Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.5.4

Chapter 6, Problem 6.5.4

3–6. 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 3x on [−π,π]

Verified step by step guidance

Step 1: Recall the formula for the arc length of a curve y = f(x) on the interval [a, b]. The arc length is given by:

\int_{a}^{b} \sqrt{1 + {\frac{d y}{d x}}^{2}} d x

Step 2: Compute the derivative of y = 2 cos(3x) with respect to x. Using the chain rule, the derivative is:

\frac{d y}{d x} = - 6 sin (3 x)

Step 3: Substitute the derivative into the arc length formula. The integrand becomes:

\sqrt{1 + {(- 6 sin (3 x))}^{2}}

Step 4: Simplify the square of the derivative. Squaring

- 6 sin (3 x)

gives:

{- 6 sin (3 x)}^{2} = 36 {sin (3 x)}^{2}

Step 5: Write the integral for the arc length. The integral becomes:

\int_{-}^{π} \sqrt{1 + 36 {sin (3 x)}^{2}} d x

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Arc Length Formula

The arc length of a curve defined by a function y = f(x) from x = a to x = b is given by the integral L = ∫[a to b] √(1 + (dy/dx)²) dx. This formula derives from the Pythagorean theorem, where the infinitesimal segments of the curve are approximated as straight lines.

Derivative of a Function

The derivative of a function, denoted as dy/dx, represents the rate of change of the function with respect to x. For the curve y = 2 cos(3x), finding the derivative is essential to apply the arc length formula, as it will be squared and added to 1 under the square root in the integral.

Definite Integral

A definite integral calculates the accumulation of quantities, such as length, over a specified interval [a, b]. In this context, the integral will be set up from -π to π, representing the total arc length of the curve y = 2 cos(3x) over that interval, without evaluating it.

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Definition of the Definite Integral

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3–6. 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 3x on [−π,π]

Verified video answer for a similar problem:

Key Concepts

Arc Length Formula

Derivative of a Function

Definite Integral

3–6. 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 3x on [−π,π]