91Ó°ÊÓ

When discussing the analysis of a curve, a common approach is to represent the coordinates of points on the curve through parametric equations. Instead of expressing one variable explicitly in terms of the other, such as \( y = f(x) \), we express both \( x \) and \( y \) as separate functions of a third variable, called the parameter, often denoted by \( t \). This is especially useful in cases involving complex curves, like ellipses.

For instance, the parametric equations \( x = 3\cos t \) and \( y = 2\sin t \) allow us to represent the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) with parameter \( t \) covering a specific range. These equations enable simplification of expressions and make differentiation more accessible by reducing a complex problem to a single-variable context.

In our scenario, parametric equations replace the given ellipse's coordinate plane description, encapsulating both dependencies within parameter \( t \). This transformation is crucial when solving for extreme values on curves.

The chain rule is a fundamental tool in calculus that helps in differentiating composite functions, functions where one function is nested inside another. In simple terms, it provides a method to break down the differentiation into simpler, more manageable parts. It works by differentiating the outer function and then multiplying by the derivative of the inner function.

For parametric equations, we often need to consider the chain rule because the function values depend indirectly on the parameter \( t \). Consider a function \( f(x, y) \), where \( x \) and \( y \) depend on \( t \). To find \( \frac{df}{dt} \), the chain rule is applied such that:

• First, partially differentiate \( f \, \text{with respect to} \, x \) and get \( \frac{\partial f}{\partial x} \).
• Differentiate \( x \, \text{with respect to} \, t \) and get \( \frac{dx}{dt} \).
• Do the same for \( y \): compute \( \frac{\partial f}{\partial y} \) and \( \frac{dy}{dt} \).

By multiplying these partial derivatives appropriately, we obtain \( \frac{df}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt} \). This step ensures that all contributions from \( x \) and \( y \) as they vary with \( t \) are considered.

Parameterizing an ellipse means expressing its points in a form that is easy to manipulate, especially for calculus applications. The traditional equation of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) can be transformed using trigonometric functions, which is helpful for simplifying calculations.

In our example, the ellipse is given by \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \). We use \( x = 3\cos t \) and \( y = 2\sin t \), where \( t \) typically varies from 0 to \( 2\pi \) to cover the complete ellipse. However, restrictions such as \( y\geq 0 \) or \( x\geq 0 \) limit \( t \) to specific intervals like \( 0 \leq t \leq \frac{\pi}{2} \) for a semi or quarter ellipse respectively.

Choosing the parameterization \( x = a\cos t \) and \( y = b\sin t \) reflects that the ellipse's radius along the \( x \)-axis is \( a \), and along the \( y \)-axis, it is \( b \). This method transforms the elliptical equation into an accessible format for further calculus operations, such as finding derivatives or critical points.

After parameterizing and differentiating the function, the next step is to find the critical points on the parametric path. Critical points occur where the derivative \( \frac{df}{dt} \) is zero or undefined. These points indicate where the function could have relative extrema or change its increasing/decreasing behavior.

In our exercise, solving \( \frac{df}{dt} = 0 \) gives us the values of \( t \) where potential maxima or minima might exist. Such values must be within the permissible range of \( t \). For the semiellipse, \( 0 \leq t \leq \pi/2 \) can lead to points like \( t = \frac{\pi}{4} \) where the tangent to the curve is horizontal or vertical.

Once critical points are obtained, they need to be cross-referenced with endpoint evaluations \( t = 0, \pi/2 \) to determine absolute extrema. By comparing the function values at these key intervals, we can ascertain where the maximum and minimum values lie, providing a comprehensive view of function behavior across the curve.