Problem 28 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(21-28 .\) The curve \(r=\sqrt{1+\sin 2 \theta}, \quad 0 \leq \theta \leq \pi \sqrt{2}\)

Step-by-Step Solution

Answer

The length of the curve is \( 2\pi \).

Step 1: Understanding the Curve's Formula

The curve is given in polar form as \( r = \sqrt{1 + \sin 2\theta} \). We need to find the length of this curve from \( \theta = 0 \) to \( \theta = \pi \sqrt{2} \). The formula for the arc length of a polar curve is \( L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \).

Step 2: Calculate \( \frac{dr}{d\theta} \)

We first need to calculate the derivative of \( r \) with respect to \( \theta \). Since \( r = \sqrt{1 + \sin 2\theta} \), by using the chain rule, we get:\[ \frac{dr}{d\theta} = \frac{1}{2\sqrt{1 + \sin 2\theta}} \cdot (2\cos 2\theta) = \frac{\cos 2\theta}{\sqrt{1 + \sin 2\theta}}. \]

Step 3: Substitute into the Arc Length Formula

Substitute \( r \) and \( \frac{dr}{d\theta} \) into the arc length formula:\[ L = \int_{0}^{\pi\sqrt{2}} \sqrt{ \left( \frac{\cos 2\theta}{\sqrt{1 + \sin 2\theta}} \right)^2 + \left( \sqrt{1 + \sin 2\theta} \right)^2 } \, d\theta. \]

Step 4: Simplify the Integrand

Simplify the expression under the square root:\[ L = \int_{0}^{\pi\sqrt{2}} \sqrt{ \frac{\cos^2 2\theta}{1 + \sin 2\theta} + 1 + \sin 2\theta } \, d\theta. \]Further simplifying gives:\[ L = \int_{0}^{\pi\sqrt{2}} \sqrt{1 + \sin 2\theta - \cos^2 2\theta + 1 + \sin 2\theta} \, d\theta. \]Noting that \( \cos^2 2\theta + \sin^2 2\theta = 1 \), the integrand reduces to:\[ L = \int_{0}^{\pi\sqrt{2}} \sqrt{2} \, d\theta. \]

Step 5: Evaluate the Integral

Since the integrand is a constant \( \sqrt{2} \), the integral simplifies to\[ L = \sqrt{2} \times \left( \pi\sqrt{2} - 0 \right) = \sqrt{2} \times \pi\sqrt{2} = 2\pi. \]

Step 6: Conclusion

Thus, the length of the curve \( r = \sqrt{1 + \sin 2\theta} \) from \( \theta = 0 \) to \( \theta = \pi\sqrt{2} \) is \( 2\pi \).

Key Concepts

Arc Length of Polar Curves

When dealing with polar coordinates, the concept of calculating the arc length of a curve is paramount. In Cartesian coordinates, curves are expressed as \( y=f(x) \), but in polar coordinates, they are defined as \( r=f(\theta) \). The arc length of a polar curve from \( \theta = a \) to \( \theta = b \) is calculated using the formula:

• \( L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \)

The formula takes into account not just the radius \( r \), but also how \( r \) changes as \( \theta \) changes, by including the derivative \( \frac{dr}{d\theta} \).
Applying this formula involves calculating this derivative correctly and manipulating it as part of an integral. This task is central to mastering the geometry of curves in polar coordinates, providing a methodical way to determine the precise path length of a curve.

Derivatives

In the context of polar coordinates, finding derivatives is a critical step in many calculations, including finding the arc length of a curve. Given that a polar curve is of the form \( r = \sqrt{1 + \sin 2\theta} \), its derivative with respect to \( \theta \) is essential.
The chain rule is employed here, which is a fundamental technique in calculus. The derivative \( \frac{dr}{d\theta} \) is:

• \( \frac{dr}{d\theta} = \frac{1}{2\sqrt{1 + \sin 2\theta}} \, (2\cos 2\theta) \)
• Simplifying, we get \( \frac{\cos 2\theta}{\sqrt{1 + \sin 2\theta}} \)

This derivative helps to determine how quickly \( r \) changes as \( \theta \) varies, and it plays a pivotal role in finding the arc length, as seen in the integrand of the arc length formula.
Understanding and calculating derivatives is crucial, as they measure the rate of change of a function, which is a core concept across calculus and its applications.

Trigonometric Functions

Trigonometric functions like sine and cosine are foundational in mathematics, especially within the realm of polar coordinates. The given curve, \( r = \sqrt{1 + \sin 2\theta} \), prominently features the sine function, which represents periodic oscillations.
These functions are vital in understanding wave behavior and rotational dynamics, often appearing in polar equations due to their inherent periodicity. Moreover, the identity \( \cos^{2}2\theta + \sin^{2}2\theta = 1 \) is exceptionally useful in simplifying expressions. For instance, during calculations,

• The expression \( 1 + \sin 2\theta - \cos^2 2\theta + 1 + \sin 2\theta \)
• Simplifies utilizing this identity, which helps to understand the geometric interpretations of trigonometric scenarios.

These functions’ properties help manipulate and simplify complex mathematical expressions, making them indispensable tools in both basic and advanced calculations.