Question

Find the arc length of the function on the given interval. \(f(x)=\frac{1}{3} x^{3 / 2}-x^{1 / 2}\) on [0,1] .

Short answer

The arc length is \(\frac{8}{3}\).

Step-by-step solution

Identify the formula for arc length

The arc length of a curve given by a function \(y = f(x)\) over the interval \([a, b]\) is calculated by the formula:\[ L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx \] First, we need to find \(f'(x)\), the derivative of the function.

Find the derivative of the function

The function is \(f(x) = \frac{1}{3} x^{3/2} - x^{1/2}\). Using the power rule, the derivative is:\[ f'(x) = \frac{1}{3} \cdot \frac{3}{2} x^{1/2} - \frac{1}{2} x^{-1/2} = \frac{1}{2} x^{1/2} - \frac{1}{2} x^{-1/2} \] Thus, \(f'(x) = \frac{1}{2} x^{1/2} - \frac{1}{2} x^{-1/2}\).

Calculate \(1 + (f'(x))^2\)

Compute \((f'(x))^2\) and add 1:\[ (f'(x))^2 = \left( \frac{1}{2} x^{1/2} - \frac{1}{2} x^{-1/2} \right)^2 \]Expanding and simplifying:\[ \left(\frac{1}{2} x^{1/2}\right)^2 - 2 \left(\frac{1}{2} x^{1/2}\right)\left(\frac{1}{2} x^{-1/2}\right) + \left(\frac{1}{2} x^{-1/2}\right)^2 = \frac{1}{4} x + \frac{1}{4}\frac{1}{x} - \frac{1}{2} \]Thus, \((f'(x))^2 = \frac{1}{4}x + \frac{1}{4}x^{-1} - \frac{1}{2}\).Now calculate \(1 + (f'(x))^2\):\[ 1 + \frac{1}{4}x + \frac{1}{4}x^{-1} - \frac{1}{2} = \frac{1}{4}x + \frac{1}{4}x^{-1} + \frac{1}{2} \].

Set up the integral for arc length

Substitute \(1 + (f'(x))^2\) into the integral:\[ L = \int_{0}^{1} \sqrt{\frac{1}{4}x + \frac{1}{4}x^{-1} + \frac{1}{2}} \, dx \].This integral needs to be solved over the interval \([0, 1]\).

Simplify and compute the integral

Simplify the expression under the square root. Notice that:\[ \sqrt{\frac{1}{4}x + \frac{1}{4}x^{-1} + \frac{1}{2}} = \frac{1}{2} \sqrt{x + \frac{1}{x} + 2}\]\[ x + \frac{1}{x} + 2 = \left(x^{1/2} + x^{-1/2}\right)^2 \]Thus:\[ \int_{0}^{1} \left(x^{1/2} + x^{-1/2}\right) \, dx \]

Evaluate the integral

Evaluate the integral:\[ \int_{0}^{1} \left(x^{1/2} + x^{-1/2}\right) \, dx \]The antiderivative of \(x^{1/2}\) is \(\frac{2}{3}x^{3/2}\) and of \(x^{-1/2}\) is \(2x^{1/2}\). Thus:\[ \frac{2}{3} x^{3/2} + 2 x^{1/2} \Bigg\vert_0^1 \] Evaluating from 0 to 1 gives:\[ \left(\frac{2}{3} \cdot 1^{3/2} + 2 \cdot 1^{1/2}\right) - \left(\frac{2}{3} \cdot 0^{3/2} + 2 \cdot 0^{1/2}\right) = \frac{8}{3}\].

Key concepts

Understanding the Derivative

To grasp the concept of finding the arc length of a function, it's essential to understand the role of derivatives. In mathematics, a **derivative** represents how a function's output changes as its input changes. Essentially, it measures the instantaneous rate of change or the slope of the function at a specific point.

Think of the derivative as the "speed" of the function. In the context of arc length, we need the derivative to determine how steep or gentle the curve is at each point.

When taking derivatives of functions involving powers, we use the **power rule**. This rule states: If you have a function of the form \(f(x) = x^n\), then its derivative is \(f'(x) = n \cdot x^{n-1}\). Using this rule, we found the derivative of \(f(x) = \frac{1}{3}x^{3/2} - x^{1/2}\), resulting in \(f'(x) = \frac{1}{2}x^{1/2} - \frac{1}{2} x^{-1/2}\).

Having the derivative is crucial, as it plays a direct role in calculating the arc length through integration.

Introduction to Integration

**Integration** is the process of finding the "whole" given its parts, which is often used to determine the area under a curve. In our case, integration helps us calculate the arc length of a function by summing up tiny line segments that approximate the curve.

You can think of integration as the opposite of differentiation. While differentiation breaks down a function into its rate of change, integration "adds up" these changes to find the total - like putting together puzzle pieces to see the entire picture.

In the arc length formula, integration is symbolized by the integral sign \(\int\). It represents adding up all the small distances calculated along the interval \[a, b\]. In this exercise, once we've manipulated the derivative and calculated \(1 + (f'(x))^2\), we integrate \(\sqrt{1 + (f'(x))^2} \) over the interval [0, 1] to find the arc length.

Dive into Integral Calculus

Understanding **integral calculus** or the "art of integration" is key to solving arc length problems. Integral calculus enables us to find exact solutions to problems involving accumulation, such as areas, volumes, and other quantities that can be approximated by summing numerous small parts.

Integral calculus consists of definite and indefinite integrals. A definite integral calculates the accumulation over a specific interval, such as [0, 1] in our arc length problem. The definite integral has limits, and its result is a numerical value.

In computing the arc length, we've used integral calculus to integrate \(\sqrt{\frac{1}{4}x + \frac{1}{4}x^{-1} + \frac{1}{2}}\). To evaluate it precisely, we simplified it to \(\frac{1}{2} \sqrt{x + \frac{1}{x} + 2}\), allowing for a more straightforward computation of the integral, which eventually led to our final result \(\frac{8}{3}\).

The Role of the Sqrt Function

The **sqrt function**, or square root, is crucial when finding arc lengths. The arc length formula includes a square root to account for the physical distance along a curve. This is because, in the traditional distance formula \(\sqrt{\Delta x^2 + \Delta y^2}\), the square root converts squared units back to their original form.

In our exercise, after computing the squared derivative, we need to incorporate the square root to compute the linear distance accurately. This square root is part of the integrand, \(\sqrt{1 + (f'(x))^2}\), reflecting the length of the curve on the interval [0, 1].

Using square roots can complicate calculations, but they are essential for obtaining true lengths, as they serve to correct the dimensions. Completing the integration over these adjusted values gives the arc length, making square roots indispensable in these problems.

• The square root ensures correct dimensionality in distance calculations.
• It prepares the integrand for evaluating the true curve length.

Understanding this allows for a deeper grasp of the relationship between derivatives, integration, and curve measurements.