Problem 62 - Calculus of a Single Variable

Question

Think About It Consider the equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) (a) Use a graphing utility to graph the equation. (b) Set up the definite integral for finding the first-quadrant arc length of the graph in part (a). (c) Compare the interval of integration in part (b) and the domain of the integrand. Is it possible to evaluate the definite integral? Is it possible to use Simpson's Rule to evaluate the definite integral? Explain. (You will learn how to evaluate this type of integral in Section \(8.8 . )\)

Step-by-Step Solution

Answer

Part (a) The graph represents an ellipse centered at the origin. Part (b) The definite integral for the first-quadrant arc length is \(\int_{0}^{3} \sqrt{1 + \frac{x^2}{9 - x^2}} dx\). Part (c) It is possible to evaluate the definite integral. However, it is not possible to use Simpson’s Rule to evaluate the definite integral because the integrand is not continuous on the closed interval [0, 3].

Step 1: Graph the given equation

The given equation is an ellipse equation of the form \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\) with \(a=3\) and \(b=2\). Using a suitable graphing utility, this equation can be graphed. The graph represents an ellipse centered at origin.

Step 2: Setup the definite integral for the first-quadrant arc length

To express the ellipse in terms of y, the equation is solved for y which gives \(y = 2\sqrt{1 - \frac{x^2}{9}}\). Using this, the first quadrant arc length of the ellipse can be expressed by the definite integral \(\int_{0}^{3} \sqrt{1 + (\frac{dy}{dx})^2} dx\). Differentiating \(y\) with respect to \(x\) gives, \(\frac{dy}{dx} = \frac{-x}{\sqrt {9 - x^2}}\). Substituting this into the integral and simplifying it, one gets \(\int_{0}^{3} \sqrt{1 + \frac{x^2}{9-x^2}} dx\).

Step 3: Compare the interval of integration and the domain of the integrand

The interval of integration is \([0, 3]\), and the domain of my integrand is \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\). Therefore, the interval of integration is included in the domain of the integrand. Hence, it is possible to evaluate the definite integral using the Fundamental Theorem of Calculus.

Step 4: Evaluate the possibility of using Simpson's Rule

Simpson's Rule could not be applied in this case because the integrand, that is \(f(x) = \sqrt{1 + \frac{x^2}{9 - x^2}}\), is not differentiable on \([0, 3]\). The derivative of the integrand is undefined at \(x = 3\). Thus, this integrand is not continuous on the closed interval \([0, 3]\), which violates one of the requirements for applying Simpson’s Rule. In conclusion, Simpson’s rule could not be used here.

Key Concepts

Ellipse Equation

An ellipse is a fascinating geometric shape that resembles a flattened circle. The equation of an ellipse is generally expressed as \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, the terms \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. In the given problem, our ellipse equation is \(\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1\), which indicates that:

• \(a = 3\)
• \(b = 2\)

This means our ellipse is wider along the x-axis, extending 3 units from the origin (center) in either direction, and it stretches 2 units along the y-axis. Understanding this equation helps us visualize the ellipse and explore further properties, like circumference or arc length. The center of the ellipse is at the origin, and the axes are aligned with the coordinate axes. Grasping the basic form of the ellipse equation is crucial for approaching problems involving geometric properties.

Definite Integral

A definite integral is a fundamental concept in calculus. It is used to find areas under curves and solve various problems in physics and engineering, such as finding arc lengths. In this exercise, the goal is to determine the arc length of one quadrant of the given ellipse. We set up a definite integral from 0 to 3, representing the x-values for one quadrant:\[\int_{0}^{3} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]To solve this, we first express the ellipse equation in terms of \(y\) and differentiate. This gives us \(\frac{dy}{dx}\) which is inserted into the integral. This expression describes the arc length formula, measuring the distance along a curve described by a function, from the beginning to the end of the interval. Understanding how to set up and interpret definite integrals is essential for solving complex calculus problems.

Simpson's Rule

Simpson’s Rule is a numerical method for approximating definite integrals. It uses parabolic arcs to approximate the curve of a function and tends to give more accurate results compared to other rules like Trapezoidal Rule. However, it has its own set of requirements; primarily, the function must be continuous and differentiable on the interval.
In our exercise, it turns out that Simpson's Rule cannot be applied. The integrand \(\sqrt{1 + \frac{x^2}{9-x^2}}\) becomes problematic because its derivative isn't defined across the full interval \([0, 3]\). Specifically, at \(x = 3\), the function isn't continuous, leading to complications in applying Simpson’s Rule. This scenario demonstrates the importance of checking function continuity and differentiability before using any numerical integration method.

Calculus

Calculus is a branch of mathematics that deals with the study of change and motion. It's divided mainly into two areas: differential calculus and integral calculus. Differential calculus focuses on rates of change (derivatives), while integral calculus deals with accumulation (integrals).
In this problem, we delve into integral calculus to determine the arc length of the ellipse in the first quadrant. We set up a definite integral, requiring us to differentiate the equation of the curve with respect to \(x\), a process rooted in differential calculus. Then we integrate over the specified interval.
Calculus provides the tools for solving complex geometric problems, including those related to arc lengths of curves. When dealing with geometric figures like ellipses, calculus allows us to transition from theoretical formulas to practical applications. Whether finding areas, volumes, or arcs, calculus is an indispensable part of higher-level mathematics.