Problem 148
Question
Given \(f(x, y)=15 x^{3}-3 x y+15 y^{3},\) find all points at which \(f_{x}(x, y)=f_{y}(x, y)=0\) simultaneously.
Step-by-Step Solution
Verified Answer
Critical points are (0, 0) and \(\left(\frac{1}{15}, \frac{1}{15}\right)\).
1Step 1: Compute Partial Derivative with respect to x
Given the function \( f(x, y) = 15x^3 - 3xy + 15y^3 \), we first find the partial derivative with respect to \( x \). The partial derivative \( f_x(x, y) \) is calculated as follows:\[f_x(x, y) = \frac{\partial}{\partial x}(15x^3 - 3xy + 15y^3) = 45x^2 - 3y\]
2Step 2: Compute Partial Derivative with respect to y
Now, find the partial derivative of \( f(x, y) \) with respect to \( y \). The partial derivative \( f_y(x, y) \) is calculated as:\[f_y(x, y) = \frac{\partial}{\partial y}(15x^3 - 3xy + 15y^3) = -3x + 45y^2\]
3Step 3: Set Partial Derivatives Equal to Zero
Since we need to find where both partial derivatives are equal to zero simultaneously, we set:\[45x^2 - 3y = 0 \quad \text{(1)} \-3x + 45y^2 = 0 \quad \text{(2)}\]
4Step 4: Solve the System of Equations
Let's solve the system of equations. From equation (1):\[45x^2 = 3y \Rightarrow y = 15x^2\]Substitute \( y = 15x^2 \) into equation (2):\[-3x + 45(15x^2)^2 = 0\]Simplify:\[-3x + 45 \cdot 225x^4 = 0 \-3x + 10125x^4 = 0 \x(10125x^3 - 3) = 0\]
5Step 5: Solve for x
From the equation \( x(10125x^3 - 3) = 0 \), we have two possible solutions: 1. \( x = 0 \)2. \( 10125x^3 - 3 = 0 \)For the second case, solve for \( x \):\[ 10125x^3 = 3 \Rightarrow x^3 = \frac{1}{3375} \Rightarrow x = \frac{1}{15} \]
6Step 6: Solve for y using x Values
For \( x = 0 \), substituting into \( y = 15x^2 \), we get:\[y = 15(0)^2 = 0\]For \( x = \frac{1}{15} \), substitute into \( y = 15x^2 \):\[y = 15 \left( \frac{1}{15} \right)^2 = 15 \cdot \frac{1}{225} = \frac{1}{15}\]
7Step 7: List the Critical Points
We have found the critical points where both partial derivatives are zero:\((x, y) = (0, 0)\) \((x, y) = \left(\frac{1}{15}, \frac{1}{15}\right)\)
Key Concepts
Critical PointsSystem of EquationsMultivariable Functions
Critical Points
Critical points are locations on a graph where the partial derivatives of a function vanish, that is, where these derivatives are equal to zero. In the context of multivariable functions, identifying critical points can be crucial for determining where local maxima, minima, or saddle points might occur.
For our given function \( f(x, y) \), critical points are found by solving the system of equations derived from the partial derivatives: \( f_x(x, y) = 45x^2 - 3y \) and \( f_y(x, y) = -3x + 45y^2 \). These equations are set to zero to determine where changes in the function start behaving regularly or are interrupted by these critical points.
Thus, critical points effectively mark the starting or stopping points of particular behaviors in the graph of a multivariable function. By setting the partial derivatives to zero, we can derive equations that aid in solving for specific \( x \) and \( y \) values, leading us to zero-in on these points of interest.
For our given function \( f(x, y) \), critical points are found by solving the system of equations derived from the partial derivatives: \( f_x(x, y) = 45x^2 - 3y \) and \( f_y(x, y) = -3x + 45y^2 \). These equations are set to zero to determine where changes in the function start behaving regularly or are interrupted by these critical points.
Thus, critical points effectively mark the starting or stopping points of particular behaviors in the graph of a multivariable function. By setting the partial derivatives to zero, we can derive equations that aid in solving for specific \( x \) and \( y \) values, leading us to zero-in on these points of interest.
System of Equations
When dealing with problems like the one provided, we often end up with a system of equations. This arises when setting the partial derivatives found in multivariable calculus to zero in order to find critical points. Such systems can consist of multiple equations that need to be solved together to find solutions that satisfy all of them simultaneously.
In our problem, the partial derivatives give us two key equations: \(45x^2 - 3y = 0\) and \(-3x + 45y^2 = 0\). Solving these equations together allows us to find the coordinates \( (x, y) \) that are critical points for the function \( f(x, y) \).
By manipulating these equations, such as by substituting one equation into another, we can often simplify or isolate specific variables. This technique is essential for finding solutions in cases where variables depend on each other. In our exercise, substituting solutions from one equation into another helped to find specific values for \( x \) and corresponding \( y \) values.
In our problem, the partial derivatives give us two key equations: \(45x^2 - 3y = 0\) and \(-3x + 45y^2 = 0\). Solving these equations together allows us to find the coordinates \( (x, y) \) that are critical points for the function \( f(x, y) \).
By manipulating these equations, such as by substituting one equation into another, we can often simplify or isolate specific variables. This technique is essential for finding solutions in cases where variables depend on each other. In our exercise, substituting solutions from one equation into another helped to find specific values for \( x \) and corresponding \( y \) values.
Multivariable Functions
Multivariable functions involve more than one variable, such as \( f(x, y) \), which is a function of both \( x \) and \( y \). These functions are mapped to a space of dimensions corresponding to the number of variables they depend on. In essence, they represent surfaces or shapes in higher dimensions, with complex behaviors.
Understanding multivariable functions is crucial in fields like physics, engineering, and economics, where systems are influenced by more than one factor. Calculus extends into these functions through the use of partial derivatives, which show how the function changes with respect to each individual variable while holding the others constant.
In examining a multivariable function like our given \( f(x, y) \), we look for points where its behavior changes, such as critical points. Partial derivatives play a key role here as they help decipher the rate and direction of change. This approach converts complex, multidimensional analysis into more manageable calculations of derivatives and equations.
Understanding multivariable functions is crucial in fields like physics, engineering, and economics, where systems are influenced by more than one factor. Calculus extends into these functions through the use of partial derivatives, which show how the function changes with respect to each individual variable while holding the others constant.
In examining a multivariable function like our given \( f(x, y) \), we look for points where its behavior changes, such as critical points. Partial derivatives play a key role here as they help decipher the rate and direction of change. This approach converts complex, multidimensional analysis into more manageable calculations of derivatives and equations.
Other exercises in this chapter
Problem 146
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
View solution Problem 147
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.
View solution Problem 149
Show that \(z=e^{x} \sin y\) satisfies the equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\).
View solution Problem 150
Show that \(f(x, y)=\ln \left(x^{2}+y^{2}\right)\) solves Laplace's equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\).
View solution