A weather map with wind arrows, the current in a river, the pull of the Earth on a satellite: each one attaches a vector to every point of a region. You already know vectors and functions of several variables. A vector field puts the two together: a function whose input is a point and whose output is a vector.
Key idea
The components $P$ and $Q$ are ordinary scalar functions, so everything you know about continuity and partial derivatives applies to each one. A field is continuous when its components are, and it is "smooth" when its components have continuous partial derivatives. Most fields in this course are smooth except perhaps at a few points, such as the origin for the field of a point charge.
Some fields to know. The rotation field $\langle -y, x \rangle$ circles the origin counterclockwise. The radial field $\langle x, y \rangle$ points straight away from the origin. A constant field such as $\langle 3, -1 \rangle$ is the same arrow everywhere, like a steady wind. Newton's law of gravitation gives an inverse-square field: a mass $M$ at the origin pulls a mass $m$ at $\mathbf{r} = \langle x, y, z \rangle$ with force $$\mathbf{F} = -\frac{GMm}{|\mathbf{r}|^3}\,\mathbf{r},$$ which points toward the origin and has length $GMm/|\mathbf{r}|^2$.
Why it works
Why the rotation field circles: $\langle -y, x \rangle \cdot \langle x, y \rangle = -xy + yx = 0$, so each arrow is perpendicular to the line from the origin, and $|\langle -y, x \rangle| = \sqrt{x^2 + y^2}$ is the distance to the origin. Arrows tangent to circles, longer farther out.
Worked examples
Worked example: Evaluating a field
- Substitute: $P(2, -1) = 2 - (-1) = 3$ and $Q(2, -1) = 2 \cdot (-1) = -2$.
- So the arrow at $(2, -1)$ is $\langle 3, -2 \rangle$, pointing right and down.
- Its length is $\sqrt{9 + 4} = \sqrt{13}$. In a velocity field, that is the speed there.
Answer: $\langle 3, -2 \rangle$, length $\sqrt{13}$
Worked example: Sketching the rotation field
- At $(1, 0)$: $\langle 0, 1 \rangle$, up. At $(0, 1)$: $\langle -1, 0 \rangle$, left. At $(-1, 0)$: $\langle 0, -1 \rangle$, down. At $(0, -1)$: $\langle 1, 0 \rangle$, right.
- At $(2, 0)$: $\langle 0, 2 \rangle$, twice as long as at $(1, 0)$.
- The arrows go around the origin counterclockwise, getting longer with distance, as the Why box shows.
Answer: counterclockwise circles, length equal to the distance from the origin
Worked example: Checking a flow line
- $\mathbf{r}'(t) = \langle -2\sin t, 2\cos t \rangle$.
- $$\mathbf{F}(\mathbf{r}(t)) = \langle -y, x \rangle = \langle -2\sin t, 2\cos t \rangle.$$
- They are equal for every $t$, so yes: a particle in this flow goes around the circle of radius ${}2$, counterclockwise.
Answer: yes
Common mistake