Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of the indicated curve. \(y=\frac{2}{3}\left(x^{2}+1\right)^{3 / 2}\) between \(x=1\) and \(x=2\)

Step-by-Step Solution

Answer

The length of the curve is \( \frac{17}{3} \).

Step 1: Recall the Formula for Arc Length

To find the length of a curve defined by a function \( y = f(x) \), the arc length formula is \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]. Here \( a = 1 \) and \( b = 2 \).

Step 2: Find the Derivative

Calculate \( \frac{dy}{dx} \) for the given function \( y = \frac{2}{3} (x^2 + 1)^{3/2} \). Applying the chain rule, we get: \[ \frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2} (x^2 + 1)^{1/2} \cdot 2x = 2x(x^2 + 1)^{1/2}. \]

Step 3: Set Up the Arc Length Integral

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

Step 4: Simplify the Expression Inside the Integral

Simplify the expression inside the square root:\[ 1 + 4x^2(x^2 + 1) = 1 + 4x^4 + 4x^2. \] This can be written as:\[ = (2x^2 + 1)^2. \] So, the integral simplifies to:\[ L = \int_{1}^{2} (2x^2 + 1) \, dx. \]

Step 5: Evaluate the Integral

Evaluate the integral:\[ L = \int_{1}^{2} (2x^2 + 1) \, dx = \left[ \frac{2}{3}x^3 + x \right]_{1}^{2}. \] Substitute the limits:\[ L = \left( \frac{2}{3}(2)^3 + 2 \right) - \left( \frac{2}{3}(1)^3 + 1 \right). \]

Step 6: Calculate the Definite Integral

Calculate the expression:\[ L = \left( \frac{16}{3} + 2 \right) - \left( \frac{2}{3} + 1 \right) = \left( \frac{16}{3} + \frac{6}{3} \right) - \left( \frac{2}{3} + \frac{3}{3} \right). \] Simplify:\[ L = \frac{22}{3} - \frac{5}{3} = \frac{17}{3}. \]

Key Concepts

Chain Rule

The Chain Rule is an essential concept in calculus, especially when dealing with composite functions. When you have a function that is composed of two or more functions, like the one in our original exercise, the chain rule helps you differentiate it properly. Imagine you have a function \( y = (u(x))^n \), where \( u(x) \) is an inner function. The chain rule states that the derivative \( \frac{dy}{dx} \) is obtained by multiplying the derivative of the outer function with respect to the inner function, by the derivative of the inner function with respect to \( x \).

• Identify the outer function and its exponent, in our case \( n = \frac{3}{2} \).
• Differentiating the outer function: \( \frac{d}{du} (u^{3/2}) = \frac{3}{2}u^{1/2} \).
• Differentiating the inner function: \( u(x) = x^2 + 1 \), results in \( \frac{du}{dx} = 2x \).
• Apply the Chain Rule: \( \frac{dy}{dx} = \frac{3}{2} (x^2 + 1)^{1/2} \cdot 2x = 2x (x^2 + 1)^{1/2} \).

This rule allows for the correct calculation of derivatives when layering multiple functions together, making it indispensable for exercises like finding the arc length of a curve.

Definite Integral

The definite integral is a powerful tool to calculate total quantities, such as the arc length of a curve between two points. After calculating the derivative and forming the expression needed under the square root, we move to set up the integral. The definite integral \( \int_{a}^{b} \, f(x) \, dx \) finds the accumulated area under the curve, precisely calculated over the interval \([a, b]\). In our arc length exercise, the definite integral also considers the smooth curve defined by \( \sqrt{1 + (\frac{dy}{dx})^2} \).

• Define the interval from \( a = 1 \) to \( b = 2 \).
• Set up the expression inside the integral using the square root result.
• Simplify as required, like rewriting \( 1 + 4x^2(x^2 + 1) \) to \((2x^2 + 1)^2\).
• Evaluate the integral \( \int_{1}^{2} (2x^2 + 1) \, dx \) within these limits.

The evaluation of this integral between the specified limits gives a precise numerical value corresponding to the length of the curve between \( x = 1 \) and \( x = 2 \).

Derivative Calculation

Calculating derivatives is a core skill in calculus, and it involves techniques like the Power Rule and Chain Rule. In the exercise, you were asked to differentiate \( y = \frac{2}{3}(x^2 + 1)^{3/2} \). This type of problem highlights the need for structured derivative calculation.

• Identify the type of derivative to calculate; here, it involves both a constant multiplier \( \frac{2}{3} \) and the Chain Rule.
• Use rules: The Power Rule applies generally to functions of the form \( x^n \).
• Combine constant multiplication with the derivative using the Chain Rule to get: \( \frac{dy}{dx} = \frac{2}{3} \times \frac{3}{2} (x^2 + 1)^{1/2} \times 2x \).
• Simplify the expression: \( \frac{dy}{dx} = 2x (x^2 + 1)^{1/2} \).

By performing these steps systematically, you ensure precise derivative calculations, which feed into further processes like integration and help solve broader calculus problems.