A vector derivative tells us how a point moves along a curve. We can now use it in familiar calculus operations: differentiate products and compositions, integrate to recover motion, and add up speed to measure distance traveled.

Differentiation rules

Let $\mathbf u(t)$ and $\mathbf v(t)$ be differentiable vector-valued functions, let $f(t)$ be a differentiable scalar-valued function, and let $c$ be a constant real number. The usual sum, product, and chain rules still work. The main extra care is deciding which kind of multiplication is involved.

Sums and scalar multiples

Differentiate sums component by component. A constant scalar stays outside the derivative:

$$\frac{d}{dt}\bigl(\mathbf u(t)+\mathbf v(t)\bigr) =\mathbf u'(t)+\mathbf v'(t),$$ $$\frac{d}{dt}\bigl(c\mathbf u(t)\bigr)=c\mathbf u'(t).$$

If the scalar changes with $t$, use the product rule instead:

$$\boxed{ \frac{d}{dt}\bigl(f(t)\mathbf u(t)\bigr) =f'(t)\mathbf u(t)+f(t)\mathbf u'(t). }$$

Each term on the right is a scalar times a vector, so the result is a vector.

Dot and cross products

A dot product is scalar-valued. Its derivative is

$$\boxed{ \frac{d}{dt}\bigl(\mathbf u(t)\cdot\mathbf v(t)\bigr) =\mathbf u'(t)\cdot\mathbf v(t) +\mathbf u(t)\cdot\mathbf v'(t). }$$

For vectors in $\mathbb R^3$, a cross product is vector-valued:

$$\boxed{ \frac{d}{dt}\bigl(\mathbf u(t)\times\mathbf v(t)\bigr) =\mathbf u'(t)\times\mathbf v(t) +\mathbf u(t)\times\mathbf v'(t). }$$

Keep the factors in their original order in both terms. Swapping the factors of a cross product changes its sign; unlike a dot product, it is not commutative.

The chain rule

In a composition $\mathbf u(f(t))$, the scalar function $f$ changes the input to the vector function $\mathbf u$. Assume the values of $f$ lie in the domain of $\mathbf u$. Then

$$\boxed{ \frac{d}{dt}\mathbf u(f(t))=\mathbf u'(f(t))f'(t). }$$

The final multiplication is scalar multiplication, not a dot product: $\mathbf u'(f(t))$ is a vector and $f'(t)$ is a real number.

To see why the rule works, write $\mathbf u(t)=\langle x(t),y(t),z(t)\rangle$. Applying the ordinary chain rule to each component gives

$$\begin{aligned} \frac{d}{dt}\mathbf u(f(t)) &=\frac{d}{dt}\langle x(f(t)),y(f(t)),z(f(t))\rangle\\ &=\langle x'(f(t))f'(t),y'(f(t))f'(t),z'(f(t))f'(t)\rangle\\ &=\mathbf u'(f(t))f'(t). \end{aligned}$$

Example: changing the parameter of a helix

Take

$$\mathbf u(t)=\langle\cos t,\sin t,t\rangle, \qquad f(t)=t^2.$$

The composition is

$$\mathbf u(f(t))=\langle\cos(t^2),\sin(t^2),t^2\rangle.$$

First differentiate the outer function, then evaluate that derivative at $t^2$, and finally multiply by $2t$:

$$\begin{aligned} \frac{d}{dt}\mathbf u(t^2) &=\langle-\sin(t^2),\cos(t^2),1\rangle(2t)\\ &=\boxed{\langle-2t\sin(t^2),2t\cos(t^2),2t\rangle.} \end{aligned}$$

Differentiating the three components directly gives the same answer. The factor $2t$ belongs in every component, including the last one.

Constant length and perpendicular velocity

If the length of a position vector stays constant, its endpoint cannot move toward or away from the origin. Its velocity must point along the sphere instead.

More precisely, if $\mathbf r$ is differentiable and $\lVert\mathbf r(t)\rVert=c$ is constant, then

$$\boxed{\mathbf r(t)\cdot\mathbf r'(t)=0.}$$

Thus $\mathbf r(t)$ and $\mathbf r'(t)$ are orthogonal. A vector can change direction even when its length does not change; constant length does not mean constant vector.

Constant distance from the origin

The dot-product rule makes the proof short. Instead of differentiating a square root, differentiate the squared length:

$$\mathbf r(t)\cdot\mathbf r(t)=c^2.$$

Because $c$ is constant,

$$\begin{aligned} 0 &=\frac{d}{dt}\bigl(\mathbf r(t)\cdot\mathbf r(t)\bigr)\\ &=\mathbf r'(t)\cdot\mathbf r(t) +\mathbf r(t)\cdot\mathbf r'(t)\\ &=2\mathbf r(t)\cdot\mathbf r'(t). \end{aligned}$$

Dividing by $2$ gives the result. When $c>0$, the curve lies on the sphere of radius $c$ centered at the origin. If $\mathbf r'(t)\ne\mathbf0$, it gives a tangent direction there. The dot-product identity still holds when the derivative is zero, but the zero vector has no direction. If $c=0$, the position vector is identically zero.

Example: a curve on the unit sphere

Consider

$$\mathbf r(t)=\langle\cos^2t,\cos t\sin t,\sin t\rangle.$$

Its components do not look like a simple circle, but its squared length simplifies:

$$\begin{aligned} \lVert\mathbf r(t)\rVert^2 &=\cos^4t+\cos^2t\sin^2t+\sin^2t\\ &=\cos^2t(\cos^2t+\sin^2t)+\sin^2t\\ &=\cos^2t+\sin^2t\\ &=1. \end{aligned}$$

So its endpoint stays on the unit sphere, and the theorem tells us immediately that $\mathbf r(t)\cdot\mathbf r'(t)=0$. There is no need to expand the derivative just to establish orthogonality.

Integrating vector-valued functions

Integration works component by component too. If $\mathbf v(t)=\langle v_1(t),v_2(t),v_3(t)\rangle$ and $V_1$, $V_2$, and $V_3$ are antiderivatives of its components, then

$$\int\mathbf v(t)\,dt =\langle V_1(t),V_2(t),V_3(t)\rangle+\mathbf C,$$

where $\mathbf C=\langle C_1,C_2,C_3\rangle$ is an arbitrary constant vector. Each component has its own constant; there is no reason for the three constants to be equal.

For continuous components on $[a,b]$, the definite integral is

$$\int_a^b\mathbf v(t)\,dt =\left\langle \int_a^b v_1(t)\,dt, \int_a^b v_2(t)\,dt, \int_a^b v_3(t)\,dt \right\rangle.$$

The Fundamental Theorem of Calculus (FTC) applies componentwise. In particular, if $\mathbf r$ is continuously differentiable on $[a,b]$, then

$$\boxed{\int_a^b\mathbf r'(t)\,dt=\mathbf r(b)-\mathbf r(a).}$$

For motion, integrating velocity gives displacement: the change from the initial position to the final position. This is a vector, not the total distance traveled along the curve.

Example: recover position from acceleration

A particle moves for $t\ge0$ with acceleration

$$\mathbf a(t)=\mathbf r''(t) =\left\langle\cos t,0,\frac1{(1+t)^2}\right\rangle.$$

Its initial velocity and position are

$$\mathbf v(0)=\langle0,1,-1\rangle, \qquad \mathbf r(0)=\langle1,1,0\rangle,$$

where $\mathbf v=\mathbf r'$ is velocity and $\mathbf a=\mathbf v'$ is acceleration. We need two integrations: one to find velocity, then another to find position. Each integration introduces a new constant vector.

First integrate acceleration:

$$\mathbf v(t) =\left\langle \sin t+C_1,\ C_2,\ -\frac1{1+t}+C_3 \right\rangle.$$

Set $t=0$ and match the components to the initial velocity:

$$\mathbf v(0)=\langle C_1,C_2,C_3-1\rangle =\langle0,1,-1\rangle,$$

so $C_1=0$, $C_2=1$, and $C_3=0$. Therefore

$$\mathbf v(t)=\left\langle\sin t,1,-\frac1{1+t}\right\rangle.$$

Now integrate velocity, using fresh constants:

$$\mathbf r(t)=\langle-\cos t+C_4,t+C_5,-\ln(1+t)+C_6\rangle.$$

Because $t\ge0$, we have $1+t>0$, so the logarithm does not need absolute-value bars on this interval. Apply the initial position:

$$\mathbf r(0)=\langle C_4-1,C_5,C_6\rangle =\langle1,1,0\rangle.$$

This gives $C_4=2$, $C_5=1$, and $C_6=0$, hence

$$\boxed{ \mathbf r(t)=\langle2-\cos t,t+1,-\ln(1+t)\rangle, \qquad t\ge0. }$$

Check the answer in both directions. At $t=0$ it gives the required initial position. Differentiating once gives the velocity above, including its initial value, and differentiating twice gives the specified acceleration. In particular, the second position component must be $t+1$: its derivative has to be the constant velocity $1$.

Arc length: add up speed

To find how far a particle travels, we need the length along the curve, not just the distance between its endpoints. A particle can return to its starting point with zero displacement after traveling a nonzero distance.

Let $\mathbf r(t)$ be a continuously differentiable curve for $a\le t\le b$, with $a<b$. Its speed is $\lVert\mathbf r'(t)\rVert$. Over a short forward interval from $t_1$ to $t_2$, set $\Delta t=t_2-t_1>0$. If speed changes little over that interval, then

$$\text{distance over the short interval} \approx\lVert\mathbf r'(t_1)\rVert\Delta t.$$

This is the familiar “speed times time” calculation. It is exact at constant speed and an approximation when we use the speed at the start of a varying-speed interval.

A short distance is approximately speed times time

Divide the full interval into $n$ equal pieces, with endpoints $t_i=a+i\Delta t$ and $\Delta t=(b-a)/n$. Adding the short-distance estimates gives

$$L\approx\sum_{i=1}^{n}\lVert\mathbf r'(t_{i-1})\rVert\Delta t.$$

As $n\to\infty$, the pieces become smaller and this Riemann sum approaches the arc length of the parametrized motion:

$$\boxed{L=\int_a^b\lVert\mathbf r'(t)\rVert\,dt.}$$

In three dimensions, this is

$$L=\int_a^b\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}\,dt.$$

The order of operations matters: take the length of the velocity vector before integrating. Compare

$$\underbrace{\int_a^b\mathbf r'(t)\,dt}_{\text{displacement vector}} \qquad\text{and}\qquad \underbrace{\int_a^b\lVert\mathbf r'(t)\rVert\,dt}_{\text{distance traveled}}.$$

If the parametrization retraces part of the curve, the length integral counts that travel again. It measures the full motion, not just the length of the set of distinct points visited.

For the helix $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$, the speed is the constant $\sqrt2$. One turn, from $t=0$ to $t=2\pi$, therefore has length

$$L=\int_0^{2\pi}\sqrt2\,dt=2\pi\sqrt2.$$

Here speed times time gives the exact answer because the speed is constant.