Problem 19 - Calculus III

Question

Find the arc length of the curve on the given interval.Set up an integral to find the circumference of the ellipse with the equation \(\mathbf{r}(t)=\cos t \mathbf{i}+2 \sin t \mathbf{j}+0 \mathbf{k}\).

Step-by-Step Solution

Answer

The integral to find the circumference of the ellipse is \( \int_0^{2\pi} \sqrt{1 + 3\cos^2 t} \, dt \), a form of the complete elliptic integral.

Step 1: Understand the Parametric Equation

We have the parametric equation of the ellipse as \(\mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k}\), where \(t\) ranges from \(0\) to \(2\pi\). The ellipse lies in the \(xy\)-plane, with semi-major axis 2 along the \(y\)-axis and semi-minor axis 1 along the \(x\)-axis.

Step 2: Calculate the Derivative \(\mathbf{r}'(t)\)

Differentiate \(\mathbf{r}(t)\) with respect to \(t\): \[ \mathbf{r}'(t) = \frac{d}{dt} (\cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k}) = -\sin t \mathbf{i} + 2 \cos t \mathbf{j} + 0 \mathbf{k} \]

Step 3: Compute the Magnitude of \(\mathbf{r}'(t)\)

The magnitude of \(\mathbf{r}'(t)\) is given by:\[ \| \mathbf{r}'(t) \| = \sqrt{(-\sin t)^2 + (2 \cos t)^2} = \sqrt{\sin^2 t + 4 \cos^2 t} \]Using the trigonometric identity \( \sin^2 t + \cos^2 t = 1 \), this simplifies to:\[ \| \mathbf{r}'(t) \| = \sqrt{1 + 3\cos^2 t} \]

Step 4: Set Up the Integral for Arc Length

The formula for the arc length \(L\) of a parametric curve \(\mathbf{r}(t)\) over \(t \in [a, b]\) is:\[ L = \int_a^b \| \mathbf{r}'(t) \| \, dt \]For the ellipse, we set up the integral as:\[ L = \int_0^{2\pi} \sqrt{1 + 3\cos^2 t} \, dt \]

Step 5: Recognize This is an Elliptic Integral

The integral \( \int_0^{2\pi} \sqrt{1 + 3\cos^2 t} \, dt \) is a form of the complete elliptic integral of the second kind. Such integrals typically require numerical methods or special functions for evaluation.

Key Concepts

Understanding Parametric Equations

When dealing with curves in mathematics, a parametric equation is a crucial concept. It allows us to express a set of equations using a parameter, typically denoted as \( t \). This approach is particularly useful for describing curves that cannot be easily expressed in a standard form.

In the given exercise, the parametric equation \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \) describes an ellipse. The parameter \( t \) represents an angle typically ranging from \( 0 \) to \( 2\pi \), corresponding to a full rotation around the ellipse.

• The term \( \cos t \mathbf{i} \) corresponds to the \( x \)-coordinate and has an amplitude of 1, which acts as the semi-minor axis.
• The term \( 2 \sin t \mathbf{j} \) corresponds to the \( y \)-coordinate with an amplitude of 2, representing the semi-major axis.
• The \( 0 \mathbf{k} \) is zero as the ellipse is restricted to the \( xy \)-plane.

This representation offers a flexible way to visualize and compute properties of curves, such as arc length, which is what this problem seeks to solve.

Exploring Elliptic Integrals

Once the arc length integral has been formulated, the complexity often rises with the result being classified as an elliptic integral. Elliptic integrals originate from calculating the arc length of an ellipse and do not resolve to simple expressions using elementary functions.

The specific integral in this problem, \( L = \int_0^{2\pi} \sqrt{1 + 3\cos^2 t} \, dt \), is a type known as a complete elliptic integral of the second kind. These integrals are beyond the scope of elementary calculus tools and usually require:

• Numerical approximation techniques, like numerical integration methods, for practical computation.
• The use of special functions, such as those available in computational software or advanced mathematics toolkits, for exact evaluation or closed-form solutions.

Understanding elliptic integrals opens doors to advanced topics in mathematics and applied sciences, providing tools for problems where the analytic solution is not straightforward.

Steps in Derivative Calculation

To find the arc length of a parametric curve, derivative calculation is a crucial step. The derivative gives insight into how the position changes with respect to the parameter \( t \). In this context, we differentiate \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \):

1. **Differentiation**: We use basic derivative rules:

• The derivative of \( \cos t \) is \( -\sin t \).
• The derivative of \( 2 \sin t \) is \( 2 \cos t \).
• Since \( 0 \mathbf{k} \) is constant, its derivative is zero.

The resulting derivative is \( \mathbf{r}'(t) = -\sin t \mathbf{i} + 2 \cos t \mathbf{j} \).

2. **Magnitude Calculation**: With derivatives, calculating the magnitude gives us the rate of change, or speed, along the curve:

• The magnitude formula \( \| \mathbf{r}'(t) \| = \sqrt{(-\sin t)^2 + (2 \cos t)^2} \) gives \( \sqrt{1 + 3\cos^2 t} \).

This expression is what is integrated from \( 0 \) to \( 2\pi \) to find the total arc length, emphasizing the importance of derivatives in obtaining and solving integrals for the arc length.