M344: Calculus III
Section 16.1
Up to this point in Calculus III we have studied
functions whose domain is $\mathbb{R}$ and codomain is $\mathbb{R}^n$ (Chapter 13), and
functions whose domain is $\mathbb{R}^n$ and codomain is $\mathbb{R}$ (Chapters 14 and 15).
The logical conclusion to this course is thus the study of functions whose domain is $\mathbb{R}^n$ and codomain is $\mathbb{R}^m$.
We'll spend almost all of our efforts on the case when $m = n$. In this case we will regard the input of the function as points in $\mathbb{R}^n$ and the output of the function as $n$-vectors in $\mathbb{V}^n$.
Vector Fields¶
Let $D \subseteq \mathbb{R}^n$. A vector field on $\mathbb{R}^n$ is a function
$$
\mathbf{F} :\mathbb{R}^n \to \mathbb{V}^n :\mathbf{x} \mapsto \mathbf{F}(\mathbf{x}),
$$
where $\mathbf{x} \in \mathbb{R}^n$ is regarded as a point in $D$, despite being written in vector notation. We will usually regard the output vector $\mathbf{F}(\mathbf{x})$ as being "attached" at $\mathbf{x}$; that is, $\mathbf{F}(\mathbf{x})$ is translated so that its initial point is $\mathbf{x}$.
Special Cases¶
Suppose $D \subseteq \mathbb{R}^2$. Then a vector field $\mathbf{F}$ on $D$ can be decomposed as
$$
\mathbf{F}(x,y) = \left\langle P(x,y), Q(x,y) \right\rangle,
$$
where $P$ and $Q$ are real-valued functions on $D$.
Suppose $D \subseteq \mathbb{R}^3$. Then a vector field $\mathbf{F}$ on $D$ can be decomposed as
$$
\mathbf{F}(x,y,z) = \left\langle P(x,y,z), Q(x,y,z), R(x,y,z) \right\rangle,
$$
where $P$, $Q$, and $R$ are real-valued functions on $D$.
Video lecture
1. Example¶
Consider the vector field on $\mathbb{R}^2$ defined by
$$
\mathbf{F}(x,y) = y\ \mathbf{i} + (x+y)\ \mathbf{j}.
$$
Plug in some points in the plane and plot the corresponding vectors to get an idea of how this vector field behaves.
Solution
We start by making a table of a few points and corresponding vectors.
| $\mathbf{x}$ | $\mathbf{F}(\mathbf{x})$ | $\mathbf{x}$ | $\mathbf{F}(\mathbf{x})$ |
|---|---|---|---|
| $(0,0)$ | $\langle 0, 0 \rangle$ | $\left(-\tfrac{1}{2},\tfrac{1}{2}\right)$ | $\left\langle -\tfrac{1}{2}, 0 \right\rangle$ |
| $(1,0)$ | $\langle 0,1 \rangle$ | $(-1,0)$ | $\langle 0,-1 \rangle$ |
| $\left(\tfrac{1}{2},\tfrac{1}{2}\right)$ | $\left\langle \tfrac{1}{2}, 1 \right\rangle$ | $\left(-\tfrac{1}{2},-\tfrac{1}{2}\right)$ | $\left\langle -\tfrac{1}{2}, -1 \right\rangle$ |
| $\left( 1,0 \right)$ | $\left\langle 1,1 \right\rangle$ |
$(0,-1)$ | $\langle -1, -1 \rangle$ |
Then we plot the vector field.
2. Example¶
Newton's Law of Gravitation states that the magnitude of the gravitational force between two objects whose masses are $m$ and $M$ (both positive constants) is given by
$$
\big\Vert\mathbf{F}(r)\big\Vert = \frac{mMG}{r^2},
$$
where $\mathbf{F}$ is the gravitational force field, $G$ is the gravitational constant, and $r$ is the distance between the centers of mass of the two objects.
Work out a formula for the gravitational field $\mathbf{F}$ exerted by an object of mass $M$ whose center of mass is the origin, on an object of mass $m$ whose center of mass is located at the point $(x,y,z)$ in $3$-dimensional space.
Check your answer
$$ \mathbf{F}(x,y,z) = \frac{-mMG}{r^3}\left\langle x,y,z \right\rangle, $$
where $r = \sqrt{x^2 + y^2 + z^2}$.
Video solution
Gradient Fields¶
One of the most important classes of examples of a vector field is one that we have already begun to study.
Suppose $f$ is a differentiable function on a domain $D \subseteq \mathbb{R}^2$. The gradient vector field along $f$ (or corresponding to $f$) is the vector field on $D$ defined by
$$
\nabla f(x,y) = \left\langle \frac{\partial f}{\partial x}(x,y), \frac{\partial f}{\partial y} (x,y) \right\rangle,
$$
or more succinctly, $\nabla f = \langle \partial_x f,\partial_y f \rangle$.
If $f$ is a function defined on a domain in $\mathbb{R}^3$, its gradient vector field is defined analogously. It is given by
$$
\nabla f = \langle \partial_x f,\partial_y f,\partial_z f \rangle.
$$
The vectors of a gradient field are orthogonal to the level curves of the function $f$.
3. Example¶
Plot the gradient field and a few level curves for the function
$$
f(x,y) = \tfrac{1}{2}\left( x^2 - y^2 \right).
$$
Solution
In the image below, the red curves are some level curves of the function $f$. The blue vectors are the gradient vectors at various points. Notice that the gradient vectors appear to be perpendicular (orthogonal) to the closest level curves at each point.
Moreover, the levels of the level curves are evenly spaced. Notice that the gradient vectors are "longer" (they have a greater magnitude) when the level curves are closer together.
4. Example¶
Suppose a particle is moving in a velocity field $\mathbf{V}$. At any point $(x,y,z)$ in the domain, the particle's velocity is given by
$$
\mathbf{V}(x,y,z) = \left\langle x^2, z + y^2, 2xyz \right\rangle.
$$
If a particle is dropped into the velocity field (with $\mathbf{0}$ acceleration) at the point $(3,2,1)$, estimate the particle's location after $0.1$ seconds.
Video solution
5. Example¶
Given the vector field
$$
\mathbf{F}(x,y) = \left\langle 3y\sin x + 3xy\cos x, 3x\sin x \right\rangle,
$$
is it possible to find a function $f$ such that $\mathbf{F} = \nabla f$? If not, why? If so, find one.
Video solution
Discussion¶
Questions? You can ask them here.