Problem 16 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C}-y^{2} d x+x y d y\), where \(C\) is given by \(x=2 t\), \(y=t^{3}, 0 \leq t \leq 2\).

Step-by-Step Solution

Answer

\(\frac{512}{7}\)

Step 1: Parameterize the Curve

The curve C is given by the parametric equations for \(x\) and \(y\): \(x = 2t\) and \(y = t^3\). We need to express \(dx\) and \(dy\) in terms of \(dt\) as well. Differentiating these with respect to \(t\), we get:\[dx = 2 \, dt\] and \[dy = 3t^2 \, dt.\]

Step 2: Substitute into the Integral

Now replace \(-y^2\) and \(xy\) in the integral with their parametric forms and multiply by \(dx\) and \(dy\), respectively. Thus the integral becomes:\[ \int_0^2 (-y^2) dx + xy dy = \int_0^2 (-(t^3)^2) (2 \, dt) + (2t)(t^3)(3t^2 \, dt). \] This simplifies to:\[ \int_0^2 -2t^6 \, dt + 6t^6 \, dt. \]

Step 3: Combine and Simplify the Integral

Combine the terms inside the integral:\[ \int_0^2 (-2t^6 + 6t^6) \, dt = \int_0^2 4t^6 \, dt. \] Simplify the expression to evaluate in the next step.

Step 4: Evaluate the Integral

The integral becomes \(\int 4t^6 \, dt\). Evaluate this from 0 to 2:\[ \int_0^2 4t^6 \, dt = [4 \cdot \frac{t^7}{7}]_0^2. \] Substitute the limits for the evaluation:\[ = \frac{4}{7}[2^7 - 0^7] = \frac{4}{7}[128]. \]

Step 5: Compute the Final Value

Multiply and simplify the evaluated expression:\[ \frac{4 \times 128}{7} = \frac{512}{7}. \] Thus, the value of the integral is \(\frac{512}{7}\).

Key Concepts

Parametric Equations

When dealing with curves and integrals, we often use parametric equations to define the curve. A parametric equation expresses a set of related quantities as explicit functions of an independent variable, usually called a parameter. In this case, the parametric variables for the curve \(C\) are given as \(x = 2t\) and \(y = t^3\), with \(t\) as the parameter.

This approach allows us to describe the path taken by moving along a curve, rather than just looking at the relationship between \(x\) and \(y\). In essence, parametric equations give you a neat way to express the coordinates of the curve point by point, allowing for a thorough exploration of its features. They are especially useful when the curve doesn't lend itself easily to a traditional \(y = f(x)\) format due to its complexity or multidimensional nature.

Differentiation

Differentiation plays a pivotal role when dealing with parametric equations in line integrals. With parametric equations, we need to find the differentials \(dx\) and \(dy\). This is done by differentiating each parametric equation with respect to its parameter.

For \(x = 2t\), differentiating with respect to \(t\) gives \(dx = 2 \, dt\). Similarly, for \(y = t^3\), differentiating gives \(dy = 3t^2 \, dt\).

• These derivatives transform the curve’s parametric form into a form usable in our integral calculation.
• They adjust the integral for the rate of change along the curve, which is crucial for accurate results.

Differentiation here provides the dynamics that allow the parametric path to be understood in terms of calculus.

Definite Integration

Definite integration is the next step after substituting the differential forms into the integral. The integral \(\int_C -y^2 dx + xy dy\) turns into \(\int_0^2 (-2t^6 + 6t^6) \, dt\) upon substitution with the parametric expressions.

In this context:

• The limits \(0\) and \(2\) represent the range of our parameter \(t\).
• States where the integral begins and ends on the curve \(C\).
• The definite integration thus offers a concrete area or "sum" under the curve between these limits.

In our example, evaluating this integral from \(t = 0\) to \(t = 2\), we compute \([4 \cdot \frac{t^7}{7}]_0^2\). We solve this and substitute the values to reach a numeric result of \(\frac{512}{7}\).

Curve Parameterization

Curve parameterization is the application of parametric equations to rename an integral path in convenient terms. It allows the representation of a curve in terms of one or more parameters, so every point on the curve is defined by some value of the parameter.

For line integrals, this method simplifies the computation by providing expressions for points along the path:

• Each coordinate \((x, y)\) becomes dependent on the parameter \(t\), translating geometrical shapes into exploreable algebraic expressions.
• This helps in managing cases where manually determining each point's formulation would be complex.

Parameterization thus assists in smooth transitioning through curves during integration, making calculus more flexible and robust when dealing with complex paths.