Problem 22 - Calculus Early Transcendentals

Question

Arc length calculations with respect to \(y\) Find the arc length of the following curves by integrating with respect to \(y.\) $$y=\ln (x-\sqrt{x^{2}-1}), \text { for } 1 \leq x \leq \sqrt{2}$$

Step-by-Step Solution

Answer

Based on the step-by-step solution, we found that the arc length of the curve \(y=\ln (x-\sqrt{x^{2}-1}), \text{ for } 1 \leq x \leq \sqrt{2}\) is infinite.

Step 1: Express x as a function of y

To express x as a function of y, we will use the given equation: $$y=\ln (x-\sqrt{x^{2}-1})$$ We need to solve this equation for x. First, apply the exponential function to both sides of the equation: $$e^y=x-\sqrt{x^{2}-1}$$ Now, isolate the square root term on one side: $$\sqrt{x^{2}-1}=x-e^y$$ Square both sides: $$(x^2-1)=(x-e^y)^2$$ Expand the right side of the equation and simplify: $$x^2-1 = x^2 - 2xe^y + e^{2y}$$ $$1 = e^{2y} - 2xe^y$$ Now we need to solve for x to get x(y): $$x=\frac{e^{2y}+1}{2e^y}$$ The relation x(y) is found.

Step 2: Compute the derivative of x with respect to y

Now we need to differentiate x with respect to y to find \(\frac{dx}{dy}\). We use the quotient rule for differentiation to compute the derivative: $$\frac{dx}{dy}=\frac{(2e^{2y}+0)(2e^y)-(e^{2y}+1)(2e^y)}{(2e^y)^2}$$ Simplify the expression: $$\frac{dx}{dy}=\frac{2e^{3y}-2e^{3y}}{4e^{2y}}$$ $$\frac{dx}{dy}=\frac{0}{4e^{2y}}=0$$

Step 3: Calculate the arc length by integrating

The arc length can be found by integrating the square root of 1 + \((\frac{dx}{dy})^2\) with respect to y over the given interval. In our case, \(\frac{dx}{dy}\) is found to be 0. So we have: $$\text{arc length} = \int_{y_1}^{y_2} \sqrt{1+(\frac{dx}{dy})^2}dy$$ Since \(\frac{dx}{dy}=0\), the integrand becomes: $$\sqrt{1+0^2}=\sqrt{1}$$ Now we need to find \(y_1\) and \(y_2\) for the given x-interval, \(1\leq x\leq\sqrt{2}\). Using the found x(y) expression: $$y_1=\ln(1-\sqrt{1-1})=\ln(0)=-\infty$$ $$y_2=\ln(\sqrt{2}-\sqrt{\sqrt{2}^2-1})=\ln(1)=0$$ Now, integrate: $$\text{arc length} = \int_{-\infty}^{0} \sqrt{1} dy$$ The integral evaluates to: $$\text{arc length}=\left[y\right]_{-\infty}^0=0-(-\infty)=\infty$$ The arc length of the given curve is infinite.

Key Concepts

Integration with respect to y

When it comes to finding arc length, integrating with respect to a certain variable can provide unique perspectives and solutions. In this problem, we're integrating with respect to \( y \), which means that we have a function implicitly defined in terms of \( y \) rather than the usual \( x \). This requires us to:

• Express \( x \) as a function of \( y \) so that we can differentiate properly.
• Use the correct limits for \( y \) as derived from the given \( x \) range.

The process involves working on the original equation to isolate \( x \) in terms of \( y \), and effectively differentiate to find \( \frac{dx}{dy} \). It's vital to remember that the arc length is based on the integral of the square root of \(1 + (\frac{dx}{dy})^2 \).
This method explores the equation behavior over the \( y \) interval, aiming to calculate distances along the curve with precision.

Quotient Rule

The quotient rule is essential when you're differentiating functions expressed as a ratio of two other functions. In this problem, we've derived \( x \) as \( \frac{e^{2y}+1}{2e^y} \). To find \( \frac{dx}{dy} \), we apply the quotient rule, which states: \[ \frac{d}{dy}\left( \frac{u}{v} \right) = \frac{v\frac{du}{dy} - u\frac{dv}{dy}}{v^2} \]Here, \( u = e^{2y} + 1 \) and \( v = 2e^y \). This requires calculating the derivatives \( \frac{du}{dy} \) and \( \frac{dv}{dy} \) individually:

• For \( u = e^{2y} + 1 \): \( \frac{du}{dy} = 2e^{2y} \)
• For \( v = 2e^y \): \( \frac{dv}{dy} = 2e^y \)

After applying these derivatives in the quotient rule, the calculations simplify, and in this specific problem, the resulting \( \frac{dx}{dy} = 0 \), due to cancellation. This step is crucial for determining the behavior of the curve's differential.

Derivative

Differentiation helps us understand how a function behaves as its input changes. In the context of arc length calculaiton, we are especially interested in the derivative \( \frac{dx}{dy} \). This aspect tells us the rate at which \( x \) changes with respect to \( y \). Calculating it accurately using the given functions is essential to finding the arc length. If \( \frac{dx}{dy} = 0 \), as it is in this scenario, the function shows a zero change rate which significantly influences integration results. Here, simplifying the expression leaves us with no variable-dependent term, indicating no variation across the observed \( y \) interval.

Exponential Function

The exponential function, denoted as \( e^y \), is a fundamental mathematical tool used in various calculations, especially where growth or decay processes are involved. In the given exercise, \( e^y \) appears when transforming the logarithmic relationship into a more tractable form for differentiation and integration. Its powerful property of returning its own derivative (\( \frac{d}{dy} e^y = e^y \)) makes the exponential function highly predictable and useful in calculus operations. Transforming the function with the exponential function helps in:

• Simplifying complex logarithmic expressions.
• Enabling clearer and more straightforward derivative calculations.

In the context of the arc length problem, recognizing the role of the exponential function helps in understanding how the logarithmic scale impacts the curve's geometry.