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Vector-Valued Functions

A progressive guide to representing curves with vector-valued functions and using limits, derivatives, integrals, and geometric quantities to analyze motion in the plane and space.

Representing Curves with Vectors

A assigns a vector to each real parameter value. In two dimensions and three dimensions, common forms are

r(t)=⟨x(t),y(t)⟩\mathbf r(t)=\langle x(t),y(t)\rangle

and

r(t)=⟨x(t),y(t),z(t)⟩.\mathbf r(t)=\langle x(t),y(t),z(t)\rangle.

The functions x(t)x(t), y(t)y(t), and z(t)z(t) are ordinary scalar component functions. As tt varies, the point represented by r(t)\mathbf r(t) traces a .

The parameter can have a geometric and a physical meaning. In motion problems it often represents time, so r(t)\mathbf r(t) gives position. Even when two parameterizations trace the same geometric path, they may move along it at different speeds or in different directions.

Takeaway: Read a component by component, while remembering that the parameter also controls how the curve is traversed.

Limits, Continuity, and Integration

Limits, continuity, derivatives, and integrals of vector-valued functions are handled componentwise. For

r(t)=⟨f(t),g(t),h(t)⟩,\mathbf r(t)=\langle f(t),g(t),h(t)\rangle,

the is

lim⁡t→ar(t)=⟨lim⁡t→af(t),lim⁡t→ag(t),lim⁡t→ah(t)⟩,\lim_{t\to a}\mathbf r(t)=\left\langle\lim_{t\to a}f(t),\lim_{t\to a}g(t),\lim_{t\to a}h(t)\right\rangle,

provided all three component limits exist. For example,

r(t)=⟨t2+1,sin⁡tt,et⟩\mathbf r(t)=\left\langle t^2+1,\frac{\sin t}{t},e^t\right\rangle

has

lim⁡t→0r(t)=⟨1,1,1⟩.\lim_{t\to 0}\mathbf r(t)=\langle 1,1,1\rangle.

A is continuous at t=at=a when

lim⁡t→ar(t)=r(a).\lim_{t\to a}\mathbf r(t)=\mathbf r(a).

This occurs exactly when every component function is continuous at aa. Thus, a function formed from familiar continuous functions is continuous wherever all of its components are defined.

Integration is also componentwise:

∫⟨f(t),g(t),h(t)⟩ dt=⟨∫f(t) dt,∫g(t) dt,∫h(t) dt⟩.\int\langle f(t),g(t),h(t)\rangle\,dt =\left\langle\int f(t)\,dt,\int g(t)\,dt,\int h(t)\,dt\right\rangle.

Takeaway: A vector limit or continuity statement succeeds only when every component satisfies the corresponding scalar condition.

Differentiating Vector-Valued Functions

The derivative is defined by the vector difference quotient

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

when the limit exists. Computing it componentwise gives

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

for r(t)=⟨f(t),g(t),h(t)⟩\mathbf r(t)=\langle f(t),g(t),h(t)\rangle.

For example, if

r(t)=⟨3t+1,t2−4t,cos⁡t⟩,\mathbf r(t)=\langle 3t+1,t^2-4t,\cos t\rangle,

then

r′(t)=⟨3,2t−4,−sin⁡t⟩\mathbf r'(t)=\langle 3,2t-4,-\sin t\rangle

and

r′′(t)=⟨0,2,−cos⁡t⟩.\mathbf r''(t)=\langle 0,2,-\cos t\rangle.

The ordinary differentiation rules extend to vector expressions. For differentiable vector-valued functions u(t)\mathbf u(t) and v(t)\mathbf v(t),

ddt[u+v]=u′+v′,\frac{d}{dt}[\mathbf u+\mathbf v]=\mathbf u'+\mathbf v',
ddt[fu]=f′u+fu′,\frac{d}{dt}[f\mathbf u]=f'\mathbf u+f\mathbf u',
ddt[u⋅v]=u′⋅v+u⋅v′,\frac{d}{dt}[\mathbf u\cdot\mathbf v]=\mathbf u'\cdot\mathbf v+\mathbf u\cdot\mathbf v',

and

ddt[u×v]=u′×v+u×v′.\frac{d}{dt}[\mathbf u\times\mathbf v]=\mathbf u'\times\mathbf v+\mathbf u\times\mathbf v'.

If r(t)⋅r(t)=c\mathbf r(t)\cdot\mathbf r(t)=c for a constant cc, differentiation gives

2r(t)⋅r′(t)=0,2\mathbf r(t)\cdot\mathbf r'(t)=0,

so r(t)\mathbf r(t) is perpendicular to r′(t)\mathbf r'(t).

Position, Velocity, Acceleration, and

For a particle with position r(t)=⟨x(t),y(t),z(t)⟩\mathbf r(t)=\langle x(t),y(t),z(t)\rangle, the is

v(t)=r′(t)=⟨x′(t),y′(t),z′(t)⟩,\mathbf v(t)=\mathbf r'(t)=\langle x'(t),y'(t),z'(t)\rangle,

and the is

a(t)=v′(t)=r′′(t)=⟨x′′(t),y′′(t),z′′(t)⟩.\mathbf a(t)=\mathbf v'(t)=\mathbf r''(t)=\langle x''(t),y''(t),z''(t)\rangle.

Consider the plane motion

r(t)=⟨t2,3t−t3⟩.\mathbf r(t)=\langle t^2,3t-t^3\rangle.

Then

v(t)=⟨2t,3−3t2⟩\mathbf v(t)=\langle 2t,3-3t^2\rangle

and

a(t)=⟨2,−6t⟩.\mathbf a(t)=\langle 2,-6t\rangle.

At t=1t=1,

r(1)=⟨1,2⟩,v(1)=⟨2,0⟩,a(1)=⟨2,−6⟩.\mathbf r(1)=\langle 1,2\rangle,\qquad \mathbf v(1)=\langle 2,0\rangle,\qquad \mathbf a(1)=\langle 2,-6\rangle.

The particle is at (1,2)(1,2), moving horizontally to the right. Its acceleration has a rightward component and a downward component.

The is the magnitude of the :

∥v(t)∥=[x′(t)]2+[y′(t)]2+[z′(t)]2.\lVert\mathbf v(t)\rVert=\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2}.

For a plane velocity ⟨−3,4⟩\langle -3,4\rangle, the is

(−3)2+42=5.\sqrt{(-3)^2+4^2}=5.

A constant does not imply zero acceleration. If is constant, then differentiating v(t)⋅v(t)=constant\mathbf v(t)\cdot\mathbf v(t)=\text{constant} yields

v(t)⋅a(t)=0.\mathbf v(t)\cdot\mathbf a(t)=0.

Therefore, acceleration is perpendicular to velocity whenever acceleration changes direction without changing .

Tangent Lines and Direction

When r′(t0)≠0\mathbf r'(t_0)\neq\mathbf 0, the derivative at t0t_0 supplies a to the curve at r(t0)\mathbf r(t_0). The tangent line can be written as

L(s)=r(t0)+sr′(t0),\mathbf L(s)=\mathbf r(t_0)+s\mathbf r'(t_0),

where ss is a new real parameter.

For a plane curve with

r(t0)=⟨x0,y0⟩andr′(t0)=⟨a,b⟩,\mathbf r(t_0)=\langle x_0,y_0\rangle \quad\text{and}\quad \mathbf r'(t_0)=\langle a,b\rangle,

the tangent line is

⟨x,y⟩=⟨x0,y0⟩+s⟨a,b⟩.\langle x,y\rangle=\langle x_0,y_0\rangle+s\langle a,b\rangle.

If a≠0a\neq 0, its slope form is

y−y0=ba(x−x0).y-y_0=\frac{b}{a}(x-x_0).

This agrees with the parametric derivative

dydx=y′(t)x′(t),x′(t)≠0.\frac{dy}{dx}=\frac{y'(t)}{x'(t)}, \qquad x'(t)\neq 0.

The removes the from the velocity:

T(t)=r′(t)∥r′(t)∥.\mathbf T(t)=\frac{\mathbf r'(t)}{\lVert\mathbf r'(t)\rVert}.

For

r(t)=⟨2cos⁡t,2sin⁡t⟩\mathbf r(t)=\langle 2\cos t,2\sin t\rangle

at t=π4t=\frac{\pi}{4},

r(π4)=⟨2,2⟩,\mathbf r\left(\frac{\pi}{4}\right)=\langle\sqrt 2,\sqrt 2\rangle,
r′(π4)=⟨−2,2⟩,\mathbf r'\left(\frac{\pi}{4}\right)=\langle-\sqrt 2,\sqrt 2\rangle,

and ∥r′(t)∥=2\lVert\mathbf r'(t)\rVert=2. Thus,

L(s)=⟨2,2⟩+s⟨−2,2⟩\mathbf L(s)=\langle\sqrt 2,\sqrt 2\rangle+s\langle-\sqrt 2,\sqrt 2\rangle

and

T(π4)=⟨−12,12⟩.\mathbf T\left(\frac{\pi}{4}\right)=\left\langle-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}\right\rangle.

Takeaway: Use the derivative for the tangent direction, divide by its magnitude for a unit tangent, and use the point-plus-direction form for the tangent line.

Recovering Motion and Solving Problems

Known acceleration can be integrated to recover velocity, and known velocity can be integrated to recover position. With initial data at t=t0t=t_0,

v(t)=v(t0)+∫t0ta(u) du\mathbf v(t)=\mathbf v(t_0)+\int_{t_0}^{t}\mathbf a(u)\,du

and

r(t)=r(t0)+∫t0tv(u) du.\mathbf r(t)=\mathbf r(t_0)+\int_{t_0}^{t}\mathbf v(u)\,du.

For example, suppose

a(t)=⟨2,6t⟩,v(0)=⟨1,−2⟩,r(0)=⟨3,4⟩.\mathbf a(t)=\langle 2,6t\rangle, \qquad \mathbf v(0)=\langle 1,-2\rangle, \qquad \mathbf r(0)=\langle 3,4\rangle.

Integrating acceleration first gives

v(t)=⟨2t+C1,3t2+C2⟩.\mathbf v(t)=\langle 2t+C_1,3t^2+C_2\rangle.

The initial velocity gives C1=1C_1=1 and C2=−2C_2=-2, so

v(t)=⟨2t+1,3t2−2⟩.\mathbf v(t)=\langle 2t+1,3t^2-2\rangle.

Integrating again,

r(t)=⟨t2+t+D1,t3−2t+D2⟩.\mathbf r(t)=\langle t^2+t+D_1,t^3-2t+D_2\rangle.

The initial position gives D1=3D_1=3 and D2=4D_2=4. Therefore,

r(t)=⟨t2+t+3,t3−2t+4⟩.\mathbf r(t)=\langle t^2+t+3,t^3-2t+4\rangle.

A reliable workflow for motion problems is:

  1. Identify the position function r(t)\mathbf r(t).

  2. Differentiate once to find v(t)\mathbf v(t).

  3. Differentiate twice to find a(t)\mathbf a(t).

  4. Evaluate the relevant vectors at the requested parameter value.

  5. Compute with ∥v(t)∥\lVert\mathbf v(t)\rVert.

  6. Use r′(t)\mathbf r'(t) as a when it is nonzero.

  7. Divide by to obtain the .

  8. Interpret negative components as decreasing motion in the corresponding coordinate direction.