Problem 16 - Thomas Calculus in SI Units

Question

a. Set up an integral for the length of the curve. b. Graph the curve to see what it looks like. c. Use your grapher's or computer's integral evaluator to find the curve's length numerically. $$x=\sqrt{1-y^{2}}, \quad-1 / 2 \leq y \leq 1 / 2$$

Step-by-Step Solution

Answer

The integral: \( L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \; dy\), length \( \approx 0.5236 \).

Step 1: Determine the Integral Form for Arc Length

The length of a curve given by a function \( x = g(y) \) from \( y = a \) to \( y = b \) can be found using the formula for arc length: \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \; dy\]In this case, \( x = \sqrt{1 - y^2} \). First, differentiate \( x \) with respect to \( y \): \[\frac{dx}{dy} = \frac{-y}{\sqrt{1 - y^2}}\]Now, substitute \( \frac{dx}{dy} \) into the arc length formula:\[L = \int_{-1/2}^{1/2} \sqrt{1 + \left(\frac{-y}{\sqrt{1 - y^2}}\right)^2} \; dy\]This simplifies to:\[L = \int_{-1/2}^{1/2} \sqrt{1 + \frac{y^2}{1 - y^2}} \; dy = \int_{-1/2}^{1/2} \sqrt{\frac{1}{1 - y^2}} \; dy\]Which results in:\[L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \; dy\]

Step 2: Graph the Curve

To understand what the curve looks like, plot the equation \( x = \sqrt{1 - y^2} \). This equation represents the right half of a circle with radius 1, centered at the origin.For the given domain of \(-1/2 \leq y \leq 1/2\), graph only this segment of the semicircle. The result is an arc that appears to be symmetric with respect to the y-axis, spanning from \( y = -1/2 \) to \( y = 1/2 \).

Step 3: Evaluate the Integral Numerically

Using a graphing calculator or a computer software capable of numerical integration, we can compute the integral we set up:\[L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \; dy\] Substituting the limits of integration into the integral gives the length of the curve. This integral evaluates to:\[L \approx 0.5236\]Thus, the numerical length of the curve for the given interval is approximately 0.5236.

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics, focusing on finding the accumulated quantity of a function. One important application is finding the arc length of a curve. In our exercise, we're tasked with finding the length of the curve given by the equation \(x = \sqrt{1 - y^2}\) over a specific interval. This involves setting up an integral with the arc length formula:\[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\]To find \(\frac{dx}{dy}\), we differentiate \(x\) with respect to \(y\), obtaining \(\frac{-y}{\sqrt{1 - y^2}}\). Substituting this into our formula, the integral simplifies to:\[L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \, dy\]This integral computes the distance along the curve from \(y = -1/2\) to \(y = 1/2\), a central concept in integral calculus used widely in physics, geometry, and engineering.

Numerical Integration

Numerical integration is a technique used to evaluate integrals that are difficult or impossible to solve analytically. In this exercise, after setting up the integral for arc length, we find it challenges our analytical skills due to its complexity or the function's form. Here is where numerical methods come into play. We use tools like graphing calculators or specialized computer software to approximate the value of the integral.For our curve, the integral we've constructed is:\[L = \int_{-1/2}^{1/2} \frac{1}{\sqrt{1 - y^2}} \, dy\]Using numerical integration, this evaluates to approximately 0.5236. Numerical methods break down the integral into small, manageable parts, calculating areas under the curve in easier chunks, such as rectangles (Riemann sums) or using advanced techniques like Simpson's rule or trapezoidal rule. This ensures that we get a close estimation of the integral's value even when analytical solutions are complex or not available.

Curve Graphing

Graphing curves helps visualize mathematical concepts effectively. In our problem, the curve given by \(x = \sqrt{1 - y^2}\) is essential to understanding the nature of the problem. By plotting this equation within its domain \(-1/2 \leq y \leq 1/2\), we see that it forms the right half of a circle, specifically an arc centered at the origin, with a range of just a segment from \(y = -1/2\) to \(y = 1/2\).Curve graphing aids in visualizing the arc. It provides insights into its symmetry and inherent properties. Most importantly, it ensures that we understand what kind of real-world shape we are dealing with. This insight can guide us in determining the direction to take when setting up equations and integrals for calculations. Understanding changes in the curve through graphing strengthens our ability to solve more complex calculus problems that involve similar principles.