Problem 19 - Thomas Calculus

Question

One leaf of a rose Find the area enclosed by one leaf of the rose \(r=12 \cos 3 \theta\)

Step-by-Step Solution

Answer

The area is \(12\pi\).

Step 1: Identify the Equation and Parameters

The equation given is the polar equation of a rose curve: \(r=12 \cos 3 \theta\). This equation represents a rose curve with petals, where \(n\) in \(r = a \cos(n\theta)\) or \(r = a \sin(n\theta)\) determines the number of petals. Here, \(a = 12\) and \(n = 3\), indicating that there are 3 petals in the curve.

Step 2: Determine the Range for One Petal

For \(r = 12 \cos 3\theta\), the range of \(\theta\) that gives one complete petal is \([0, \pi/3]\). This is because each petal of a cosine rose curve \(r = a \cos(n\theta)\) forms when the angle varies from \(0\) to \(\pi/n\).

Step 3: Use the Polar Area Formula

The area \(A\) enclosed by a curve \(r(\theta)\) from \(\theta = \alpha\) to \(\theta = \beta\) is given by the formula: \[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \ d\theta\]Plug in the values: \[\begin{align*}A &= \frac{1}{2} \int_{0}^{\pi/3} (12 \cos 3\theta)^2 \ d\theta \&= \frac{1}{2} \int_{0}^{\pi/3} 144 \cos^2 3\theta \ d\theta\end{align*}\]

Step 4: Simplify Using Trigonometric Identity

The identity \(\cos^2 3\theta = \frac{1+\cos 6\theta}{2}\) will simplify the integral. Substitute it into the integral:\[\begin{align*}A &= \frac{1}{2} \int_{0}^{\pi/3} 144 \left(\frac{1+\cos 6\theta}{2}\right)\ d\theta \&= 72 \int_{0}^{\pi/3} \left(\frac{1}{2} + \frac{1}{2}\cos 6\theta\right) d\theta\end{align*}\]

Step 5: Evaluate the Integral

Split the integral into two parts and evaluate:\[\begin{align*}A &= 72 \int_{0}^{\pi/3} \frac{1}{2} d\theta + 72 \int_{0}^{\pi/3} \frac{1}{2} \cos 6\theta \ d\theta \&= 36 \left[\theta\right]_{0}^{\pi/3} + 36 \left[\frac{1}{6} \sin 6\theta\right]_{0}^{\pi/3}\end{align*}\]Calculating each:\[\begin{align*}36 \left[\theta\right]_{0}^{\pi/3} &= 36 \cdot \frac{\pi}{3} - 36 \cdot 0 = 12\pi \36 \left[\frac{1}{6} \sin 6\theta\right]_{0}^{\pi/3} &= 36 \cdot 0 - 36 \cdot 0 = 0 \\end{align*}\]Thus, \(A = 12\pi\).

Step 6: Conclusion

The area enclosed by one petal of the rose curve \(r=12 \cos 3 \theta\) is \(12\pi\).

Key Concepts

Rose Curve

A rose curve is a fascinating mathematical concept commonly encountered in polar coordinates. They are captivating because they form a petal-like pattern on a graph. The general equations for a rose curve are either \( r = a \cos(n\theta) \) or \( r = a \sin(n\theta) \), where:

• \(a\) is a coefficient that determines the length of each petal.
• \(n\) is the number of petals if \(n\) is odd, or twice the number if \(n\) is even.

In our exercise, the rose curve described by \( r = 12 \cos 3\theta \) represents a rose with three petals because the value of \(n\) is 3, which is an odd number. The most intriguing aspect of rose curves is their symmetry and how they repeat their pattern in a circular manner.

Integral Calculus

Integral calculus is a critical branch of mathematics that deals with the accumulation of quantities and the areas under curves. In our polar coordinate exercise, we use integrals to find the area within one petal of the rose curve. Polar coordinates add a layer of complexity to integration because the curves are not in the traditional \((x, y)\) Cartesian format. Instead, they are defined in terms of \( (r, \theta) \), where \(r\) is the radius from the origin and \(\theta\) is the angle.
To integrate in polar coordinates, we use specific formulas that account for this curvature, ensuring we accurately find areas or accumulate sums along the curve. Integral calculus is not only central to mathematics but also to physics, engineering, and other sciences, making it an essential tool for understanding the behavior of different systems.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for all values of the variables involved. One of the key identities used in our exercise is \(\cos^2\theta = \frac{1 + \cos(2\theta)}{2}\). This identity simplifies the integration process significantly. It helps transform complex trigonometric integrals into simpler forms that are easier to solve.
Such identities are vital because they link different trigonometric expressions, providing elegant solutions to otherwise complicated problems. By employing identities like these, we can simplify problems, which makes finding solutions more manageable. These simplifications greatly impact not just mathematics, but also signal processing, engineering, and data analysis where trigonometric functions are prevalent.

Area Calculation

Area calculation in polar coordinates uses a unique approach compared to Cartesian coordinates. The formula for finding the area enclosed by a polar curve from \(\theta = \alpha\) to \(\theta = \beta\) is \( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \). This formula accounts for the circular nature of the curve.
In our specific exercise, we calculated the area of one petal of the rose curve \(r=12 \cos 3 \theta\). We determined the bounds as \([0, \pi/3]\) because this range captures one complete petal. By substituting \(r = 12 \cos 3\theta\) into the formula, we could compute the area accurately. Area calculation in polar coordinates often requires these thoughtful steps to determine precise boundaries and the appropriate form of \(r\) to ensure an accurate measurement.