Problem 3 - Calculus

Question

Find the lengths of the curves. $$ y=\frac{1}{3}\left(x^{2}+2\right)^{3 / 2} \text { from } x=0 \text { to } x=1 $$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{4}{3} \).

Step 1: Identify the Curve Length Formula

To find the length of the curve, we will use the formula \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \), where \( y \) is the function given and \( \frac{dy}{dx} \) is its derivative. The limits of integration are from \( x = 0 \) to \( x = 1 \).

Step 2: Differentiate the Function

Find the derivative of \( y = \frac{1}{3}(x^2 + 2)^{3/2} \). Using the chain rule, \( \frac{dy}{dx} = (x^2+2)^{1/2} \cdot x \).

Step 3: Substitute and Simplify

Substitute \( \frac{dy}{dx} = (x^2+2)^{1/2} \cdot x \) into the integral formula: \( L = \int_{0}^{1} \sqrt{1 + \left(x \cdot (x^2+2)^{1/2}\right)^2} \, dx \). This simplifies to \( L = \int_{0}^{1} \sqrt{1 + x^2(x^2+2)} \, dx \).

Step 4: Simplify the Expression Under the Square Root

Simplify the expression \( 1 + x^2(x^2+2) = 1 + x^4 + 2x^2 \).

Step 5: Evaluate the Integral

Now evaluate \( L = \int_{0}^{1} \sqrt{x^4 + 2x^2 + 1} \, dx \). Notice that \( x^4 + 2x^2 + 1 = (x^2+1)^2 \), so \( \sqrt{(x^2+1)^2} = |x^2+1| \), which is \( x^2+1 \) over the interval [0,1].

Step 6: Integrate the Simplified Expression

Now integrate \( \int_{0}^{1} (x^2 + 1) \, dx \). This resolves to \( \left[ \frac{x^3}{3} + x \right]_{0}^{1} = \frac{1}{3} + 1 = \frac{4}{3} \).

Key Concepts

Integration

Integration is a fundamental concept in calculus, acting as the reverse process of differentiation. By integrating a function, we determine the accumulation of quantities, such as finding area under a curve or, as in this case, the length of a curve. The key to integration is understanding its notation and limits. In the curve length problem, the integral \[ L = \int_{0}^{1} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]provides a means to compute the length within specified limits of 0 to 1. This process requires applying specific techniques or formulas aimed at simplifying the integrand's expression under the square root, enabling straightforward computation or estimation.- Integration helps in summing infinitely small parts.- The definite integral has upper and lower limits.- Integration can involve substitution or transformation to simplify complex parts, like squaring an expression.

Differentiation

Differentiation is the method used to find the derivative, or the rate at which a function changes at any point. In the context of our problem, we differentiated the function to determine the slope component needed for the curve length formula.For the curve given, \[ y = \frac{1}{3}(x^2 + 2)^{3/2} \]applying differentiation helps find \( \frac{dy}{dx} \) by using rules such as the power rule and the chain rule, which simplifies the expression. Differentiation is important because the derived function gives rise to the squared term within the integral calculation, essential for evaluating the arc length.Key aspects include:- Differentiation provides the slope or rate of change.- It involves applying mathematical rules to function components.- Crucial for deriving expressions further simplified in integration.

Chain Rule

The chain rule plays a crucial role in both differentiation and integration, particularly when dealing with composite functions. A composite function consists of nested functions, where one function is inside another.In our problem, the function\[ y = \frac{1}{3}(x^2 + 2)^{3/2} \]is a composite function involving the inner function \( u = x^2 + 2 \) raised to a power. The chain rule is necessary for differentiating such functions efficiently, allowing for the proper handling of inner modifications.The chain rule can be summarized as:- Derive the outer function while keeping the inner unchanged.- Multiply by the derivative of the inner function.- Essential in differentiating composite functions in complex integrals.

Integral Calculus

Integral Calculus extends the simple concepts of integration and differentiation into more comprehensive analyses of functions. It covers more advanced applications like determining the total sum of infinite series portions, as seen in arc length problems.In this exercise, we've applied integral calculus by first identifying the proper function derivative, then simplifying it into an easier-to-integrate form:\[ x^4 + 2x^2 + 1 = (x^2+1)^2 \] recognizing this as \[ \int_{0}^{1} (x^2 + 1) \, dx \]avoids the complexity of integrating roots, showcasing seamless integration steps.Important elements of integral calculus:- Extends integration for obtaining sums of entire functions.- Involves recognizing and transforming expressions.- Links differentiated outputs to computed integrals for precise quantities like curve lengths.