Problem 5 - Calculus Single Variable

Question

In each of Exercises \(5-12,\) calculate the arc length \(L\) of the graph of the given function over the given interval. $$ f(x)=2+x^{3 / 2} \quad I=[1,4] $$

Step-by-Step Solution

Answer

The arc length \(L\) is approximately 15.206 units.

Step 1: Understand the Arc Length Formula

The arc length of a curve defined by the function \( f(x) \) over the interval \([a,b]\) is given by the formula: \[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \] Identify \( f(x) = 2 + x^{3/2} \) and the interval \([1, 4]\).

Step 2: Find the Derivative of the Function

Calculate the derivative of the function \( f(x) = 2 + x^{3/2} \).Use the power rule: \[ f'(x) = \frac{d}{dx}(2 + x^{3/2}) = \frac{3}{2}x^{1/2} \]

Step 3: Set Up the Arc Length Integral

Substitute \( f'(x) = \frac{3}{2}x^{1/2} \) into the arc length formula: \[ L = \int_{1}^{4} \sqrt{1 + \left( \frac{3}{2}x^{1/2} \right)^2} \, dx \]Simplify the term inside the square root: \[ L = \int_{1}^{4} \sqrt{1 + \frac{9}{4}x} \, dx \]

Step 4: Simplify the Integral

Express the square root in its simplified integral form: \[ L = \int_{1}^{4} \sqrt{\frac{4+9x}{4}} \, dx \]Which can be written as: \[ L = \int_{1}^{4} \frac{\sqrt{4 + 9x}}{2} \, dx \]

Step 5: Evaluate the Integral

Solve the integral: 1. Substitute \( u = 4 + 9x \), thus \( du = 9dx \) or \( dx = \frac{du}{9} \).2. Change limits of integration accordingly: when \( x = 1, u = 13 \); when \( x = 4, u = 40 \).3. Substitute into the integral: \[ L = \int_{13}^{40} \frac{\sqrt{u}}{18} \, du \]4. Evaluate the integral: \[ L = \left[ \frac{2}{27}u^{3/2} \right]_{13}^{40} \]Which yields: \[ L = \frac{2}{27}((40)^{3/2} - (13)^{3/2}) \]

Step 6: Compute the Final Result

Calculate the values of the fractional powers and plug them into the evaluated integral expression:1. \( 40^{3/2} = \sqrt{40^3} = 40 \times \sqrt{40} \approx 253.02 \).2. \( 13^{3/2} = \sqrt{13^3} = 13 \times \sqrt{13} \approx 46.87 \).3. Substitute these values back into the expression for \( L \):\[ L = \frac{2}{27}((253.02) - (46.87)) \approx 15.206 \].

Key Concepts

Integral Calculus

Integral calculus plays a critical role in calculating the arc length of a curve. It involves finding the integral of a function over a specific interval. This is crucial when we want to determine the total length of a curve between two points.
In the context of the exercise, we are dealing with the curve defined by the function \( f(x) = 2 + x^{3/2} \). To find the arc length, we apply the formula for arc length which is:

• \( L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \)

Here, \( [a, b] = [1, 4] \). Integral calculus allows us to sum up an infinite number of infinitesimally small segments along the curve. Each segment is approximated by an arc, and the entire length is the integral of these arcs. By evaluating the integral, we get the exact arc length.

Derivatives

Derivatives are the foundation of understanding how a function changes at any given point. In terms of arc length calculation, derivatives help us find the rate of change of the function vertically with respect to its horizontal change. This rate of change is encapsulated in the derivative \( f'(x) \).
For our given function \( f(x) = 2 + x^{3/2} \), we find the derivative using the power rule. It results in:

• \( f'(x) = \frac{3}{2}x^{1/2} \)

This derivative forms a vital part of the arc length formula. Essentially, \( f'(x) \) represents the slope of the tangent to the curve at any point \( x \). When we square this derivative and add 1 under the square root in the arc length formula, it reflects how the curve deviates from a straight path, thus affecting its total length.

Substitution Method

The substitution method is a practical technique used to simplify integrals, especially when dealing with complex functions. It involves changing the variable of integration to transform the integral into a more manageable form.
In this exercise, to handle the integral \( \int \frac{\sqrt{4+9x}}{2} \, dx \), we use substitution:

• We set \( u = 4 + 9x \), which simplifies the square root term.
• The derivative \( du = 9\,dx \) indicates that \( dx = \frac{du}{9} \).
• We adjust the limits of integration: When \( x = 1, \ u = 13 \) and when \( x = 4, \ u = 40 \).

After substitution, the integral transforms to \( \int_{13}^{40} \frac{\sqrt{u}}{18} \, du \). This new integral is simpler to evaluate. The substitution method effectively reduces the complexity, making the calculation of the arc length feasible.