Question

Find the area of the described region. Enclosed by one petal of the rose curve \(r=\cos (n \theta),\) where \(n\) is a positive integer.

Short answer

The area of one petal of the curve is \(\frac{\pi}{4n}\).

Step-by-step solution

Identify the form of the rose curve

The rose curve is described by the polar equation \(r = \cos(n\theta)\). Since \(n\) is a positive integer, the total number of petals is given by \(n\) if \(n\) is odd and \(2n\) if \(n\) is even.

Determine bounds for a single petal

To find the bounds for a single petal, determine the range of \(\theta\) for one petal. Since the rose curve is symmetric, a single petal appears as \(\theta\) ranges from 0 to \(\pi/n\).

Set up the integral for the area

The area \(A\) of a single petal in polar coordinates is given by the integral: \[A = \frac{1}{2} \int_{a}^{b} (r^2) \, d\theta\]Plugging in the values from the rose curve \(r = \cos(n\theta)\) and the bounds \(\theta = 0\) to \(\theta = \pi/n\), we have:

Compute the integral

Substitute \(r = \cos(n\theta)\) into the area formula:\[A = \frac{1}{2} \int_{0}^{\pi/n} (\cos^2(n\theta)) \, d\theta\]Using the identity \(\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\), the integral becomes:\[A = \frac{1}{2} \int_{0}^{\pi/n} \frac{1}{2} (1 + \cos(2n\theta)) \, d\theta = \frac{1}{4} \int_{0}^{\pi/n} (1 + \cos(2n\theta)) \, d\theta\]

Solve the integral

Compute the integral:\[A = \frac{1}{4} \left[\theta + \frac{1}{2n} \sin(2n\theta) \right]_0^{\pi/n}\]Evaluating, we get:\[A = \frac{1}{4} \left( \frac{\pi}{n} + \frac{1}{2n}(\sin(2\pi) - \sin(0)) \right) = \frac{1}{4} \cdot \frac{\pi}{n} = \frac{\pi}{4n}\]

Conclusion

Thus, the area of one petal of the rose curve \(r = \cos(n\theta)\) is \(\frac{\pi}{4n}\).

Key concepts

Integrals

Integrals are a fundamental concept in calculus. They are used to calculate the area under a curve, among other things. In the case of polar coordinates, like rose curves, integrals help us find the area enclosed by the curves. Integrals can be thought of as a sum of infinitesimally small areas under the curve.
In polar coordinates, we often use the formula \[A = \frac{1}{2} \int_{a}^{b} (r^2) \, d\theta\]to find the area. Here, \( r \) is the distance from the origin to a point on the curve, and \( \theta \) measures the angle. This formula is adapted from the rectangular coordinates formula, considering the sector of a circle.
The process involves setting up an integral with the correct bounds and function for \( r \). Then, you evaluate the integral to find the exact area. This method is particularly useful for complex shapes like rose curves, where direct geometric formulas may not suffice.

Rose Curves

Rose curves are a type of polar graph, described by equations like \( r = \cos(n \theta) \) or \( r = \sin(n \theta) \). They are called rose curves because they look like petals of a rose. Depending on the integer \( n \), the number of petals changes.

• If \( n \) is odd, the curve will have \( n \) petals.
• If \( n \) is even, it will have \( 2n \) petals.

These curves are symmetric, making them visually appealing and mathematically interesting. To find the area of a single petal, it is crucial to understand the symmetry and equation of the curve. For a single petal, the range for \( \theta \) is typically from 0 to \( \pi/n \). This range ensures the integral only covers one complete petal.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variable. They are vital tools in simplifying and solving trigonometry problems. One common identity used in calculus is the double angle formula for cosine: \[\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\]This identity helps in converting a square of a cosine function into a form that is easier to integrate.
In the context of calculating areas in polar coordinates, especially for equations like \( r = \cos(n\theta) \), using this identity transforms a challenging integral into a manageable one. By substituting the identity, you simplify the integrand and can find the integral of trigonometric functions efficiently. These identities are powerful, making complex expressions more straightforward and integrations doable.