10.2 Parametric curves and vector functions

A moving point carries more information than its path

An equation such as $x^2+y^2=R^2$ describes a collection of points. It does not say where a motion starts, which direction it takes, or how quickly it moves. A parametrization adds this information by assigning a point to each value of a parameter. When the parameter represents time, the same geometric path can support several different motions.

We begin with continuity, which prevents the point from jumping as the parameter changes. Differentiability is an additional requirement, used when we want an instantaneous velocity. A curve need not be differentiable merely because it is continuous. Keeping the definition separate from this later requirement avoids excluding legitimate continuous examples such as the graph of an arbitrary continuous function.

Curves, components, and traces

Definition

Parametric curve and trace

Let $I$ be an interval and $n$ a positive integer. A parametric curve in $\mathbb R^n$ is a continuous function

$$\gamma:I\longrightarrow\mathbb R^n,\qquad \gamma(t)=(x_1(t),\ldots,x_n(t)).$$

The functions $x_i$ are its coordinate components. The trace, or image, is

$$\gamma(I)=\{\gamma(t):t\in I\}.$$

Thus $\gamma$ is the function specifying the parametrization, while $\gamma(I)$ is the set of points it visits.

The parameter interval is part of the definition. Restricting the interval may restrict the trace, or may remove repeated visits without changing the trace. The parameter is a single real number even when the output has many coordinates. A vector function is another name for this vector-valued function; it is not a separate collection of unrelated scalar motions.

Theorem

Continuity can be checked componentwise

The function $\gamma:I\to\mathbb R^n$ is continuous at $t_0\in I$ if and only if every component $x_i$ is continuous there. Limits at an endpoint are taken within the interval.

For any vector $\mathbf z=(z_1,\ldots,z_n)$, the Euclidean norm satisfies

$$|z_i|\le\|\mathbf z\|\le\sqrt n\max_{1\le j\le n}|z_j|.$$

The first inequality follows because $z_i^2$ is one nonnegative term of the sum defining the squared norm. For the second, each of the $n$ squared components is at most the square of the largest absolute component. Apply these inequalities to $\gamma(t)-\gamma(t_0)$. If its norm tends to zero, every coordinate difference does. Conversely, finitely many coordinate differences tending to zero make their maximum tend to zero, so the norm does too. This proves both directions rather than assuming that vector limits automatically behave coordinatewise.

Worked example

Eliminate a parameter without losing its interval

Let $\gamma:[0,1]\to\mathbb R^2$ be $\gamma(t)=(t,2t)$. Its components are continuous, and

$$\gamma(0)=(0,0),\qquad \gamma(1)=(1,2).$$

Since $x=t$ and $y=2t$, elimination gives $y=2x$. But $t\in[0,1]$ also gives $0\le x\le1$. The trace is the line segment joining the stated endpoints, not the entire line. Conversely, every point $(x,2x)$ with $0\le x\le1$ occurs by taking $t=x$, so the description is exact. Increasing the parameter moves from $(0,0)$ to $(1,2)$.

Concept lensStructural

The parametrization and the image answer different questions

The function $\gamma$ answers “where is the point at this parameter value?” Its image answers “which points occur at least once?” Passing to the image forgets the order of visits, their timing, and how often they occur. In the segment example, the equation $y=2x$ with its interval describes the image; the two component formulas also specify the motion along it. To identify a parametric curve, record the map and its domain, not only an eliminated coordinate equation.

Closed curves and simple curves

Definition

Closedness and simplicity on a closed interval

Let $\gamma:[a,b]\to\mathbb R^n$ with $a\lt b$. It is closed if $\gamma(a)=\gamma(b)$. We call it simple if

$$\gamma(t_1)\ne\gamma(t_2) \qquad a\le t_1\lt t_2\le b,$$

except that the pair $t_1=a$, $t_2=b$ is allowed to have equal images. A simple closed curve satisfies both conditions.

Under this convention a simple curve with distinct endpoints is injective. A simple closed curve is injective apart from its single allowed endpoint identification. The exception permits one complete circuit without declaring that its meeting endpoints form a self-intersection. It does not permit an interior point to be visited twice. These definitions concern the chosen parametrization, not just the appearance of its trace.

For example, traversing a circle twice produces the same circular trace as one traversal, but the double traversal is not simple. Closedness alone says only that the final point is the initial point; it does not limit what happens between them. A constant curve on a nondegenerate closed interval is closed but not simple, because many distinct interior parameters have the same image.

A circle with a specified orientation

Worked example

One anticlockwise circuit

For $R>0$, define

$$\gamma(t)=(R\cos t,R\sin t),\qquad 0\le t\le2\pi.$$

Its trace lies on $x^2+y^2=R^2$, by the identity $\cos^2t+\sin^2t=1$. Conversely every point of that circle has a polar angle in this interval, so every point is visited. The initial and final points are $(R,0)$, making the curve closed.

If two parameter values give the same coordinates, their sine and cosine both agree, so their difference is an integer multiple of $2\pi$. Within this interval the only distinct such pair is $0,2\pi$. The curve is therefore simple under the stated endpoint convention. From $(R,0)$ it initially moves into the upper half-plane; the parameter is the increasing polar angle, so the traversal is anticlockwise.

Translating the center gives the general circle parametrization:

$$\gamma(t)=(h+R\cos t,k+R\sin t),\qquad 0\le t\le2\pi,$$

where $h,k\in\mathbb R$ and $R>0$. Subtracting the center coordinates shows that its trace is $(x-h)^2+(y-k)^2=R^2$. Translation changes the position of every visited point without changing their order or the parameter interval. The radius condition excludes the collapsed constant curve, whose simplicity properties would differ.

Derivative, velocity, and speed

Definition

Derivative of a vector function

At an interior point $t_0$ of the parameter interval, define

$$\gamma'(t_0)=\lim_{h\to0}\frac{\gamma(t_0+h)-\gamma(t_0)}{h},$$

if this vector limit exists. If $t$ represents time, $\gamma'(t)$ is the velocity vector and $\|\gamma'(t)\|$ is the speed.

Here the limit uses nonzero $h$ for which both parameters belong to the domain. At an included endpoint we use the corresponding one-sided derivative when needed. A differentiable curve in this note has derivatives at all interior points, with endpoint derivatives understood one-sided when stated. This convention prevents a formula on a closed interval from silently requesting values outside its domain.

The numerator is displacement; division by elapsed parameter gives average velocity over that interval. Its limit is instantaneous velocity. Speed retains only the length of this vector and therefore has no sign or direction. If the parameter is not physical time, these quantities measure change per unit of parameter; calling them a physical velocity requires that additional interpretation.

Theorem

Componentwise differentiation

The derivative exists at $t_0$ if and only if all component derivatives exist, and then

$$\gamma'(t_0)=(x_1'(t_0),\ldots,x_n'(t_0)),\qquad \|\gamma'(t_0)\|=\sqrt{\sum_{i=1}^n(x_i'(t_0))^2}.$$

For the proof, apply the same norm bounds used for continuity to the vector of difference quotients minus a proposed limiting vector. A vector limit forces each scalar quotient to converge to its corresponding component. Conversely, convergence of all the scalar quotients bounds the vector error by $\sqrt n$ times their largest error, which tends to zero. The finite number of components is essential to this argument. The speed formula then follows from the definition of the Euclidean norm. Exactly the same proof works for a one-sided limit at an endpoint.

For the preceding circle,

$$\gamma'(t)=(-R\sin t,R\cos t),\qquad \|\gamma'(t)\|=R.$$

The velocity changes direction while its norm stays constant. Its dot product with the radius vector $(R\cos t,R\sin t)$ is zero, so it is tangent to the circle. The value $R$ is the speed for this particular parameter choice, not an intrinsic speed possessed by the circular set of points.

The same trace can have different velocities

For $\omega>0$, put

$$\xi(t)=(R\cos(\omega t),R\sin(\omega t)), \qquad 0\le t\le\frac{2\pi}{\omega}.$$

The angle $\omega t$ runs from $0$ to $2\pi$, so this is again exactly one anticlockwise traversal. Differentiation gives

$$\xi'(t)=(-R\omega\sin(\omega t),R\omega\cos(\omega t)), \qquad \|\xi'(t)\|=R\omega.$$

For any nonzero real $\omega$, use the interval $[0,2\pi/|\omega|]$ instead. The speed is then $R|\omega|$; a negative $\omega$ makes the angle decrease and reverses the orientation. The value $\omega=0$ would give a constant map, not a full circle, so the stated one-circuit interval is not defined in that case.

Counterexample mode

The same circular trace does not determine the curve

Compare $\gamma(t)=(R\cos t,R\sin t)$ on $[0,2\pi]$ with $\xi(t)=(R\cos(2t),R\sin(2t))$ on $[0,\pi]$. They have the same trace, but different domains and speeds $R$ and $2R$. Even on their shared parameter interval they are not the same map: at $t=\pi/2$ their values are $(0,R)$ and $(-R,0)$. Equality of traces is therefore weaker than equality of parametric curves.

If the second map were instead given the larger domain $[0,2\pi]$, it would traverse twice. Then $\xi(0)=\xi(\pi)$ is a forbidden repetition for simplicity, because $\pi$ is an interior parameter rather than the final endpoint. Thus the domain also controls multiplicity of traversal.

Worked example

A clockwise unit circle starting at the top

The requested start and end point is $(0,1)$. Choose

$$\eta(t)=(\sin t,\cos t),\qquad 0\le t\le2\pi.$$

This is $(\cos(\pi/2-t),\sin(\pi/2-t))$, so its polar angle decreases from $\pi/2$ by one full turn. Both endpoints are $(0,1)$, and

$$\eta'(t)=(\cos t,-\sin t),\qquad \|\eta'(t)\|=1.$$

At the top, the initial velocity $(1,0)$ points right, confirming clockwise motion. Merely negating both coordinates of the usual circle formula would move the start point to the opposite side without reversing the direction; check the changing angle, rather than judging orientation from one position.

Space curves: the helix

Worked example

Circular motion together with vertical motion

For $R,h>0$, define the space curve

$$\gamma(t)=(R\cos t,R\sin t,ht),\qquad t\in\mathbb R.$$

The first two components stay on $x^2+y^2=R^2$, while the third increases linearly. This is a helix around the vertical axis. After the parameter increases by $2\pi$, its horizontal position repeats but its vertical coordinate has increased by $2\pi h$. It does not return to the same point. In fact, equality of two curve values forces $ht_1=ht_2$, and $h>0$ then forces $t_1=t_2$.

Its velocity and speed are

$$\gamma'(t)=(-R\sin t,R\cos t,h),\qquad \|\gamma'(t)\|=\sqrt{R^2+h^2}.$$

The horizontal and vertical velocity components are perpendicular, so their squares add in the speed. Every restriction to a nondegenerate closed interval has distinct endpoints and is simple. Closedness, as defined above, is a property stated for such a closed parameter interval; the domain $\mathbb R$ itself has no starting and ending parameter to compare.

Segments and graphs as vector functions

Worked example

The complete segment between two points

Let the position vectors of $A,B\in\mathbb R^n$ be $\mathbf a,\mathbf b$. Then

$$\gamma(t)=\mathbf a+t(\mathbf b-\mathbf a) =(1-t)\mathbf a+t\mathbf b,\qquad 0\le t\le1$$

parametrizes their segment. The endpoints follow by substituting $0$ and $1$; the weights are nonnegative and sum to one throughout the interval. Conversely each point of the segment has such a weight, so the whole segment is covered. Its derivative is $\mathbf b-\mathbf a$, with constant speed $\|\mathbf b-\mathbf a\|$.

If $A\ne B$, equality at two parameters implies $(t_1-t_2)(\mathbf b-\mathbf a)=\mathbf0$, hence $t_1=t_2$. It is simple with distinct endpoints. If $A=B$, it is constant and closed, but not simple on $[0,1]$. Extending the parameter to all real values gives the whole line only in the nonconstant case.

If $f:[a,b]\to\mathbb R$ is continuous with $a\lt b$, its graph has parametrization

$$\gamma(t)=(t,f(t)),\qquad t\in[a,b].$$

Continuity follows componentwise. Every graph point occurs at its own first coordinate, and different parameters have different first coordinates, so this curve is simple and has distinct endpoints. If $f$ is differentiable at an interior point, then $\gamma'(t)=(1,f'(t))$ and its speed is $\sqrt{1+(f'(t))^2}$. Without that extra hypothesis, the graph remains a continuous curve but no derivative assertion is justified.

A nonconstant curve can nevertheless have zero velocity at one parameter. For instance, the segment parametrization $\rho(t)=(t^3,2t^3)$ on $[-1,1]$ has derivative $(3t^2,6t^2)$, which is zero at $t=0$. The map is nonconstant and injective because $t^3$ is strictly increasing. A stationary instant therefore does not by itself establish a repeated point or a constant curve.

Quick checks

Checkpoint

What information is lost when a parametrization is replaced by its trace?

Compare a function with its image set.

Solution · Answer

The trace forgets timing, direction, order of visits, and repeated traversals. It retains only the points visited.

Checkpoint

Is one complete traversal of the standard circle simple despite its coincident endpoints?

Use the exact endpoint exception in the definition.

Solution · Answer

Yes. The only distinct parameters with equal images are $0$ and $2\pi$, the permitted endpoint pair. It is simple and closed.

Checkpoint

Does zero speed at one parameter force the whole curve to be constant?

Inspect the cubic segment parametrization.

Solution · Answer

No. $\rho(t)=(t^3,2t^3)$ is nonconstant but has $\rho'(0)=\mathbf0$.

Exercises

1. Describe the exact trace and speed of $\gamma(t)=(t,2t)$ on $[0,1]$.
2. For $R>0$, parametrize the circle $(x-h)^2+(y-k)^2=R^2$ once anticlockwise from $(h+R,k)$, and compute its speed for that parameter choice.
3. Give a clockwise unit-circle parametrization from $(0,1)$ back to itself.
4. For $R>0$ and $\xi(t)=(R\cos(\omega t),R\sin(\omega t))$ with $\omega\ne0$, specify a nonnegative-time interval for one circuit and compute its speed.
5. For the helix with $R,h>0$, find the vertical rise after one revolution and explain why no two parameter values give the same point.
6. For distinct points $A,B$, prove the segment parametrization is injective.
7. Explain which assumptions on $f$ ensure that $(t,f(t))$ is a curve, and which additional assumptions justify its velocity formula.

Guided solutions

Solution · Model solution 1

The trace is $\{(x,2x):0\le x\le1\}$. The derivative is $(1,2)$, so the speed is $\sqrt5$.

Solution · Model solution 2

Use $(h+R\cos t,k+R\sin t)$ for $0\le t\le2\pi$. Its derivative is $(-R\sin t,R\cos t)$ and its speed is $R$.

Solution · Model solution 3

Use $(\sin t,\cos t)$ for $0\le t\le2\pi$; its angle $\pi/2-t$ decreases, and both endpoint values are $(0,1)$.

Solution · Model solution 4

Use $[0,2\pi/|\omega|]$. Differentiation gives speed $R|\omega|$. Positive $\omega$ gives anticlockwise motion; negative $\omega$ gives clockwise.

Solution · Model solution 5

The rise is $2\pi h$. Equality of positions implies $ht_1=ht_2$; division by nonzero $h$ gives $t_1=t_2$.

Solution · Model solution 6

Subtract equal curve values to obtain $(t_1-t_2)(\mathbf b-\mathbf a)=\mathbf0$. At least one component of $\mathbf b-\mathbf a$ is nonzero, so its scalar coefficient is zero.

Solution · Model solution 7

Continuity of $f$ on its interval makes both components continuous. Differentiability at an interior point gives velocity $(1,f'(t))$ there; at included endpoints, any derivative claim uses the corresponding one-sided derivative.