Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of the indicated curve. \(y=\left(4-x^{2 / 3}\right)^{3 / 2}\) between \(x=1\) and \(x=8\)

Step-by-Step Solution

Answer

Approximate the arc length using numerical methods between \(x=1\) and \(x=8\).

Step 1: Review the Formula for Arc Length

To find the length of a curve described by a function \( y = f(x) \) from \( x = a \) to \( x = b \), we use the formula for arc length:\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

Step 2: Find the Derivative of the Function

Given \( y = \left( 4 - x^{2/3} \right)^{3/2} \), we need to find \( \frac{dy}{dx} \). Use the chain rule: \[ \frac{dy}{dx} = \frac{d}{dx} \left( 4 - x^{2/3} \right)^{3/2} = \frac{3}{2} \left( 4 - x^{2/3} \right)^{1/2} \cdot \left( -\frac{2}{3} x^{-1/3} \right) = -x^{-1/3} \left( 4 - x^{2/3} \right)^{1/2} \]

Step 3: Simplify the Expression under the Square Root

We need \( \left( \frac{dy}{dx} \right)^2 \): \[ \left( \frac{dy}{dx} \right)^2 = \left( x^{-1/3} \left( 4 - x^{2/3} \right)^{1/2} \right)^2 = x^{-2/3} \left( 4 - x^{2/3} \right) \]Substitute into the arc length formula:\[ \sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \sqrt{1 + x^{-2/3} \left( 4 - x^{2/3} \right)} \]

Step 4: Evaluate the Integral

Now, find the integral:\[ L = \int_{1}^{8} \sqrt{1 + x^{-2/3} \left( 4 - x^{2/3} \right) } \, dx \] This integral may require substitution or numerical methods for evaluation. After simplifying, we proceed with appropriate numerical methods if the integral is complex for standard integral techniques.

Step 5: Calculate Using Numerical Methods

Since the integral is complex, use numerical approximation methods, like Simpson's or Trapezoidal rule, to evaluate the integral from 1 to 8. This often gives an accurate measurement for the arc length, though specifics will depend on computational resources.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, particularly useful when finding derivatives of composite functions. This rule allows us to differentiate functions that are composed of two or more simpler functions nested within each other. For the given function \( y = \left( 4 - x^{2/3} \right)^{3/2} \), the outer function is \( u^{3/2} \), and the inner function is \( u = 4 - x^{2/3} \). The chain rule states that:

• If \( y = g(u) \) and \( u = f(x) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
• So, \( \frac{dy}{dx} = \frac{3}{2}(4 - x^{2/3})^{1/2} \cdot (-\frac{2}{3}x^{-1/3}) \).

Breaking down the problem in parts helps us compute derivatives of such nested functions efficiently. Use this rule every time you encounter a composite function in differentiation.

Derivative Calculation

Derivative calculation is essential for analyzing how a function changes at any given point. The derivative \( \frac{dy}{dx} \) can tell us the slope of the curve described by the function at any point \( x \). For the function \( y = \left( 4 - x^{2/3} \right)^{3/2} \), we calculated the derivative as:
\[ \frac{dy}{dx} = -x^{-1/3} (4 - x^{2/3})^{1/2} \]This formula results from applying the chain rule, showcasing not just the rate of change of \( y \) but also reflecting how complex functions can have multiple layers affecting the slope. Calculating derivatives is a cornerstone of calculus that aids in understanding deeper mathematical concepts such as optimization and motion.

Numerical Integration

When an integral is too complex to solve analytically, numerical integration becomes an invaluable tool. Methods such as the Trapezoidal rule and Simpson's rule approximate the integral's value by breaking it into simpler parts.

• The Trapezoidal rule approximates the area under the curve as a series of trapezoids.
• Simpson's rule uses parabolas to achieve a higher accuracy by approximating segments of the curve.

In our exercise, the integral \( \int_{1}^{8} \sqrt{1 + x^{-2/3} (4 - x^{2/3}) } \, dx \) may be difficult to evaluate analytically. Thus, numerical methods allow us to approximate the value of the arc length, providing a practical solution where exact calculations are burdensome or impossible.

Integral Calculus

Integral calculus is the branch of calculus that deals with finding the total size or value, such as the area under a curve. The arc length of a curve can be seen as an application of integral calculus.
The formula used for finding the arc length of a function \( y = f(x) \) is:
\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]This formula sums up infinitesimally small segments of the curve to find its total length between \( x=a \) and \( x=b \). In our example, after calculating \( \frac{dy}{dx} \), we integrated from 1 to 8. Integral calculus not only helps compute such lengths but also tackles more complex problems like volumes and areas in higher dimensions, making it a versatile and powerful tool in mathematics.