91Ó°ÊÓ

Parametrizing a curve is an important step when dealing with line integrals, particularly when you need to simplify the process for integration. In our exercise, the curve given is described by the equation \(y = \frac{x^2}{2}\). Parameterization involves expressing both \(x\) and \(y\) as functions of a new variable, commonly denoted as \(t\), which is known as the parameter. This makes the variables easier to handle during integration.
To parameterize this curve, we can set \(x = t\). Immediately, we derive that \(y = \frac{t^2}{2}\). This means the curve in a parametric form can be expressed as \((t, \frac{t^2}{2})\). This representation is crucial for integrating along the curve and helps in transforming our integral according to the path taken by the curve.

Determining the integration limits is key to evaluating an integral properly along a curve. In our specific problem, the curve starts at the point \((1, \frac{1}{2})\) and ends at \((0, 0)\). Initially, you must check what parameter \(t\) values correspond to these spatial coordinates.
Initially, when the point is \((1, \frac{1}{2})\), substituting into our parameterization \((t, \frac{t^2}{2})\), it implies that \(t = 1\). Similarly, when at point \((0,0)\), \(t = 0\). Therefore, the range of integration is from \(t = 1\) to \(t = 0\), moving along the curve from start to end. This backward integration suggests you must be careful about such direction changes when performing computations.

Calculating the derivatives of parametric equations is necessary for setting up the line integral correctly. Here, since we have parameterized the curve with \(x = t\) and \(y = \frac{t^2}{2}\), we need to find the derivatives of \(x\) and \(y\) with respect to \(t\).
For our case:

• \(dx = dt\) because \(x = t\)
• \(dy = \frac{d}{dt}(\frac{t^2}{2}) = t \, dt\) since the derivative of \(\frac{t^2}{2}\) is \(t\)

These derivatives play a pivotal role in transforming the original integral \ \(\int_C f(x, y) \, dx\ \) into the parametric form as they replace differential elements \(dx\) and \(dy\) during the integration process.

The substitution method is a powerful technique used during integration, simplifying expressions by changing variables. In our line integral, once the curve has been parameterized, substitution helps in expressing the function \(f(x, y)\) entirely in terms of \(t\).
Consider the function given: \(f(x, y) = \frac{x + y^2}{\sqrt{1 + x^2}}\). By substituting \(x = t\) and \(y = \frac{t^2}{2}\), this function transforms into:

• \(f(t, \frac{t^2}{2}) = \frac{t + \frac{t^4}{4}}{\sqrt{1 + t^2}}\)

This substitution simplifies the expression, making it easier to integrate over the parameter \(t\).
Overall, substitution transforms an integration problem from its original variables, often leading to simpler integrands or allowing the use of standard integrals, easing the evaluation.