Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{C}(2 x+9 z) d s ; C\) is the curve \(x=t, y=t^{2}, z=t^{3}\), \(0 \leq t \leq 1\).

Step-by-Step Solution

Answer

Approximate the integral using numeric integration methods.

Step 1: Parametrize the Curve

The curve \( C \) is given by the parametric equations \( x = t \), \( y = t^2 \), and \( z = t^3 \). Since these are already in terms of \( t \), the curve is well parameterized.

Step 2: Compute the Differential Arc Length

The differential arc length \( ds \) is given by \( ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \). Compute the derivatives: \( \frac{dx}{dt} = 1 \), \( \frac{dy}{dt} = 2t \), \( \frac{dz}{dt} = 3t^2 \). Therefore, \( ds = \sqrt{1 + (2t)^2 + (3t^2)^2} \, dt = \sqrt{1 + 4t^2 + 9t^4} \, dt \).

Step 3: Set Up the Integral

Substitute the parametric equations and \( ds \) into the integral. The integral becomes: \[\int_{0}^{1} (2x + 9z) \, ds = \int_{0}^{1} (2t + 9t^3) \sqrt{1 + 4t^2 + 9t^4} \, dt.\]

Step 4: Evaluate the Integral

Evaluating this integral requires either numerical methods or advanced techniques, as its integrand is not straightforwardly elementary. If manual evaluation is not required, use a calculator to approximate the definite integral. For example, using numeric integration methods on a calculator, approximate the integral to find a value.

Key Concepts

Arc Length

Arc length is a fundamental concept in calculus, especially when dealing with curves in three-dimensional space. It gives us the precise length of a curve over a specified interval. When we have a curve defined parametrically by equations like \(x = f(t)\), \(y = g(t)\), and \(z = h(t)\), the arc length between two points, say from \(t = a\) to \(t = b\), is calculated using the formula:

• \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \)

This equation gives a measure of how much space the curve occupies. The expression inside the square root is akin to finding the length of a tiny segment of the curve, derived by the Pythagorean theorem. By integrating these segments over the interval from \(a\) to \(b\), we get the complete length of the curve.

Parametric Equations

Parametric equations are a powerful tool in mathematics to describe a curve or surface. Instead of describing the relationship between \(x\) and \(y\) directly, parametric equations express each variable as a function of an independent parameter, typically \(t\). For a curve, we often have:

• \(x = f(t)\)
• \(y = g(t)\)
• \(z = h(t)\)

This approach is quite advantageous, particularly when dealing with complex shapes, such as loops or spirals, where the \(x\) and \(y\) coordinates cannot be described with a simple function. Using parameters provides more flexibility in controlling how the curve behaves. In the given exercise, the parametric equations \(x = t\), \(y = t^2\), and \(z = t^3\) easily describe the path the curve takes as \(t\) changes. This simplicity helps when setting up integrals for line integrals or evaluating properties like arc length.

Numerical Integration

Numerical integration is a collection of algorithms used to approximate the value of definite integrals. This method becomes crucial when integrals cannot be solved analytically due to the integrand's complexity or lack of elementary functions. In our example, after setting up the integral for the curve, it results in an expression that is often too complex for standard analytical techniques:

• \(\int_{0}^{1} (2t + 9t^3) \sqrt{1 + 4t^2 + 9t^4} \, dt\)

Tools like numerical integration techniques—such as the Trapezoidal Rule, Simpson's Rule, or computational algorithms like the Romberg method—become indispensable. These tools approximate the integral's area by dividing it into tiny segments or using functional approximations, providing a practical solution when theoretical calculations are harsh or impossible. This approach ensures students and professionals alike can tackle complex real-world problems efficiently.