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.

Example: one turn of a helix

For the helix $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$, one turn corresponds to $0\le t\le2\pi$. Its velocity is $\mathbf r'(t)=\langle-\sin t,\cos t,1\rangle$, so

$$\lVert\mathbf r'(t)\rVert =\sqrt{(-\sin t)^2+(\cos t)^2+1} =\sqrt2.$$

The speed is constant, so speed times time gives the exact length:

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

Example: tracing a circle twice

Consider

$$\mathbf r(t)=\langle\cos(2t),\sin(2t)\rangle, \qquad 0\le t\le2\pi.$$

The points lie on the unit circle, but the angle $2t$ runs from $0$ to $4\pi$. The particle makes two full turns, not one.

One circle, two trips around it

Differentiating and taking the norm gives

$$\mathbf r'(t)=\langle-2\sin(2t),2\cos(2t)\rangle,$$ $$\lVert\mathbf r'(t)\rVert =\sqrt{(-2\sin(2t))^2+(2\cos(2t))^2} =2.$$

Therefore

$$L=\int_0^{2\pi}2\,dt=4\pi.$$

The circle has circumference $2\pi$, but the distance traveled is $4\pi$. The same geometric curve can give different travel distances when the parameter interval includes different numbers of trips around it.

Example: the circumference of an ellipse

Let $a>0$ and $b>0$ be the horizontal and vertical semiaxis lengths of the ellipse

$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1.$$

One complete traversal is

$$\mathbf r(t)=\langle a\cos t,b\sin t\rangle, \qquad 0\le t\le2\pi.$$

The two semiaxes of an ellipse

Its derivative is $\mathbf r'(t)=\langle-a\sin t,b\cos t\rangle$, so its circumference is

$$\boxed{L=\int_0^{2\pi}\sqrt{a^2\sin^2t+b^2\cos^2t}\,dt.}$$

Setting up an arc-length integral is often easier than evaluating it. For unequal positive semiaxes, this integrand has no elementary antiderivative. The integral is an elliptic integral; we can still evaluate it numerically or express it using special functions. An exact integral is a valid answer even when ordinary integration techniques do not simplify it.

If $a=b$, the ellipse is a circle. The speed simplifies to $a$, and the circumference becomes $2\pi a$.

The arc-length function

The total length $L$ uses a fixed endpoint. To track how much distance has accumulated as the endpoint moves, define the arc-length function

$$\boxed{s(t)=\int_0^t\lVert\mathbf r'(u)\rVert\,du.}$$

Here $0$ is the chosen starting parameter and $u$ is a dummy integration variable. Assume $\mathbf r$ is continuously differentiable on the interval being considered. For $t\ge0$, $s(t)$ is the distance traveled along the curve from parameter $0$ to parameter $t$.

The Fundamental Theorem of Calculus gives

$$\boxed{s'(t)=\lVert\mathbf r'(t)\rVert.}$$

In words, the rate at which distance accumulates is speed. For the helix above,

$$s(t)=\int_0^t\sqrt2\,du=\sqrt2\,t.$$

If we allow $t<0$, the same integral is a signed arc-length coordinate relative to the starting point, not a nonnegative distance. For this helix, the distance along the curve between parameters $0$ and $t$ is $\sqrt2\,|t|$.

Parametrizing by arc length

A parameter usually tells us when a point is reached, not how far along the curve it lies. We can instead choose a parameter that increases by one whenever we travel one unit of length.

A curve is parametrized by arc length, or has a unit-speed parametrization, when the norm of its derivative with respect to its parameter is $1$. The name of the parameter does not matter: if it is still called $t$, the condition is

$$\lVert\mathbf r'(t)\rVert=1.$$

With starting parameter $0$, its arc-length function is then

$$s(t)=\int_0^t\lVert\mathbf r'(u)\rVert\,du =\int_0^t1\,du=t.$$

More generally, the length traveled from parameter $a$ to parameter $b$, with $a<b$, is simply

$$\boxed{L=\int_a^b1\,dt=b-a.}$$

Changing to an arc-length parameter

Starting from a continuously differentiable curve with positive speed throughout the interval:

1. Compute the arc-length function $s(t)$.
2. Solve $s=s(t)$ for $t$ as a function of $s$. Write the inverse as $t=t(s)$.
3. Substitute this inverse into the original curve to get $\mathbf R(s)=\mathbf r(t(s))$.

Positive speed makes $s'(t)>0$, so $s(t)$ is strictly increasing and can be inverted on the interval. Finding an explicit formula for that inverse may still be difficult.

The chain rule explains why this change produces unit speed. Since $ds/dt=\lVert\mathbf r'(t)\rVert$, we have

$$\frac{dt}{ds}=\frac1{\lVert\mathbf r'(t(s))\rVert}, \qquad \mathbf R'(s) =\frac{\mathbf r'(t(s))}{\lVert\mathbf r'(t(s))\rVert}.$$

The new derivative is the original velocity divided by its length, so $\lVert\mathbf R'(s)\rVert=1$. The curve's points and direction of travel stay the same; its speed with respect to the new parameter is now $1$.

Example: a unit-speed helix

For $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$, we already found $s(t)=\sqrt2\,t$. Invert this relation:

$$s=\sqrt2\,t \quad\Longrightarrow\quad t=\frac{s}{\sqrt2}.$$

Substitute into every component of the original function:

$$\boxed{ \mathbf R(s)=\left\langle \cos\left(\frac{s}{\sqrt2}\right), \sin\left(\frac{s}{\sqrt2}\right), \frac{s}{\sqrt2} \right\rangle. }$$

To check the speed directly, differentiate with respect to $s$:

$$\mathbf R'(s)=\left\langle -\frac1{\sqrt2}\sin\left(\frac{s}{\sqrt2}\right), \frac1{\sqrt2}\cos\left(\frac{s}{\sqrt2}\right), \frac1{\sqrt2} \right\rangle.$$

Each component carries a factor $1/\sqrt2$ from the chain rule. Its norm is

$$\begin{aligned} \lVert\mathbf R'(s)\rVert &=\sqrt{ \frac12\sin^2\left(\frac{s}{\sqrt2}\right) +\frac12\cos^2\left(\frac{s}{\sqrt2}\right) +\frac12}\\ &=\sqrt{\frac12+\frac12}=1. \end{aligned}$$

For $t\in\mathbb R$ and $s\in\mathbb R$, the two parametrizations describe the same entire helix, but the original has speed $\sqrt2$ and the new one has speed $1$. One turn uses $0\le t\le2\pi$ in the original parameter and $0\le s\le2\pi\sqrt2$ in the arc-length parameter. The latter interval's length is exactly the length of that turn.