Problem 4 - Calculus Early Transcendentals

Question

Setting up arc length integrals Write and simplify, but do not evaluate, an integral with respect to \(x\) that gives the length of the following curves on the given interval. $$y=2 \cos 3 x \text { on }[-\pi, \pi]$$

Step-by-Step Solution

Answer

Question: Set up the integral expression for the arc length of the curve defined by the function \(y = 2\cos(3x)\) on the interval \([-\pi, \pi]\). Answer: The integral expression for the arc length of the curve is \(L=\int_{-\pi}^{\pi}\sqrt{1+36\sin^2(3x)} dx\).

Step 1: Find the derivative of the function

We are given the function \(y=2cos(3x)\). First, we need to find its derivative with respect to \(x\). $$\frac{dy}{dx} = \frac{d}{dx} (2 \cos(3x))$$ Apply the chain rule: $$\frac{dy}{dx} = -6 \sin(3x)$$

Step 2: Set up the arc length integral formula

Now that we have the derivative of the function, we can set up the integral for the arc length using the formula mentioned before: $$L = \int_{a}^{b} \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$ Substitute the given interval \([-\pi, \pi]\) and the derivative we found: $$L=\int_{-\pi}^{\pi}\sqrt{1+(-6\sin(3x))^2}dx$$

Step 3: Simplify the integral

Now we will simplify the integral by squaring the derivative: $$L=\int_{-\pi}^{\pi}\sqrt{1+(36\sin^2(3x))} dx$$ Now we have the required simplified integral expression for the length of the curve on the given interval: $$L=\int_{-\pi}^{\pi}\sqrt{1+36\sin^2(3x)} dx$$

Key Concepts

Chain Rule

In calculus, the chain rule is an essential tool for finding the derivative of composite functions. The chain rule allows us to differentiate functions composed of other functions, which is important when dealing with more complex expressions. The basic idea is to take the derivative of the outer function and multiply it by the derivative of the inner function.

For example, if you have a function like \( y = 2 \, \cos(3x) \), you see that \( \cos(3x) \) is the inner function and the entire expression is the outer function multiplied by 2. To differentiate, follow these steps:

• Identify the outer function (\( 2 \cos(u) \) where \( u = 3x \)) and the inner function (\( u = 3x \)).
• Differentiate the outer function with respect to the inner function: \( \frac{d}{du}(2 \cos(u)) = -2 \sin(u) \).
• Differentiate the inner function with respect to \( x \): \( \frac{du}{dx} = 3 \).
• Apply the chain rule: \( \frac{dy}{dx} = \frac{d}{du}(2 \cos(u)) \cdot \frac{du}{dx} = -2 \sin(3x) \cdot 3 = -6 \sin(3x) \).

This chain rule process lets you handle derivatives of complex functions step-by-step.

Derivative of Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, especially since they often appear in periodic phenomena. Knowing their derivatives helps analyze their behavior in different situations.

The derivatives of the basic trigonometric functions are as follows:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

These rules are crucial when differentiating expressions involving trigonometric functions. In our case, the original function is \( y = 2 \cos(3x) \). When finding its derivative, we used the rule that the derivative of \( \cos(x) \) is \( -\sin(x) \), and applied it alongside the chain rule. This combination helps us convert the trigonometric function into a form ready for integration or other operations.

Integral Simplification

Simplifying integrals is a vital step in both solving and understanding calculus problems. For the arc length calculation, we initially transform the integrand to make the integral easier to handle.

Starting with the expression obtained through differentiation, \( \frac{dy}{dx} = -6 \sin(3x) \), we plug this into the arc length formula:\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]Upon substituting the derivative into this formula, we square \( -6 \sin(3x) \), resulting in:\[ 1 + 36 \sin^2(3x) \]The key here is to recognize that simplifying an integral often involves rewriting it in a form that is less complex, possibly using trigonometric identities or algebraic manipulations.

By simplifying the inside of the square root, we make potential evaluations or numerical approximations more straightforward, even though we do not evaluate the integral in this problem. In other problems, this approach can reveal patterns or insights into the function's behavior over the interval.