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Critical points are essential locations where a function's rate of change might either stop or change direction, making them key candidates for identifying maxima and minima. To find these critical points for a function of two variables, like our example function \( f(x, y) = x^2 + y^2 + 4x - 1 \), we need to first compute its partial derivatives.

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x + 4 \).
• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2y \).

Setting these derivatives equal to zero helps us find where the gradient is zero, leading us to critical points. Solving \( 2x + 4 = 0 \) gives \( x = -2 \), and \( 2y = 0 \) gives \( y = 0 \). Thus, the critical point is \((-2, 0)\). This is a crucial step as it identifies candidate points for absolute extrema within the domain or boundary.

When searching for absolute maxima or minima within a defined region, evaluating the behavior of the function on the boundary is often necessary. For this problem, the boundary is defined by the circle \(x^2 + y^2 = 9\). To assess the function along this boundary, we must express the function using the parameters that describe the boundary. By substituting these boundary conditions into our function, we explore how the function behaves at the perimeter of the region.

Parameterization involves expressing boundary conditions with a simpler form. For a circle, this is done using trigonometric functions:

• \( x = 3\cos(\theta) \)
• \( y = 3\sin(\theta) \)

Here, \( \theta \) represents an angle that represents every point on the boundary, ranging from \( 0 \) to \( 2\pi \). This method transforms the problem into a simpler form. In our example, substituting these into the function yields \( f(3\cos(\theta), 3\sin(\theta)) = 8 + 12\cos(\theta) \). This new function is easier to analyze because it varies simply with \( \theta \), allowing straightforward determination of extremal values.

Partial derivatives play a crucial role in multivariable calculus and are instrumental in identifying critical points of a function involving more than one variable. In our example, partial derivatives were used to determine how changes in each variable individually affect the function. For the function \( f(x, y) = x^2 + y^2 + 4x - 1 \), we calculated:

• \( \frac{\partial f}{\partial x} = 2x + 4 \)
• \( \frac{\partial f}{\partial y} = 2y \)

Setting these derivatives to zero helped us find the location where the function doesn’t change its slope, i.e., the critical point \((-2, 0)\). This method not only provides insight into function behavior at critical points but also simplifies the analysis of boundary conditions when parameterizing curves.