9–20. Arc length calculations Find the arc length of the following curves on the given interval.
x = 2e^√2y + 1/16e^−√2y, for 0 ≤ y ≤ ln²/√2

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Identify the given function: \(x = 2e^{\sqrt{2}y} + \frac{1}{16}e^{-\sqrt{2}y}\), with the interval \(0 \leq y \leq \frac{(\ln 2)^2}{\sqrt{2}}\).

Recall the formula for the arc length of a curve defined as \(x = f(y)\) over an interval \([a, b]\): \[L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\]

Compute the derivative \(\frac{dx}{dy}\): Differentiate each term of \(x\) with respect to \(y\) using the chain rule, noting that \(\frac{d}{dy} e^{k y} = k e^{k y}\).

Square the derivative \(\left(\frac{dx}{dy}\right)^2\) and add 1 inside the square root to form the integrand: \[\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\]

Set up the definite integral for the arc length over the given interval and prepare to evaluate: \[L = \int_0^{\frac{(\ln 2)^2}{\sqrt{2}}} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\]

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주요 개념

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Arc Length Formula for Parametric and Explicit Curves

The arc length of a curve defined by x = f(y) over an interval [a, b] is found using the integral L = ∫ from a to b √(1 + (dx/dy)²) dy. This formula measures the length of the curve by summing infinitesimal line segments along the curve.

Differentiation of Exponential Functions

To find dx/dy for functions involving exponentials like e^(√2 y), apply the chain rule. The derivative of e^(g(y)) is e^(g(y)) * g'(y), where g(y) is the exponent function. Accurate differentiation is essential for setting up the arc length integral.

Evaluating Definite Integrals with Logarithmic Limits

The interval involves limits with logarithmic expressions, such as y from 0 to (ln 2)/√2. Understanding how to handle these limits and simplify expressions involving logarithms is important for correctly evaluating the definite integral for arc length.

9–20. Arc length calculations Find the arc length of the following curves on the given interval.x = 2e^√2y + 1/16e^−√2y, for 0 ≤ y ≤ ln²/√2