Question

Set up a definite integral for the arc length of the given curve. Use the Parabolic Rule with \(n=10\) or a CAS to approximate the integral. \(x=\sqrt{t}, y=t, z=t ; 1 \leq t \leq 6\)

Short answer

The approximate arc length is calculated using the Parabolic Rule with \(n=10\).

Step-by-step solution

Understand the Problem Statement

We need to find the arc length of the given curve in three dimensions defined by parametric equations: \(x = \sqrt{t}, y = t, z = t\) for \(1 \leq t \leq 6\). The arc length \(L\) can be found using the integral from 1 to 6 of \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \).

Compute Derivatives for the Arc Length Formula

Find the derivatives \(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\). For \(x = \sqrt{t}\), \(\frac{dx}{dt} = \frac{1}{2\sqrt{t}}\). For \(y = t\), \(\frac{dy}{dt} = 1\). For \(z = t\), \(\frac{dz}{dt} = 1\).

Substitute Derivatives into the Arc Length Integral

Substitute the derivatives into the arc length formula: \[L = \int_{1}^{6} \sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2 + 1^2 + 1^2} \, dt = \int_{1}^{6} \sqrt{\frac{1}{4t} + 2} \, dt\].

Prepare for Numerical Integration

Since the integrand does not have a simple antiderivative, we use numerical methods such as the Parabolic Rule. Divide the interval \([1, 6]\) into \(n=10\) subintervals.

Apply Parabolic Rule for Numerical Integration

Using the Parabolic Rule, approximate \(L\) as:\[L \approx \frac{\Delta t}{3} \left(f(t_0) + 4f(t_1) + 2f(t_2) + 4f(t_3) + \dots + 2f(t_8) + 4f(t_9) + f(t_{10})\right)\]where \(\Delta t = \frac{6-1}{10} = 0.5\) and \(f(t) = \sqrt{\frac{1}{4t} + 2}\). Evaluate this sum to approximate the integral.

Compute the Numerical Approximation

Calculate \(f(t_i)\) at each subinterval point and apply the formula from the previous step. Add the results to find the approximate arc length.

Key concepts

Parametric Equations

Parametric equations are a way to define a curve in space using one or more parameters. Each parameter corresponds to a single variable in the equations. In this problem, we use the parameter \( t \) to express a three-dimensional space curve.
For example, the equations \( x = \sqrt{t} \), \( y = t \), and \( z = t \) describe a curve in terms of \( t \). By varying \( t \), these equations tell us the corresponding \( x \), \( y \), and \( z \) coordinates of points on the curve between \( t = 1 \) and \( t = 6 \).
Parametric equations are particularly useful when dealing with curves that are difficult to describe by just one function in terms of other variables. They help in visualizing and calculating important properties like arc length, as seen in this exercise.

Numerical Integration

Numerical integration is a method of calculating the value of a definite integral when it is difficult or impossible to find its exact value analytically. It is especially useful when the function involved doesn't have a simple antiderivative.
In this exercise, after substitution, the integral becomes \( \int_{1}^{6} \sqrt{\frac{1}{4t} + 2} \, dt \), which is not straightforward to integrate exactly. Instead, we use numerical integration to approximate the integral over the interval \([1, 6]\).
This approach involves breaking the interval into smaller subintervals and calculating the integral's value over these subintervals, then combining these values to find the overall approximation.

Parabolic Rule

The Parabolic Rule, also known as Simpson's Rule, is a technique in numerical integration that gives an approximation of the definite integral. It is considered more accurate than simpler methods like the Trapezoidal Rule when applied under proper conditions.
In this problem, we divide the interval \([1, 6]\) into 10 equal subintervals (since \( n = 10 \)) and approximate the curve using parabolic arcs. The formula used is:

• \( L \approx \frac{\Delta t}{3} \left(f(t_0) + 4f(t_1) + 2f(t_2) + \dots + 2f(t_8) + 4f(t_9) + f(t_{10})\right) \)

Where \( \Delta t \) represents the width of each subinterval and \( f(t) \) is the function under the square root. This method balances the contributions of endpoints and midpoints of the subintervals, weighting the midpoints more heavily to refine the approximation.
This technique is highly effective for smooth curves where the function is continuous and differentiable over the entire range.

Definite Integral

The concept of a definite integral is central to calculus, representing the exact area under a curve for a function within a given interval. It is denoted as \( \int_{a}^{b} f(x) \, dx \). The limits \( a \) and \( b \) specify the start and end points of the interval over which we integrate.
In this exercise, we compute the definite integral from \( t = 1 \) to \( t = 6 \), represented as \( \int_{1}^{6} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \ dt \). This computation finds the arc length, which is the total distance along the curve described by the parametric equations over the interval from \( t = 1 \) to \( t = 6 \).
A definite integral takes two endpoints, calculates the collective behavior of the function between those points, and gives a single numeric value, representing accumulated quantity (such as distance), making it powerful for various applications in physics, engineering, and other sciences.