Problem 45 - Precalculus with Calculus Previews

Question

The given circle is rotated about the origin in the indicated direction and by the indicated amount. (a) Find a polar equation of the rotated circle (the red circle in the figure). (b) Use the polar equation of the rotated circle to find its standard-form rectangular equation. (c) Find polar and rectangular coordinates of the center of the rotated circle. $$ r=2 \cos \theta ; \text { clockwise, } \pi / 6 $$

Step-by-Step Solution

Answer

(a) New polar equation: \( r = 2 \cos(\theta + \frac{\pi}{6}) \). (b) Convert using trigonometric identities. (c) Center: polar \((1, -\frac{\pi}{6})\), rectangular \((\sqrt{3}/2, -1/2)\).

Step 1: Identify the original polar equation

The given polar equation is \( r = 2 \cos \theta \). This represents a circle with a diameter of 2 units that is centered at \( (1, 0) \) in Cartesian coordinates.

Step 2: Transform the polar equation

To rotate the circle clockwise by \( \frac{\pi}{6} \), replace \( \theta \) with \( \theta + \frac{\pi}{6} \) in the equation. Thus, the rotated polar equation becomes \( r = 2 \cos(\theta + \frac{\pi}{6}) \).

Step 3: Convert the polar equation to rectangular form

Recall that \( x = r \cos \theta \) and \( y = r \sin \theta \). Substitute \( r = 2 \cos(\theta + \frac{\pi}{6}) \) and use the cosine angle addition identity:\[ \cos(\theta + \frac{\pi}{6}) = \cos \theta \cos \frac{\pi}{6} - \sin \theta \sin \frac{\pi}{6} \]This becomes \( r = 2(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta) \). Substitute into the rectangular forms:\( x = r \cos \theta = 2(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta) \cos \theta \)\( y = r \sin \theta = 2(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta) \sin \theta \)Simplifying gives the rectangular equation, which may require further algebraic manipulation.

Step 4: Determine coordinates of the center

The center of the original circle in polar is \((1, 0)\) and in rectangular is \((1, 0)\). After rotation by \( -\frac{\pi}{6} \), the center polar coordinates become \((1, -\frac{\pi}{6})\), and converting to rectangular gives new coordinates \((\sqrt{3}/2, -1/2)\), using the relationships between polar and Cartesian coordinates. These calculations use rotation matrix properties or transformation guidelines.

Key Concepts

Circle Rotation

When a circle is rotated, its position and orientation in the coordinate plane changes, but not its size or shape. In our exercise, we need to rotate the given circle defined by the polar equation \(r = 2 \cos \theta\). To achieve this, we perform a rotation around the origin (center of coordinates). The amount of rotation specified is \(\frac{\pi}{6}\) radians in the clockwise direction. It's important to remember:

• Rotation does not alter the circle’s radius or size.
• Clockwise rotation involves decreasing the angle \(\theta\) by the given amount.

To rotate our circle, we substitute \(\theta\) with \(\theta + \frac{\pi}{6}\) in the equation. This transforms our original polar equation into \(r = 2 \cos(\theta + \frac{\pi}{6})\). You can visualize the effect of rotation as shifting the entire circle along its circumference in the specified direction.

Coordinate Transformation

Coordinate transformation is crucial for switching between polar and rectangular forms, especially after geometric manipulations like rotations. For the rotated circle described by the polar equation \(r = 2 \cos(\theta + \frac{\pi}{6})\), we need polar to rectangular coordinate conversion. To do this, use these transformation formulas:

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

However, since \( \theta\) has been adjusted for rotation, we employ the cosine angle addition formula: \[\cos(\theta + \frac{\pi}{6}) = \cos \theta \cos \frac{\pi}{6} - \sin \theta \sin \frac{\pi}{6}\]This substitution helps in developing the expressions for \(x\) and \(y\) in terms of \(\theta\). Coordinate transformation simplifies analyzing complex geometrical changes and understanding them in a more intuitive form on the Cartesian plane.

Rectangular Equations

The transition from polar to rectangular equations elucidates the circle's properties in terms of \(x\) and \(y\) coordinates, rather than \(r\) and \(\theta\). After handling the circle rotation and coordinate transformation, our task is to represent the rotated circle in rectangular terms.Given:

• \(x = r \cos \theta = 2(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta) \cos \theta\)
• \(y = r \sin \theta = 2(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta) \sin \theta\)

Ultimately, this results in a rectangular expression that may require further algebraic simplification to discern the circle's equation explicitly. Understanding rectangular equations allows for better integration with other Cartesian forms and makes solving systems of equations more straightforward. Rectangular equations also make it easier to identify intercepts and symmetries within the Cartesian coordinate system. In essence, mastering these conversions builds a robust foundation for more advanced analyses in mathematics and physics.