Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of the curve with the given parametric equations (a) \(x=\sin \theta, y=\cos \theta\) for \(0 \leq \theta \leq 2 \pi\) (b) \(x=\sin 3 \theta, y=\cos 3 \theta\) for \(0 \leq \theta \leq 2 \pi\) (c) Explain why the lengths in parts (a) and (b) are not equal.

Step-by-Step Solution

Answer

(a) Length is \(2\pi\); (b) Length is \(6\pi\); Lengths differ since part (b) traces the circle 3 times.

Step 1: Identify the Parametric Equations for Part (a)

For part (a), the parametric equations are given as \(x = \sin \theta\) and \(y = \cos \theta\) for the range \(0 \leq \theta \leq 2\pi\). This describes a standard circle parameterized by angle \(\theta\).

Step 2: Calculate the Derivatives for Part (a)

Compute the derivatives: \(\frac{dx}{d\theta} = \cos \theta\) and \(\frac{dy}{d\theta} = -\sin \theta\).

Step 3: Use the Formula for Arc Length in Part (a)

The formula for the arc length \(L\) of a parametric curve is \(L = \int_{a}^{b} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \ d\theta\). Substitute the derivatives: \(L = \int_{0}^{2\pi} \sqrt{(\cos \theta)^2 + (-\sin \theta)^2} \ d\theta = \int_{0}^{2\pi} \sqrt{\cos^2 \theta + \sin^2 \theta} \ d\theta\).

Step 4: Simplify and Integrate for Part (a)

Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), the integral becomes \(L = \int_{0}^{2\pi} 1 \ d\theta\). Evaluating this integral gives \(L = \theta \bigg|_{0}^{2\pi} = 2\pi\).

Step 5: Identify the Parametric Equations for Part (b)

For part (b), the parametric equations become \(x = \sin 3\theta\) and \(y = \cos 3\theta\) for the range \(0 \leq \theta \leq 2\pi\). This describes a circle traced 3 times.

Step 6: Calculate the Derivatives for Part (b)

Compute the derivatives: \(\frac{dx}{d\theta} = 3\cos 3\theta\) and \(\frac{dy}{d\theta} = -3\sin 3\theta\).

Step 7: Use the Formula for Arc Length in Part (b)

Substitute the derivatives into the arc length formula: \(L = \int_{0}^{2\pi} \sqrt{(3\cos 3\theta)^2 + (-3\sin 3\theta)^2} \ d\theta = \int_{0}^{2\pi} \sqrt{9(\cos^2 3\theta + \sin^2 3\theta)} \ d\theta\).

Step 8: Simplify and Integrate for Part (b)

Using the identity \(\cos^2 3\theta + \sin^2 3\theta = 1\), the integral becomes \(L = \int_{0}^{2\pi} 3 \ d\theta\). Evaluating this integral gives \(L = 3\theta \bigg|_{0}^{2\pi} = 6\pi\).

Step 9: Compare the Lengths and Explain the Difference

In part (a), the length of the circle traced once is \(2\pi\), whereas in part (b), the path is traced three times resulting in a length of \(6\pi\). The difference is because in part (b) the circle is wrapped around three times, multiplying the total length by 3.

Key Concepts

Arc Length

When working with curves in calculus, determining the arc length is a crucial concept. Arc length is essentially the distance along the curved line between two points. For parametric equations, the formula to find the arc length \( L \) is given by:

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

This formula involves integrating the square root of the sum of the squares of the derivatives of the parametric functions \(x(\theta)\) and \(y(\theta)\). It's akin to finding the hypotenuse of a right triangle for a very short segment of the curve.
The given parameter \( \theta \) can vary from a starting point \(a\) to an endpoint \(b\). In simple terms, you are summing up all these tiny hypotenuse values to get the full length of the curve. Understanding this formula is fundamental because it allows you to calculate the precise length of any curve described by parametric equations.

Circle Parameterization

Parameterization is a technique of describing a curve using a set of equations. When we talk about circles, a common parameterization involves using trigonometric functions of a parameter, typically \( \theta \). For a unit circle, the equations are:

• \( x = \sin \theta \)
• \( y = \cos \theta \)

These equations are derived from the unit circle in trigonometry where a circle with radius 1 is centered at the origin in the coordinate plane.
This parameterization travels around the circle once as \( \theta \) goes from 0 to \( 2\pi \).
In more complex scenarios, such as in part (b) of the exercise, we have:

• \( x = \sin 3\theta \)
• \( y = \cos 3\theta \)

This indicates the circle is being traced three times as \( \theta \) varies from 0 to \( 2\pi \). Each increase in the coefficient (like 3 here) wraps the circle that many times, effectively increasing the total arc length proportionally.

Derivative Calculation

Calculating derivatives is a pivotal step in finding the arc length of a parametric curve. The derivative helps determine how the curve changes with respect to the parameter \( \theta \). For part (a) of the exercise, the derivatives of the parametric equations \( x = \sin \theta \) and \( y = \cos \theta \) are:

• \( \frac{dx}{d\theta} = \cos \theta \)
• \( \frac{dy}{d\theta} = -\sin \theta \)

These derivatives tell us how fast the x and y coordinates change as \( \theta \) changes. This rate of change is crucial because it is used in the arc length formula to consider the direction and distance traveled by the curve.
For part (b), we have more complex derivatives because the equations include \( 3\theta \) instead of \( \theta \):

• \( \frac{dx}{d\theta} = 3\cos 3\theta \)
• \( \frac{dy}{d\theta} = -3\sin 3\theta \)

Here, the additional factor of 3 stems from the chain rule, reflecting how the wrapping of the circle three times speeds up the rate of change proportionally.

Trigonometric Identities

Trigonometric identities play a pivotal role in simplifying expressions when working with parametric curves. One of the essential identities used in these calculations is the Pythagorean identity:

• \( \cos^2 \theta + \sin^2 \theta = 1 \)

In our exercise, this identity simplifies the expression within the square root in the arc length integrals. For both scenarios—whether the parameterization is a single wrap or a triple wrap—this identity helps reduce the expression to a constant:

• For part (a): \( \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt{1} = 1 \)
• For part (b): \( \sqrt{9(\cos^2 3\theta + \sin^2 3\theta)} = \sqrt{9} = 3 \)

Using these identities not only simplifies the calculation process but also directly shows how the integrals evaluate to simple multiplication, making the connection clear between the parameterization and the arc length.