91Ó°ÊÓ

To identify the range of θ values for one leaf, we need to find the range of values between consecutive points where r = 0. Solve for θ: $$ r = \cos 5\theta $$ $$ 0 = \cos 5\theta $$ $$ 5\theta = \frac{(2n+1) \pi}{2} $$ (for integer values of n) $$ \theta = \frac{(2n+1) \pi}{10} $$ So, the θ values are: $$ \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, ... $$ For one leaf, θ ranges between consecutive values. Therefore, the range of θ for one leaf is: $$ \frac{\pi}{10} \le \theta \le \frac{3\pi}{10} $$

To easily sketch the curve, we can convert the polar equation to a Cartesian equation: $$ r = \cos 5\theta $$ Using the conversion formulas: $$ x = r \cos \theta $$ $$ y = r \sin \theta $$ We get: $$ x = \cos 5\theta \cos \theta $$ $$ y = \cos 5\theta \sin \theta $$ To eliminate θ, we can employ the trigonometric identity \(\sin 2 \alpha = 2 \sin \alpha \cos \alpha\). First, let \(2\alpha = 5\theta\). Then, $$ x = \frac{1}{2} \sin(10\theta) $$ $$ y = \frac{1}{2} \sin(10\theta) \cot \theta $$ Now, you can easily sketch the bounding curve using a graphing calculator or software.

To find the area of the region inside one leaf, we can use the formula for the area in polar coordinates: $$ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta $$ As we determined in Step 1, the range of θ for one leaf is between \(\frac{\pi}{10}\) and \(\frac{3\pi}{10}\). Substituting these values and the equation \(r = \cos 5\theta\) into the area formula, we get: $$ A = \frac{1}{2} \int_{\frac{\pi}{10}}^{\frac{3\pi}{10}} (\cos 5\theta)^2 d\theta $$ Now, we can evaluate the integral: $$ A = \frac{1}{2} \int_{\frac{\pi}{10}}^{\frac{3\pi}{10}} \frac{1}{2}(1 + \cos(10\theta)) d\theta $$ $$ A = \frac{1}{4} \left[ \int_{\frac{\pi}{10}}^{\frac{3\pi}{10}} d\theta + \int_{\frac{\pi}{10}}^{\frac{3\pi}{10}} \cos(10\theta) d\theta \right] $$ $$ A = \frac{1}{4} \left[ \theta \Big|_{\frac{\pi}{10}}^{\frac{3\pi}{10}} + \frac{1}{10} \sin(10\theta) \Big|_{\frac{\pi}{10}}^{\frac{3\pi}{10}} \right] $$ $$ A = \frac{1}{4} \left[ \frac{2\pi}{10} + \frac{1}{10} (\sin(3\pi) - \sin(\pi)) \right] $$ $$ A = \frac{1}{4} \left[ \frac{2\pi}{10} \right] $$ $$ A = \frac{1}{4} \cdot \frac{\pi}{5} $$ $$ A = \frac{\pi}{20} $$ So, the area of the region inside one leaf of the rose \(r = \cos 5 \theta\) is \(\frac{\pi}{20}\).