9–20. Arc length calculations Find the arc length of the following curves on the given interval.
y = ln (x−√x²−1), for 1 ≤ x ≤ √2(Hint: Integrate with respect to y.)

Verified step by step guidance

1. Recognize that the problem asks for the arc length of the curve defined by \(y = \ln \left(x - \sqrt{x^2 - 1}\right)\) on the interval \$1 \leq x \leq \sqrt{2}\(. The hint suggests integrating with respect to \)y\(, so consider expressing \)x\( as a function of \)y\( instead of \)y\( as a function of \)x$.
2. Start by rewriting the given function to find \(x\) in terms of \(y\). Since \(y = \ln \left(x - \sqrt{x^2 - 1}\right)\), exponentiate both sides to get \(e^y = x - \sqrt{x^2 - 1}\). Then solve this equation for \(x\).
3. After isolating \(x\), differentiate \(x\) with respect to \(y\) to find \(\frac{dx}{dy}\). This derivative will be used in the arc length formula when integrating with respect to \(y\).
4. Recall the arc length formula when integrating with respect to \(y\): \(L = \int_{y_1}^{y_2} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\). Determine the new limits of integration \(y_1\) and \(y_2\) by substituting the original \(x\)-values into the original function to find corresponding \(y\)-values.
5. Set up the integral for the arc length using the expression for \(\frac{dx}{dy}\) and the limits \(y_1\) and \(y_2\). The final step is to evaluate this integral to find the arc length, but per instructions, do not compute the final value here.

Key Concepts

Arc Length Formula

The arc length of a curve between two points is found by integrating the square root of 1 plus the derivative squared. When the curve is given as y = f(x), the formula is L = ∫√(1 + (dy/dx)²) dx. Alternatively, if integrating with respect to y, the formula becomes L = ∫√(1 + (dx/dy)²) dy.

Inverse Hyperbolic Functions and Logarithms

The function y = ln(x − √(x² − 1)) is related to inverse hyperbolic functions, specifically the inverse hyperbolic cosine. Recognizing this helps simplify derivatives and integrals. Understanding the relationship between logarithmic and inverse hyperbolic functions aids in manipulating the expression for easier integration.

Integration with Respect to y

Sometimes it is easier to compute arc length by integrating with respect to y instead of x, especially when the function or its inverse is simpler in terms of y. This requires expressing x as a function of y, finding dx/dy, and applying the arc length formula accordingly. This approach can simplify the integral and make it more manageable.