) and \(\theta = \frac{\pi}{2}\), since it lies on the positive y-axis.

Step 3: Convert the equation to polar coordinates

The general conversion from Cartesian to polar coordinates is \(x = r \cos \theta\) and \(y = r \sin \theta\). Substituting these into the equation gives: \(x^2 + (y-1)^2 = r^2 \cos^2 \theta + (r \sin \theta - 1)^2\).

Step 4: Simplify the polar equation

Expanding and simplifying \((r \sin \theta - 1)^2\) gives \((r\sin\theta)^2 - 2r\sin\theta + 1\). The equation becomes \(r^2 \cos^2 \theta + r^2 \sin^2 \theta - 2r \sin \theta + 1 = 1\).

Step 5: Use the Pythagorean identity

Since \(r^2 (\cos^2 \theta + \sin^2 \theta) = r^2\) by the identity \(\cos^2 \theta + \sin^2 \theta = 1\), simplify to \(r^2 - 2r \sin \theta + 1 = 1\).

Step 6: Solve for \(r\)

Subtract 1 from both sides to get \(r^2 - 2r \sin \theta = 0\). Rearrange to factor and solve, giving \(r(r - 2\sin \theta) = 0\), and hence \(r = 0\) or \(r = 2\sin \theta\).

Step 7: Select the appropriate polar equation

Since we're describing a circle, the solution \(r = 2\sin \theta\) represents the circle in polar form, as it captures the center and radius appropriately.

Key Concepts

Circle Equation

A circle equation is a geometric representation of a circle on the coordinate plane. In Cartesian coordinates, a circle centered at the point \((h, k)\) with radius \(r\) is given by the equation:

• \( (x - h)^2 + (y - k)^2 = r^2 \)

The given equation \( x^2 + (y-1)^2 = 1 \) describes a circle centered at \((0, 1)\) with a radius of 1.
This is a standard form of a circle equation, making it easy to identify the circle's center and radius. The equation indicates that every point \((x, y)\) on the circumference is equidistant from the center \((0, 1)\), maintaining a constant radius of 1.
Understanding how a circle equation works helps you translate it into different coordinate systems, such as polar coordinates, which can further simplify calculations in various contexts.

Cartesian to Polar Conversion

To convert between Cartesian and polar coordinates, we use the relations between the two systems. While Cartesian coordinates are based on \(x\) and \(y\)-axes, polar coordinates depend on an angle \(\theta\) and radius \(r\) from the origin.

• The equations for conversion are:
• \(x = r \cos\theta\)
• \(y = r \sin\theta\)

To translate the given circle equation into polar form, we substitute these expressions:

• \(x^2 = (r\cos\theta)^2\)
• \( (y-1)^2 = (r\sin\theta - 1)^2 \)

These steps will align circular geometries with angular and radial measurements, often leading to simplifications and clearer geometric interpretations, especially in cases dealing with symmetry or radial positions.

Pythagorean Identity

The Pythagorean identity is a fundamental building block in trigonometry, which holds that

• \( \cos^2 \theta + \sin^2 \theta = 1 \)

This identity expresses the inherent relationship between the trigonometric functions \(\cos\theta\) and \(\sin\theta\) of an angle \(\theta\). It is derived from the Pythagorean theorem applied to the unit circle.
In the context of converting a circle equation, this identity simplifies expressions involving \(r^2\) when the equation has both sine and cosine components. For example, recognizing that \(r^2(\cos^2\theta + \sin^2\theta) = r^2\) allows you to simplify polar equations directly, as it confirms that the square of the hypotenuse (radius, or \(r\)) fully describes any point on the circle.
This simplification is key to more effectively manipulating and solving polar equations, converting them into simpler forms for further mathematical processing.

Trigonometric Simplification

Trigonometric simplification involves reducing complex trigonometric equations or expressions into simpler forms. During the conversion of a circle equation from Cartesian to polar coordinates, we encounter terms like \((r \sin\theta - 1)^2\). Expanding this leads to:

• (\(r \sin \theta\))^2 - 2\(r \sin \theta\) + 1

Recognizing trigonometric identities and factoring techniques in part facilitates this simplification. It leads to results like:

• \((r^2 - 2r \sin \theta + 1) = 1 \)

By applying a factorization, we reveal key solutions such as \(r = 0\) or \(r = 2 \sin \theta\). These reflect changes in equations from complex expressions back into simpler, familiar forms.
Understanding these simplifications enables solving broader mathematical problems by reducing complexity, making equations particularly useful in engineering and physics applications involving circular or wave motions.