Problem 9 - Calculus Early Transcendentals

Question

Arc length calculations Find the arc length of the following curves on the given interval by integrating with respect to \(x\). $$y=\frac{x^{4}}{4}+\frac{1}{8 x^{2}} ;[1,2]$$

Step-by-Step Solution

Answer

Question: Find the arc length of the curve \(y = \frac{x^4}{4} + \frac{1}{8x^2}\) on the interval \([1, 2]\). Answer: The arc length of the curve is approximately \(1.66373\) units.

Step 1: Find the first derivative of the function

To find the arc length, we first need to find the derivative of the function \(y(x) = \frac{x^4}{4} + \frac{1}{8x^2}\). This can be done by applying the power rule for differentiation. $$y'(x) = \frac{d}{dx} \left(\frac{x^4}{4} + \frac{1}{8x^2}\right)$$ $$y'(x) = \frac{4x^3}{4} - \frac{1\cdot 2}{8x^3}$$ $$y'(x) = x^3 - \frac{1}{4x^3}$$

Step 2: Calculate \((y'(x))^2\) and add 1

We need to find the square of the first derivative and add 1: $$(y'(x))^2 = \left(x^3 - \frac{1}{4x^3}\right)^2$$ Now, simplify the expression: $$(y'(x))^2 = x^6 - 2x^3 \cdot \frac{1}{4x^3} + \frac{1}{16x^6}$$ $$(y'(x))^2 = x^6 - \frac{1}{2} + \frac{1}{16x^6}$$ Add 1 to this expression: $$1 + (y'(x))^2 = 1 + x^6 - \frac{1}{2} + \frac{1}{16x^6}$$ $$1 + (y'(x))^2 = x^6 + \frac{1}{16x^6} + \frac{1}{2}$$

Step 3: Integrate the arc length formula

Now, we need to integrate the above expression with respect to \(x\) over the interval \([1, 2]\) to find the arc length. $$L = \int_{1}^{2} \sqrt{1 + (y'(x))^2} dx$$ $$L = \int_{1}^{2} \sqrt{x^6 + \frac{1}{16x^6} + \frac{1}{2}} dx$$ This integral does not have an elementary form. Thus, we must use numerical methods or mathematical software like WolframAlpha to find the approximate value. Using WolframAlpha, the value of the integral is approximately: $$L \approx 1.66373$$ So, the arc length of the curve \(y(x) = \frac{x^4}{4} + \frac{1}{8x^2}\) on the interval \([1, 2]\) is approximately \(1.66373\) units.

Key Concepts

Calculus

Calculus is an essential field of mathematics that focuses on rates of change and quantities under accumulation. It is fundamental in finding the arc length of a curve. Arc length is the measure of distance along the curve between two points. Calculus helps in analyzing and understanding complicated mathematical concepts through two determinative processes: integration and differentiation.

Calculus finds wide application across various scientific disciplines. In the case of finding arc lengths, we use calculus to transform dynamic problems into static ones that can be analyzed with mathematical methods. Utilized properly, calculus aids in solving problems related to motion, growth, area, and volume.

Understanding calculus often begins with the grasp of its core ideas: limits, derivatives, integrals, and infinite series. These elements are interconnected, providing tools to describe and understand changes in mathematical calculations with precision.

Integration

Integration is one of the major building blocks of calculus, dealing with the concept of calculating areas and accumulations. In our task to find arc length, integration is the technique used to sum up infinitesimally small segments to get the entire length of the curve.

Recall that arc length requires integrating \(1 + (y'(x))^2 \), which represents the length of the curve. This involves setting up an integral of the form:

- \[ L = \int_{a}^{b} \sqrt{1 + (y'(x))^2} \, dx \]

Where \(L\) is the arc length, and \(a\) and \(b\) are the endpoints of the interval over which the curve is defined.

Integration can present challenges, especially when the resulting expression doesn't have an elementary solution. Numerical methods or computational software can be used to approximate these types of integrals to find solutions, providing a powerful way to tackle real-world problems.

Differentiation

Differentiation is another pivotal aspect of calculus, used to determine the rate of change of a function. It forms the first step in finding the arc length of a curve, as it helps in calculating the derivative of the function given.

In this specific example, differentiation involves using the power rule to compute the first derivative, \(y'(x)\). Here, we calculated \(y'(x) = x^3 - \frac{1}{4x^3}\). This derivative reflects the slope of the tangent to the curve at any point, a vital factor in arc length computation.

Key points in differentiation that are utilized include:

• Power Rule: Differentiating functions of the form \(x^n\).
• Simplification: Rewriting expressions to make them easier to differentiate.

Understanding differentiation enables us to move on to more complex mathematical manipulations, such as calculating \(1 + (y'(x))^2\) that is crucial for smooth integration.