Math Core

Lesson 2.2 · Vector-Valued Functions

Derivatives and integrals of vector functions

Once a curve is described by a vector function, calculus tells you which way the curve is heading at each point and how to rebuild a path from its rate of change. The good news is that almost nothing new is needed: you differentiate and integrate one component at a time, and the geometry comes along for free.

The derivative as a limit of secant vectors

The derivative of a vector function is defined by the familiar difference quotient:

r′(t)=lim⁡h→0r(t+h)−r(t)h.\mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h}.

Geometrically, r(t+h)−r(t)\mathbf{r}(t + h) - \mathbf{r}(t) is the vector from one point of the curve to a nearby point, a secant vector. Dividing by hh rescales it (and keeps it pointing forward even when hh is negative). As h→0h \to 0, the secant vector swings into the direction the curve is heading. So when r′(t)\mathbf{r}'(t) exists and is not 0\mathbf{0}, it is a tangent vector to the curve at the tip of r(t)\mathbf{r}(t), pointing in the direction of increasing tt.

Since subtraction, scaling and limits all happen componentwise, so does the derivative.

Differentiate each component

If r(t)=⟨f(t),g(t),h(t)⟩\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle with ff, gg, hh differentiable, then

r′(t)=⟨f′(t),  g′(t),  h′(t)⟩.\mathbf{r}'(t) = \langle f'(t),\; g'(t),\; h'(t) \rangle.

When r′(t)≠0\mathbf{r}'(t) \ne \mathbf{0}, the unit tangent vector is T(t)=r′(t)∣r′(t)∣\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}, and the tangent line at t=at = a is L(s)=r(a)+s r′(a)\mathbf{L}(s) = \mathbf{r}(a) + s\,\mathbf{r}'(a).

A curve is called smooth on an interval when r′\mathbf{r}' is continuous and never equal to 0\mathbf{0} there. At a point where r′(t)=0\mathbf{r}'(t) = \mathbf{0} the curve can have a sharp corner or cusp, so there is no well-defined tangent direction. For example, ⟨t2,t3,0⟩\langle t^2, t^3, 0 \rangle has a cusp at the origin, where r′(0)=0\mathbf{r}'(0) = \mathbf{0}.

Worked example: A derivative and a unit tangent

Let r(t)=⟨t2,  2t,  ln⁡t⟩\mathbf{r}(t) = \langle t^2,\; 2t,\; \ln t \rangle. Find r′(t)\mathbf{r}'(t) and T(1)\mathbf{T}(1).

Differentiate each component: r′(t)=⟨2t,  2,  1t⟩\mathbf{r}'(t) = \left\langle 2t,\; 2,\; \dfrac{1}{t} \right\rangle. At t=1t = 1, r′(1)=⟨2,2,1⟩\mathbf{r}'(1) = \langle 2, 2, 1 \rangle, whose length is 4+4+1=3\sqrt{4 + 4 + 1} = 3. So

T(1)=⟨23,  23,  13⟩.\mathbf{T}(1) = \left\langle \tfrac{2}{3},\; \tfrac{2}{3},\; \tfrac{1}{3} \right\rangle.

Worked example: Tangent line to a helix

Find the tangent line to the helix r(t)=⟨2cos⁡t,  2sin⁡t,  3t⟩\mathbf{r}(t) = \langle 2\cos t,\; 2\sin t,\; 3t \rangle at t=π2t = \tfrac{\pi}{2}.

The point is r ⁣(π2)=⟨0,  2,  3π2⟩\mathbf{r}\!\left(\tfrac{\pi}{2}\right) = \left\langle 0,\; 2,\; \tfrac{3\pi}{2} \right\rangle. The derivative is r′(t)=⟨−2sin⁡t,  2cos⁡t,  3⟩\mathbf{r}'(t) = \langle -2\sin t,\; 2\cos t,\; 3 \rangle, so r′ ⁣(π2)=⟨−2,0,3⟩\mathbf{r}'\!\left(\tfrac{\pi}{2}\right) = \langle -2, 0, 3 \rangle. The tangent line is

x=−2s,y=2,z=3π2+3s.x = -2s, \qquad y = 2, \qquad z = \tfrac{3\pi}{2} + 3s.

Notice the tangent line never changes yy: at this point the helix is moving across the front of its cylinder while it climbs.

Higher derivatives are defined the same way: r′′(t)=⟨f′′(t),g′′(t),h′′(t)⟩\mathbf{r}''(t) = \langle f''(t), g''(t), h''(t) \rangle. In the next lessons, r′\mathbf{r}' and r′′\mathbf{r}'' will become velocity and acceleration.

Direction versus length

It helps to separate the two pieces of information packed into r′(t)\mathbf{r}'(t). Its direction, captured by T(t)\mathbf{T}(t), belongs to the curve itself: any parametrization that traces the curve the same way gives the same unit tangent at a given point. Its length ∣r′(t)∣|\mathbf{r}'(t)| depends on the parametrization. If you replace tt by 2t2t, the curve is traced twice as fast, and the chain rule doubles every tangent vector without changing a single direction.

That's why tangent lines are safe to compute from any parametrization: the direction vector might be scaled differently, but the line is the same. It's also why the unit tangent will be the starting point for curvature, which should depend only on the shape of the curve and not on how quickly you move along it.

Differentiation rules

The single-variable rules carry over. If u\mathbf{u} and v\mathbf{v} are differentiable vector functions, cc is a scalar and ff is a real-valued function:

ddt[u+v]=u′+v′ddt[f(t) u(t)]=f′(t) u(t)+f(t) u′(t)ddt[u⋅v]=u′⋅v+u⋅v′ddt[u×v]=u′×v+u×v′ddt[u(f(t))]=f′(t) u′(f(t))\begin{aligned} \frac{d}{dt}\big[\mathbf{u} + \mathbf{v}\big] &= \mathbf{u}' + \mathbf{v}' \\ \frac{d}{dt}\big[f(t)\,\mathbf{u}(t)\big] &= f'(t)\,\mathbf{u}(t) + f(t)\,\mathbf{u}'(t) \\ \frac{d}{dt}\big[\mathbf{u} \cdot \mathbf{v}\big] &= \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}' \\ \frac{d}{dt}\big[\mathbf{u} \times \mathbf{v}\big] &= \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}' \\ \frac{d}{dt}\big[\mathbf{u}(f(t))\big] &= f'(t)\,\mathbf{u}'(f(t)) \end{aligned}

Each can be proved by writing out components and using the ordinary product or chain rule.

Common mistake

In the cross product rule, keep the order of the factors. Because a×b=− b×a\mathbf{a} \times \mathbf{b} = -\,\mathbf{b} \times \mathbf{a}, writing v×u′\mathbf{v} \times \mathbf{u}' instead of u′×v\mathbf{u}' \times \mathbf{v} flips the sign of that term and gives a wrong answer.

The dot product rule has a striking consequence.

Worked example: Constant length means perpendicular derivative

Suppose ∣r(t)∣=c|\mathbf{r}(t)| = c for all tt, where cc is a constant. Show that r′(t)\mathbf{r}'(t) is perpendicular to r(t)\mathbf{r}(t).

Since r(t)⋅r(t)=∣r(t)∣2=c2\mathbf{r}(t) \cdot \mathbf{r}(t) = |\mathbf{r}(t)|^2 = c^2 is constant, its derivative is zero:

0=ddt[r⋅r]=r′⋅r+r⋅r′=2 r⋅r′.0 = \frac{d}{dt}\big[\mathbf{r} \cdot \mathbf{r}\big] = \mathbf{r}' \cdot \mathbf{r} + \mathbf{r} \cdot \mathbf{r}' = 2\,\mathbf{r} \cdot \mathbf{r}'.

So r(t)⋅r′(t)=0\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0. Geometrically: a curve on a sphere centered at the origin always moves perpendicular to the radius, just as the tangent to a circle is perpendicular to the radius.

Integrals of vector functions

Integrals are componentwise too:

∫abr(t) dt=⟨∫abf(t) dt,  ∫abg(t) dt,  ∫abh(t) dt⟩.\int_a^b \mathbf{r}(t)\,dt = \left\langle \int_a^b f(t)\,dt,\; \int_a^b g(t)\,dt,\; \int_a^b h(t)\,dt \right\rangle.

The Fundamental Theorem of Calculus still holds: if R′(t)=r(t)\mathbf{R}'(t) = \mathbf{r}(t), then ∫abr(t) dt=R(b)−R(a)\displaystyle\int_a^b \mathbf{r}(t)\,dt = \mathbf{R}(b) - \mathbf{R}(a). An indefinite integral has a constant vector C\mathbf{C}, which amounts to one constant per component. A single initial condition like r(0)=⟨1,2,3⟩\mathbf{r}(0) = \langle 1, 2, 3 \rangle pins down all three constants at once.

Worked example: Recovering a curve from its derivative

Find r(t)\mathbf{r}(t) if r′(t)=⟨3t2,  cos⁡t,  e−t⟩\mathbf{r}'(t) = \langle 3t^2,\; \cos t,\; e^{-t} \rangle and r(0)=⟨1,−2,4⟩\mathbf{r}(0) = \langle 1, -2, 4 \rangle.

Integrate each component:

r(t)=⟨t3,  sin⁡t,  −e−t⟩+C.\mathbf{r}(t) = \left\langle t^3,\; \sin t,\; -e^{-t} \right\rangle + \mathbf{C}.

At t=0t = 0 this gives ⟨0,0,−1⟩+C=⟨1,−2,4⟩\langle 0, 0, -1 \rangle + \mathbf{C} = \langle 1, -2, 4 \rangle, so C=⟨1,−2,5⟩\mathbf{C} = \langle 1, -2, 5 \rangle. Therefore

r(t)=⟨t3+1,  sin⁡t−2,  5−e−t⟩.\mathbf{r}(t) = \left\langle t^3 + 1,\; \sin t - 2,\; 5 - e^{-t} \right\rangle.

Check: differentiating gives back ⟨3t2,cos⁡t,e−t⟩\langle 3t^2, \cos t, e^{-t} \rangle, and r(0)=⟨1,−2,4⟩\mathbf{r}(0) = \langle 1, -2, 4 \rangle.

Tip

After finding C\mathbf{C}, plug t=0t = 0 (or whatever the initial time is) back into your final answer. A sign slip in one component shows up immediately.

Practice

Practice 1

Let r(t)=⟨t3,  2t2,  5t⟩\mathbf{r}(t) = \langle t^3,\; 2t^2,\; 5t \rangle. Find r′(1)\mathbf{r}'(1).

Enter a point like (2, -3)

Practice 2

Let r(t)=⟨e2t,  cos⁡3t,  ln⁡(1+t)⟩\mathbf{r}(t) = \langle e^{2t},\; \cos 3t,\; \ln(1 + t) \rangle. Find r′(0)\mathbf{r}'(0).

Enter a point like (2, -3)

Practice 3

Find the unit tangent vector T(2)\mathbf{T}(2) for r(t)=⟨2t,  t2,  13t3⟩\mathbf{r}(t) = \left\langle 2t,\; t^2,\; \tfrac{1}{3}t^3 \right\rangle.

Enter a point like (2, -3)

Practice 4

Find the point where the tangent line to r(t)=⟨t2,  3t,  t3+1⟩\mathbf{r}(t) = \langle t^2,\; 3t,\; t^3 + 1 \rangle at t=1t = 1 crosses the xyxy-plane.

Enter a point like (2, -3)

Practice 5

Evaluate ∫01⟨4t3,  11+t,  πcos⁡πt2⟩dt\displaystyle\int_0^1 \left\langle 4t^3,\; \frac{1}{1 + t},\; \pi\cos\frac{\pi t}{2} \right\rangle dt. Enter ln⁡2\ln 2 as ln(2).

Enter a point like (2, -3)

Practice 6

Suppose r′(t)=⟨2t,  6t2,  et⟩\mathbf{r}'(t) = \langle 2t,\; 6t^2,\; e^t \rangle and r(0)=⟨1,−1,3⟩\mathbf{r}(0) = \langle 1, -1, 3 \rangle. Find r(1)\mathbf{r}(1).

Enter a point like (2, -3)

Practice 7

Suppose u(0)=⟨1,2,−1⟩\mathbf{u}(0) = \langle 1, 2, -1 \rangle, u′(0)=⟨3,0,4⟩\mathbf{u}'(0) = \langle 3, 0, 4 \rangle, v(0)=⟨2,1,1⟩\mathbf{v}(0) = \langle 2, 1, 1 \rangle and v′(0)=⟨−1,5,2⟩\mathbf{v}'(0) = \langle -1, 5, 2 \rangle. Find ddt[u(t)⋅v(t)]\dfrac{d}{dt}\big[\mathbf{u}(t) \cdot \mathbf{v}(t)\big] at t=0t = 0.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

With the same u\mathbf{u} and v\mathbf{v} as the previous problem, find ddt[u(t)×v(t)]\dfrac{d}{dt}\big[\mathbf{u}(t) \times \mathbf{v}(t)\big] at t=0t = 0.

Enter a point like (2, -3)