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 \(m ?\)

Short answer

Answer: For the \(4m\)-leaf case (even values of \(m\)), the total area enclosed is given by the expression: \(A_{\text{total}} = 2m \int_{\frac{\pi}{4m}}^{\frac{3\pi}{4m}} \cos^2(2m\theta) d\theta\) For the \((2m+1)\)-leaf case (odd values of \(m\)), the total area enclosed is given by the expression: \(A_{\text{total}} = \frac{1}{2}(2m+1) \int_{\frac{\pi}{(2m+1)}}^{\frac{3\pi}{(2m+1)}} \cos^2((2m+1)\theta) d\theta\)

Step-by-step solution

Find the area enclosed by the rose curves using polar integration

We can find the area enclosed by the rose curve by using the following formula: \(A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta\) For both (a) and (b), we will find the area by integrating over the region that covers one leaf of the rose curve. Once we find the area of one leaf, we can multiply it by the number of leaves to get the total area.

Finding area enclosed for (a)

First, we need to find the limits of integration for one leaf. The bounds of integration occur when \(r=0\). Therefore, we can set \(r=\cos(2m\theta)=0\) and solve for \(\theta\). For the even case, we can find \(\theta\) values for which the cosine function equals 0: \(\cos(2m\theta)=0 \Rightarrow 2m\theta = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow \theta = \frac{\pi}{4m}, \frac{3\pi}{4m}\) Now, we can calculate the area of one leaf: \(A_{\text{one leaf}} = \frac{1}{2} \int_{\frac{\pi}{4m}}^{\frac{3\pi}{4m}} \cos^2(2m\theta) d\theta\) For even \(m\), there are \(4m\) leaves. Hence, the total area enclosed by the \(4m\)-leaf rose is: \(A_{\text{total}} = 4m A_{\text{one leaf}} = 2m \int_{\frac{\pi}{4m}}^{\frac{3\pi}{4m}} \cos^2(2m\theta) d\theta\)

Finding area enclosed for (b)

Similarly to step 2, we will find the bounds of integration for one leaf in the odd case by solving \(r=\cos((2m+1)\theta)=0\): \((2m+1)\theta = \pi, 3\pi \Rightarrow \theta = \frac{\pi}{(2m+1)}, \frac{3\pi}{(2m+1)}\) Now, we can find the area of one leaf: \(A_{\text{one leaf}} = \frac{1}{2} \int_{\frac{\pi}{(2m+1)}}^{\frac{3\pi}{(2m+1)}} \cos^2((2m+1)\theta) d\theta\) For odd \(m\), there are \(2m+1\) leaves. Thus, the total area enclosed by the \((2m+1)\)-leaf rose is: \(A_{\text{total}} = (2m+1) A_{\text{one leaf}} = \frac{1}{2}(2m+1) \int_{\frac{\pi}{(2m+1)}}^{\frac{3\pi}{(2m+1)}} \cos^2((2m+1)\theta) d\theta\) Now we have found the relationship between the enclosed areas for both types of rose curves and the integer \(m\).

Key concepts

Rose Curve

The rose curve is a famous type of polar graph, named for its flower-like pattern with symmetrical 'leaves' that emanate from the origin. It is represented by the equation \(r = \text{cos}(k\theta)\) or \(r = \text{sin}(k\theta)\), where \(k\) is a constant that determines the number of 'leaves' or 'petals' the curve has. If \(k\) is an integer, the curve will have either \(k\) leaves if \(k\) is odd, or \(2k\) leaves if \(k\) is even.

Understanding the structure of a rose curve is crucial when calculating its area, particularly as it relates to symmetry and periodicity. The curves exhibit repetitive behavior, making it possible to calculate the area of one 'leaf' and then multiply by the total number of leaves to find the entire area enclosed by the curve. A characteristic of rose curves is they intersect themselves, and the area calculations need to take into account the overlapping regions to avoid counting any region more than once.

Area of Polar Curves

Calculating the area enclosed by a curve in polar coordinates involves integrating with respect to the angle \( \theta \). For polar curves, the formula to find the area \( A \) within a region bounded by \( \theta = \theta_1 \) and \( \theta = \theta_2 \) is given by:
\[ A = \frac{1}{2} \text{integrate} \big( r^2 \big) d\theta \]
where \( r \) is the polar radius as a function of \( \theta \). The integral of \( r^2 \) with respect to \( \theta \) gives the area of a sector, and the factor of \(\frac{1}{2}\) is applied since the actual area is half of what the integral calculates. To find the total area enclosed by recurring patterns, such as the leaves of a rose curve, one typically computes the area of a single region and then multiplies it by the number of identical regions. It is important to correctly identify the bounds of integration, \( \theta_1 \) and \( \theta_2 \), to ensure the integration is performed over exactly one region or pattern of the curve.

Calculus

Calculus plays an instrumental role in understanding and analyzing continuous phenomena, which is particularly evident in the study of polar curves. Fundamental tools of calculus, namely differentiation and integration, are extensively used to describe and quantify the properties of these curves. In the context of polar graphs, integration is used to find areas and distances, whereas differentiation can help in determining slopes and tangent lines.

In the context of polar graphs, such as the rose curve, calculus not only measures simple one-dimensional distances but also calculates two-dimensional areas confined by the curves. Identifying the integration limits and setting up the integral properly are key steps in calculus that lead to accurate results. Moreover, integration techniques, such as substitution or trigonometric identities, often simplify the integrals that arise in the computation of areas enclosed by polar curves. These concepts reflect the real-world applications of calculus, from physics to engineering, by providing the tools to model and solve problems with precision.