3.2 Calculus of Vector-Valued Functions

Big idea

The difference quotient that defines the derivative of an ordinary function makes sense without modification for a vector-valued function. Subtracting two vectors is defined, dividing a vector by a small number is defined, and the limit of a vector-valued expression is defined componentwise. So the derivative of $\mathbf{r}$ is the limit of $\left[\mathbf{r}(t+h) - \mathbf{r}(t)\right]/h$, and nothing in that formula had to be invented for this chapter.

Working out what the limit equals produces the result that makes the subject practical: differentiate each component separately. A vector-valued function is three ordinary functions in a container, and the derivative reaches into the container and differentiates each one. That means every technique from single-variable calculus is immediately available, and the only genuinely new content is what the answer means.

What it means is a direction. The numerator $\mathbf{r}(t+h) - \mathbf{r}(t)$ is a chord of the curve, running from one position to a nearby one. Dividing by $h$ rescales it without turning it. As $h$ shrinks, the chord pivots toward the tangent line, so $\mathbf{r}'(t)$ points along the curve in the direction of increasing $t$. Its length measures how fast the point is moving, which is why the next two sections can treat $\mathbf{r}'$ as a velocity without further explanation.

The differentiation rules follow the same pattern as before, with one addition. There are now two ways to multiply vectors, so there are two product rules, and the cross product version must keep its factors in order because the cross product does not commute. Integration is equally componentwise, and an indefinite integral of a vector-valued function carries an arbitrary constant vector, not an arbitrary constant.

Decoder

The derivative of a vector-valued function is obtained by differentiating each component, and when it is nonzero it is a vector tangent to the curve, pointing in the direction of increasing parameter.

The word “tangent” is the payoff. For a real-valued function the derivative is a slope, a number attached to a picture only by convention. Here the derivative is itself an arrow you can draw at the point $\mathbf{r}(t)$, lying along the curve. Dividing it by its own length gives the unit tangent vector, which records the direction of travel with the speed stripped out.

The qualifier “when it is nonzero” is not decoration. A parametrization can slow to a stop and turn a corner, and at such a parameter value the derivative is $\mathbf{0}$ and determines no direction at all. Curves whose derivative is continuous and never zero are called smooth, and the results of the next two sections assume smoothness for exactly this reason.

The classic mistake is carrying over the wrong intuition about magnitudes. Differentiating does not commute with taking length: $\|\mathbf{r}(t)\|$ and $\|\mathbf{r}'(t)\|$ are unrelated quantities, and the derivative of the first is not the second. A point moving on a sphere has constant $\|\mathbf{r}\|$, so the derivative of its length is zero, while its velocity is certainly not. What is true in that situation is more interesting: constant length forces $\mathbf{r}$ and $\mathbf{r}'$ to be orthogonal.

Definitions and results

The derivative. For a vector-valued function $\mathbf{r}$,

$$ \mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h} $$

when the limit exists. If $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ with all three components differentiable, then

$$ \mathbf{r}'(t) = \langle f'(t), \; g'(t), \; h'(t) \rangle $$

Tangent vectors. If $\mathbf{r}'(t_0) \ne \mathbf{0}$, it is tangent to the curve at $\mathbf{r}(t_0)$. The tangent line there is the line through that point with direction $\mathbf{r}'(t_0)$, so it has vector equation $\mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0)$.

Smooth curves. A parametrization is smooth on an interval when $\mathbf{r}'$ is continuous and never $\mathbf{0}$ there. Smoothness rules out corners and stopping points, and it guarantees that the unit tangent vector below is defined everywhere on the interval.

Unit tangent vector. For a smooth curve,

$$ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} $$

It has length $1$ at every parameter value and depends only on the direction of travel, not on the speed.

Basic rules. For differentiable vector-valued $\mathbf{u}$, $\mathbf{v}$, a constant $c$, a differentiable real-valued $f$, and a constant vector $\mathbf{c}$:

$$ \frac{d}{dt}\mathbf{c} = \mathbf{0}, \qquad \frac{d}{dt}\left[ c\,\mathbf{u} \right] = c\,\mathbf{u}', \qquad \frac{d}{dt}\left[ \mathbf{u} \pm \mathbf{v} \right] = \mathbf{u}' \pm \mathbf{v}' $$

$$ \frac{d}{dt}\left[ f(t)\,\mathbf{u}(t) \right] = f'(t)\,\mathbf{u}(t) + f(t)\,\mathbf{u}'(t), \qquad \frac{d}{dt}\, \mathbf{u}\big(f(t)\big) = f'(t)\, \mathbf{u}'\big(f(t)\big) $$

The dot product rule. The derivative of a dot product is a scalar function:

$$ \frac{d}{dt}\left[ \mathbf{u}(t) \cdot \mathbf{v}(t) \right] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t) $$

The cross product rule. The derivative of a cross product is a vector function, and the order of the factors must be preserved in both terms:

$$ \frac{d}{dt}\left[ \mathbf{u}(t) \times \mathbf{v}(t) \right] = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t) $$

Constant magnitude. If $\|\mathbf{r}(t)\|$ is constant, then $\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0$ for all $t$. The proof is one line: differentiate $\mathbf{r} \cdot \mathbf{r} = \text{constant}$ with the dot product rule to get $2\,\mathbf{r} \cdot \mathbf{r}' = 0$. Applied to $\mathbf{T}$, which always has length $1$, this says $\mathbf{T}'$ is orthogonal to $\mathbf{T}$, the fact the next section builds on.

Integrals. Antiderivatives and definite integrals are also computed componentwise:

$$ \int \mathbf{r}(t)\,dt = \left\langle \int f(t)\,dt, \; \int g(t)\,dt, \; \int h(t)\,dt \right\rangle + \mathbf{C} $$

where $\mathbf{C}$ is an arbitrary constant vector, and the definite integral over $[a,b]$ is the vector of the three definite integrals. An initial condition determines $\mathbf{C}$ completely.

Worked examples

A derivative and a unit tangent vector

Let $\mathbf{r}(t) = \langle t^2, \; 2t, \; \ln t \rangle$ for $t > 0$. Differentiating each component,

$$ \mathbf{r}'(t) = \left\langle 2t, \; 2, \; \frac{1}{t} \right\rangle $$

The middle component is never zero, so $\mathbf{r}'$ never vanishes and the curve is smooth on $t > 0$. At $t = 1$,

$$ \mathbf{r}'(1) = \langle 2, 2, 1 \rangle, \qquad \|\mathbf{r}'(1)\| = \sqrt{4 + 4 + 1} = 3, \qquad \mathbf{T}(1) = \left\langle \tfrac{2}{3}, \tfrac{2}{3}, \tfrac{1}{3} \right\rangle $$

The unit tangent has length $1$, since $\tfrac{4}{9} + \tfrac{4}{9} + \tfrac{1}{9} = 1$.

The tangent line to a space curve

Using the same curve, the point at $t = 1$ is $\mathbf{r}(1) = \langle 1, 2, 0 \rangle$, since $\ln 1 = 0$. With direction $\langle 2,2,1 \rangle$, the tangent line is

$$ x = 1 + 2s, \qquad y = 2 + 2s, \qquad z = s $$

Check that the line touches the curve where it should: $s = 0$ gives $(1,2,0)$, which is the point on the curve. Any nonzero multiple of $\mathbf{r}'(1)$ would serve equally well as the direction, so using $\mathbf{T}(1)$ instead simply reparametrizes the same line by arc length.

Verifying the dot product rule

Let $\mathbf{u}(t) = \langle t, 1, t^2 \rangle$ and $\mathbf{v}(t) = \langle t^2, -2t, 3 \rangle$.

Compute the dot product first and differentiate afterward:

$$ \mathbf{u} \cdot \mathbf{v} = t^3 - 2t + 3t^2, \qquad \frac{d}{dt}\left[ \mathbf{u} \cdot \mathbf{v} \right] = 3t^2 + 6t - 2 $$

Now use the rule instead. With $\mathbf{u}' = \langle 1, 0, 2t \rangle$ and $\mathbf{v}' = \langle 2t, -2, 0 \rangle$,

$$ \mathbf{u}' \cdot \mathbf{v} = t^2 + 0 + 6t, \qquad \mathbf{u} \cdot \mathbf{v}' = 2t^2 - 2 + 0 $$

Adding gives $3t^2 + 6t - 2$, the same answer. The rule is not a shortcut here, since the product was easy to expand, but for a cross product or for functions you cannot expand in closed form it is the only route.

Recovering a function from its derivative

Suppose $\mathbf{r}'(t) = \langle 4t, \; 3t^2, \; e^{-t} \rangle$ and $\mathbf{r}(0) = \langle 1, -2, 0 \rangle$. Antidifferentiate componentwise:

$$ \mathbf{r}(t) = \left\langle 2t^2 + C_1, \; t^3 + C_2, \; -e^{-t} + C_3 \right\rangle $$

The initial condition forces $C_1 = 1$, $C_2 = -2$ and $-1 + C_3 = 0$, so $C_3 = 1$. Therefore

$$ \mathbf{r}(t) = \left\langle 2t^2 + 1, \; t^3 - 2, \; 1 - e^{-t} \right\rangle $$

Differentiating this back gives $\langle 4t, 3t^2, e^{-t} \rangle$, and substituting $t = 0$ gives $\langle 1, -2, 0 \rangle$, so both conditions hold. A definite integral works the same way; for instance

$$ \int_0^1 \left\langle 2t, \; 3t^2, \; \frac{1}{1+t^2} \right\rangle dt = \left\langle 1, \; 1, \; \frac{\pi}{4} \right\rangle $$

since each component is evaluated by the fundamental theorem on its own.

Practice

These problems exercise the calculus of this section: differentiating componentwise, finding tangent vectors, unit tangent vectors and tangent lines, checking smoothness, applying the scalar, dot and cross product rules, and integrating vector-valued functions with and without initial conditions.

Quiz

Five items on componentwise differentiation, tangent and unit tangent vectors, the product rules, and antiderivatives and definite integrals of vector-valued functions.