Problem 45 - Calculus Early Transcendentals

Question

\(45-48\) Find the exact length of the polar curve. $$ r=3 \sin \theta, \quad 0 \leqslant \theta \leqslant \pi / 3 $$

Step-by-Step Solution

Answer

The length of the curve is \( \pi \).

Step 1: Understand the Polar Curve

The polar equation given is \( r = 3 \sin \theta \). This represents a circle where the radius varies with \( \theta \). The problem asks us to find the length of this curve from \( \theta = 0 \) to \( \theta = \pi/3 \).

Step 2: Use the Formula for Arc Length

The formula for the arc length \( L \) of a polar curve \( r(\theta) \) from \( \theta = a \) to \( \theta = b \) is: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta \]. In this case, \( a = 0 \) and \( b = \pi/3 \), and \( r = 3 \sin \theta \).

Step 3: Find Derivative of r

Calculate \( \frac{dr}{d\theta} \) where \( r = 3 \sin \theta \). Thus, \( \frac{dr}{d\theta} = 3 \cos \theta \).

Step 4: Substitute in the Arc Length Formula

Substitute \( r \) and \( \frac{dr}{d\theta} \) into the arc length formula: \[ L = \int_{0}^{\pi/3} \sqrt{(3 \cos \theta)^2 + (3 \sin \theta)^2} \, d\theta \].

Step 5: Simplify the Expression

Simplify the expression under the square root: \[ (3 \cos \theta)^2 + (3 \sin \theta)^2 = 9 \cos^2 \theta + 9 \sin^2 \theta = 9(\cos^2 \theta + \sin^2 \theta) = 9 \]. Therefore, the integral becomes \[ L = \int_{0}^{\pi/3} \sqrt{9} \, d\theta = \int_{0}^{\pi/3} 3 \, d\theta \].

Step 6: Evaluate the Integral

Evaluate the integral: \[ L = \int_{0}^{\pi/3} 3 \, d\theta = 3\left[\theta\right]_{0}^{\pi/3} = 3\left(\frac{\pi}{3} - 0\right) = \pi \].

Step 7: Conclusion

Thus, the exact length of the polar curve \( r = 3 \sin \theta \) from \( \theta = 0 \) to \( \theta = \pi/3 \) is \( \pi \).

Key Concepts

Arc Length of Polar Curves

When dealing with polar curves, finding the arc length can involve a bit more complexity compared to Cartesian curves. The length of a polar curve is measured between two values of \(\theta\), given by the formula: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta \]Here, \(a\) and \(b\) are the bounds for \(\theta\). This formula incorporates both the rate at which the radius changes with \(\theta\) (\(\frac{dr}{d\theta}\)) and the radius itself (\(r\)), capturing the intricacies of polar curves.
It's important to properly set up the integral, by substituting the polar equation \(r(\theta)\) and its derivative into the formula, as shown in our example. Although polar coordinates are beautiful and elegant, the algebra involved can sometimes be cumbersome when finding arc lengths. However, this approach provides a precise measurement, essential in applications such as engineering and physics.

Polar Coordinates

Polar coordinates offer a unique way of describing a point in a plane using two values: the radius and the angle. Unlike Cartesian coordinates, where a location is defined by \((x, y)\), polar coordinates extend from the origin outwards with \(r\) (radius) and \(\theta\) (angle measured from the positive x-axis).
Key features of polar coordinates:

• The radial coordinate \(r\), shows the distance from the origin.
• The angular coordinate \(\theta\), indicates the direction.
• Useful in scenarios where relationships are radial, such as circles or spirals.

Understanding polar coordinates is crucial for analyzing curves that are naturally described this way, such as the curve \(r = 3 \sin \theta\) used in our exercise. Here, the curve traces out a portion of a circle as \(\theta\) ranges from \(0\) to \(\pi/3\). This system is advantageous when handling rotations or circular motions, simplifying the expression of geometry compared to traditional Cartesian methods.

Integral Calculus in Polar Coordinates

Integral calculus allows us to sum infinitesimally small quantities, like slices of a curve, to measure total quantities such as area or arc length. In the realm of polar coordinates, integration becomes a powerful tool for determining such properties.
When dealing with polar curves, integration is used via the specific formula for arc length, requiring us to:

• Determine the function \(r(\theta)\).
• Differentiate \(r(\theta)\) to obtain \(\frac{dr}{d\theta}\).
• Substitute into the integral formula for arc length.

In the example given, the integral calculation allows us to find the curve's length without having to physically measure each segment. This underscores the flexibility and power of integral calculus in mathematical problem-solving, providing exact analytical answers.

Differentiation in Polar Coordinates

Differentiation in polar coordinates involves computing the rate of change of the radius function \(r(\theta)\) as \(\theta\) varies. It's crucial for understanding how the curve behaves as \(\theta\) increases.
Steps in the differentiation process:

• Identify \(r(\theta)\), the radius as a function of \(\theta\).
• Compute \(\frac{dr}{d\theta}\), the derivative of \(r\), which tells us how \(r\) changes per unit increase in \(\theta\).
• The derivative \(\frac{dr}{d\theta}\) is then used in the arc length formula.

For \(r = 3 \sin \theta\), differentiation gives us \(\frac{dr}{d\theta} = 3 \cos \theta\). This result captures the change in radial distance as the angle sweeps out a section of a circle. Without this crucial step of differentiation, determining curve lengths and further insights into curve behavior would be significantly more challenging.