Problem 146 - Calculus 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 where both partial derivatives are zero is \((-1, 1)\).

Step 1: Calculate Partial Derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate the function \( f(x, y) = 2x^2 + 2xy + y^2 + 2x - 3 \) with respect to \( x \). This gives: \[ \frac{\partial f}{\partial x} = 4x + 2y + 2. \]

Step 2: Set Partial Derivative with respect to x to 0

Set the expression obtained from the partial derivative with respect to \( x \) equal to zero and solve for \( y \): \[ 4x + 2y + 2 = 0. \] Simplifying, we get: \[ 2y = -4x - 2. \]

Step 3: Calculate Partial Derivative with respect to y

Find \( \frac{\partial f}{\partial y} \) by differentiating \( f(x, y) = 2x^2 + 2xy + y^2 + 2x - 3 \) with respect to \( y \): \[ \frac{\partial f}{\partial y} = 2x + 2y. \]

Step 4: Set Partial Derivative with respect to y to 0

Set the expression for \( \frac{\partial f}{\partial y} \) equal to zero and solve for \( x \): \[ 2x + 2y = 0. \] Simplifying, we find: \[ y = -x. \]

Step 5: Solve the System of Equations

Substitute \( y = -x \) into the equation from Step 2: \[ 2(-x) = -4x - 2. \] Simplifying gives: \[ -2x = -4x - 2 \] \[ 2x = -2 \] \[ x = -1. \] Now substitute \( x = -1 \) back into \( y = -x \) to find \( y \): \[ y = -(-1) = 1. \] So the point is \( (-1, 1) \).

Key Concepts

Function Partial Derivative

A partial derivative is a key concept from multivariable calculus that measures how a function changes as one of its input variables changes, while keeping the other variables constant. When dealing with functions of two variables, such as \( f(x, y) \), we take the partial derivative with respect to one variable, like \( x \), while treating \( y \) as a constant. The process reveals the rate at which the function \( f \) changes as \( x \) changes along a slice parallel to the \( x \)-axis.

For the example function \( f(x, y) = 2x^2 + 2xy + y^2 + 2x - 3 \), calculating its partial derivative with respect to \( x \) involves using familiar derivative rules, yielding \( \frac{\partial f}{\partial x} = 4x + 2y + 2 \). This represents the tangent slope in the direction of the \( x \)-axis, providing insight into the function's behavior at specific \( x \) values while ignoring changes in \( y \).

Understanding partial derivatives is crucial in determining how multi-dimensional functions behave locally, which forms the foundation for further analysis like finding critical points or optimizing functions.

Simultaneous Equations

Simultaneous equations involve finding values for variables that satisfy multiple equations at once. In multivariable calculus, dealing with simultaneous equations often arises when working with partial derivatives. We set these derivatives, like \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), to zero to find critical points of a function.

In our example, after computing these partial derivatives and setting them equal to zero, we have the equations:

• \( 4x + 2y + 2 = 0 \)
• \( 2x + 2y = 0 \)

The goal is to find values of \( x \) and \( y \) that simultaneously satisfy both equations, revealing the points where the function's surface is "flat" along both axes. This process of solving creates a system of equations that must be simplified using algebraic techniques, like substitution or elimination, to find a common solution.

This step is essential in identifying points of interest in a function's graph, particularly for optimization or finding regions with special properties.

Critical Points

Critical points are locations on a graph where a function's derivative equals zero. Beyond simple functions, in multivariable calculus, finding critical points involves ensuring that all partial derivatives equal zero at the same time. These points can indicate maxima, minima, or saddle points in a multi-dimensional graph.

The critical point of a function is where its rate of change flattens out. In our example, after solving the simultaneous equations formed by \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \), we find \( x = -1 \) and \( y = 1 \).

This critical point \( (-1,1) \) suggests a location where the function may achieve a local extremum or possibly act as a saddle point, meaning neither a maximum nor minimum, but a point of inflection. Analyzing critical points helps in understanding the larger characteristics of the function's surface.

Multivariable Calculus

Multivariable calculus extends basic calculus concepts to functions with more than one variable. Functions in this realm often depend on multiple inputs, like \( x \) and \( y \), and are studied using techniques that account for their larger dimensionality.

Through concepts like partial derivatives, we explore how functions change in different directions and dimensions. This is vital for disciplines like physics, engineering, and economics where multi-faceted systems are common.

For instance, in the function \( f(x, y) \), we observe how it behaves as we tweak \( x \) while holding \( y \) constant, or vice versa, aiding in defining paths over its surface. Multivariable calculus also helps in defining and finding optimization points and solving real-world problems that come with many variables.

By understanding multivariable calculus, we gain powerful tools to mathematically model and investigate interactions across complex systems, making it indispensable in mathematics and applied sciences.