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Ch. 6 - Applications of Definite Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 6 - Applications of Definite IntegralsProblem 6.PE.21
Chapter 6, Problem 6.PE.21

Find the lengths of the curves in Exercises 19–22.
y = (5/12) x⁶/⁵ ― (5/8)x⁴/⁵ , 1 ≤ x ≤ 32

Verified step by step guidance
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Recall the formula for the length of a curve defined by a function \(y = f(x)\) from \(x = a\) to \(x = b\): \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]
Identify the function given: \[y = \frac{5}{12} x^{\frac{6}{5}} - \frac{5}{8} x^{\frac{4}{5}}\] and the interval: \[1 \leq x \leq 32\]
Find the derivative \(\frac{dy}{dx}\) by differentiating each term using the power rule: \[\frac{dy}{dx} = \frac{5}{12} \cdot \frac{6}{5} x^{\frac{6}{5} - 1} - \frac{5}{8} \cdot \frac{4}{5} x^{\frac{4}{5} - 1}\] Simplify the coefficients and exponents carefully.
Substitute \(\frac{dy}{dx}\) into the arc length formula under the square root: \[\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\] This expression will be the integrand for the length integral.
Set up the definite integral for the length of the curve: \[L = \int_{1}^{32} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\] At this point, you can simplify the integrand if possible, then evaluate the integral using appropriate methods (analytical or numerical).

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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 y = f(x) from x = a to x = b is found using the integral L = ∫_a^b √(1 + (dy/dx)²) dx. This formula calculates the distance along the curve by summing infinitesimal line segments, accounting for both horizontal and vertical changes.
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Arc Length of Parametric Curves

Derivative of a Function

The derivative dy/dx represents the rate of change or slope of the function y with respect to x. For the arc length, computing dy/dx is essential to determine how steep the curve is at each point, which affects the length calculation.
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Derivatives of Other Trig Functions

Integration Techniques for Power Functions

Since the function involves fractional exponents, integrating the arc length expression requires familiarity with integrating power functions and possibly simplifying the integrand. Understanding how to handle fractional powers and algebraic manipulation is key to solving the integral.
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Representing Functions as Power Series
Related Practice
Textbook Question

Work


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a. stretch the spring 4 ft from its natural length?

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