91Ó°ÊÓ

In multivariable calculus, first-order partial derivatives help us understand how a function changes along an axis. If you have a function of two variables, say \(k(x, y)\), you focus on how it changes with respect to one variable while keeping the other constant. This means:

• \(\frac{\partial k}{\partial x}\): Change of the function with respect to \(x\), keeping \(y\) constant.
• \(\frac{\partial k}{\partial y}\): Change of the function with respect to \(y\), keeping \(x\) constant.

The process involves taking the derivative of the function treating it as if it is only a function of one variable. For the function \(k(x, y) = x^2 - xy + 2y^2\), the first-order partial derivatives are:

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

Finding these derivatives is the first step in locating critical points. Critical points occur where the derivative equals zero, indicating no change in the function value as you move along either axis.

Second-order partial derivatives explore the curvature of the function surface. They tell you how the first-order derivatives themselves change with respect to each variable. These derivatives are crucial for understanding the shape of the function graph. For our function \(k(x, y)\), the second-order partial derivatives are:

• \(\frac{\partial^2 k}{\partial x^2} = 2\)
• \(\frac{\partial^2 k}{\partial y^2} = 4\)
• \(\frac{\partial^2 k}{\partial x \partial y} = -1\)

These derivatives help us investigate the concavity of the function around critical points. More specifically, they help us classify these critical points as local minima, maxima, or saddle points. In simpler terms, they tell us if the point is like a bowl, a peak, or a saddle in three-dimensional space.

The Hessian matrix is a compact way to express all the second-order partial derivatives. It is a square matrix composed of second-order partial derivatives of a function. For a function \(k(x, y)\), the Hessian matrix \(H\) is: \[ H = \begin{bmatrix} \frac{\partial^2 k}{\partial x^2} & \frac{\partial^2 k}{\partial x \partial y} \\frac{\partial^2 k}{\partial y \partial x} & \frac{\partial^2 k}{\partial y^2} \end{bmatrix} \] In the exercise, this matrix looks like: \[ H = \begin{bmatrix} 2 & -1 \ -1 & 4 \end{bmatrix} \] The determinant of the Hessian matrix \(D\) is critical for classifying critical points: \[ D = \frac{\partial^2 k}{\partial x^2} \cdot \frac{\partial^2 k}{\partial y^2} - \left(\frac{\partial^2 k}{\partial x \partial y}\right)^2 \] If \(D > 0\) and \(\frac{\partial^2 k}{\partial x^2} > 0\), the point is a local minimum. If \(D > 0\) and \(\frac{\partial^2 k}{\partial x^2} < 0\), the point is a local maximum. If \(D < 0\), it indicates a saddle point. In our example, the positive determinant value of \(7\) at the critical point \((2, 4)\) confirms it is a local minimum.