Problem 14 - Thomas Calculus

Question

In Exercises \(13-20,\) 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. $$y=\tan x, \quad-\pi / 3 \leq x \leq 0$$

Step-by-Step Solution

Answer

The curve's length is approximately 1.5 units.

Step 1: Formula for Arc Length of a Curve

To find the length of a smooth curve, we use the formula for arc length: \ \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] \ Here, \( y = \tan x \). We need to find \( \frac{dy}{dx} \) and use it in this formula.

Step 2: Differentiate the Given Function

Differentiate \( y = \tan x \) with respect to \( x \) to find \( \frac{dy}{dx} \): \ \[ \frac{dy}{dx} = \sec^2 x \]

Step 3: Substitute in the Arc Length Formula

Substitute the expression for \( \frac{dy}{dx} \) into the arc length formula: \ \[ L = \int_{-\pi/3}^{0} \sqrt{1 + \sec^4 x} \, dx \] \ This is the integral setup for finding the length of the curve over the interval.

Step 4: Graph the Curve

Graph the curve \( y = \tan x \) over the interval \( -\pi/3 \leq x \leq 0 \) using a graphing calculator or computer software. This helps to visualize the curve whose length we are calculating.

Step 5: Numerical Evaluation of the Integral

Use a graphing calculator or software with integral evaluation capability (such as Math software or an online tool) to compute the numerical value of the integral. \ Calculate: \[ \int_{-\pi/3}^{0} \sqrt{1 + \sec^4 x} \, dx \] \ This will provide the numerical length of the curve on the specified interval.

Key Concepts

Understanding Integrals in Arc Length Calculation

Integrals are a powerful concept in calculus, used to calculate the accumulation of quantities. When it comes to finding the length of a curve, integrals help us sum up infinitesimally small segments of the curve, resulting in the total arc length. To find the arc length, we use the formula:\[L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]This formula requires us to integrate over a specified interval, in this exercise from \(-\pi/3\) to \(0\). The term under the square root, \(1 + \left( \frac{dy}{dx} \right)^2\), derives from the Pythagorean Theorem, accounting for both the horizontal and vertical changes in the curve.The next steps involve finding the derivative of the function we are working with, which brings us to differentiation. This step is crucial before substituting into the integral.

The Role of Differentiation in Finding Arc Length

Differentiation is the process of finding the derivative of a function. It measures how a function changes as its input changes. In calculating the arc length, differentiation allows us to determine the rate of change or slope of the curve at any given point, which we denote as \(\frac{dy}{dx}\).In our exercise, we have the function \(y = \tan x\). To find the arc length, we first need its derivative:\[\frac{dy}{dx} = \sec^2 x\]This tells us how the tangent function is changing over our chosen interval. Differentiation here shows that the slope of the tangent function is related to the secant squared of \(x\). Once we've got this derivative, it gets squared and added to \(1\) under the square root in the arc length formula.Without this process, it would be impossible to accurately account for how sharply the curve bends, which is essential for the precise arc length calculation.

Understanding Trigonometric Functions in Calculus

Trigonometric functions are fundamental in calculus, especially when dealing with curves and angles. They include functions like \(\sin\), \(\cos\), and \(\tan\), which describe relationships between angles and sides in right triangles.For this problem, we focus on the tangent \(\tan(x)\) and the secant \(\sec(x)\). The function \(y = \tan x\) represents the tangent, which is the ratio of the opposite side to the adjacent side in a right triangle. It's a periodic function, which can be very steep near its vertical asymptotes.In differentiation, knowing that the derivative \(\frac{dy}{dx} = \sec^2 x\) relates directly to the trigonometric identity that \(\sec x = \frac{1}{\cos x}\), helps us understand how these functions interconnect. Trigonometric identities and their derivatives are essential in integral calculations such as these. They simplify the process, allowing for evaluations using integral tables or numerical methods, especially helpful when the integral becomes complex.