Vector Field Formula for Graph in [-2,2] x [-2,2]

[carl123]

Give an example of a formula for a vector field whose graph would closely resemble the one shown. The box for this figure is [−2, 2] x [−2, 2].View attachment 4930

Not sure where to start.

 

[I like Serena]

Give an example of a formula for a vector field whose graph would closely resemble the one shown. The box for this figure is [−2, 2] x [−2, 2].
Not sure where to start.

Hi carl123! (Smile)

Typically we can write a vector field as the gradient of some $f(x,y)$.
From the graph given, there are 2 singular points at $x=\pm 1$.
Can you think of a $f(x,y)$ such that those are indeed singular points? (Wondering)

 

[carl123]

Will it be f (x,y) = f (-2,2)?

 

[I like Serena]

Will it be f (x,y) = f (-2,2)?

Huh? That's not a formula for f(x,y). (Worried)

Consider that for instance $\frac{1}{x-1}$ is a formula where $x=1$ is a singular point.

 

[carl123]

Huh? That's not a formula for f(x,y). (Worried)

Consider that for instance $\frac{1}{x-1}$ is a formula where $x=1$ is a singular point.

where x = 1 is a singular point, the denominator will be 0 then.

 

[I like Serena]

where x = 1 is a singular point, the denominator will be 0 then.

Exactly! (Nod)

 

[carl123]

Exactly! (Nod)

if the denominator is 0, the formula will be undefined. What does that say about the graph?

Also, when I found the cross product of [-2,2] and [-2,2], I got 0. Does that mean, there's no field that resembles the graph?

Thanks.

 

[I like Serena]

if the denominator is 0, the formula will be undefined. What does that say about the graph?

Suppose we pick $f(x,y)=\frac{y}{x-1}$.
Then the gradient will be:
$$\begin{pmatrix}\pd f x\\\pd f y\end{pmatrix} =\begin{pmatrix}-\frac{y}{(x-1)^2}\\\frac{1}{x-1}\end{pmatrix}$$

We can see the resulting vector plot here.

Also, when I found the cross product of [-2,2] and [-2,2], I got 0. Does that mean, there's no field that resembles the graph?

How would that cross product be related?

 

[carl123]

Suppose we pick $f(x,y)=\frac{y}{x-1}$.

How would that cross product be related?

Their cross product is equal

 

[topsquark]

Also, when I found the cross product of [-2,2] and [-2,2], I got 0. Does that mean, there's no field that resembles the graph?

Thanks.

[-2,2] x [-2,2] is not the cross product of two vectors. (How can you take the cross product of two vectors in 2D space??) It represents the domain of the function: \(\displaystyle x \in [ -2,2 ] \) and \(\displaystyle y \in [-2,2] \).

-Dan