Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that \(r=a \sin \theta+b \cos \theta\) represents a circle and find its center and radius.

Step-by-Step Solution

Answer

The equation represents a circle with center \((\frac{b}{2}, \frac{a}{2})\) and radius \(\frac{\sqrt{a^2 + b^2}}{2}\).

Step 1: Recognize the Polar Equation

The given equation is in polar form: \(r = a \sin \theta + b \cos \theta\). We need to transform this polar equation into a Cartesian form to identify the geometric shape.

Step 2: Express Polar Coordinates in Cartesian Form

Recall that in polar coordinates, \(r\) is the radial distance, and \(\theta\) is the angle. The relationships between polar and Cartesian coordinates are \(x = r \cos \theta\) and \(y = r \sin \theta\). We can use these to substitute for \(x\) and \(y\).

Step 3: Substitute to Find Cartesian Equation

Multiply both sides of the equation by \(r\) to get rid of the division in polar coordinates: \[r^2 = r(a \sin \theta + b \cos \theta)\] Using \(x = r \cos \theta\) and \(y = r \sin \theta\), we have: \[x^2 + y^2 = ya + xb\]

Step 4: Rearrange into Circle Equation Form

Rearrange the equation to have all terms on one side: \[x^2 + y^2 - ya - xb = 0\] This can be rewritten as: \[(x - \frac{b}{2})^2 + (y - \frac{a}{2})^2 = \left(\frac{a^2 + b^2}{4}\right)\] by completing the square.

Step 5: Identify the Center and Radius

After completing the square, the equation is now in the standard form of a circle: \[(x - h)^2 + (y - k)^2 = r^2\] Here, the center \((h, k)\) is \(\left(\frac{b}{2}, \frac{a}{2}\right)\) and the radius \(r\) is \(\sqrt{\frac{a^2 + b^2}{4}}\), which simplifies to \(\frac{\sqrt{a^2 + b^2}}{2}\).

Key Concepts

Circle Equation

To understand the shape represented by the polar equation \(r = a \sin \theta + b \cos \theta\), we need to delve into the concept of a circle's equation.
A circle in the Cartesian coordinate system is characterized by the equation \((x - h)^2 + (y - k)^2 = r^2\). Here:

• \((h, k)\) represents the center of the circle.
• \(r\) denotes the radius.

In our exercise, by transforming the given polar equation into a format that resembles this standard circle equation, we verify that it indeed sketches a circle in the Cartesian plane.
The equation reveals that the location of the circle and its size (radius) can be identified after the transformation and reformation steps.

Completing the Square

Completing the square is a technique used to rewrite quadratic equations, like those for circles, in a more interpretable form. This method allows us to convert a general quadratic equation into its standard form.
For example:

• Start with a general quadratic like \(x^2 + y^2 - ya - xb = 0\).
• Reorganize it into a perfect square structure, such as \((x - \frac{b}{2})^2 + (y - \frac{a}{2})^2 = \left(\frac{a^2 + b^2}{4}\right)\).

The technique involves adding and subtracting terms to both sides, ensuring that both \(x\) and \(y\) terms are perfect squares. This method redefines the equation into a recognizable circle equation, revealing the circle's center and radius.
It's a crucial skill in algebra, particularly when dealing with conic sections like circles.

Polar Form Equation

The polar form equation expresses curves like circles in terms of the radial distance \(r\) and the angle \(\theta\).
In polar coordinates:

• \(r\) signifies the distance from the origin to a point.
• \(\theta\) is the angle measured from the positive x-axis.

For our specific equation \(r = a \sin \theta + b \cos \theta\), it indicates a dependence of \(r\) on the trigonometric functions of the angle \(\theta\), reflecting how a point moves around a potential circle.
Recognizing and transforming this into a Cartesian framework allows us to identify the conic section precisely and aids greatly in graphing and understanding its parameters.

Cartesian Form Transformation

Transforming polar equations into Cartesian form involves using relationships between these coordinate systems: \(x = r \cos \theta\) and \(y = r \sin \theta\).
This transformation facilitates:

• Understanding complex shapes in more familiar Cartesian terms.
• Plugging into the equations \(x\) and \(y\) directly to identify shapes and descriptors.

By multiplying the entire polar equation by \(r\), we eliminate the division inherent in polar coordinates, converting it into something workable in Cartesian terms.
This step bridges the gap between polar and Cartesian systems, allowing for a comprehensive analysis of the shape, often simplifying understanding and ensuring connectivity between different mathematical perspectives.