91Ó°ÊÓ

When dealing with line integrals, parametrization of the path is a crucial step. In the given exercise, we are provided with a path \((t, t^2)\), which beautifully demonstrates parametric equations. These equations describe a curve by defining both x and y as functions of a third variable, often denoted as t. This variable t is known as the parameter, tracing the position on the curve as it varies.

• For this problem, the parametric equations are \(x(t) = t\) and \(y(t) = t^2\).
• The parameter t ranges from 0 to 1.
• By taking the derivatives, we find \(dx = dt\) and \(dy = 2t \, dt\).

These derivatives are crucial because they enable the transformation of the original integral along the path into a standard form suitable for integration. Parametric equations simplify the evaluation by allowing us to focus on a single variable, t, thus transforming a complex integral into a solvable one.

Integral computation is a core concept in calculus. It involves finding a function whose derivative is the given function. In line integrals, we compute the integral along a path in the field, considering both the function and the path.For this problem, because we have parametrized the function, the integral \[\int_{C} \cos x \cos y \,dx - \sin x \sin y \,dy\]is transformed into a single-variable integral:\[\int_{0}^{1} \left(\cos(t) \cos(t^2) - 2t \sin(t) \sin(t^2)\right) dt.\]
It is important to note that the simplification of the integral, as shown above, prepares it for further evaluation. The terms \(\cos(t) \cos(t^2)\) and \(2t \sin(t) \sin(t^2)\) arise from substituting the parametric equations and their differentials. With this simplified form, the integral is usually addressed using numerical methods or symbolic computation if it does not resolve to a familiar elementary function.

The substitution method is a powerful tool for simplifying integrals, especially in the context of line integrals with parametric equations. Essentially, substitution allows us to transform a complicated integral with multiple variables into a simpler one with a single variable.Here's how it was applied in this exercise:

• We substituted \(x = t\) and \(y = t^2\)
• This transformation allowed us to replace \(dx\) with \(dt\) and \(dy\) with \(2t \, dt\)

This substitution helps us to further transform the integral into the expression\[\int_{0}^{1} \left( \cos(t) \cos(t^2) - 2t \sin(t) \sin(t^2) \right) dt,\]which is much easier to handle.Substitution method in integrals leads to converting a multivariable problem to a more manageable single-variable problem. This technique is widely used in calculus to locate solutions that aren't straightforwardly apparent from the original problem format.