Problem 12 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The arc length is computed numerically due to complex integrals.

Step 1: Understand the formula for arc length

The formula to find the arc length of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is given by \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]. For the function \( f(x) = \frac{1}{3}x^{3/2} - x^{1/2} \) on the interval \([0, 1]\), we need to first find the derivative \( \frac{dy}{dx} \), and then plug it into this formula.

Step 2: Find the derivative of the function

The function is given as \( f(x) = \frac{1}{3}x^{3/2} - x^{1/2} \). To find its derivative, use the power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \). Thus, the derivative \( f'(x) \) is: \( f'(x) = \frac{1}{3} \cdot \frac{3}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \). Simplifying, we have \( f'(x) = \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \).

Step 3: Set up the integral for arc length

Substitute \( f'(x) \) into the arc length formula: \[ L = \int_{0}^{1} \sqrt{1 + \left( \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \right)^2} \, dx \]. Simplify the expression inside the square root first:

Step 4: Simplify the expression inside the integral

Expand the expression: \( \left( \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \right)^2 = \frac{1}{4}x + \frac{1}{4}x^{-1} - \frac{1}{2} \). Adding \(1\), we get \( 1 + \frac{1}{4}x + \frac{1}{4}x^{-1} - \frac{1}{2} = \frac{1}{2} + \frac{1}{4}x + \frac{1}{4}x^{-1} \).

Step 5: Solve the integral

Now, we need to calculate \( \int_{0}^{1} \sqrt{\frac{1}{2} + \frac{1}{4}x + \frac{1}{4}x^{-1}} \, dx \), which is generally complex and may involve substitution or numerical methods. The exact solution requires advanced techniques such as using elliptic integrals, which is typically done using a calculator or software.

Key Concepts

Derivative Calculation

One of the first steps in finding the arc length of a curve is determining the derivative of the function. This is because the derivative represents the slope of the tangent line at any given point on the curve. In our exercise, we have the function:
\[ f(x) = \frac{1}{3}x^{3/2} - x^{1/2} \]
To find its derivative, we use the Power Rule, which states that for any function of the form \( x^n \), the derivative is \( nx^{n-1} \).
Applying this rule to our function:

• The term \( \frac{1}{3}x^{3/2} \) becomes \( \frac{1}{3} \times \frac{3}{2}x^{1/2} = \frac{1}{2}x^{1/2} \).
• The term \( -x^{1/2} \) becomes \( -\frac{1}{2}x^{-1/2} \).

Therefore, the derivative of \( f(x) \) is:
\[ f'(x) = \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \]
This derivative is crucial for setting up the integral needed to find the arc length.

Integral Setup

Once we have the derivative, the next step is to set up the integral. The formula for the arc length \( L \) of a curve described by \( y = f(x) \) from \( x = a \) to \( x = b \) is:
\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
In our specific exercise, after calculating the derivative, we need to substitute \( f'(x) = \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \) into the arc length formula. This results in the following integral:
\[ L = \int_{0}^{1} \sqrt{1 + \left( \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \right)^2} \, dx \]
Before integrating, it's usually necessary to simplify the expression under the square root. In this case, simplifying gives us:
\[ \sqrt{\frac{1}{2} + \frac{1}{4}x + \frac{1}{4}x^{-1}} \]
At this point, we can set up the integral to calculate the arc length.

Power Rule

The Power Rule is a fundamental tool in calculus used to find the derivative of a polynomial function. It's straightforward and very handy, particularly when dealing with functions where a variable is raised to a power.
The rule is stated as: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
Here's how you can use the Power Rule effectively:

• If the exponent is a positive integer, find the derivative by multiplying by the exponent and subtracting one from it.
• If the exponent is a fraction (as in our exercise), the same rule applies. For example, \( x^{3/2} \) becomes \( \frac{3}{2}x^{1/2} \).
• Even when the exponent is negative, the rule holds true. Take \( x^{-1/2} \) which becomes \( -\frac{1}{2}x^{-3/2} \).

The Power Rule speeds up the process of differentiation, providing a quick way to find derivatives and hence aiding in problems involving arc lengths of curves.

Elliptic Integrals

Elliptic integrals are a class of integrals that cannot be expressed in terms of elementary functions, which makes them complex and often require numerical or software-based solutions.
In our arc length problem, the integral we derived:
\[ \int_{0}^{1} \sqrt{\frac{1}{2} + \frac{1}{4}x + \frac{1}{4}x^{-1}} \, dx \]
is not easily solvable by simple analytical methods. This is where elliptic integrals come into play. They appear in calculations involving arc lengths of ellipses or other complex curves.
To solve this, you typically:

• Use numerical methods or software tools like Matlab or Mathematica which are equipped to approximate elliptic integrals efficiently.
• Understand that these integrals often involve parameterizing the curve and reformulating the integral in terms of the elliptic form.
• Recognize that while they may seem daunting, proper resources and tools make them manageable.

Elliptic integrals bridge the gap between elementary calculus techniques and more advanced calculations, highlighting the richness and depth of mathematical studies.