Problem 33 - Thomas Calculus in SI Units

Question

Find the arc length function for the graph of \(f(x)=2 x^{3 / 2}\) using (0, 0) as the starting point. What is the length of the curve from (0,0) to (1,2)\(?\)

Step-by-Step Solution

Answer

The arc length is approximately 2.268 units.

Step 1: Find the Derivative of the Function

The given function is \( f(x) = 2x^{3/2} \). To find the derivative, use the power rule. The derivative \( f'(x) \) is calculated as: \[ f'(x) = \frac{d}{dx}(2x^{3/2}) = 2 \times \frac{3}{2} x^{3/2 - 1} = 3x^{1/2} \]

Step 2: Set Up the Arc Length Integral

The formula for the arc length \( L \) of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is: \[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \] For the function \( f(x) \), the derivative is \( f'(x) = 3x^{1/2} \). Thus, the integral becomes: \[ L = \int_{0}^{1} \sqrt{1 + (3x^{1/2})^2} \, dx \] \[ L = \int_{0}^{1} \sqrt{1 + 9x} \, dx \]

Step 3: Integrate the Arc Length Function

To solve \( \int_{0}^{1} \sqrt{1 + 9x} \, dx \), use a substitution method. Let \( u = 1 + 9x \), which gives \( du = 9 \, dx \), or \( dx = \frac{1}{9} \, du \). The limits of integration change from \( x = 0 \) to \( u = 1+9\times0 = 1 \) and from \( x = 1 \) to \( u = 1+9\times1 = 10 \). The integral becomes: \[ L = \int_{1}^{10} \sqrt{u} \cdot \frac{1}{9} \, du \] \[ L = \frac{1}{9} \int_{1}^{10} u^{1/2} \, du \] Integrating, you get: \[ L = \frac{1}{9} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{10} \] \[ L = \frac{2}{27} \left[ 10^{3/2} - 1^{3/2} \right] \]

Step 4: Evaluate the Integral for Arc Length

Finish evaluating by computing the numerical values: \[ L = \frac{2}{27} (31.6227766017 - 1) \] \[ L \approx \frac{2}{27} \times 30.6227766017 \] \[ L \approx 2.268 \] Thus, the length of the curve from \((0,0)\) to \((1,2)\) is approximately 2.268 units.

Key Concepts

Derivative Calculation

To find the arc length of a curve defined by a function, the first step often involves calculating the derivative of the function. In the exercise, the given function is \( f(x) = 2x^{3/2} \). Calculating derivatives is fundamental in mathematics, as they tell us about the rate of change at any point on the curve. Here, we utilized the power rule for derivatives, which states that if \( f(x) = ax^n \), then the derivative \( f'(x) = anx^{n-1} \). Applying this rule to our function:- Multiply the coefficient (2) by the exponent (\(\frac{3}{2}\)): 2 \(\times\frac{3}{2} = 3\)- Decrease the exponent by 1: \(\frac{3}{2} - 1 = \frac{1}{2}\)Putting it all together, the derivative is \( f'(x) = 3x^{1/2} \). This is crucial for setting up the arc length formula.

Arc Length Integral

The arc length integral is a tool used to determine the length of a curve between two points. Once you have the derivative, you can set up the arc length integral. For a function \( y = f(x) \), the arc length \( L \) from \( x = a \) to \( x = b \) is given by the integral:\[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \]Using our derivative \( f'(x) = 3x^{1/2} \), the integral becomes:- Calculate \((f'(x))^2 = (3x^{1/2})^2 = 9x\),- So, \(\sqrt{1 + (f'(x))^2} = \sqrt{1 + 9x}\).Thus, the integral for arc length from 0 to 1 is:\[ L = \int_{0}^{1} \sqrt{1 + 9x} \, dx \]

Substitution Method

The substitution method is commonly used to simplify complex integrals, allowing them to be calculated more easily. In our arc length problem, let \( u = 1 + 9x \). This substitution helps to simplify the integral:- Differentiate the substitution: \( du = 9 \, dx \) or \( dx = \frac{1}{9} \, du \).- Change the limits of integration: - When \( x = 0 \), \( u = 1 + 9 \times 0 = 1 \). - When \( x = 1 \), \( u = 1 + 9 \times 1 = 10 \).Substituting these into the integral gives:\[ L = \int_{1}^{10} \sqrt{u} \cdot \frac{1}{9} \, du \]This integral is simpler to evaluate, as it now involves a straightforward power of \( u \).

Definite Integral Evaluation

After organizing the integral into a simplified form using substitution, it is time to evaluate it. The integral becomes:\[ L = \frac{1}{9} \int_{1}^{10} u^{1/2} \, du \]Evaluating \( \int u^{1/2} \, du \) yields \( \frac{2}{3} u^{3/2} \). When evaluating definite integrals, calculate the antiderivative at the upper limit and subtract the value at the lower limit:- Compute \( \left[ \frac{2}{3} u^{3/2} \right]_{1}^{10} \) which represents \( \frac{2}{3} \times 10^{3/2} - \frac{2}{3} \times 1^{3/2} \).- Substitute these back into the main expression:\[ L = \frac{1}{9} \left( \frac{2}{3} (31.6227766017 - 1) \right) \]After calculating, the numerical result is approximately \( L \approx 2.268 \). This represents the arc length from \((0, 0)\) to \((1, 2)\) on the curve defined by the function.