Let's calculate values for \( r \), \( x \), and \( y \) for the given polar equation: \[ r = 6 - 6\cos\theta \] We need to pick enough points for \(\theta\) (e.g., \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\)). Recall the conversions: - \( x = r\cos\theta \) - \( y = r\sin\theta \) Let's calculate for: - \(\theta = 0\) - \(\theta = \frac{\pi}{2}\) - \(\theta = \pi\) - \(\theta = \frac{3\pi}{2}\) --- ### Calculations #### 1. \(\theta = 0\) - \( r = 6 - 6\cos(0) = 6 - 6(1) = 0 \) - \( x = 0 \cdot \cos(0) = 0 \) - \( y = 0 \cdot \sin(0) = 0 \) #### 2. \(\theta = \frac{\pi}{2}\) - \( r = 6 - 6\cos(\frac{\pi}{2}) = 6 - 6(0) = 6 \) - \( x = 6 \cdot \cos(\frac{\pi}{2}) = 6 \cdot 0 = 0 \) - \( y = 6 \cdot \sin(\frac{\pi}{2}) = 6 \cdot 1 = 6 \) #### 3. \(\theta = \pi\) - \( r = 6 - 6\cos(\pi) = 6 - 6(-1) = 6 + 6 = 12 \) - \( x = 12 \cdot \cos(\pi) = 12 \cdot (-1) = -12 \) - \( y = 12 \cdot \sin(\pi) = 12 \cdot 0 = 0 \) #### 4. \(\theta = \frac{3\pi}{2}\) - \( r = 6 - 6\cos(\frac{3\pi}{2}) = 6 - 6(0) = 6 \) - \( x = 6 \cdot \cos(\frac{3\pi}{2}) = 6 \cdot 0 = 0 \) - \( y = 6 \cdot \sin(\frac{3\pi}{2}) = 6 \cdot (-1) = -6 \) --- ### Table | | 1 (\(\theta = 0\)) | 2 (\(\theta = \frac{\pi}{2}\)) | 3 (\(\theta = \pi\)) | 4 (\(\theta = \frac{3\pi}{2}\)) | |-----|---------------------|-------------------------------|----------------------|----------------------------------| | R | 0 | 6 | 12 | 6 | | X | 0 | 0 | -12 | 0 | | Y | 0 | 6 | 0 | -6 | You can use these points to sketch the rough graph of the equation \( r = 6 - 6\cos\theta \). The graph will look like a limaçon with an inner loop.