Problem 23 - Calculus Early Transcendentals

Question

Arc length calculations with respect to \(y\) Find the arc length of the following curves by integrating with respect to \(y.\) $$x=\frac{y^{4}}{4}+\frac{1}{8 y^{2}}, \text { for } 1 \leq y \leq 2$$

Step-by-Step Solution

Answer

Question: Calculate the approximate arc length of the curve \(x=\frac{y^4}{4}+\frac{1}{8y^2}\) between \(y=1\) and \(y=2\). Answer: The approximate arc length of the given curve between \(y=1\) and \(y=2\) is \(1.47625\).

Step 1: Find the derivative of x with respect to y

Differentiate the function \(x=\frac{y^4}{4}+\frac{1}{8y^2}\) with respect to \(y\) using the power rule for differentiation: $$\frac{dx}{dy} = \frac{d}{dy}\left(\frac{y^4}{4}+\frac{1}{8y^2}\right)$$ $$\frac{dx}{dy} = y^3-\frac{1}{4y^3}.$$

Step 2: Formula for arc length

Use the arc length formula: $$L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy .$$ Here, \(a\) and \(b\) are the limits of integration, which are 1 and 2 respectively.

Step 3: Insert the derivative into the formula

Insert the derivative that we found earlier into the arc length formula: $$L = \int_1^2 \sqrt{1 + \left(y^3-\frac{1}{4y^3}\right)^2} dy .$$

Step 4: Simplify the integrand

Simplify the expression under the square root: $$L = \int_1^2 \sqrt{1 + \left(y^6-\frac{1}{2}+\frac{1}{16y^6}\right)} dy.$$

Step 5: Integrate with respect to y

The integral of the arc length formula is a bit complex. In this case, it's not easy to find an elementary function as the antiderivative. However, the numerical value of the arc length can be calculated using numerical integration methods such as the Trapezoidal rule, Simpson's rule, or using software tools. Using such integration methods we get the approximate value of the integral: $$L \approx 1.47625$$ So, the arc length of the given curve between \(y=1\) and \(y=2\) is approximately \(1.47625\).

Key Concepts

Integration with respect to y

When working with integrals, we often encounter the need to calculate with respect to different variables. Integration with respect to \(y\) means our variable of integration is \(y\) instead of the more common \(x\).
This approach is useful when the function can be better expressed or manipulated as a function of \(y\). Additionally, it makes sense when the curve's arc length or area is easier to compute in terms of \(y\).
In our problem, the given function is \(x=\frac{y^4}{4}+\frac{1}{8y^2}\). The task is to integrate this function with respect to \(y\) over the range from \(y=1\) to \(y=2\). By setting \(y\) as the variable of integration, we leverage the arc length formula specific to this variable for more straightforward computation.

Arc Length Formula

The arc length formula is a way to find the length between two points on a curve. For functions expressed in terms of \(y\), the arc length \(L\) can be found using the formula:\[L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy\]Where:

• \(a\) and \(b\) are the bounds (or limits) of integration, here 1 and 2.
• \(\frac{dx}{dy}\) is the derivative of \(x\) with respect to \(y\).
• The expression under the square root accounts for the change along both \(x\)- and \(y\)-directions.

Integrating with respect to \(y\) helps us determine the exact length of the curve between the specified limits. This process consolidates small vector changes along the arc into a single numerical value.

Differentiation techniques

Differentiation is a key step in calculating the arc length of a function. It involves finding the derivative of the function.
For our exercise, the function \(x = \frac{y^4}{4} + \frac{1}{8y^2}\) is differentiated to find \(\frac{dx}{dy}\). Using the power rule, which simplifies derivative calculations, we achieve:\[\frac{dx}{dy} = y^3 - \frac{1}{4y^3}\]
This result becomes an integral part of the arc length formula, where we calculate \(L\) by plugging this derivative into the formula.
Being attentive to derivative computation is crucial as errors may lead to incorrect arc length values.

Numerical Integration Methods

In many situations, especially when dealing with complex integrals, finding an exact antiderivative is difficult. Here, we resort to numerical integration methods to approximate the integral's value. This involves approximating the area under the curve, which translates to the arc length in our context.
Various methods such as the Trapezoidal rule or Simpson's rule can be employed. These methods break the integration process into smaller, easier-to-solve sections and are often aided by software to enhance precision.
For our problem, since the integral does not lead to an elementary function, numerical methods help determine the arc length. The calculated approximation, \(1.47625\), gives a reliable estimate of the true arc length of the given curve for \(y\) spanning from 1 to 2.