Problem 11 - Calculus Early Transcendentals

Question

Find the arc length of the following curves on the given interval by integrating with respect to \(x\) $$y=\frac{1}{3} x^{3 / 2} \text { on }[0,60]$$

Step-by-Step Solution

Answer

The arc length of the curve given by the function \(y = \frac{1}{3}x^{3/2}\) on the interval \([0, 60]\) is 168.

Step 1: Find the derivative of y with respect to x

To find the derivative of \(y\) with respect to \(x\), we can use the power rule: $$ \frac{d}{dx} x^n = nx^{n-1} $$ Applying the power rule to our function, we get $$ y' = \frac{d}{dx} \left(\frac{1}{3}x^{3/2}\right) = \frac{1}{2}x^{1/2} $$

Step 2: Plug the derivative into the arc length formula

Now that we have the derivative \(y'\), we can plug it into the arc length formula: $$ L = \int_{0}^{60} \sqrt{1 + \left(\frac{1}{2}x^{1/2}\right)^2}\, dx $$ Simplify the expression inside the square root before proceeding: $$ L = \int_{0}^{60} \sqrt{1 + \frac{1}{4}x}\, dx $$

Step 3: Compute the integral

To compute the integral, we can use substitution. Let \(u = 1 + \frac{1}{4}x\), so \(\frac{du}{dx} = \frac{1}{4}\). Thus, \(dx = 4\,du\). The limits of integration will also change accordingly: when \(x = 0\), \(u = 1\) and when \(x = 60\), \(u = 16\). Now, substitute in the integral: $$ L = \int_{1}^{16} \sqrt{u}\cdot 4\, du = 4 \int_{1}^{16} u^{1/2}\, du $$ Now, integrate with respect to u: $$ L = 4\left[\frac{2}{3}u^{3/2}\right]_{1}^{16} $$ Finally, evaluate the integral at the limits of integration: $$ L = 4\left[\frac{2}{3}(16)^{3/2} - \frac{2}{3}(1)^{3/2}\right] = 4\left[\frac{2}{3}(64) - \frac{2}{3}\right] = 4\left[\frac{2}{3}(63)\right] = 168 $$ Therefore, the arc length of the curve \(y = \frac{1}{3}x^{3/2}\) on the interval \([0, 60]\) is \(168\).

Key Concepts

Integration

In mathematics, integration is a fundamental concept that represents the process of finding the whole from its parts. It's like reverse differentiation. When you integrate a function, you're essentially finding the area under its curve over a certain interval. This is particularly useful in calculating the arc length of a curve. In our exercise, we integrated the expression \( \sqrt{1 + (\frac{1}{2}x^{1/2})^2} \) over the interval \([0, 60]\). This involves transforming a complex expression into a simpler, aggregate form that represents the total arc length of the given curve

• The aim is to sum an infinite number of infinitesimally small quantities.
• Integration can be definite or indefinite. Here it is definite since it is over a specific interval.
• It’s essentially finding an antiderivative.

By integrating step-by-step and simplifying the expression, we find the arc length, a crucial concept when dealing with curves.

Power Rule

The power rule is a basic principle in calculus that simplifies the process of taking derivatives. It states that if you have a function of form \( x^n \), its derivative is \( nx^{n-1} \). This rule makes it easier to derive polynomial functions efficiently. In the given problem, we used the power rule to find the derivative of \( y \), where \( y = \frac{1}{3}x^{3/2} \).

• Subtract one from the exponent.
• Multiply the term by the old exponent.
• In our problem, this became \( y' = \frac{1}{2}x^{1/2} \).

The power rule is one of the first rules learned in calculus due to its simplicity and broad applicability for polynomial expressions, making it foundational for understanding derivatives deeply.

Derivative

A derivative represents the rate of change of a function with respect to a variable. Think of it as finding the speed of a moving object. Derivatives help determine the slope of the curve at a specific point and uncover critical features of functions, like maxima and minima.

• In our exercise, the derivative helped us find the slope of the curve \( y = \frac{1}{3}x^{3/2} \).
• It is necessary for calculating arc length as it factors into the arc length formula.
• Derivatives can be computed using rules like the power rule, chain rule, product rule, etc.

The derivative is a core tool in calculus that allows us to understand dynamic and changing systems, positioning it as a fundamental concept in solving calculus problems.

Substitution Method

The substitution method is a technique used in integration to simplify complex integrals. It involves substituting a part of the integral, usually the inner function, with a new variable. This converts a tough integral into an easier one.

In our exercise, we chose \( u = 1 + \frac{1}{4}x \). This new variable \( u \) transformed the original integral into a more manageable form.

• Step 1: Choose a substitution, \( u \).
• Step 2: Express \( dx \) in terms of \( du \).
• Step 3: Change limits of integration if needed.
• Step 4: Rewrite the integral.

This method is particularly handy in solving integrals that are difficult to integrate using standard methods or when a direct antiderivative is hard to find, allowing for more straightforward calculations.