Problem 27 - Thomas Calculus

Question

Find the lengths of the curves. The curve \(r=\sqrt{1+\cos 2 \theta}, \quad 0 \leq \theta \leq \pi \sqrt{2}\)

Step-by-Step Solution

Answer

The arc length requires evaluating the integral numerically due to its complexity.

Step 1: Understand the Formula for Arc Length in Polar Coordinates

To find the length of a curve in polar coordinates, we use the formula: \[ L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta \]where \( r \) is given by the equation of the curve and \( \alpha \) and \( \beta \) are the bounds of \( \theta \). Here, \( r = \sqrt{1 + \cos 2\theta} \) and \( \theta \) ranges from 0 to \( \pi \sqrt{2} \).

Step 2: Calculate the Derivative of \( r \) with Respect to \( \theta \)

The function given is \( r = \sqrt{1 + \cos 2\theta} \). We need to find \( \frac{dr}{d\theta} \). Using the chain rule,\[ \frac{dr}{d\theta} = \frac{1}{2\sqrt{1 + \cos 2\theta}} \cdot (-2\sin 2\theta) \cdot 2 \]Thus,\[ \frac{dr}{d\theta} = \frac{-2\sin 2\theta}{\sqrt{1 + \cos 2\theta}} \]

Step 3: Substitute into the Arc Length Formula

Now substitute \( r = \sqrt{1 + \cos 2\theta} \) and \( \frac{dr}{d\theta} = \frac{-2\sin 2\theta}{\sqrt{1 + \cos 2\theta}} \) into the arc length formula:\[ L = \int_0^{\pi\sqrt{2}} \sqrt{\left(\frac{-2\sin 2\theta}{\sqrt{1 + \cos 2\theta}}\right)^2 + (\sqrt{1 + \cos 2\theta})^2} \, d\theta \]

Step 4: Simplify the Expression Inside the Integral

Simplify the expression inside the square root:\[ \left(\frac{-2\sin 2\theta}{\sqrt{1 + \cos 2\theta}}\right)^2 + (\sqrt{1 + \cos 2\theta})^2 \]This becomes:\[ \frac{4\sin^2 2\theta}{1 + \cos 2\theta} + 1 + \cos 2\theta \]Adding these fractions, \[ \frac{4\sin^2 2\theta + (1 + \cos 2\theta)^2}{1 + \cos 2\theta} \]

Step 5: Integrate the Expression

Evaluate the integral. This may require trigonometric identities or numerical methods, as closed-form solutions might not be easily available without simplification. Compute using known integration techniques or numerical approximations:\[ L = \int_0^{\pi\sqrt{2}} \frac{4\sin^2 2\theta + (1 + \cos 2\theta)^2}{1 + \cos 2\theta} \, d\theta \]

Step 6: Provide the Result

After evaluating the integral, the arc length of the curve \( r = \sqrt{1 + \cos 2\theta} \) from \( \theta = 0 \) to \( \theta = \pi\sqrt{2} \) is found. This calculation would typically involve numerical integration due to complexity.

Key Concepts

Polar Coordinates

Polar coordinates offer a different way to represent points in a plane compared to Cartesian coordinates. Instead of using x and y coordinates, polar coordinates describe a point using a distance from the origin, r, and an angle, \( \theta \), from the positive x-axis. For example, in the case of the curve given in the exercise, \( r = \sqrt{1 + \cos 2\theta} \), \( r \) is not a constant but varies with \( \theta \).

The advantage of polar coordinates is that they can simplify the graphing of certain curves, like spirals and circles, which may be complex to express in Cartesian coordinates. When dealing with curves in polar form, every set of \( (r, \theta) \) values corresponds uniquely to a point in the plane, making it easier to describe curves that have rotational symmetry.
To find the length of curves expressed in this form, a special arc length formula is used, which factors in both the varying radius and the angle change, accounting for the unique way these curves are parameterized.

Trigonometric Identities

Trigonometric identities are essential tools that simplify the process of working with expressions involving trigonometric functions. In the context of finding arc lengths in polar coordinates, identities such as \( \sin^2 x + \cos^2 x = 1 \) and the double angle formulas become particularly useful.

For our exercise, simplifying terms like \( \sin 2\theta \) and \( \cos 2\theta \) involves applying these identities. For instance, knowing that \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \), we can simplify expressions and integrate them more easily.

• Helps in transforming complex trigonometric expressions into simpler forms for easier integration.
• Facilitates finding derivatives and integrating polar functions by rewriting terms.

Understanding these identities not only aids in performing operations on trigonometric functions but also in resolving terms during multiplication, division, and within combined integrals.

Numerical Integration

Numerical integration becomes necessary especially when dealing with integrals that are difficult to solve analytically. In our problem, even after simplification, evaluating the integral of the expression might not yield a straightforward antiderivative. In such situations, techniques like the Trapezoidal Rule, Simpson's Rule, or numerical software can approximate the value of the integral closely.

The process involves dividing the range of integration into small intervals, calculating the area within these intervals, and summing it up to estimate the total value. Though typically less direct than algebraic methods, numerical integration provides:

• A practical means to solve complex or non-standard integrals.
• Flexibility to handle curves and equations that don't simplify easily.

It's a powerful tool in calculus, ensuring solutions are feasible even when algebraic tools fall short.

Derivative Calculation

Calculating derivatives in polar coordinates is a critical skill when working with arc lengths and understanding the dynamics of curves. Unlike Cartesian coordinates, where the derivative represents the slope of the tangent line, in polar coordinates, it helps in determining the rate of change of the radius with respect to the angle.

For instance, in this exercise, the derivative \( \frac{dr}{d\theta} \) tells us how the radius changes as \( \theta \) changes, which directly factors into the arc length formula. Here's a brief breakdown of the derivative calculation process:

• Apply the chain rule: when \( r = \sqrt{1 + \cos 2\theta} \), the derivative is found by differentiating the inside function first, then the outer function, accounting for chain effects.
• Ensure to simplify: this leads to expressions that integrate more easily.

Understanding these calculations is key to working with curves in polar coordinates and provides insights into their behavior as \( \theta \) varies.