Vector fields

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We introduce the idea of a vector at every point in space.

A review of functions

When we started on our journey exploring calculus, we investigated functions: $f:\R \to \R$ Typically, we interpert these functions as being curves in the $(x,y)$ -plane:

We’ve also studied vector valued functions: $\vec {f}:\R \to \R ^n$ We can interpert these functions as parametric curves in space:

We’ve studied functions of several variables $F:\R ^n \to \R$ We can interpert these functions as surfaces in $\R ^{n+1}$ ,

Now we study vector-valued functions of several variables $\vec {F}:\R ^n\to \R ^n$ and we intererpert these as vector fields, meaning for each point in the $(x,y)$ -plane we have a vector.

Let’s be explicit

A vector field in $\R ^n$ is a function $\vec {F}: \R ^n\to \R ^n$ where for every point in the domain, we assign a vector to the range.

Consider the following table describing a vector field $\vec {F}$ :

What is $\vec {F}(1,7)$ ? $\vec {F}(1,7) = \vector {\answer {1},\answer {-1}}$

What is $\vec {F}(3,5)$ ? $\vec {F}(3,5) = \vector {\answer {0},\answer {2}}$

What is $\vec {F}(2,6)$ ? $\vec {F}(2,6) = \vector {\answer {0},\answer {0}}$

Consider the following vector field:

Which expression is best described by this vector field?

$\vec {F}(x,y)=\vector {x,y/2}$ $\vec {F}(x,y)=1/2$ $\vec {F}(x,y)=x+y/2$ $\vec {F}(x,y)=\vector {1,1/2}$

Note, $\grad (x+y/2) = \vector {1,1/2}$ .

Now we give you a small sampling of some important vector fields.

Radial fields

Let’s see some examples of radial vector fields:

Here we see $\vec {F}(x,y) = \vector {\frac {x}{\sqrt {x^2+y^2}},\frac {y}{\sqrt {x^2+y^2}}}:$

Here we see $\vec {G}(x,y) = \vector {\frac {-x}{x^2+y^2},\frac {-y}{x^2+y^2}}:$

Here we see $\vec {H}(x,y,z) = \vector {\frac {x}{(x^2+y^2+z^2)^2},\frac {y}{(x^2+y^2+z^2)^2},\frac {z}{(x^2+y^2+z^2)^2}}:$

Each of the vector fields above is a radial vector field.

A radial vector field is a field of the form $\vec {F}:\R ^n\to \R ^n$ where $\vec {F}(\vec {x}) = \frac {\pm \vec {x}}{|\vec {x}|^p}$ where $p$ is a real number.

Is $\vec {F}(x,y,z) = \vector {x,y,z}$ a radial vector field?

yes no

Absolutely! This vector field can be rewritten as: $\vector {\frac {x}{(\sqrt {x^2+y^2+z^2})^p},\frac {y}{(\sqrt {x^2+y^2+z^2})^p},\frac {z}{(\sqrt {x^2+y^2+z^2})^p}}$ where $p=\answer {0}$ .

Rotational fields

Vector fields can easily exhibit what looks like “rotation” to the human eye.

Here we see $\vec {F}(x,y) = \vector {-y,x}:$

Here we see $\vec {F}(x,y) = \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}}:$

At this point, we’re going to give some “spoilers.” It turns out that from a local perspective, meaning all points very very close each other, only the first example exhibits “rotation.” While the second example does look like it is “rotating,” as we will see, it does not exhibit “local rotation.”

Gradient fields

We’re going to start with the definition.

Consider any differentiable function $F:\R ^n\to \R$ . A gradient field is a vector field $\vec {G}:\R ^n\to \R ^n$ where $\vec {G} = \grad F.$

Since we are assuming $F$ is differentiable, we are also assuming that $\vec {G}$ is defined for all points in $\R ^n$ . This is important to note, as we will see.

Consider $F(x,y) = \frac {\sin (3x)+\sin (2y)}{1+x^2+y^2}$ . A plot of this function looks like this:

The gradient field of $F$ looks like this:

Note we can see the vector pointing in the initial direction of greatest increase. Let’s see a plot of both together:

What is the connection between gradient vectors and level curves?

Gradient vectors are orthogonal to level curves. Gradient vectors point in the direction of level curves. Gradient vectors are independent of level curves.

Now consider $F(x,y) = \frac {1}{\sqrt {x^2+y^2}}$ . A plot of this function looks like this:

The gradient field of $F$ $\grad F(x,y) = \vector {\frac {-x}{(x^2+y^2)^{3/2}},\frac {-y}{(x^2+y^2)^{3/2}}}$ looks like this:

Note we can see the vector pointing in the initial direction of greatest increase. Let’s see a plot of both together:

The Clairaut gradient test

Now we give a method to determine if a field is a gradient field.

Clairaut A vector field $\vec {F}(x,y) = \vector {M(x,y),N(x,y)}$ , where $M$ and $N$ have continuous partial derivatives, is a gradient field if and only if $\pp [N]{x} -\pp [M]{y} = 0$ for all $x$ and $y$ .

If $\vec {G} = \grad F$ , then $F$ is called a potential function for $\vec {G}$ .

Is $\vec {G} = \vector {2x+y^2,2y+x^2}$ a gradient field? If so find a potential function.

yes no

Is $\vec {G} = \vector {x^3,-y^4}$ a gradient field? If so find a potential function.

yes no

Find a potential function $F$ such that $F(\vec {0}) = 0$ . $F(x,y) = \answer {x^4/4-y^5/5}.$

Is $\vec {G} = \vector {y\cos (x),\sin (x)}$ a gradient field? If so find a potential function.

yes no

Find a potential function $F$ such that $F(\vec {0}) = 0$ . $F(x,y) = \answer {y\sin (x)}.$

Is $\vec {G} = \vector {\frac {-y}{x^2+y^2},\frac {x}{x^2+y^2}}$ a gradient field? If so find a potential function.

yes no

What happens at $\vec {0}$ ?