Problem 14 - Thomas Calculus

Question

In Exercises \(13-20,\) do the following. \begin{equation}\begin{array}{l}{\text { a. Set up an integral for the length of the curve. }} \\ {\text { b. Graph the curve to see what it looks like. }} \\\ {\text { c. Use your grapher's or computer's integral evaluator to find }} \\\ {\text { the curve's length numerically. }}\end{array}\end{equation} $$y=\tan x, \quad-\pi / 3 \leq x \leq 0$$

Step-by-Step Solution

Answer

Use a calculator to find the arc length numerically for \(y=\tan x\) from \(-\frac{\pi}{3}\) to \(0\).

Step 1: Define the Curve

The given curve is described by the function \(y = \tan x\) over the interval \(-\frac{\pi}{3} \leq x \leq 0\).

Step 2: Set Up the Integral

The formula for the arc length \(L\) of a curve defined by \(y = f(x)\) from \(x = a\) to \(x = b\) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx \]First, find the derivative \(f'(x)\) of \(y = \tan x\), which is \(f'(x) = \sec^2 x\).So, the integral becomes: \[ L = \int_{-\frac{\pi}{3}}^{0} \sqrt{1 + \sec^4 x} \, dx \]

Step 3: Graph the Curve

Graph the curve \(y = \tan x\) from \(-\frac{\pi}{3}\) to \(0\). The graph will show a segment of the tangent curve, starting from the point where \(x = -\frac{\pi}{3}\) to \(x = 0\), which is near the origin, showing a vertical asymptote.

Step 4: Numerically Evaluate the Integral

Use a graphing calculator or computer software to evaluate the integral \[ \int_{-\frac{\pi}{3}}^{0} \sqrt{1 + \sec^4 x} \, dx \] This will provide the numerical value for the arc length of the curve.

Key Concepts

Integration

When tasked with finding the arc length of a curve, integration is our go-to mathematical tool. It involves summing up infinitely small segments to get an entire length, area, volume, or other quantities. In the case of arc length, the continuous sum—or integral—is calculated using a specific formula that incorporates the derivative of the function.

For a function defined as \( y = f(x) \), the arc length \( L \) between two points \( a \) and \( b \) is determined by the integral:

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

This formula integrates the square root of one plus the square of the derivative over the interval. Essentially, it accounts for every infinitesimally small slope change along the curve. In practical applications without a simple analytical solution, tools like computer software or advanced calculators are often used to evaluate these integrals numerically.

Tangent Function

The tangent function \( y = \tan x \) is one of the basic trigonometric functions that describe a wave-like curve with periodicity and vertical asymptotes at certain points. When examining its graph between certain bounds, like \(-\frac{\pi}{3}\) and 0 in this instance, you will notice how the curve moves upwards rapidly as it approaches the vertical lines called asymptotes.

These asymptotes represent points where the function approaches infinity. Visually, the tangent curve repeats itself every \( \pi \) units but is undefined at odd multiples of \( \frac{\pi}{2} \). Thus, for understanding the arc length specifically, knowing the behavior of the tangent function over the defined interval is critical as it informs what part of this repeating curve we're analyzing.

Derivative

A derivative represents the instantaneous rate of change of a function and is fundamental in calculating the slope of the tangent line at any given point on a curve.

For our specific function \( y = \tan x \), the derivative \( f'(x) \) is \( \sec^2 x \), which tells us how quickly and in what way the tangent curve is changing at each point. The secant squared function itself is derived from the tangent function and has its own unique properties, including being always positive or zero, which is crucial when used in the integral formula for arc length.

Understanding derivatives is essential for multiple areas in calculus, especially in contexts like motion, where they indicate velocity or acceleration, or in our case, contributing to the accurate calculation of arc length.

Numerical Evaluation

Numerical evaluation allows us to approximate the value of integrals or other calculations that cannot be easily solved analytically.

By using technology such as graphing calculators or computer software, we can compute complex integrals like \( \int_{-\frac{\pi}{3}}^{0} \sqrt{1 + \sec^4 x} \, dx \).

Tools might use methods such as the trapezoidal rule or Simpson's rule to estimate integrals by breaking the area under the curve into simpler shapes (like trapezoids or parabolas) that approximate the region more closely than simple rectangles could.
By adjusting the number of subdivisions and refining calculations, we can achieve an increasingly precise estimate for the desired arc length, making numerical evaluation an essential skill for applied mathematics.