Problem 18 - Calculus Early Transcendentals

Question

a. Write and simplify the integral that gives the arc length of the following curves on the given interval. b. If necessary, use technology to evaluate or approximate the integral. $$y=\sin x \text { on }[0, \pi]$$

Step-by-Step Solution

Answer

Answer: The approximate arc length is $$3.820$$.

Step 1: Find the derivative of y with respect to x

Differentiating y with respect to x, we get: $$\frac{dy}{dx} = \frac{d}{dx} (\sin x) = \cos x$$

Step 2: Square the derivative and add 1

Now, square the derivative and add 1: $$1 + (\frac{dy}{dx})^2 = 1 + (\cos x)^2 = 1 + \cos^2 x$$

Step 3: Simplify the integral expression

Now we will find the arc length using the formula: $$L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx = \int_0^{\pi} \sqrt{1 + \cos^2 x} dx$$

Step 4: Evaluate or approximate the integral

This integral doesn't have a simple closed-form solution, so we'll need to use technology (like a calculator or online tools) to approximate the integral. Here we can use Wolfram Alpha, Desmos, or a graphing calculator. Using technology, we get: $$L \approx \int_0^{\pi} \sqrt{1 + \cos^2 x} dx \approx 3.820$$ So, the arc length of the curve $$y = \sin x$$ on the interval $$[0, \pi]$$ is approximately $$3.820$$.

Key Concepts

Integral Calculus

Integral calculus is one of the two main branches of calculus, with the other being differential calculus. It deals with the accumulation of quantities, such as areas under curves or the net change of a quantity over a period. In our exercise, we used integral calculus to find the arc length of a curve.

For the curvature of a function represented by \( y = f(x) \), the arc length from \( x = a \) to \( x = b \) is given by the integral formula \( L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx \). This formula stems from the Pythagorean theorem, applied infinitesimally along the curve. Simplifying or evaluating this integral gives us the total length of the curve between those two points.

To solve such problems effectively, we need to establish the derivative, square it, add one, and then integrate the resulting expression over the desired interval. In certain cases, like the one in our exercise, the integral does not have an elementary closed-form solution and numerical methods or technology are used for approximation.

Derivative of Trigonometric Functions

The derivative of a function at any point measures the rate at which the function's value is changing at that point. Derivatives of trigonometric functions are a fundamental concept in calculus as they often appear in a variety of applications including physics, engineering, and other sciences.

For instance, the derivative of \( \text{sin}(x) \) with respect to \( x \) is \( \text{cos}(x) \), which we expressed symbolically as \( \frac{d}{dx} (\text{sin} x) = \text{cos} x \). Understanding how to obtain and use these derivatives is crucial when finding the length of curves like the sine wave in our exercise. This derivative helps us establish the formula under the square root in the integral needed to find the arc length.

Numerical Integration Methods

When the integrals do not have a simple closed-form expression, or are too complex to be integrated analytically, numerical integration methods come into play. These methods approximate the value of an integral using a finite process, and are indispensable in applied mathematics and various scientific fields.

Numerical integration typically involves algorithms like the Trapezoidal Rule, Simpson's Rule, or more complex ones like Gaussian quadrature. For computational purposes, technology such as computer software or graphing calculators execute these algorithms, handling the tedious calculations involved.

In our curve's arc length problem, we saw that the integral \( \int_0^{\pi} \sqrt{1 + \text{cos}^2 x} dx \) had no straightforward analytical solution. Hence, numerical methods or technology—like Wolfram Alpha, Desmos, or a graphing calculator—were suggested to approximate the integral's value, showing the practical application of these methods when dealing with real-world mathematical problems.