This chapter is concerned with applying calculus in the context of vector fields. A two-dimensional vector field is a function $f$ that maps each point $(x,y)$ in $\R^2$ to a two-dimensional vector $\langle u,v\rangle$, and similarly a three-dimensional vector field maps $(x,y,z)$ to $\langle u,v,w\rangle$. Since a vector has no position, we typically indicate a vector field in graphical form by placing the vector $f(x,y)$ with its tail at $(x,y)$. Figure 16.1.1 shows a representation of the vector field $f(x,y)=\langle x/\sqrt{x^2+y^2+4},-y/\sqrt{x^2+y^2+4}\rangle$. For such a graph to be readable, the vectors must be fairly short, which is accomplished by using a different scale for the vectors than for the axes. Such graphs are thus useful for understanding the sizes of the vectors relative to each other but not their absolute size.
| Figure 16.1.1. fig:vector field |
Vector fields have many important applications, as they can be used to represent many physical quantities: the vector at a point may represent the strength of some force (gravity, electricity, magnetism) or a velocity (wind speed or the velocity of some other fluid).
We have already seen a particularly important kind of vector field---the gradient. Given a function $f(x,y)$, recall that the gradient is $\langle f_x(x,y),f_y(x,y)\rangle$, a vector that depends on (is a function of) $x$ and $y$. We usually picture the gradient vector with its tail at $(x,y)$, pointing in the direction of maximum increase. Vector fields that are gradients have some particularly nice properties, as we will see. An important example is $$F= \left \langle {-x\over (x^2+y^2+z^2)^{3/2}},{-y\over (x^2+y^2+z^2)^{3/2}},{-z\over (x^2+y^2+z^2)^{3/2}}\right\rangle,$$ which points from the point $(x,y,z)$ toward the origin and has length $${\sqrt{x^2+y^2+z^2}\over(x^2+y^2+z^2)^{3/2}}= {1\over(\sqrt{x^2+y^2+z^2})^2},$$ which is the reciprocal of the square of the distance from $(x,y,z)$ to the origin---in other words, $F$ is an "inverse square law". The vector $\bf F$ is a gradient: $$F = \nabla {1\over\sqrt{x^2+y^2+z^2} }, \eqrdef{eq:inverse square field as gradient} \eqno{(\xrefn{eq:inverse square field as gradient})} $$ which turns out to be extremely useful.
Exercises 16.1
Sketch the vector fields; check your work with Sage's \vb|plot_vector_field| function.
Ex 16.1.1 $\langle x,y\rangle$
Ex 16.1.2 $\langle -x, -y\rangle$
Ex 16.1.3 $\langle x,-y\rangle$
Ex 16.1.4 $\langle \sin x,\cos y\rangle$
Ex 16.1.5 $\langle y,1/x\rangle$
Ex 16.1.6 $\langle x+1,x+3\rangle$
Ex 16.1.7 Verify equation MISSING XREFN(eq:inverse square field as gradient(.