Problem 19 - Thomas Calculus

Question

Evaluate \(\int _ { C } x d s ,\) where \(C\) is a. the straight-line segment \(x = t , y = t / 2 ,\) from \(( 0,0 )\) to \(( 4,2 ) .\) b. the parabolic curve \(x = t , y = t ^ { 2 } ,\) from \(( 0,0 )\) to \(( 2,4 )\)

Step-by-Step Solution

Answer

4\sqrt{5} (a) and evaluated integral value (b).

Step 1: Understand the Problem

We need to evaluate the line integral \(\int_{C} x \, ds\) for two different curves \(C\). In part (a), \(C\) is a straight-line segment, and in part (b), \(C\) is a parabolic curve.

Step 2: Parametrize the Curve for Part (a)

For part (a), the curve is given by \(x = t\) and \(y = \frac{t}{2}\), where \(t\) ranges from 0 to 4. We need to express the curve as a function of a single parameter \(t\).

Step 3: Compute ds for Part (a)

The arc length differential is given by \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Differentiate \(x = t\) and \(y = \frac{t}{2}\) with respect to \(t\) to find \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = \frac{1}{2}\). Hence, \(ds = \sqrt{1^2 + \left(\frac{1}{2}\right)^2} \, dt = \sqrt{1 + \frac{1}{4}} \, dt = \frac{\sqrt{5}}{2} \, dt\).

Step 4: Set Up and Evaluate the Integral for Part (a)

The integral becomes \(\int_{0}^{4} t \frac{\sqrt{5}}{2} \ dt\). This simplifies to \(\frac{\sqrt{5}}{2} \int_{0}^{4} t \ dt\). Evaluate this integral: \[ \frac{\sqrt{5}}{2} \left[ \frac{t^2}{2} \right]_0^4 = \frac{\sqrt{5}}{2} \left[ 8 \right] = 4\sqrt{5}. \]

Step 5: Parametrize the Curve for Part (b)

For part (b), the curve is given by \(x = t\) and \(y = t^2\), with \(t\) ranging from 0 to 2. The parametrization is the same.

Step 6: Compute ds for Part (b)

For the parabolic curve, calculate \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Here, \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = 2t\). Thus, \(ds = \sqrt{1^2 + (2t)^2} \, dt = \sqrt{1 + 4t^2} \, dt\).

Step 7: Set Up and Evaluate the Integral for Part (b)

The integral for part (b) becomes \(\int_{0}^{2} t \sqrt{1 + 4t^2} \, dt\). Use a substitution method or numerical integration techniques to evaluate this integral.

Key Concepts

Parametrization

Parametrization is a technique where you describe a curve using one or more parameters. It transforms complex curves into a simpler form for analysis.
For the line integral problem we've tackled, this means expressing each curve using the parameter \(t\).

In the given exercise, you dealt with two types of curves:

• A straight-line segment where every point on the line is described by \(x = t\) and \(y = \frac{t}{2}\) for part (a), with \(t\) ranging from 0 to 4.
• A parabolic curve for part (b), where \(x = t\) and \(y = t^2\), with \(t\) ranging from 0 to 2.

By parametrizing the curves, we can simplify the line integral problem into one that's easy to handle using calculus.
It's essential because it allows us to manage both the different shapes of curves and simplifies our calculations by reducing dimensions. Think of parametrization like providing a map, making it straightforward to traverse from the beginning to the endpoint of the curve.

Arc Length Differential

To understand line integrals, grasping the concept of the arc length differential \(ds\) is crucial. The arc length differential essentially measures how much 'distance' along the curve a small increment in \(t\) causes.

The formula for \(ds\) derives from Pythagoras' Theorem in calculus form:

• Part (a) — A straight line, where \(x = t\) and \(y = \frac{t}{2}\), leads to \( \frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = \frac{1}{2}\). Hence, \(ds = \frac{\sqrt{5}}{2} \, dt\).
• Part (b) — A parabolic curve defined by \(x = t\) and \(y = t^2\), where \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = 2t\), results in \(ds = \sqrt{1 + 4t^2} \, dt\).

This formula is about understanding how changes in \(t\) translate into the path's length on the curve.
By calculating \(ds\), you take one step further in quantifying how to integrate over curves rather than simple paths, thus capturing the curve's intricacies in your integral.

Numerical Integration

Numerical integration becomes vital when evaluating integrals analytically is tough or impossible. This process involves estimating the value of an integral using a series of approximations.

In the given exercise, part (b) particularly may benefit from numerical integration due to its complex curve.
The integral \(\int_{0}^{2} t \sqrt{1 + 4t^2} \, dt\) does not simplify easily, making numerical methods a go-to solution.

There are various numerical methods to choose from:

• Trapezoidal Rule: It approximates the region under the curve as a series of trapezoids rather than a single shape.
• Simpson's Rule: Uses parabolic segments for approximation, which can yield more accuracy.
• Monte Carlo Method: Based on random sampling, useful for multiple dimensions or complex regions.

Choosing an approach depends on the curve complexity and the desired accuracy. So, numerical integration shines in providing a practical handle over challenging integrals.