91Ó°ÊÓ

Partial derivatives are a fundamental tool in multivariable calculus that help us analyze functions of multiple variables, like the one given in the exercise. If you have a function, such as \[ z = x^3 - xy + y^2 - 1, \] we take its partial derivative with respect to each variable independently. The partial derivative with respect to \( x \) is computed by treating \( y \) as a constant, and vice versa. This process allows us to understand how the function changes or slopes in different directions.For the function from the exercise:

• The partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = 3x^2 - y \).
• The partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = -x + 2y \).

To find critical points where the function's slope is zero, we set these partial derivatives to zero and solve the resulting system of equations. This helps us determine where the function has no directional slope, optimal points, or potential extremes.

The second derivative test is a method used to classify critical points obtained from partial derivatives. After finding a critical point, you compute second-order partial derivatives to construct the Hessian matrix, which aids in determining the nature of these points.For a function \( f(x, y) \) with critical points, the key components in the Hessian matrix are:

• \( \frac{\partial^2 f}{\partial x^2} \) - the second partial derivative with respect to \( x \).
• \( \frac{\partial^2 f}{\partial y^2} \) - the second partial derivative with respect to \( y \).
• \( \frac{\partial^2 f}{\partial x \partial y} \) - the mixed partial derivative.

The determinant, \( D \), of the Hessian is calculated as: \[ D = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2. \]This determinant helps determine the classification:- If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), it's a local minimum.- If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), it's a local maximum.- If \( D < 0 \), it's a saddle point.Using these criteria, you can classify the critical points you found in the previous step.

A saddle point represents a critical point that is neither a local maximum nor a local minimum. It is named for its resemblance to the shape of a saddle on horseback; it can appear as a minimum in one cross-section and a maximum in another. This unique feature is identified when the determinant of the Hessian is negative (\( D < 0 \)).In the context of our exercise, the critical point \((0, 0)\) was classified as a saddle point. The calculation:- \( \frac{\partial^2 z}{\partial x^2} = 0 \),- \( D = (0)(2) - (-1)^2 = -1. \)Since \( D < 0 \), it confirms that this point is a saddle point. Such a classification shows how the function behaves differently along various paths around this critical point.

A local minimum is a critical point where the function value is lower than the function values at surrounding points. This concept indicates a small "valley" or dip relative to its immediate environment.For the function in question, once the second derivative test is applied to the critical point \( \left( \frac{1}{6}, \frac{1}{12} \right) \), it is classified as a local minimum. The test involves:- Evaluating the second partial derivative: \( \frac{\partial^2 z}{\partial x^2} = 1 \),- Determinant: \( D = (1)(2) - (-1)^2 = 1. \)Since \( D > 0 \) and \( \frac{\partial^2 z}{\partial x^2} > 0 \), this point is verified to be a local minimum.Local minima are crucial in fields like optimization, where you might want to find the lowest point in a dataset or physical process. Understanding such points in a function can help predict outcomes and plan further analysis.