Problem 45 - Thomas Calculus

Question

Replace the polar equations in Exercises \(23-48\) by equivalent Cartesian equations. Then describe or identify the graph. $$ r=2 \cos \theta+2 \sin \theta $$

Step-by-Step Solution

Answer

The Cartesian equation is \((x-1)^2 + (y-1)^2 = 2\), which represents a circle centered at (1, 1) with a radius of \(\sqrt{2}\).

Step 1: Express r in terms of x and y

In polar coordinates, \( r = \sqrt{x^2 + y^2} \), \( \cos \theta = \frac{x}{r} \), and \( \sin \theta = \frac{y}{r} \). Substitute these into the given polar equation \( r = 2 \cos \theta + 2 \sin \theta \):\[\sqrt{x^2 + y^2} = 2\left(\frac{x}{\sqrt{x^2 + y^2}}\right) + 2\left(\frac{y}{\sqrt{x^2 + y^2}}\right)\]

Step 2: Eliminate the square roots

Multiply both sides of the equation by \( \sqrt{x^2 + y^2} \) to clear the denominators:\[x^2 + y^2 = 2x + 2y\]

Step 3: Rearrange the equation into standard form

Rearrange terms to bring the equation to a standard quadratic form:\[x^2 + y^2 - 2x - 2y = 0\]

Step 4: Complete the square

Complete the square for both \(x\) and \(y\). For \(x\): \[x^2 - 2x = (x-1)^2 - 1\]For \(y\):\[y^2 - 2y = (y-1)^2 - 1\]Substitute back into the equation:\[(x-1)^2 - 1 + (y-1)^2 - 1 = 0\]

Step 5: Simplify the completed square

Combine constants and simplify:\[(x-1)^2 + (y-1)^2 = 2\]

Step 6: Identify the graph

The equation \((x-1)^2 + (y-1)^2 = 2\) represents a circle with center at \((1, 1)\) and radius \(\sqrt{2}\).

Key Concepts

Completing the Square

Completing the square is a key algebraic technique used to transform a quadratic expression into a more convenient form. This method helps identify geometric features like vertices or centers in equations more easily. When dealing with an equation such as \[ x^2 - 2x + y^2 - 2y = 0, \]we can apply completing the square for both the \( x \) and \( y \) terms.

• For \(x^2 - 2x\), we add and subtract 1 to form \((x-1)^2 - 1\).
• For \(y^2 - 2y\), we similarly add and subtract 1, resulting in \((y-1)^2 - 1\).

By substituting these values back, we simplify to:\[(x-1)^2 + (y-1)^2 = 2.\]This transformation clarifies the equation, making it easier to interpret graphically.

Circle Equation

The equation obtained from the completing the square process reveals a fundamental shape in geometry: a circle. The standard form for a circle's equation is:\[(x-h)^2 + (y-k)^2 = r^2,\]where \( (h, k) \) is the center of the circle and \( r \) is the radius.

In our case, the circle equation \[(x-1)^2 + (y-1)^2 = 2\]indicates the circle has:

• Center at the point \((1, 1)\)
• Radius of \(\sqrt{2}\)

This form allows an easy identification of the circle's location and size, which is critical for graphing and analysis.

Graph Identification

Once the equation is in an understandable form, identifying the graph becomes straightforward. Knowing that \[ (x-1)^2 + (y-1)^2 = 2 \]is a circle, we can predict significant features about its graph.

• The circle is centered at \((1, 1)\).
• It has a radius of \(\sqrt{2}\), which means any point on the circle is \(\sqrt{2}\) units away from the center.

While plotting, you start by marking the center and then measure \(\sqrt{2}\) units in all directions to outline the perimeter. This simple process aids in visualizing the equation's geometric representation in the coordinate plane.