Problem 16 - Thomas Calculus

Question

In Exercises \(13-20,\) do the following. \begin{equation}\begin{array}{l}{\text { a. Set up an integral for the length of the curve. }} \\ {\text { b. Graph the curve to see what it looks like. }} \\\ {\text { c. Use your grapher's or computer's integral evaluator to find }} \\\ {\text { the curve's length numerically. }}\end{array}\end{equation} $$x=\sqrt{1-y^{2}}, \quad-1 / 2 \leq y \leq 1 / 2$$

Step-by-Step Solution

Answer

The curve's length is approximately 0.5236.

Step 1: Identify the Function Type

We need to recognize the type of the function given. Here, the function is given in implicit form relating \(x\) and \(y\) through the equation \(x = \sqrt{1-y^2}\). This relates to a part of a circle since it resembles the equation \(x^2 + y^2 = 1\), i.e., a circle with radius 1 centered at the origin.

Step 2: Derive the Expression for the Arc Length

The general formula for the length of a curve given by \( x = f(y) \) over the interval \([c, d]\) is \( L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \). First, differentiate \( x = \sqrt{1 - y^2} \) with respect to \( y \): \( \frac{dx}{dy} = \frac{-y}{\sqrt{1 - y^2}} \). Substitute into the formula: \( L = \int_{-1/2}^{1/2} \sqrt{1 + \left(-\frac{y}{\sqrt{1 - y^2}}\right)^2} \, dy \).Simplify this to:\( L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \, dy \).

Step 3: Graph the Curve

The graph of the function \(x = \sqrt{1 - y^2}\) for \(-1/2 \leq y \leq 1/2\) is a semicircle with radius 1 in the first quadrant, specifically the top half of the semicircle. The interval restrictions mean we only plot part of this semicircle within these \(y\) bounds.

Step 4: Evaluate the Integral Numerically

Use a graphing calculator or computer software to evaluate the integral: \( L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \, dy \).Using numerical integration tools, calculate the length. The evaluated length is approximately 0.5236.

Key Concepts

Understanding the Implicit Function

An implicit function is one where the relationship between variables is given in a more indirect manner. Unlike explicit functions, which clearly define one variable in terms of another (like \( y = f(x) \)), implicit functions involve a relationship where the variables are mixed together in one equation. In this exercise, the equation \( x = \sqrt{1 - y^2} \) is part of an implicit function because it involves both \( x \) and \( y \) simultaneously.

Implicit functions are particularly useful for describing curves with more complicated relationships. When dealing with implicit functions, one operates with both variables to explore the behavior and characteristics of the curve. Understanding these functions involves:

• Identifying pairs of \( x \) and \( y \) that satisfy the given equation.
• Graphing the function to visualize its shape and extent.
• Using calculus, like derivatives, to find properties such as the slope or arc length.

This approach is fundamental when it comes to geometric shapes like circles, where the standard equation \( x^2 + y^2 = 1 \) represents a complete circle.

Connecting to the Circle

A circle is one of the simplest and most familiar geometric shapes. It is defined as the set of all points in a plane that are equidistant from a given point, the center. In standard form, a circle's equation is written as \( x^2 + y^2 = r^2 \), where \( r \) is the radius.

In this problem, the function \( x = \sqrt{1 - y^2} \) describes the upper half of a circle with a radius of 1, centered at the origin. This is because if squared, it resembles the circle equation: \( x^2 + y^2 = 1 \). The range \(-1/2 \leq y \leq 1/2\) just captures a segment of this semicircle, giving a clearer perspective of the part of the whole circle not fulfilled in the problem.

With circles:

• The complete curvature is captured when considering the full equation.
• Parts of a circle, such as halves or quarters, can be specified by restricting the range of \( x \) or \( y \).
• Geometry and calculus intersect as we explore measures like perimeter - here, the arc length.

Understanding this connection helps students visualize implicit functions in context.

The Role of Numerical Integration

Numerical integration is a mathematical tool used to approximate the value of an integral. This process is crucial when finding exact solutions analytically is too complex or impossible. Numerical methods provide a way to estimate these values with high accuracy by using a set of points over the interval of integration.

In this exercise, numerical integration provides a solution to compute the arc length of the curve represented by \( x = \sqrt{1 - y^2} \) from \( y = -1/2 \) to \( y = 1/2 \). It's essential because this integral cannot be directly solved by standard methods taught in elementary calculus.

Key aspects about numerical integration include:

• Using calculators or software to handle complex calculations.
• The output here is an approximation - in our case, approximately 0.5236 for the arc length.
• It's applicable for real-world problems where exact math is either impractical or inexplicable by hand.

By understanding numerical integration, students gain an important skill for solving advanced mathematical and engineering problems.