Solving Multivariate Problem: Critical Points for ##y=-x##

[squenshl]

Homework Statement

Consider the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f(x,y) := \ln{\left(3+(x+y)^2\right)}$.

1. Find $\nabla f$, the gradient of $f$, and determine at which points $\nabla f$ is zero.

2. Determine whether the critical points of $f$ are local minima, local maxima, or saddle points by considering the level curves of $f$.

3. Calculate all the second order mixed partial derivatives of $f$.

Relevant Equations

Gradient vector $\nabla f$.

There are sets of the form $\left\{(x,y)\in \mathbb{R}^2: f(x,y) = \ln{\left(3+(x+y)^2\right)} = c\right\}$ where $c$ is some fixed number $> 1$. Let's see what happens for a few values of $c$.
Suppose $c = 2$, then $\ln{\left(3+(x+y)^2\right)} = 2 \Longleftrightarrow (x+y)^2 = e^2-3$, or $x+y = \pm\sqrt{e^2-3}$ which are straight lines of gradient $-1$ through $\left(\pm\sqrt{e^2-3},0\right)$ and $\left(0,\pm\sqrt{e^2-3}\right)$.
For $c = 1$, then $\ln{\left(3+(x+y)^2\right)} = 3 \Longleftrightarrow (x+y)^2 = e^3-3$, or $x+y = \pm\sqrt{e^3-3}$ which are straight lines of gradient $-1$ through $\left(\pm\sqrt{e^3-3},0\right)$ and $\left(0,\pm\sqrt{e^3-3}\right)$.

1. We find the partial derivatives of $f$ with respect to $x$ and $y$ to get $\frac{\partial f}{\partial x} = f_x = \frac{2(x+y)}{3+(x+y)^2}$ and $\frac{\partial f}{\partial y} = f_y = f_x = \frac{2(x+y)}{3+(x+y)^2}$. This makes the gradient vector
$$\nabla{f} = \begin{bmatrix}
f_x \\
f_y
\end{bmatrix} = \begin{bmatrix}
\frac{2(x+y)}{3+(x+y)^2} \\[6pt]
\frac{2(x+y)}{3+(x+y)^2}
\end{bmatrix}.$$

Equating these to zero gives the critical points and doing so gives $2(x+y) = 0$, or $y = -x$. Hence, the critical point is the line $y = -x$.

2. From the figure above we see that the level curves seem to increase the further we get away from the critical points which suggests that they are local minima.

3. We have $f_x = \frac{2(x+y)}{3+(x+y)^2}$ which means $f_{xx} = \frac{\left(2(3+(x+y)^2\right)-4(x+y)^2}{\left(3+(x+y)^2\right)^2}$. Similarly we have $f_y = 2ye^{x^2+y^2}3$ which means $f_{yy} = \frac{\left(2(3+(x+y)^2\right)-4(x+y)^2}{\left(3+(x+y)^2\right)^2}$. Finally, we know that $f_{xy} = f_{yx} = \frac{\left(2(3+(x+y)^2\right)-4(x+y)^2}{\left(3+(x+y)^2\right)^2}$ since the (Hessian) matrix of second order partial derivatives is symmetric.

My question is what are the critical points if $\nabla f = 0$ implies $y = -x$??

 

[HallsofIvy]

I did not go through the entire post but if "$$\nabla f= 0$$ implies y= -x" is the only condition then the critical points are all the points on the line y= -x.

 

[pasmith]

Problem Statement: Consider the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f(x,y) := \ln{\left(3+(x+y)^2\right)}$.

1. Find $\nabla f$, the gradient of $f$, and determine at which points $\nabla f$ is zero.

2. Determine whether the critical points of $f$ are local minima, local maxima, or saddle points by considering the level curves of $f$.

3. Calculate all the second order mixed partial derivatives of $f$.
Relevant Equations: Gradient vector $\nabla f$.

There are sets of the form $\left\{(x,y)\in \mathbb{R}^2: f(x,y) = \ln{\left(3+(x+y)^2\right)} = c\right\}$ where $c$ is some fixed number $> 1$.

Where does ">1" come from? I agree that there are no points where $f(x,y) = 1$, but the actual minimum of $f$ is slightly higher than this.

Lets see what happens for a few values of $c$.

Or you could solve the general case...

Suppose $c = 2$, then $\ln{\left(3+(x+y)^2\right)} = 2 \Longleftrightarrow (x+y)^2 = e^2-3$, or $x+y = \pm\sqrt{e^2-3}$ which are straight lines of gradient $-1$ through $\left(\pm\sqrt{e^2-3},0\right)$ and $\left(0,\pm\sqrt{e^2-3}\right)$.

For $c = 1$, then $\ln{\left(3+(x+y)^2\right)} = 3 \Longleftrightarrow (x+y)^2 = e^3-3$, or $x+y = \pm\sqrt{e^3-3}$ which are straight lines of gradient $-1$ through $\left(\pm\sqrt{e^3-3},0\right)$ and $\left(0,\pm\sqrt{e^3-3}\right)$.

Is this for $c = 1$ as stated, or $c = 3$ which is the calculation you have actually done?

1. We find the partial derivatives of $f$ with respect to $x$ and $y$ to get $\frac{\partial f}{\partial x} = f_x = \frac{2(x+y)}{3+(x+y)^2}$ and $\frac{\partial f}{\partial y} = f_y = f_x = \frac{2(x+y)}{3+(x+y)^2}$. This makes the gradient vector
$$\nabla{f} = \begin{bmatrix}
f_x \\
f_y
\end{bmatrix} = \begin{bmatrix}
\frac{2(x+y)}{3+(x+y)^2} \\[6pt]
\frac{2(x+y)}{3+(x+y)^2}
\end{bmatrix}.$$

Equating these to zero gives the critical points and doing so gives $2(x+y) = 0$, or $y = -x$. Hence, the critical point is the line $y = -x$.

"Hence every point on the line $y = -x$ is critical."

2. From the figure above we see that the level curves seem to increase the further we get away from the critical points which suggests that they are local minima.

Is this suggestion correct? Are they minima or not?

3. We have $f_x = \frac{2(x+y)}{3+(x+y)^2}$ which means $f_{xx} = \frac{\left(2(3+(x+y)^2\right)-4(x+y)^2}{\left(3+(x+y)^2\right)^2}$. Similarly we have $f_y = 2ye^{x^2+y^2}3$ which means $f_{yy} = \frac{\left(2(3+(x+y)^2\right)-4(x+y)^2}{\left(3+(x+y)^2\right)^2}$. Finally, we know that $f_{xy} = f_{yx} = \frac{\left(2(3+(x+y)^2\right)-4(x+y)^2}{\left(3+(x+y)^2\right)^2}$ since the (Hessian) matrix of second order partial derivatives is symmetric.

Note that all elements of the Hessian are equal. This requires that the Hessian have a zero eigenvalue.

My question is what are the critical points if $\nabla f = 0$ implies $y = -x$??

You have a line of degenerate critical points.