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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 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]

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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: ∫ab1+dydx2dx.
Step 2: Compute the derivative of y = 2 cos(3x) with respect to x. Using the chain rule, the derivative is: dydx=-6sin(3x).
Step 3: Substitute the derivative into the arc length formula. The integrand becomes: 1+(-6sin(3x))2.
Step 4: Simplify the square of the derivative. Squaring -6sin(3x) gives: -6sin(3x)2=36sin(3x)2.
Step 5: Write the integral for the arc length. The integral becomes: ∫-ππ1+36sin(3x)2dx.

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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.
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Arc Length of Parametric Curves

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.
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Derivatives of Other Trig Functions

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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