91Ó°ÊÓ

In calculus, parameterization is a powerful technique used to deal with curves and surfaces by expressing the position along a curve using a single variable, typically denoted as \(t\). This is particularly useful when working with line integrals, where the path of integration is along a curve. In this problem, we have a curve defined by the equation \(x = y^3\). To parameterize the curve from \((-1,-1)\) to \((1,1)\), we substitute a parameter \(t\) for \(y\). Therefore, we set \(y = t\), which naturally leads to \(x = t^3\) by the curve equation.

• The parameter \(t\) ranges from \(-1\) to \(1\).
• The parameterized equations are \(x = t^3\) and \(y = t\).

As a result, the parameterization of the curve can be expressed through the vector function \(\mathbf{r}(t) = (t^3, t)\). This form makes it easier to substitute variables and evaluate integrals.

Calculating an antiderivative is finding a function whose derivative produces the original function. In this context, when working with line integrals, finding the antiderivative of a transformed integral can simplify the evaluation. Here, the line integral of interest becomes \(\int_{-1}^{1} 3t^2 e^{t^3} \, dt\). We employ the substitution technique, where we let \(u = t^3\). Consequently, the derivative \(du = 3t^2 \, dt\) clearly matches the terms in our integral, allowing a straightforward substitution.

• The integral transforms to \(\int e^u \, du\).
• The antiderivative of \(e^u\) is simply \(e^u\).

Now, apply the limits of integration: as \(t\) changes, \(u\) moves from \(-1^3 = -1\) to \(1^3 = 1\), so you evaluate \(\left[ e^u \right]_{-1}^{1}\), which results in \(e^1 - e^{-1} = e - \frac{1}{e}\). Understanding this step is crucial for evaluating complex line integrals neatly.

Curve parameterization is integral in calculus for converting complex curves into manageable, calculable forms. The idea is to express the coordinates of points on a curve as functions of a single parameter. In this example, we parameterized the curve \(x = y^3\) for \(y\) in the interval \([-1, 1]\).

• The original curve is rewritten as functions of \(t\), providing \(x = t^3\) and \(y = t\).
• This allows step-by-step transformation of the integral into a simpler form.

Why is parameterizing helpful? It simplifies the mathematical expression of curves, especially when integrating or differentiating along a curve. It also allows endpoints to be straightforwardly specified in terms of the parameter \(t\), simplifying limit evaluations. The deriving power of parameterization lies in its ability to simplify complex geometrical figures into familiar functional forms, making them easier to analyze and calculate.