Problem 1 - Calculus for Biology and Medicine

Question

The functions are defined for all \((x, y) \in R^{2} .\) Find all candidates for local extrema, and use the Hessian matrix to determine the type (maximum, minimum, or saddle point). $$ f(x, y)=x^{2}+y^{2}+2 y $$

Step-by-Step Solution

Answer

The local minimum occurs at \((0, -1)\).

Step 1: Find the critical points

To find the critical points, we first need to compute the partial derivatives of the function. The partial derivative with respect to \(x\) is given by:\[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 + y^2 + 2y) = 2x. \]The partial derivative with respect to \(y\) is:\[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 + y^2 + 2y) = 2y + 2. \]Setting these partial derivatives equal to zero to find critical points:\[ 2x = 0 \quad \text{and} \quad 2y + 2 = 0. \]Solving these equations gives:\[ x = 0 \quad \text{and} \quad y = -1. \]Thus, the critical point is \((0, -1)\).

Step 2: Form the Hessian matrix

The Hessian matrix is a square matrix of second-order partial derivatives. For the function \( f(x, y) = x^2 + y^2 + 2y \), it's given by:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy}\end{bmatrix}. \]First, compute the second partial derivatives:- \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \)- \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \)- \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \) (and \( f_{yx} = f_{xy} = 0 \) because mixed partial derivatives are continuous and equal)Thus, the Hessian matrix is:\[H = \begin{bmatrix} 2 & 0 \ 0 & 2\end{bmatrix}. \]

Step 3: Evaluate the Hessian at the critical point

Now, we evaluate the Hessian matrix at the critical point \((0, -1)\). Since the Hessian does not depend on \( (x, y) \), it remains:\[H = \begin{bmatrix} 2 & 0 \ 0 & 2\end{bmatrix}. \]The determinant of \(H\) is computed as:\[\det(H) = (2)(2) - (0)(0) = 4.\]Since \( \det(H) > 0\) and \( f_{xx} > 0 \), the critical point \((0, -1)\) is a local minimum.

Key Concepts

Local Extrema

Local extrema are significant points on a function where the function reaches a minimum or maximum value in a nearby region. Imagine a local hilltop or a valley; these would represent local maximum and minimum points, respectively.

In the context of functions with multiple variables, local extrema are analyzed using the first and second derivative tests. For a function like \( f(x, y) \), local extrema occur where the partial derivatives (slopes in each direction) equal zero. This indicates neither increase nor decrease initially, setting the stage for a potential extremum.

Knowing this, let's see how you can find these points in two-dimensional functions.

• Use the first partial derivatives to find critical points where they both equal zero.
• Use the determinant of the Hessian matrix, a matrix of second partial derivatives, to determine the nature of each critical point.

This approach helps distinguish between maxima, minima, and saddle points, which can appear as upward or downward curves like hills or valleys, or neither.

Partial Derivatives

Partial derivatives measure how a function changes as one variable changes while the other remains constant. Consider them like slopes in specific directions. When dealing with functions of two variables such as \( f(x, y) \), we use partial derivatives to identify potential changes when shifting slightly in the \(x\)-direction or \(y\)-direction.

To find a partial derivative with respect to \(x\), \( f_x(x, y) \), hold \(y\) constant and see how the function changes as \(x\) changes. Likewise, for \(y\), find \( f_y(x, y) \) by maintaining \(x\) constant.

Here's why they are essential:

• Identify critical points by setting partial derivatives to zero.
• Determine function behavior around those points.

In our example, the partial derivatives lead us to find that \( f_x = 2x \) and \( f_y = 2y + 2 \). These expressions provide us with the slopes of the function in each direction and help locate potential critical points like \((0, -1)\).

Critical Points

Critical points are crucial as they are potential candidates for local extrema. They occur where the partial derivatives of a function are both equal to zero, meaning the slope in each direction (\(x\) and \(y\)) is flat. This absence of upward or downward trend suggests the likelihood of a peak, trough, or saddle point.

To locate critical points, begin by calculating the partial derivatives and equate them to zero. This will yield equations you can solve for \(x\) and \(y\).

Consideration for critical points includes:

• Setting both \( f_x(x, y) = 2x = 0 \) and \( f_y(x, y) = 2y + 2 = 0 \).
• Solving these will result in \(x = 0\) and \(y = -1\).

These points are valuable as they act as checkpoints to further investigate the nature of extrema through the Hessian matrix, deciding whether the point is a maximum, minimum, or saddle point based on the behavior of the function around it.