Question

Set up a definite integral that gives the arc length of the given curve. Approximate the integral using the Para. bolic Rule with \(n=8\). \(x=t \ln t, y=t-1 ; 1 \leq t \leq 3\)

Short answer

The arc length is approximately the result of the integral using Parabolic Rule with calculated values.

Step-by-step solution

Understand the Problem

We need to find the arc length of the parametric curve defined by the equations \(x = t \ln t\) and \(y = t - 1\) over the interval \(1 \leq t \leq 3\). The arc length formula for parametric equations \(x = f(t)\) and \(y = g(t)\) is given by the integral \(\int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\).

Find Derivatives

First, we find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). For \(x = t \ln t\), we have \(\frac{dx}{dt} = \ln t + 1\) (using the product rule). For \(y = t - 1\), we simply have \(\frac{dy}{dt} = 1\).

Set Up the Arc Length Integral

Insert the derivative components into the arc length formula: \[ L = \int_{1}^{3} \sqrt{(\ln t + 1)^2 + 1^2} \, dt = \int_{1}^{3} \sqrt{(\ln t + 1)^2 + 1} \, dt \]

Approximate Using the Parabolic Rule

Since we are using the Parabolic Rule with \(n = 8\), divide the interval \([1, 3]\) into 8 equal parts of width \(\Delta t = \frac{3-1}{8} = 0.25\). Calculate the values of \(t\) as \(t_0 = 1, t_1 = 1.25, \dots, t_8 = 3\). Then apply the Parabolic Rule:\[ L \approx \frac{\Delta t}{3} \left[f(t_0) + 4f(t_1) + 2f(t_2) + 4f(t_3) + \, \text{...} \, + 4f(t_7) + f(t_8)\right] \] Where \(f(t) = \sqrt{(\ln t + 1)^2 + 1}\).

Perform the Calculation

Calculate the function \(f(t)\) at each \(t_i\): - \(f(t_0 = 1)\), \(f(t_1 = 1.25)\), ..., \(f(t_8 = 3)\).Use these values in the Parabolic Rule expression from the previous step to approximate \(L\).

Key concepts

Parametric Equations

Parametric equations are a powerful mathematical tool used to express curves in terms of a parameter, usually denoted as \( t \). This is different from the standard Cartesian equations, which rely on \( x \) and \( y \) coordinates. With parametric equations, each coordinate is a function of \( t \), giving more flexibility in defining curves. For example, in our exercise, we have the parametric equations \( x = t \ln t \) and \( y = t - 1 \). These describe a curve in the plane where \( t \) runs from 1 to 3.

The beauty of parametric equations is their ability to represent complex curves that are difficult to describe using standard equations. This method can connect curves that loop back on themselves or create spirals and waves, which might be impossible in a standard \( y = f(x) \) form.

Definite Integral

A definite integral is a fundamental concept in calculus that calculates the accumulation of quantities, such as areas under curves. It is denoted as \( \int_{a}^{b} f(x) \; dx \), where \( a \) and \( b \) are the limits of integration. In our exercise, the definite integral determines the total arc length of the curve described parametrically.

Here, we utilize the arc length formula for parametric curves, which integrates the square root of the sum of the squares of the derivatives of \( x \) and \( y \), with respect to \( t \). This is expressed as \[ L = \int_{1}^{3} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \; dt \]. By setting the appropriate integral limits \( (1 \) to \( 3) \), we encapsulate the segment of the curve we're interested in.

Partial Derivatives

The concept of partial derivatives comes into play when functions depend on several variables. In our exercise, each part of a parametric equation requires differentiation, but with respect to a single variable, \( t \). For the parametric equations \( x = t \ln t \) and \( y = t - 1 \), we compute the derivatives \( \frac{dx}{dt} = \ln t + 1 \) and \( \frac{dy}{dt} = 1 \).

It's critical in the calculation of arc lengths because these derivatives represent the rate of change of each coordinate. By plugging these derivatives into the arc length formula, you can proficiently handle the integral that specifies the curve's arc length.

Numerical Integration Methods

Numerical integration methods are used when an integral is too complex to solve analytically or when we need an approximate value quickly. One such method is the Parabolic Rule, a type of Simpson’s Rule that approximates the area under a curve using parabolas. In this exercise, the function is integrated numerically over the interval from 1 to 3.

By dividing the interval into 8 equal segments, each of width \( \Delta t = 0.25 \), and calculating the function \( f(t) = \sqrt{(\ln t + 1)^2 + 1} \) at each node, we use the Parabolic Rule:
\[ L \approx \frac{\Delta t}{3} \left[f(t_0) + 4f(t_1) + 2f(t_2) + \ldots + 4f(t_{n-1}) + f(t_n)\right] \].
This approach averages the values at these points to give a precise estimate of the integral, perfect when handling non-linear and complex functions.