Problem 20 - Thomas Calculus

Question

In Exercises \(17-24,\) do the following. 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 curve's length is approximately 1.0472.

Step 1: Understand the Curve's Equation

We have the parametric equation of the curve: \( x = \sqrt{1-y^{2}} \) over the interval \( -\frac{1}{2} \leq y \leq \frac{1}{2} \). This is a portion of the circle \( x^{2} + y^{2} = 1 \).

Step 2: Find the Derivative

To find the length of the curve, we will need \( \frac{dx}{dy} \). Differentiating \( x = \sqrt{1-y^{2}} \) with respect to \( y \), we get \( \frac{dx}{dy} = -\frac{y}{\sqrt{1-y^{2}}} \).

Step 3: Setup the Integral for Arc Length

The formula for the arc length \( L \) of a curve given as \( x = f(y) \) is \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^{2}} \, dy \). Substitute \( \frac{dx}{dy} = -\frac{y}{\sqrt{1-y^{2}}} \) into the formula:\[L = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1 + \left( -\frac{y}{\sqrt{1-y^{2}}} \right)^{2}} \, dy \]Simplifying under the square root gives:\[L = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1 + \frac{y^{2}}{1-y^{2}}} \, dy = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\frac{1}{1-y^{2}}} \, dy = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{1-y^{2}}} \, dy \]

Step 4: Graph the Curve

Recognize that \( x = \sqrt{1-y^{2}} \) represents the upper half of a circle with radius 1 centered at the origin. On the interval \( -\frac{1}{2} \leq y \leq \frac{1}{2} \), this is an arc centered around the x-axis.

Step 5: Compute the Integral Numerically

Using software or a graphing calculator, evaluate the integral:\[\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{1-y^{2}}} \, dy.\] The result of this integral is approximately the length of the curve on the given interval, which is approximately 1.0472.

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus that allows us to find the total size or value of a given quantity. In the context of arc length, integral calculus helps us determine the length of curves. Whenever you hear about calculating areas, volumes, or in this case, arc lengths, integral calculus is the tool used.

In this exercise, we used the concept of integration to calculate the length of a curve defined parametricall by the equation: \[ x = \sqrt{1-y^2}. \]This integral is set up to account for all tiny, infinitesimal lengths along the arc, summing them up into a complete value. To set up the integration for the arc length, we use the formula:\[L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy, \]where \(a\) and \(b\) are the boundaries of the parameter \(y\).

• The derivative \( \frac{dx}{dy} \) is calculated to understand how sharp or gentle the curve is at every point.
• This ensures that every tiny segment of the arc is accounted for in the total length.

Parametric Curves

Parametric curves are a way to represent geometric figures using parameters. In many cases, parameterization simplifies complex equations and allows us to analyze curves that are difficult to express using only \( x \) or \( y \). Here, the parametric form \( x = \sqrt{1-y^2} \) describes a segment of a circle.

A circle is naturally described by a relationship between \( x \) and \( y \), typically in the form of \( x^2 + y^2 = 1 \). However, by using parametric equations, we isolate one variable (\( y \)) and express another (\( x \)) in terms of this parameter.

• This is particularly useful when you want to pinpoint sections of the curve, like in this case, the upper half of the circle between \( y = -\frac{1}{2} \) and \( y = \frac{1}{2} \).
• The parameter \( y \) runs through a specified interval, allowing us to slice the entire circle into manageable arcs or segments.

Numerical Integration

Numerical integration is an approach used when a direct computation of an integral is difficult or impossible to find analytically. Instead of solving an integral by hand, we use computers or calculators to get an approximate value.

In this exercise, after setting up the integral:\[L = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{1-y^{2}}} \, dy,\]we used numerical integration to compute the length of the curve.

• Numerical methods, like the Trapezoidal Rule or Simpson's Rule, are common techniques to approximate these integrals.
• They work by breaking the area under the curve into small trapezoids or parabolic slices and summing them for a total area estimation.
• This method is particularly advantageous for complex curves where exact integration would require complicated calculations.
• Our result, approximately 1.0472, tells us the length of the curve with a high degree of precision.