Problem 2 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C} G(x, y) d x, \int_{C} G(x, y) d y\), and \(\int_{C} G(x, y) d s\) on the indicated curve \(C\). $$ G(x, y)=x^{3}+2 x y^{2}+2 x ; x=2 t, y=t^{2}, 0 \leq t \leq 1 $$

Step-by-Step Solution

Answer

\( \int_C G(x,y) \, dx = 10 \), \( \int_C G(x,y) \, dy = \frac{104}{15} \), \( \int_C G(x,y) \, ds \) requires numerical methods.

Step 1: Set Up Parameterization of Curve C

The curve is parameterized by the functions given in the problem: \( x(t) = 2t \) and \( y(t) = t^2 \), with the parameter \( t \) ranging from 0 to 1.

Step 2: Compute \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \)

Differentiate the expressions for \( x \) and \( y \) with respect to \( t \) to obtain: \( \frac{d x}{d t} = 2 \) and \( \frac{d y}{d t} = 2t \).

Step 3: Evaluate \( \int_C G(x,y) \, dx \)

Substitute \( x(t) \), \( y(t) \) into \( G(x,y) \), resulting in \( G(2t, t^2) = (2t)^3 + 2(2t)(t^2) + 2(2t) \). Simplify to get \( G(2t, t^2) = 8t^3 + 4t^3 + 4t = 12t^3 + 4t \). Now compute \( \int_0^1 (12t^3 + 4t) \cdot 2 \, dt \). This simplifies to \( \int_0^1 (24t^3 + 8t) \, dt \). Evaluate this integral to get \( \left[ 6t^4 + 4t^2 \right]_0^1 = 6 + 4 = 10 \).

Step 4: Evaluate \( \int_C G(x,y) \, dy \)

Use \( G(x, y) = 12t^3 + 4t \), obtained from the previous step. Compute \( \int_0^1 (12t^3 + 4t) (2t) \, dt \). This simplifies to \( \int_0^1 (24t^4 + 8t^2) \, dt \). Evaluate this integral to get \( \left[ \frac{24}{5}t^5 + \frac{8}{3}t^3 \right]_0^1 = \frac{24}{5} + \frac{8}{3} = \frac{104}{15} \approx 6.9333 \).

Step 5: Evaluate \( \int_C G(x,y) \, ds \)

Find \( ds \) using \( ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt = \sqrt{4 + (2t)^2} \, dt = \sqrt{4 + 4t^2} \, dt = 2\sqrt{1 + t^2} \, dt \). Evaluate \( \int_0^1 (12t^3 + 4t) \times 2\sqrt{1 + t^2} \, dt \) which is more complex and will require numerical methods to solve. The focus here is understanding the setup, involving solving through integration after substitution.

Key Concepts

Parameterization of Curves

When discussing line integrals, a key step is the parameterization of curves. Parameterization involves defining a curve using a set of equations that express the coordinates as functions of a parameter, usually denoted as \(t\). In our example, the curve \(C\) is defined by \(x(t) = 2t\) and \(y(t) = t^2\). Here, \(t\) acts as the parameter that allows us to traverse the curve from its start to end as \(t\) varies within a given range.
By expressing a curve in this manner, it becomes easier to evaluate integrals with respect to the curve, as everything is dependent on a single variable, \(t\). This simplification is crucial for subsequent integration processes, especially when dealing with complex curves in multiple dimensions.

Integration with Respect to a Parameter

Integration with respect to a parameter involves computing integrals after substituting the parameterized expressions of a curve into the given function. For the integral \(\int_C G(x,y) \, dx\), you substitute the expressions \(x(t) = 2t\) and \(y(t) = t^2\) into the function \(G(x,y)\). This converts the integral into an integral over a single variable, \(t\), which is \(\int_0^1 (24t^3 + 8t) \, dt\).
The main idea is to express every component of the integral in terms of \(t\), including the limits of integration, the function itself, and the differential \(dx\) or \(dy\). This complete transformation helps streamline the calculation process and makes it feasible to apply standard techniques of calculus on the problem.

Differentiation of Parametric Equations

Differentiation of parametric equations is a technique used to compute derivatives when a curve is given in a parametric form. This typically involves differentiating each parametric equation with respect to the parameter \(t\). For the given problem, we have \(x(t) = 2t\) and \(y(t) = t^2\). Differentiating these gives \(\frac{dx}{dt} = 2\) and \(\frac{dy}{dt} = 2t\).
These derivatives are important because they help compute differentials like \(dx\) and \(dy\), which are needed for evaluating line integrals. Additionally, these derivatives are used to compute the arc length differential, \(ds\), and play a role in transforming the integrals into a more manageable form.

Arc Length Differential

The arc length differential, denoted as \(ds\), is an element that represents an infinitesimally small distance along the curve. It is pivotal in line integrals to find the integration path's length. For a parameterized curve \((x(t), y(t))\), \(ds\) is computed using the formula \(ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\).
In our problem, this results in \(ds = \sqrt{4 + (2t)^2} \, dt = 2\sqrt{1 + t^2} \, dt\). This expression accounts for changes in both \(x\) and \(y\) coordinates as we move along the curve, thus providing a direct way to integrate over the path length. Although some integrals involving \(ds\) can be more complicated, the setup is crucial for numerical methods or further analytical approaches.