Problem 19 - 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} $$y=\int_{0}^{x} \tan t d t, \quad 0 \leq x \leq \pi / 6$$

Step-by-Step Solution

Answer

The curve's length is approximately 0.64.

Step 1: Understand the Curve Function

The curve is defined by the function \( y = \int_{0}^{x} \tan t \, dt \). This represents the area under the curve \( \tan t \) from \( t = 0 \) to \( t = x \). To find the length of this curve, we need to first establish the limits for \( y \) as our function ranges from \( 0 \leq x \leq \frac{\pi}{6} \).

Step 2: Set Up the Integral for Arc Length

To find the length of the curve represented by \( y = \int_{0}^{x} \tan t \, dt \), we need the arc length formula: \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \). First, find \( \frac{dy}{dx} \), which is the derivative of \( y \). Since \( y = \int_{0}^{x} \tan t \, dt \), \( \frac{dy}{dx} = \tan x \). Thus, the integral for the arc length becomes \[ L = \int_{0}^{\frac{\pi}{6}} \sqrt{1 + \tan^2 x} \, dx \].

Step 3: Simplify the Integral Expression

Recall the trigonometric identity \( 1 + \tan^2 x = \sec^2 x \). Substitute into the integral \( L = \int_{0}^{\frac{\pi}{6}} \sec x \, dx \). This is a standard integral and simplifies our computation of the arc length.

Step 4: Graph the Curve

To graph \( y = \int_{0}^{x} \tan t \, dt \), we consider the fact that \( y \) represents the area under the curve \( \tan t \) from \( 0 \) to \( x \). The graph will have increasing \( y \)-values as \( x \) moves from \( 0 \) to \( \frac{\pi}{6} \). Plot the curve to visualize how it behaves over this interval.

Step 5: Numerically Evaluate the Integral

Using a graphing calculator or computer software, evaluate the integral \( L = \int_{0}^{\frac{\pi}{6}} \sec x \, dx \). Numerically solving this integral will provide the length of the curve from \( x = 0 \) to \( x = \frac{\pi}{6} \). The computation gives a length of approximately \( 0.64 \).

Key Concepts

Arc Length

In calculus, finding the length of a curve, known as arc length, is essential for understanding many geometric properties. To compute the arc length, we use a specific formula. This involves calculating an integral that takes the derivative of the curve into account.

The formula for arc length for a function \( y = f(x) \) from \( a \) to \( b \) is given by:

• \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \).

This formula considers the change in \( y \) with respect to \( x \) and calculates the incremental segments of the curve. Each segment's length is determined by the Pythagorean theorem, forming the integral's core.

To solve an arc length problem, you first need to find the derivative \( \frac{dy}{dx} \), then substitute it into the formula, and solve the definite integral, which gives the total length of the curve over the interval \([a, b]\).

Definite Integrals

Definite integrals are used to calculate the accumulation of quantities, like areas under curves, and play a crucial role in Integral Calculus. In the context of arc length, definite integrals help us sum up the small distances along a curve to find its total length.

A definite integral is written as:

• \( \int_{a}^{b} f(x) \, dx \)

where \( f(x) \) is the function being integrated over the interval \([a, b]\). The \/...\/ in the integral represents the area under the curve from point \( a \) to point \( b \). In this case, evaluating the integral gives us a numerical value representing the total change accumulated by the function.

To apply this in integrating for arc length, we calculate the integral of the expression involving the derivative of the given function. This allows us to measure how much the curve 'stretches out' over the specified interval.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the variables within their domain. They are vital in simplifying integration problems, particularly those involving arc length.

For example, in the process of finding the arc length of a curve defined by \( y = \int_{0}^{x} \tan t \, dt \), the derivative \( \frac{dy}{dx} = \tan x \) leads us to the integral \( L = \int_{0}^{\pi/6} \sqrt{1 + \tan^2 x} \, dx \).

Using the trigonometric identity:

• \( 1 + \tan^2 x = \sec^2 x \)

This substitution simplifies the integral to \( L = \int_{0}^{\pi/6} \sec x \, dx \). Much simpler than the original problem, this makes calculations more tractable. Understanding and applying these identities is key to efficiently solving integral problems in calculus.