Problem 19 - 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. $$ x=\sin y, \quad 0 \leq y \leq \pi $$

Step-by-Step Solution

Answer

The arc length of the curve is approximately 3.8202 units.

Step 1: Formula for Arc Length

To find the length of a curve expressed in parametric form \(x = f(y)\), the formula for the arc length \(L\) is \[L = \int_a^b \sqrt{ 1 + \left( \frac{dx}{dy} \right)^2 } \, dy,\] where \(\frac{dx}{dy}\) is the derivative of \(x\) with respect to \(y\).

Step 2: Differentiate x with respect to y

Given, \(x = \sin y\). Differentiate \(x\) with respect to \(y\): \[\frac{dx}{dy} = \cos y.\]

Step 3: Set Up the Integral

Substitute \(\frac{dx}{dy} = \cos y\) into the arc length formula: \[L = \int_0^\pi \sqrt{1 + (\cos y)^2} \, dy.\]

Step 4: Graph the Curve

Plot the curve \(x = \sin y\) for \(0 \leq y \leq \pi\). It describes a wave-like arc ranging from \((0,0)\) to \((0,\pi)\), effectively starting and ending at height zero and peaking in the middle.

Step 5: Numerical Evaluation of the Integral

Use a graphing calculator or computer software to evaluate the integral \(\int_0^\pi \sqrt{1 + \cos^2 y} \, dy\). The numerical approximation of the length of the curve gives approximately 3.8202.

Key Concepts

Parametric Equations

Parametric equations are a way of defining a curve using parameters. For each parameter, typically represented by a variable like \(t\), there are equations that give the coordinates of points on the curve. In this exercise, we used the equation \(x = \sin y\) to describe the shape of the curve with respect to \(y\). This approach allows us to describe complex curves in a compact way.

• Each point on the parametric curve is given by \((x(t), y(t))\).
• Gives more flexibility in representing curves than standard Cartesian equations.
• Commonly used in calculus to find derivatives and integrals of curves.

In our example, while \(y\) acts as the parameter, the parametric equation gives a clear picture of how \(x\) changes with \(y\). This approach is particularly beneficial when modeling cyclic or trigonometric patterns across specified intervals like \([0, \pi]\).

Numerical Integration

Numerical integration is a mathematical tool used to calculate the integral of functions when an analytical solution may be difficult or impossible. In our particular exercise, we used numerical integration to find the length of the curve defined by the parametric equation \(x = \sin y\).

• Substitute the derivative \(\frac{dx}{dy} = \cos y\) into the arc length formula.
• Evaluate the integral \(\int_0^\pi \sqrt{1 + \cos^2 y}\, dy\).
• Use technology like graphing calculators for simplifying complex calculations.

Numerical methods, such as Simpson's rule or the trapezoidal rule, can provide us the approximate length of the curve. In practice, you will often rely on graphing calculators or computational software for these calculations, just as we did to arrive at the length of approximately 3.8202. This illustrates how numerical integration plays a pivotal role in practical applications of calculus.

Graphing Calculators

Graphing calculators are powerful tools that often come in handy when dealing with complex mathematical problems like the one we've solved here. They can not only assist in plotting curves but also help in solving integrals numerically.

• Allow visualization of curves represented by parametric equations.
• Numerically compute definite integrals, aiding in finding quantities like arc lengths.
• Ease the process of checking calculations and ensuring accuracy.

In this exercise, the graphing calculator was used for both graphing the function \(x = \sin y\) across the interval \([0, \pi]\) and for numerically evaluating the integral to find the arc length. Do not underestimate a graphing calculator, as it is equipped to handle intricate and time-consuming calculations quickly, making it indispensable in both education and professional mathematics.