Problem 147 - Calculus Volume 3

Question

Given \(f(x, y)=y^{3}-3 y x^{2}-3 y^{2}-3 x^{2}+1,\) find all points on \(f\) at which \(f_{x}=f_{y}=0\) simultaneously.

Step-by-Step Solution

Answer

The points are (0, 0), (0, 2), (\( \sqrt{3} \), -1), (-\( \sqrt{3} \), -1).

Step 1: Find Partial Derivative with respect to x

To find the points where both partial derivatives are zero, we start by finding the partial derivative of \( f(x, y) = y^{3} - 3 y x^{2} - 3 y^{2} - 3 x^{2} + 1 \) with respect to \( x \). The derivative is:\[ f_{x} = \frac{\partial}{\partial x} (y^{3} - 3 y x^{2} - 3 y^{2} - 3 x^{2} + 1) = -6yx - 6x \]Simplifying this, we get:\[ f_{x} = -6x(y + 1) \]

Step 2: Find Partial Derivative with respect to y

Next, find the partial derivative of \( f(x, y) \) with respect to \( y \). The derivative is:\[ f_{y} = \frac{\partial}{\partial y} (y^{3} - 3 y x^{2} - 3 y^{2} - 3 x^{2} + 1) = 3y^2 - 3x^2 - 6y \]

Step 3: Set Partial Derivatives Equal to Zero

To find stationary points, set the partial derivatives equal to zero:1. \( f_{x} = -6x(y + 1) = 0 \)2. \( f_{y} = 3y^2 - 3x^2 - 6y = 0 \)From equation (1), two cases arise:- \( x = 0 \)- \( y = -1 \)

Step 4: Solve when x = 0

Substitute \( x = 0 \) into the equation for \( f_{y} \):\[ 3y^2 - 6y = 0 \]Factor this equation:\[ 3y(y - 2) = 0 \]This gives the solutions:- \( y = 0 \)- \( y = 2 \)Therefore, when \( x = 0 \), the points are \( (0, 0) \) and \( (0, 2) \).

Step 5: Solve when y = -1

Substitute \( y = -1 \) into the equation for \( f_{y} \):\[ 3(-1)^2 - 3x^2 - 6(-1) = 0 \]Simplify and solve this:\[ 3 - 3x^2 + 6 = 0 \]\[ 9 = 3x^2 \]\[ x^2 = 3 \]\[ x = \pm \sqrt{3} \]Therefore, when \( y = -1 \), the points are \( (\sqrt{3}, -1) \) and \( (-\sqrt{3}, -1) \).

Step 6: List All Critical Points

Combine the solutions from the above cases. The points where both partial derivatives are zero are:1. \( (0, 0) \)2. \( (0, 2) \)3. \( (\sqrt{3}, -1) \)4. \( (-\sqrt{3}, -1) \)

Key Concepts

Critical Points

Critical points in multivariable calculus are the locations on the function where the gradient, or the vector of its partial derivatives, is equal to zero. To find these points, we determine where both the partial derivative with respect to each variable equals zero simultaneously. These points are of great interest because they often indicate where maxima, minima, or saddle points occur in a function.

In our exercise, we're working with a function of two variables, and the critical points are found by solving:

• The partial derivative with respect to x, given as: \[ f_{x} = -6x(y + 1) = 0 \]
• The partial derivative with respect to y, given as: \[ f_{y} = 3y^2 - 3x^2 - 6y = 0 \]

This process involves setting up and solving these equations to find the values of x and y that satisfy both conditions. When you solve these equations, you determine the points where the function can change its behavior. These include \((0, 0)\), \((0, 2)\), \((\sqrt{3}, -1)\), and \((-\sqrt{3}, -1)\) for this specific example.

Stationary Points

Stationary points are a specific type of critical point where the derivative itself becomes zero. This concept is closely connected to critical points but emphasized in scenarios where we are particularly interested in the flatness of a curve at these points.

In the given function, stationary points occur where the slope of tangent lines to the graph is zero. For functions of several variables like our two-variable example, the stationary points satisfy both:

• \( f_{x} = 0 \), meaning the slope in the x-direction is zero.
• \( f_{y} = 0 \), meaning the slope in the y-direction is zero.

Both the partial derivatives must simultaneously be zero. By examining the conditions \( x = 0 \) and \( y + 1 = 0 \), we find that the character of the function doesn't change. We derived solutions like \( y = 0 \) or \( y = -1 \), and these facilitate finding corresponding \( x \) values. Such analysis directly ties into exploring the geometry and behavior of multivariable functions at the stationary points derived, like \((0, 2)\) or \((\sqrt{3}, -1)\).

Multivariable Calculus

Multivariable calculus explores functions with multiple inputs, typically in the form \( f(x, y, z, ...) \). It extends the concepts from single-variable calculus, such as differentiation and integration, into higher dimensions. Critical skills in this field include understanding how functions behave concerning their input variables and the geometric interpretations of these functions.

A key concept is the partial derivative, which measures how the function changes as one variable changes, holding the others constant. In our given function \( f(x, y) = y^{3} - 3 y x^{2} - 3 y^{2} - 3 x^{2} + 1 \), partial derivatives help us unravel where the slopes are zero, especially across multiple dimensions. This aids in identifying where—and how—a function might achieve its max, min, or even neither (saddle points).

Topics such as critical and stationary points are fundamental as they allow you to sketch the behavior of functions accurately, analyze surfaces, and solve optimization problems. Mastery of multivariable calculus opens doors to further studies and applications, such as in engineering, physics, economics, and various fields where multiple variables influence outcomes.