A vector function ${\bf r}(t)=\langle f(t),g(t),h(t)\rangle$ is a function of one variable—that is, there is only one "input'' value. What makes vector functions more complicated than the functions $y=f(x)$ that we studied in the first part of this book is of course that the "output'' values are now three-dimensional vectors instead of simply numbers. It is natural to wonder if there is a corresponding notion of derivative for vector functions. In the simpler case of a function $y=s(t)$, in which $t$ represents time and $s(t)$ is position on a line, we have seen that the derivative $s'(t)$ represents velocity; we might hope that in a similar way the derivative of a vector function would tell us something about the velocity of an object moving in three dimensions.

One way to approach the question of the derivative for vector functions is to write down an expression that is analogous to the derivative we already understand, and see if we can make sense of it. This gives us \eqalign{ {\bf r}'(t)&=\lim_{\Delta t\to0}{{\bf r}(t+\Delta t)-{\bf r}(t)\over \Delta t}\cr &=\lim_{\Delta t\to0}{\langle f(t+\Delta t)-f(t),g(t+\Delta t)-g(t), h(t+\Delta t)-h(t)\rangle\over \Delta t}\cr &=\lim_{\Delta t\to0}\langle {f(t+\Delta t)-f(t)\over\Delta t}, {g(t+\Delta t)-g(t)\over\Delta t}, {h(t+\Delta t)-h(t)\over \Delta t}\rangle\cr &=\langle f'(t),g'(t),h'(t)\rangle,\cr } if we say that what we mean by the limit of a vector is the vector of the individual coordinate limits. So starting with a familiar expression for what appears to be a derivative, we find that we can make good computational sense out of it—but what does it actually mean?

We know how to interpret ${\bf r}(t+\Delta t)$ and ${\bf r}(t)$—they are vectors that point to locations in space; if $t$ is time, we can think of these points as positions of a moving object at times that are $\Delta t$ apart. We also know what $\Delta {\bf r}= {\bf r}(t+\Delta t)-{\bf r}(t)$ means—it is a vector that points from the head of ${\bf r}(t)$ to the head of ${\bf r}(t+\Delta t)$, assuming both have their tails at the origin. So when $\Delta t$ is small, $\Delta {\bf r}$ is a tiny vector pointing from one point on the path of the object to a nearby point. As $\Delta t$ gets close to 0, this vector points in a direction that is closer and closer to the direction in which the object is moving; geometrically, it approaches a vector tangent to the path of the object at a particular point.

Figure 13.2.1. Approximating the derivative.

Unfortunately, the vector $\Delta{\bf r}$ approaches 0 in length; the vector $\langle 0,0,0\rangle$ is not very informative. By dividing by $\Delta t$, when it is small, we effectively keep magnifying the length of $\Delta{\bf r}$ so that in the limit it doesn't disappear. Thus the limiting vector $\langle f'(t),g'(t),h'(t)\rangle$ will (usually) be a good, non-zero vector that is tangent to the curve.

What about the length of this vector? It's nice that we've kept it away from zero, but what does it measure, if anything? Consider the length of one of the vectors that approaches the tangent vector: $$\left|{{\bf r}(t+\Delta t)-{\bf r}(t)\over \Delta t}\right|={|{\bf r}(t+\Delta t)-{\bf r}(t)|\over|\Delta t|}$$ The numerator is the length of the vector that points from one position of the object to a "nearby'' position; this length is approximately the distance traveled by the object between times $t$ and $t+\Delta t$. Dividing this distance by the length of time it takes to travel that distance gives the average speed. As $\Delta t$ approaches zero, this average speed approaches the actual, instantaneous speed of the object at time $t$.

So by performing an "obvious'' calculation to get something that looks like the derivative of ${\bf r}(t)$, we get precisely what we would want from such a derivative: the vector ${\bf r}'(t)$ points in the direction of travel of the object and its length tells us the speed of travel. In the case that $t$ is time, then, we call ${\bf v}(t)={\bf r}'(t)$ the velocity vector. Even if $t$ is not time, ${\bf r}'(t)$ is useful—it is a vector tangent to the curve.

Example 13.2.1 We have seen that ${\bf r}=\langle \cos t,\sin t,t\rangle$ is a helix. We compute ${\bf r}'=\langle -\sin t,\cos t,1\rangle$, and $|{\bf r}'|=\sqrt{\sin^2 t+\cos^2 t+1}=\sqrt2$. So thinking of this as a description of a moving object, its speed is always $\sqrt2$; see figure 13.2.2. $\square$