Problem 117 - Calculus Volume 3

Question

Show that the total length of the ellipse \(x=4 \sin \theta, y=3 \cos \theta\) is \(L=16 \int_{0}^{\pi / 2} \sqrt{1-e^{2} \sin ^{2} \theta} d \theta, \quad\) where \(\quad e=\frac{c}{a} \quad\) and \(c=\sqrt{a^{2}-b^{2}}\).

Step-by-Step Solution

Answer

The total length of the ellipse is given by the integral \( L = 16 \int_{0}^{\pi / 2} \sqrt{1-e^{2} \sin ^{2} \theta} \, d \theta \).

Step 1: Identify the Parametric Equations

The given parametric equations for the ellipse are:\[ x = 4 \sin \theta, \quad y = 3 \cos \theta \]Here, the coefficients in front of \(\sin\) and \(\cos\) represent the semi-major axis \(a\) and the semi-minor axis \(b\):\[ a = 4, \quad b = 3 \]

Step 2: Calculate the Eccentricity

The eccentricity \(e\) of an ellipse is defined as \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus, given by \(c = \sqrt{a^2 - b^2}\). We first calculate \(c\):\[ c = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7} \]Then, the eccentricity is:\[ e = \frac{\sqrt{7}}{4} \]

Step 3: Use Ellipse Arc Length Formula

The arc length \(L\) of an ellipse in terms of \(\theta\) can be expressed by the integral \[ L = \int_0^{2\pi} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \ d\theta \]Calculating the derivatives:\[ \frac{dx}{d\theta} = 4 \cos \theta, \quad \frac{dy}{d\theta} = -3 \sin \theta \]

Step 4: Simplify the Integral for the Ellipse Arc Length

Substitute the derivatives into the arc length formula:\[ L = \int_0^{2\pi} \sqrt{(4 \cos \theta)^2 + (-3 \sin \theta)^2} \ d\theta \]This simplifies to:\[ L = \int_0^{2\pi} \sqrt{16 \cos^2 \theta + 9 \sin^2 \theta} \ d\theta \]

Step 5: Transform the Integral

Factor and simplify inside the square root:\[ L = \int_0^{2\pi} \sqrt{16 (\cos^2 \theta + \frac{9}{16} \sin^2 \theta)} \ d\theta \]This can be rewritten using \(e = \frac{\sqrt{7}}{4}\):\[ L = 4 \int_0^{2\pi} \sqrt{1 - e^2 \sin^2 \theta} \ d\theta \]

Step 6: Adjust Limits of Integration for One Quadrant

Since the ellipse is symmetric, we only need to integrate over one quadrant and multiply by 4. The integral then becomes:\[ L = 16 \int_0^{\pi/2} \sqrt{1 - e^2 \sin^2 \theta} \ d\theta \]

Step 7: Final Step: Conclusion

We have shown that the total length of the ellipse is given by the formula:\[ L = 16 \int_{0}^{\pi/2} \sqrt{1-e^2 \sin^2 \theta} \ d\theta \] where \(e = \frac{\sqrt{7}}{4}\), which concludes the derivation.

Key Concepts

Parametric Equations

Parametric equations are a way to express the coordinates of a curve as functions of a parameter. In the case of an ellipse, these equations separately define the x and y coordinates based on a parameter, typically represented by \( \theta \). For the given ellipse, we have:

• \( x = 4 \sin \theta \)
• \( y = 3 \cos \theta \)

Here, \( \theta \) plays the role of the parameter that varies to trace the complete shape of the ellipse. The equations reveal a simple yet effective way to describe curves that are not functions in the traditional sense due to their rotational symmetry.
By representing the ellipse through these parametric forms, we can easily calculate various properties, such as arc length.

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a perfect circle. It is defined as \( e = \frac{c}{a} \), where \( c \) is the distance from the center to a focus and \( a \) is the length of the semi-major axis. For our ellipse, we calculate \( c \) using the relation:\[ c = \sqrt{a^2 - b^2} = \sqrt{4^2 - 3^2} = \sqrt{7} \]Therefore, the eccentricity is:\[ e = \frac{\sqrt{7}}{4} \]A smaller eccentricity means the ellipse is closer to a circle, while a larger eccentricity indicates a more elongated shape. For this ellipse, \( e \) is relatively small, suggesting it is not too far from being circular.

Integral Calculus

In determining the arc length of an ellipse, integral calculus comes into play. Calculating the arc length involves integrating a function over a specific interval. The formula used here, \[ L = \int_0^{2\pi} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta \]requires finding derivatives of the parametric equations to find the infinitesimal changes in \( x \) and \( y \). The integral represents a continuous summation of infinitesimally small segments of the curve.
The integration limits can be adjusted by symmetry to simplify calculations, as shown in the solution where only one quadrant is considered.

Semi-Major Axis

The semi-major axis is the longest radius of an ellipse, stretching from the center to the furthest edge. It is crucial not only in defining the ellipse's size but also in its equation and properties. For this particular ellipse, identified through parametric equations, the semi-major axis \( a \) is 4, as seen from the coefficient of \( \sin \theta \) in \( x = 4 \sin \theta \).
The semi-major axis helps determine other critical characteristics such as the eccentricity and is a part of the integral for arc length calculation.

Semi-Minor Axis

The semi-minor axis is the shortest radius of an ellipse, connecting the center to one of its closest points. In terms of the parametric representation of the ellipse \( y = 3 \cos \theta \), the semi-minor axis is 3. The semi-minor axis, together with the semi-major axis, defines the shape and extent of the ellipse.
Understanding these axes is essential not just for the geometry of the ellipse but also for calculations like the eccentricity and arc length, as they both factor heavily into integral calculus solutions for these shapes.