Problem 10 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The function has saddle points at \((n\pi, 0)\) for \(n \in \mathbb{Z}\).

Step 1: Compute Partial Derivatives

To find the candidates for local extrema, we first calculate the partial derivatives of the function \( f(x, y) = y \sin x \) with respect to \(x\) and \(y\).Partial derivative with respect to \(x\):\[\frac{\partial f}{\partial x} = y \cos x\]Partial derivative with respect to \(y\):\[\frac{\partial f}{\partial y} = \sin x\]

Step 2: Find Critical Points

Set the partial derivatives equal to zero to find the critical points.For \( \frac{\partial f}{\partial x} = 0 \):\[ y \cos x = 0 \]This implies either \(y = 0\) or \(\cos x = 0\). If \(\cos x = 0\), \(x = \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\).For \( \frac{\partial f}{\partial y} = 0 \):\[ \sin x = 0 \]This implies \(x = n\pi\) for \(n \in \mathbb{Z}\).From both conditions, we get the critical points at \((x, y) = (n\pi, 0)\).

Step 3: Construct the Hessian Matrix

The Hessian matrix \(H\) is used to determine the nature of critical points. It is formed with the second partial derivatives of \(f\):\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]Calculate each second partial derivative:\[ \frac{\partial^2 f}{\partial x^2} = -y \sin x \]\[ \frac{\partial^2 f}{\partial x \partial y} = \cos x \]\[ \frac{\partial^2 f}{\partial y \partial x} = \cos x \]\[ \frac{\partial^2 f}{\partial y^2} = 0 \]Thus, the Hessian is:\[H = \begin{bmatrix} -y \sin x & \cos x \ \cos x & 0 \end{bmatrix}\]

Step 4: Determine the Nature of Critical Points

Evaluate the Hessian at the critical points \((n\pi, 0)\).At \(x = n\pi\), \(\sin x = 0\) and \(\cos x = (-1)^n\), the Hessian becomes:\[H(n\pi, 0) = \begin{bmatrix} 0 & (-1)^n \ (-1)^n & 0 \end{bmatrix}\]The determinant of \(H\) is given by:\[ \det(H) = 0 \times 0 - ((-1)^n)^2 = -1 < 0 \]Since the determinant is negative, each \((n\pi, 0)\) is a saddle point.

Key Concepts

Calculus

Calculus is an essential branch of mathematics that studies continuous change. It provides tools for analyzing and understanding functions, slopes, areas, and other phenomena related to change, which are not easily handled by basic algebra. The subject is split primarily into two fields: differential calculus and integral calculus. Differential calculus involves computing derivatives, which measure how a function changes at any given point. Integral calculus, on the other hand, is concerned with summation, or finding the total accumulation of values, usually resulting in area calculation under a curve.

In this particular exercise, we focus on differential calculus to find where a function has local extrema. Extrema refer to points where a function reaches a maximum or minimum value, which are pivotal in many applications such as optimization problems. This involves using derivatives to determine function behavior. Calculus provides the foundational approach, and when extended to multiple variables, it becomes a gateway to multidimensional analyses, such as this function of two variables, where aspects like the Hessian matrix become important.

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus, a critical extension of calculus that deals with functions of several variables. When a function depends on multiple variables, like in our exercise where the function is defined as \(f(x, y) = y \sin x\), each variable can change independently. With partial derivatives, we examine how a function changes as one particular variable changes while keeping the others constant.

To find the partial derivative of \(f(x, y) = y \sin x\):

• The partial derivative with respect to \(x\), denoted as \(\frac{\partial f}{\partial x}\), focuses on how \(f\) changes as \(x\) changes, yielding \(y \cos x\).
• The partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y}\), observes changes in \(f\) with changes in \(y\), resulting in \(\sin x\).

These derivatives form the basis for finding critical points by setting each derivative to zero, revealing where the rate of change halts momentarily. This is instrumental not only in locating but analyzing critical points for further understanding of local behavior around a point.

Critical Points

In calculus, critical points of a function are where its derivative is zero or undefined, indicating potential local maxima, minima, or saddle points. For functions of multiple variables, like our function \(f(x, y)\), critical points arise from setting its partial derivatives to zero, revealing where these directional changes pause.

The exercise identified critical points at \((n\pi, 0)\) by:

• Equating \(\frac{\partial f}{\partial x} = y \cos x = 0\), thus implying \(y=0\) or \(cos x = 0\), leading to \(x=\frac{\pi}{2} + n\pi\).
• Solving \(\frac{\partial f}{\partial y} = \sin x = 0\), giving \(x=n\pi\).

These calculations reveal that all critical points occur along the y-axis at multiples of \(\pi\) for \(x\), with y fixed at zero. The subsequent step involved calculating the Hessian matrix to determine the nature of these critical points. Here, the matrix's negative determinant indicated each critical point is a saddle point, a scenario where the point isn't a max or min but might look like both. Thus, understanding these points is vital for grasping function dynamics.