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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.2.81

81. Find the lengths of the following curves.
a. y = (x²/8) - ln(x), 4≤x≤8

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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: \[y = \frac{x^2}{8} - \ln(x)\] and the interval: \[4 \leq x \leq 8\]
Compute the derivative \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^2}{8} - \ln(x) \right) = \frac{2x}{8} - \frac{1}{x} = \frac{x}{4} - \frac{1}{x}\]
Square the derivative: \[\left( \frac{dy}{dx} \right)^2 = \left( \frac{x}{4} - \frac{1}{x} \right)^2\] Expand this expression carefully to simplify the integrand.
Set up the integral for the length of the curve: \[L = \int_{4}^{8} \sqrt{1 + \left( \frac{x}{4} - \frac{1}{x} \right)^2} \, dx\] This integral can then be evaluated (analytically or numerically) to find the length.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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 changes in both x and y.
추천 영상:
가이드 코스
06:29
Arc Length of Parametric Curves

Derivative of the Given Function

To apply the arc length formula, you need the derivative dy/dx of the function y = (x²/8) - ln(x). Differentiating each term separately helps find the slope of the curve at any point, which is essential for computing the integrand √(1 + (dy/dx)²).
추천 영상:
06:30
Derivatives of Other Trig Functions

Definite Integration over the Interval

After finding the integrand, evaluate the definite integral from x = 4 to x = 8. This process sums the infinitesimal lengths along the curve within the specified interval, yielding the total length of the curve segment.
추천 영상:
가이드 코스
05:43
Definition of the Definite Integral