Problem 146 - Calculus Volume 3

Question

Given \(f(x, y)=2 x^{2}+2 x y+y^{2}+2 x-3,\) find all points at which \(\frac{\partial f}{\partial x}=0\) and \(\frac{\partial f}{\partial y}=0\) simultaneously.

Step-by-Step Solution

Answer

The point is \((-1, 1)\).

Step 1: Find Partial Derivative with Respect to x

To find \(\frac{\partial f}{\partial x}\), take the derivative of \(f(x, y)\) with respect to \(x\), treating \(y\) as a constant. Start by differentiating each term separately.\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2x^2) + \frac{\partial}{\partial x}(2xy) + \frac{\partial}{\partial x}(y^2) + \frac{\partial}{\partial x}(2x) - \frac{\partial}{\partial x}(3)\]. This simplifies to \[\frac{\partial f}{\partial x} = 4x + 2y + 2\].

Step 2: Find Partial Derivative with Respect to y

Now, compute \(\frac{\partial f}{\partial y}\) by differentiating \(f(x, y)\) with respect to \(y\), treating \(x\) as a constant. Differentiate each term: \[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2x^2) + \frac{\partial}{\partial y}(2xy) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial y}(2x) - \frac{\partial}{\partial y}(3)\]. This simplifies to \[\frac{\partial f}{\partial y} = 2x + 2y\].

Step 3: Set Partial Derivatives to Zero

Set each of the partial derivatives equal to zero to find the points where they vanish. This gives the system of equations: \[4x + 2y + 2 = 0\] and \[2x + 2y = 0\].

Step 4: Solve the System of Equations

Solve the system of equations: \[4x + 2y + 2 = 0\] and \[2x + 2y = 0\]. Simplify the second equation: \[x + y = 0\]. Substitute \(y = -x\) into the first equation: \[4x + 2(-x) + 2 = 0\], which becomes \[2x + 2 = 0\]. Solving gives \[x = -1\]. Substituting \(x = -1\) into \(y = -x\) gives \(y = 1\).

Step 5: Identify the Solution Point

The solution to the system is the point \((-1, 1)\), which satisfies both \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\) simultaneously.

Key Concepts

Calculus Problem Solving

When tackling calculus problems, it's essential to break down the process into manageable steps. This makes it easier to understand and solve complex problems like the one involving partial derivatives.
In our exercise, we start with a function of two variables,

• Identify what you need to compute: here, we focus on partial derivatives.
• Calculate the partial derivatives separately with respect to each variable.
• Set the partial derivatives to zero to find critical points.

Viewing each step individually and solving systematically ensures that no critical part of the problem is overlooked. This methodical approach to calculus problem-solving allows you to tackle seemingly daunting equations confidently.

Critical Points

In calculus, critical points occur where the derivative of a function equals zero or is undefined. For functions of two variables, partial derivatives are used to find these points.
In our exercise, we calculate the partial derivatives and set them to zero, producing a system of equations. Solving these equations helps us find where the function has critical points. Critical points can tell us where the function reaches minimum, maximum, or saddle points, depending on the problem's context.

• Critical points are necessary for understanding the behavior of multi-variable functions.
• They help to determine where changes occur in the slope of the function, which can indicate peaks or troughs.

Comprehending critical points' role in calculus provides deeper insights into analyzing functions and their behavior over a particular domain.

System of Equations

When partial derivatives for a multi-variable function are determined, it's common to end up with a system of equations to solve. These equations arise from equating each partial derivative to zero.
In the current example,

• Two equations were obtained: \(4x + 2y + 2 = 0\) and \(2x + 2y = 0\).
• Simplify the equations wherever possible; here, the latter simplifies to \(x + y = 0\).
• Substitute one equation into another to find the variables' values, helping locate the critical point \((-1, 1)\).

Solving the system of equations is a crucial step that solidifies understanding of how partial derivatives work and interacts, providing the solutions needed for identifying where these derivatives vanish. This lays the groundwork for further analysis of the function's behavior at these points.