Problem 1 - Calculus: Early Transcendentals

Question

Find the arc length of \(f(x)=x^{3 / 2}\) on [0,2] .

Step-by-Step Solution

Answer

The arc length of \( f(x) = x^{3/2} \) on [0,2] is approximately 2.5317 units.

Step 1: Formula for Arc Length

The formula for the arc length of a function \( f(x) \) from \( a \) to \( b \) is given by \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx \]. We need to find the derivative \( \frac{df}{dx} \) for our function \( f(x) = x^{3/2} \).

Step 2: Taking the Derivative

Differentiate the function \( f(x) = x^{3/2} \). Using the power rule, \( \frac{d}{dx}[x^{n}] = nx^{n-1} \), we get: \( \frac{df}{dx} = \frac{3}{2}x^{1/2} \).

Step 3: Insert Derivative into the Arc Length Formula

Substitute \( \frac{df}{dx} = \frac{3}{2}x^{1/2} \) into the arc length formula: \[ L = \int_{0}^{2} \sqrt{1 + \left( \frac{3}{2}x^{1/2} \right)^2} \, dx \]. This simplifies to \[ L = \int_{0}^{2} \sqrt{1 + \frac{9}{4}x} \, dx \].

Step 4: Simplify and Evaluate the Integral

This integral requires a substitution method. Let \( u = 1 + \frac{9}{4}x \), then \( du = \frac{9}{4}dx \) or \( dx = \frac{4}{9}du \). Change the limits: when \( x = 0, u = 1 \) and when \( x = 2, u = 1 + \frac{9}{4}(2) = \frac{17}{4} \). The integral becomes \[ L = \int_{1}^{\frac{17}{4}} \sqrt{u} \times \frac{4}{9} \, du \].

Step 5: Solve the Integral

Compute the integral: \[ L = \frac{4}{9} \int_{1}^{\frac{17}{4}} u^{1/2} \, du \]. The antiderivative of \( u^{1/2} \) is \( \frac{2}{3}u^{3/2} \). Evaluate it from 1 to \( \frac{17}{4} \): \[ L = \frac{4}{9} \left[ \frac{2}{3} \left( \left( \frac{17}{4} \right)^{3/2} - 1^{3/2} \right) \right] \].

Step 6: Final Calculation

Calculate the values: \( \left( \frac{17}{4} \right)^{3/2} = \left( \sqrt{\frac{17}{4}} \right)^{3} = \left( \frac{\sqrt{17}}{2} \right)^{3} \). Continuing, the arc length \( L \) is \( \frac{4}{9} \times \frac{2}{3} \times \left( \left( \frac{\sqrt{17}}{2} \right)^{3} - 1 \right) \). Solve the numerical value to get the approximate measure of the arc.

Key Concepts

Arc Length

Understanding the concept of arc length is crucial when dealing with curves in calculus. The arc length measures the distance along the curve of a function from one point to another. It differs from simply measuring the distance between two points in a straight line. This is particularly useful in practical applications, such as landscape design or engineering.
The formula for the arc length of a function, denoted as \( L \), between two points \( a \) and \( b \) can be expressed as:
\[L = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dx} \right)^2} \, dx\]This equation accounts for the slope of the function at every point, calculated through its derivative, adapting the regular integration process for the curved line.
By applying this formula, you can find the length of any smooth curve represented by a function, which is essential for more complex geometry and analysis situations.

Integral Calculus

Integral calculus is the branch of calculus that deals with integrals and their applications. It is the reverse process of differentiation and focuses on computing the accumulating quantities, such as areas under a curve or total accumulated change.
In the context of finding arc length, integral calculus is used to sum up infinitely small sections of a curve to find its total length. The challenge is integrating the function under the square root of the arc length formula, often requiring more advanced techniques like substitution to simplify and solve.
Integral calculus is widely considered a powerful tool for analysis in a variety of scientific fields, including physics, engineering, and economics, as it allows us to model and solve real-world problems involving changing quantities.

Power Rule

The power rule is a basic derivative rule important in calculus, especially when dealing with polynomial functions. It states that for any real number \( n \), the derivative of \( x^n \) with respect to \( x \) is given by:

• \( \frac{d}{dx}[x^n] = nx^{n-1} \)

In our arc length problem for the function \( f(x) = x^{3/2} \), we utilize the power rule to differentiate the function and find:

• \( \frac{df}{dx} = \frac{3}{2}x^{1/2} \)

The power rule allows us to handle the derivatives of polynomial functions efficiently, as it simplifies the process of finding how a function's value changes as its input changes.

Substitution Method

The substitution method is a common technique used in integral calculus to simplify integrals. It is particularly useful in more complex integrals that are difficult to solve directly.
In essence, substitution involves changing the variable of integration to make the integral more manageable. You do this by letting a new variable replace an expression within the original integral. The trick is to choose a substitution that simplifies the integral significantly.
For example, in our arc length problem, we made the substitution \( u = 1 + \frac{9}{4}x \) to replace a complex expression under the square root. This step transforms the integral into a more straightforward form, facilitating easier integration. Once the integration is complete, we convert back to the original variable to find the final solution.
The substitution method can be a powerful tool for solving complicated integrals and is an essential skill for anyone studying calculus.