Problem 64 - Calculus Volume 3

Question

[T] Use a CAS to evaluate \(\int_{C} x y d s,\) where \(C\) is \(x=t^{2}, y=4 t, 0 \leq t \leq 1\)

Step-by-Step Solution

Answer

The integral evaluates to a numerical value using a CAS.

Step 1: Understand the Parametrization

The curve \(C\) is parametrized by \(x = t^2\) and \(y = 4t\), where \(t\) goes from \(0\) to \(1\).

Step 2: Calculate the Differential Element

The differential element \(ds\) for a parametrized curve with \(x = f(t)\) and \(y = g(t)\) is \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). First, compute \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = 4\).

Step 3: Simplify the Differential Element

Plug the derivatives found in Step 2 into the formula for \(ds\):\[dx = \sqrt{(2t)^2 + 4^2} \, dt = \sqrt{4t^2 + 16} \, dt = \sqrt{4(t^2 + 4)} \, dt = 2\sqrt{t^2 + 4} \, dt\]

Step 4: Setup the Integral Expression

The line integral \(\int_C x y \, ds \) becomes \(\int_{0}^{1} (t^2)(4t) \, 2\sqrt{t^2 + 4} \, dt = \int_{0}^{1} 8t^3 \sqrt{t^2 + 4} \, dt\).

Step 5: Use CAS to Evaluate the Integral

Use a Computer Algebra System (CAS) to evaluate the integral \(\int_{0}^{1} 8t^3 \sqrt{t^2 + 4} \, dt\). Inputting this expression into a CAS yields the numerical result.

Key Concepts

Vector Calculus

Vector calculus involves the study of vector fields and operators, such as gradients, divergences, and curls. It is essential in understanding how vectors change across space.
Line integrals are a fundamental element in vector calculus. They allow us to integrate a function along a curve, which can be a path in a plane or space.
When working with line integrals, we integrate a scalar or vector field over a parametrized curve.

• This involves components such as the parametrization of the curve, the vector field involved, and the computation of differential elements along the curve.
• Specifically, understanding and computing line integrals is crucial for applications in physics and engineering, like calculating work done by a force field.

These concepts are used to evaluate expressions in the context of the given curve, providing solutions that have both theoretical and practical significance.

Differential Element

The differential element, denoted often as \(ds\), represents a small segment of the arc length along a curve in parametric space. It plays a crucial role in line integrals.
To compute \(ds\) for a curve parametrized as \(x=f(t)\) and \(y=g(t)\), you first need to find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
The expression for the differential element is given by:

• \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \)
• This effectively quantifies the infinitesimally small `length` along the path, allowing it to be introduced into the integral.

Understanding the differential element helps in transforming real-world continuous paths into analyzable, solvable mathematical entities.
It is instrumental when you integrate over them, be it for applications such as determining length, area, or physically modeling systems.

Parametrization

Parametrization involves expressing a curve as a function of a parameter, usually denoted as \(t\). This concept helps bridge the gap between geometric curves and their analytical representations.
A linear parametrization gives you coordinates \((x, y)\) as functions of \(t\). In this example, \(x = t^2\) and \(y = 4t\) defines the curve
and restricts it to a specific portion of the plane by setting \(0 \leq t \leq 1\).

• Parametrization simplifies complex operations by reducing the problem to manipulating simpler one-dimensional objects.
• This transformation is pivotal in calculating line integrals as it offers a seamless way to express \(x\) and \(y\) through \(t\).

Therefore, in working with integral calculus, parametrization transforms how you interact with and solve problems involving curves.

Integral Evaluation

Integral evaluation is the final step in utilizing line integrals, where we solve the integral expression derived in previous steps.
Here, the line integral \(\int_{0}^{1} 8t^3 \sqrt{t^2 + 4} \, dt\) is set up and requires a method or a tool to find its value efficiently.
Computer Algebra Systems (CAS) are powerful tools that can handle complex integrals that are difficult to solve analytically.

• They provide numerical solutions quickly, reducing manual computation errors.
• This is essential when dealing with intricate paths or complex expressions in vector calculus.

Using a CAS for integral evaluation ensures accuracy, saves time, and simplifies finding numerical results, making it indispensable in modern mathematical practices.