Problem 52 - Calculus Early Transcendentals

Question

Assume \(m\) is a positive integer. a. Even number of leaves: What is the relationship between the total area enclosed by the \(4 m\) -leaf rose \(r=\cos (2 m \theta)\) and \(m ?\) b. Odd number of leaves: What is the relationship between the total area enclosed by the \((2 m+1)\) -leaf rose \(r=\cos ((2 m+1) \theta)\) and \(m ?\)

Step-by-Step Solution

Answer

Answer: The relationship between the total area enclosed by rose curves and the parameter m can be expressed as follows: a. For even numbers of leaves (4m-leaf rose): $$A_{even}=\frac{\pi m}{2}$$ b. For odd numbers of leaves ((2m+1)-leaf rose): $$A_{odd}=\frac{\pi}{2(2m+1)}$$

Step 1: Area of polar curves formula

In order to find the area enclosed by a polar curve given in the form \(r=f(\theta)\), we can use the following formula: $$A=\frac{1}{2}\int_{\alpha}^{\beta}r^2 d\theta$$ where \(\alpha\) and \(\beta\) are the limits of integration that we'll determine based on the symmetry properties of the rose curves.

Step 2: Limits of integration for even number of leaves (\(4m\)-leaf rose)

For the \(4m\)-leaf rose, the equation is given as \(r=\cos(2m\theta)\). Observe that this curve has a period of \(\frac{\pi}{m}\) due to the \(2m \theta\) term. Moreover, one period will cover two leaves of the curve. Therefore, to capture all \(4m\) leaves, we integrate over \(0 \leq \theta \leq 2\pi m\). So, the limits of integration are \(\alpha = 0\) and \(\beta = 2\pi m\).

Step 3: Calculate the area for even number of leaves

Now, we can plug our limits of integration and the function into our formula to calculate the area for the even number of leaves case: $$A_{even}=\frac{1}{2}\int_{0}^{2\pi m}\cos^2(2m\theta)d\theta$$ To simplify this integral, we can use the double-angle formula for cosine: $$\cos^2(2m\theta) = \frac{1+\cos(4m\theta)}{2}$$ Now, our integral becomes: $$A_{even}=\frac{1}{2}\int_{0}^{2\pi m} \frac{1+\cos(4m\theta)}{2}d\theta$$ We can separate the integral into two parts: $$A_{even}=\frac{1}{4}\int_{0}^{2\pi m}d\theta + \frac{1}{4}\int_{0}^{2\pi m}\cos(4m\theta)d\theta$$ The first integral evaluates to \(\frac{1}{4}(2\pi m) = \frac{\pi m}{2}\). The second integral evaluates to zero since it represents the integral of a single period of a cosine function over its entire period. Thus, the total area enclosed for the even number of leaves is: $$A_{even}=\frac{\pi m}{2}$$

Step 4: Limits of integration for odd number of leaves (\((2m+1)\)-leaf rose)

For the \((2m+1)\)-leaf rose, the equation is given as \(r=\cos((2m+1)\theta)\). Observe that this curve has a period of \(\frac{2\pi}{(2m+1)}\). Since one period corresponds to all the \((2m+1)\) leaves, our limits of integration are \(\alpha = 0\) and \(\beta = \frac{2\pi}{2m+1}\).

Step 5: Calculate the area for odd number of leaves

Now, we can plug our limits of integration and the function into our formula to calculate the area for the odd number of leaves case: $$A_{odd}=\frac{1}{2}\int_{0}^{\frac{2\pi}{2m+1}}\cos^2((2m+1)\theta)d\theta$$ We also use the double-angle formula for cosine to simplify this integral: $$\cos^2((2m+1)\theta) = \frac{1+\cos(2(2m+1)\theta)}{2}$$ Now, our integral becomes: $$A_{odd}=\frac{1}{2}\int_{0}^{\frac{2\pi}{2m+1}} \frac{1+\cos(4m\theta+2\theta)}{2}d\theta$$ We can separate the integral into two parts like before: $$A_{odd}=\frac{1}{4}\int_{0}^{\frac{2\pi}{2m+1}}d\theta + \frac{1}{4}\int_{0}^{\frac{2\pi}{2m+1}}\cos(4m\theta+2\theta)d\theta$$ The first integral evaluates to \(\frac{1}{4}\left(\frac{2\pi}{2m+1}\right) = \frac{\pi}{2(2m+1)}\). Similar to the even case, the second integral evaluates to zero. Thus, the total area enclosed for the odd number of leaves is: $$A_{odd}=\frac{\pi}{2(2m+1)}$$ So, the relationship between the total area enclosed and \(m\) is given as follows: a. Even number of leaves (\(4m\)-leaf rose): $$A_{even}=\frac{\pi m}{2}$$ b. Odd number of leaves (\((2m+1)\)-leaf rose): $$A_{odd}=\frac{\pi}{2(2m+1)}$$