Question

Use Simpson's Rule, with \(n=4,\) to approximate the arc length of the function on the given interval. \(f(x)=\sec x\) on \([-\pi / 4, \pi / 4]\)

Short answer

Using Simpson's Rule results in an approximate arc length.

Step-by-step solution

Understand Simpson's Rule for Arc Length

Simpson's Rule is used to approximate the integral of a function. For arc length, the integral is a bit special. The formula for arc length of a curve \(y = f(x)\) from \(a\) to \(b\) is given by:\[L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx\]Simpson's Rule approximates \( \int_a^b f(x) \, dx\) as \[\frac{b-a}{3n} \left( f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right)\] where \(n\) must be even and each \(x_i = a + i\Delta x\) with \(\Delta x = \frac{b-a}{n}\).

Define the Integral and Derivatives

The task is to approximate \[L = \int_{-\pi/4}^{\pi/4} \sqrt{1 + (f'(x))^2} \, dx\] where \( f(x) = \sec x \). First, find the derivative \(f'(x)\):\[ f'(x) = \sec x \tan x \]So now:\[1 + (f'(x))^2 = 1 + \sec^2 x \tan^2 x\]

Simplify the Integral Expression

Further simplify the expression inside the square root:\[1 + \sec^2 x \tan^2 x = \sec^2 x (1 + \tan^2 x)\]Knowing \(1 + \tan^2 x = \sec^2 x\), we get:\[\sec^2 x (1 + \tan^2 x) = \sec^4 x\]Thus, \[L = \int_{-\pi/4}^{\pi/4} \sec^2 x \, dx\]

Apply Simpson's Rule

To apply Simpson's Rule, we choose \(n = 4\), so \[\Delta x = \frac{\pi/2}{4} = \frac{\pi}{8}\].The points are:\[x_0 = -\pi/4, \, x_1 = -\pi/8, \, x_2 = 0, \, x_3 = \pi/8, \, x_4 = \pi/4\].Calculate \(f(x_i) = \sec^2(x_i)\) and apply:\[L \approx \frac{\pi/8}{3} [\sec^2(-\pi/4) + 4\sec^2(-\pi/8) + 2\sec^2(0) + 4\sec^2(\pi/8) + \sec^2(\pi/4)]\].

Compute Function Values

Calculate each \(\sec(x_i)\) value:- \(\sec(-\pi/4) = \sqrt{2}\)- \(\sec(-\pi/8) = \frac{1}{\cos(-\pi/8)}\)- \(\sec(0) = 1\)- \(\sec(\pi/8) = \frac{1}{\cos(\pi/8)}\)- \(\sec(\pi/4) = \sqrt{2}\)Square these to find \( \sec^2(x_i)\).

Evaluate the Approximation

Substitute the calculated values:- \(\sec^2(-\pi/4) = 2\)- \(\sec^2(-\pi/8) = \left(\frac{1}{\cos(-\pi/8)}\right)^2\)- \(\sec^2(0) = 1\)- \(\sec^2(\pi/8) = \left(\frac{1}{\cos(\pi/8)}\right)^2\)- \(\sec^2(\pi/4) = 2\)Then,\[L \approx \frac{\pi/24} [2 + 4(\sec^2(-\pi/8)) + 2 + 4(\sec^2(\pi/8)) + 2]\]Calculate using known trigonometric values or approximations.

Key concepts

Arc Length

Arc length helps us determine the distance along a curve over a given interval. This is essential in many fields, such as engineering and physics, where measuring the distance along a curved path is necessary. To find the arc length, we utilize calculus and integration. The general formula for the arc length of a curve defined by a function \(y = f(x)\) is

• \[L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx\]

The integrand \(\sqrt{1 + (f'(x))^2}\) ensures that we account for the slope of the curve when measuring distances. Differentiating the function gives us \(f'(x)\), and squaring it before adding one helps adjust our calculations to the curve's geometry.
In the current exercise, \(f(x) = \sec x\), and its derivative \(f'(x) = \sec x \tan x\) is used to find the arc length across the interval \([\-\pi/4, \pi/4]\).

Numerical Integration

Numerical integration is a technique used to approximate the value of integrals, especially when an exact analytical solution is difficult or impossible to determine. This technique becomes powerful when dealing with functions that have complex forms or are known only at discrete points.
Simpson's Rule is one of the most popular methods for numerical integration. This rule assumes that the function can be approximated by quadratic polynomials over intervals, providing a good balance between complexity and accuracy. The formula is:

• \[ \frac{b-a}{3n} \left( f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right) \]

In applying Simpson's Rule, the interval \([a, b]\) is divided into \(n\) equal subintervals. Despite being an approximation, it offers a reasonably accurate estimate when \(n\) is even. In our exercise, with \(n=4\), we use this rule to approximate the arc length.

Trigonometric Functions

Trigonometric functions, like sine, cosine, and secant, are essential in understanding periodic phenomena, such as sound waves and circular motion. They relate angles of a right triangle to the ratios of its sides. Here, the secant function \(\sec x = \frac{1}{\cos x}\) plays a pivotal role.
Trigonometric functions often lead to complex integrals when applied in calculus. In this exercise, we need the derivative of the secant function, which is \(f'(x) = \sec x \tan x\). Simplifying expressions using trigonometric identities like \(1 + \tan^2 x = \sec^2 x\) is key to handle complex integrals. Trigonometric functions introduce unique behaviors into calculus problems, which may require special approaches for both differentiation and integration.

Calculus

Calculus, the mathematical study of change and motion, covers differentiation and integration as its core components. Differentiation finds the rate of change, or the slope of a curve, and integration aggregates these changes to find areas or other aggregate measures.

• Differentiation: Using derivatives, we understand how a function continuously changes, such as finding \(f'(x)\).
• Integration: Involves summing these continuous changes, useful for finding areas under curves and quantities like arc length.

In this particular exercise, calculus helps find the arc length of \(f(x) = \sec x\). Differentiation provides us with \(f'(x)\), and integration aggregates this data into a single arc length measure. While calculus can often be challenging, it’s a powerful tool for uncovering insights into the natural world through rigorous mathematical analysis.