Problem 329 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=y^{2}+x y+3 y+2 x+3 $$

Step-by-Step Solution

Answer

The critical point \((1, -2)\) is a saddle point.

Step 1: Find the First Partial Derivatives

To find the critical points of the function \( f(x, y) = y^2 + xy + 3y + 2x + 3 \), we first compute the partial derivatives \( f_x \) and \( f_y \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial}{\partial x}(y^2 + xy + 3y + 2x + 3) = y + 2 \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial}{\partial y}(y^2 + xy + 3y + 2x + 3) = 2y + x + 3 \).

Step 2: Set Partial Derivatives to Zero

Set \( f_x = 0 \) and \( f_y = 0 \) to find the critical points. From \( f_x = y + 2 = 0 \), we get \( y = -2 \). Substitute \( y = -2 \) into \( f_y = 2y + x + 3 = 0 \) to find \( x \). This gives \( 2(-2) + x + 3 = 0 \), or \( -4 + x + 3 = 0 \), which simplifies to \( x = 1 \). So, the critical point is \( (1, -2) \).

Step 3: Find the Second Partial Derivatives

Next, compute the second partial derivatives necessary for the second derivative test. Find \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 0 \), \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \), and \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 1 \).

Step 4: Apply the Second Derivative Test

Use the second derivative test, which involves calculating \( D = f_{xx}f_{yy} - (f_{xy})^2 \) at the critical point \((1, -2)\). Compute \( D = (0)(2) - (1)^2 = -1 \). When \( D < 0 \), the critical point is a saddle point.

Key Concepts

Critical Points

In calculus, critical points are where the first partial derivatives of a function are zero. For the function \( f(x, y) = y^2 + xy + 3y + 2x + 3 \), we find critical points by setting both partial derivatives equal to zero.
Critical points are important because they can indicate potential locations of local maxima, minima, or saddle points.

• First partial derivative with respect to \( x \): \( f_x = y + 2 \)
• First partial derivative with respect to \( y \): \( f_y = 2y + x + 3 \)

We solve the equations \( f_x = 0 \) and \( f_y = 0 \) to find the critical points. In this example, \( y + 2 = 0 \) leads to \( y = -2 \), and substitution into \( 2y + x + 3 = 0 \) gives \( x = 1 \), producing the critical point \((1, -2)\). This process helps in finding places where the function could potentially change behavior.

Partial Derivatives

Partial derivatives are a way to find the rate of change of a function with respect to one variable while keeping others constant. They are critical in multivariable calculus and transformational analyses.

• To find \( f_x \) or \( \frac{\partial f}{\partial x} \), differentiate by holding \( y \) constant.
• For \( f_y \) or \( \frac{\partial f}{\partial y} \), differentiate by holding \( x \) constant.

In our function \( f(x, y) = y^2 + xy + 3y + 2x + 3 \), partial derivatives help us understand how the function changes as each variable changes independently. These derivatives are the first step to identifying critical points and applying further tests for characteristics of the function.

Saddle Point

A saddle point is a critical point where the function does not have a maximum or a minimum. It's analogous to a mountain pass: not at the top of a hill nor the bottom of a valley.
For the function \( f(x, y) = y^2 + xy + 3y + 2x + 3 \), using the second derivative test, we calculate:

• \( f_{xx} = 0 \)
• \( f_{yy} = 2 \)
• \( f_{xy} = 1 \)

With these, the determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 = (0)(2) - (1)^2 = -1 \). Since \( D < 0 \), the function has a saddle point at \( (1, -2) \). Understanding saddle points is essential in optimizing and characterizing functions beyond basic maxima and minima.

Calculus 3

Calculus 3 extends the concepts of Calculus 1 and 2 into multivariable functions. Concepts like partial derivatives and critical points come alive in this context.

These tools help us analyze and understand multi-dimensional systems, crucial in fields like physics and engineering. Finding and examining critical points of multivariable functions is a concept commonly explored in Calculus 3. This involves partial derivatives and the second derivative test, key elements of multivariable optimization.

• Expands derivative concepts into higher dimensions.
• Introduces partial derivatives and multiple integrals.
• Applies mathematical principles to real-world phenomena.

This level of calculus provides the foundation to analyze dynamic systems where variables interact in complex ways, enabling predictions and optimizations in various scientific and engineering domains.