Problem 145 - Calculus Volume 3

Question

Given \(f(x, y)=x^{2}+x-3 x y+y^{3}-5,\) find all points at which \(f_{x}=f_{y}=0\) simultaneously.

Step-by-Step Solution

Answer

The points are \((1, 1)\) and \((\frac{1}{4}, \frac{1}{2})\)."

Step 1: Find partial derivatives

To find critical points, we need the partial derivatives of the function \( f(x, y) \). For \( f(x, y) = x^{2} + x - 3xy + y^{3} - 5 \), calculate \( f_{x} \) and \( f_{y} \):1. \( f_{x} = \frac{\partial}{\partial x}(x^{2} + x - 3xy + y^{3} - 5) = 2x + 1 - 3y \).2. \( f_{y} = \frac{\partial}{\partial y}(x^{2} + x - 3xy + y^{3} - 5) = -3x + 3y^{2} \).

Step 2: Set partial derivatives to zero

We need to find \((x, y)\) such that both partial derivatives are zero:1. \( 2x + 1 - 3y = 0 \).2. \( -3x + 3y^{2} = 0 \).Solve these equations simultaneously.

Step 3: Solve the system of equations

Solve the two equations from Step 2:1. From \(-3x + 3y^{2} = 0\), simplify to \(x = y^{2}\).2. Substitute \(x = y^{2}\) into \(2x + 1 - 3y = 0\): \( 2(y^{2}) + 1 - 3y = 0 \) becomes \( 2y^{2} - 3y + 1 = 0 \).3. Solve the quadratic equation \(2y^{2} - 3y + 1 = 0\) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) where \(a = 2\), \(b = -3\), and \(c = 1\). This yields \( y = 1 \) and \( y = \frac{1}{2} \).

Step 4: Find corresponding x-values

Using \(x = y^{2}\), find \(x\) for the values of \(y\) found:1. For \(y = 1\), \(x = 1^{2} = 1\).2. For \(y = \frac{1}{2}\), \(x = (\frac{1}{2})^{2} = \frac{1}{4}\).

Step 5: Identify points

Combine \((x, y)\) solutions from Step 4:1. For \(y = 1\), the point is \((1, 1)\).2. For \(y = \frac{1}{2}\), the point is \((\frac{1}{4}, \frac{1}{2})\).

Key Concepts

Critical Points

In the study of multivariable calculus, critical points are the locations where the gradient of a function is zero or undefined. For the function \(f(x, y) = x^2 + x - 3xy + y^3 - 5\), we seek points where the partial derivatives \(f_x\) and \(f_y\) both equal zero. These are potential points where a function might have local extrema or saddle points.

Finding critical points involves locating spots where either one or more changes in the function's growth rate occur. We take the derivatives with respect to all variables, set them to zero, and solve for the variables. These critical points give valuable insights into the behavior of a function on a graph.

System of Equations

A system of equations consists of multiple equations that are solved together, where the solution satisfies all equations simultaneously. In the context of finding critical points, our task is to solve:

• \(2x + 1 - 3y = 0\)
• \(-3x + 3y^2 = 0\)

For these two equations, our goal is to find values of \(x\) and \(y\) where both equations are true. These equations can be interpreted graphically as two curves on a plane, and any solutions are intersection points of these curves. Solving the system gives us solutions that help in identifying the critical points of the function.

Quadratic Formula

The quadratic formula is a tool for solving quadratic equations that take the form \(ax^2 + bx + c = 0\). The formula is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our problem, one of the critical steps is solving the quadratic equation \(2y^2 - 3y + 1 = 0\). Here, we apply the quadratic formula with \(a = 2\), \(b = -3\), and \(c = 1\) to find the roots. Solving gives \(y = 1\) and \(y = \frac{1}{2}\). It is fundamental to understand how each parameter influences the solution, and taking the square root of \(b^2 - 4ac\) helps us in identifying the nature of the roots.

• This results in real solutions, which hints at potential critical points.

Simultaneous Equations

Solving simultaneous equations is about finding a common solution to a set of equations. In other words, it's about finding the values that satisfy all equations in a system at once. In our specific example, the equations derived from setting partial derivatives equal to zero are solved together.

To solve:

• From \(-3x + 3y^2 = 0\), we get \(x = y^2\).
• Substituting \(x = y^2\) into the first equation \(2x + 1 - 3y = 0\) results in a quadratic equation \(2y^2 - 3y + 1 = 0\).

This process seamlessly integrates solving, substituting, and simplifying to find values for \(x\) and \(y\). Mastering simultaneous equations is a handy skill, particularly in calculus, as it allows us to find intersections or overlapping points essential for thorough analysis of a function’s behavior.