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Section 3.2 Parametric curves in \(\mathbb{R}^n\)

In the previous section we showed how one-forms can be integrated over intervals in \(\mathbb{R}\text{.}\) Our goal is to generalize this construction to curves in \(\mathbb{R}^n\) (we will focus on \(\mathbb{R}^2\) and \(\mathbb{R}^3\) in this course). To this end, we must first study in more details parametric curves in \(\mathbb{R}^n\text{,}\) and the concept of orientation of a parametric curve.

Subsection 3.2.1 Parametric curves

We start by recalling the definition of parametric curves.

Definition 3.2.1. Parametric curves.

A parametric curve in \(\mathbb{R}^n\) is a vector-valued function

\begin{align*} \alpha : [a,b] \amp\to \mathbb{R}^n\\ t \amp \mapsto \alpha(t) = (x_1(t), \ldots, x_n(t)), \end{align*}

such that:

  1. \(\alpha\) can be extended to a \(C^1\)-function 1  on an open set containing \([a,b]\text{;}\)

  2. \(\alpha'(t) \neq \mathbf{0} \) for all \(t \in [a,b]\text{;}\)

  3. if \(\alpha(s) = \alpha(t)\) for any two distinct \(s,t \in [a,b]\text{,}\) then \(s,t \in \{a,b \} \text{.}\)  2  In other words, the map \(\alpha\) is injective everywhere except possibly at the end points \(a\) and \(b\text{.}\)

The image of \(\alpha\text{,}\) which we will denote by \(C = \alpha([a,b])\text{,}\) is a one-dimensional subspace of \(\mathbb{R}^n\text{,}\) which is the curve itself. We say that the parametric curve is smooth if \(\alpha\) can be extended to a smooth (\(C^\infty\)) function on an open set containing \([a,b]\text{.}\)

A \(C^1\)-function is a differentiable function whose derivative is continuous.
Here, the standard notation \(\{ a,b \}\) means the set with two elements \(a\) and \(b\text{.}\) This should not be confused with \([a,b]\text{,}\) which means the closed interval from \(a\) to \(b\text{,}\) and \((a,b)\text{,}\) which means the open interval from \(a\) to \(b\text{.}\)

In words, what we are doing is mapping the interval \([a,b]\) to a one-dimensional subspace \(C \subset \mathbb{R}^n\text{,}\) which is the curve we are interested in. The map gives us a way of doing calculus on the curve, by mapping points on the interval, where we can do calculus easily, to points on the curve, which is a more difficult object to work with. This will prove important to define integration along curves: we will pullback using the parametrization to rephrase integration along a curve into integration along the interval \([a,b]\text{,}\) which we know how to do.

Properties 1 and 2 in Definition 3.2.1 are important to ensure that the image curve \(C\) does not have kinks or corners (see for instance Exercise 3.2.6.6). Property 2 also ensures that a parametrization induces a well defined orientation on the curve, as we will see.

Property 3 in Definition 3.2.1 imposes that the map \(\alpha\) is injective everywhere except possibly at the endpoints \(a\) and \(b\) of the interval, which could be mapped to the same point. This ensures that the image curve does not cross itself, and that the parametrization does not go through the same path multiple times (for instance, going around a circle two times).

Because of Property 3, we can distinguish between two types of parametric curves, depending on whether the image curve has endpoints or not.

Definition 3.2.2. Closed parametric curves.

Let \(\alpha:[a,b] \to \mathbb{R}^n\) be a parametric curve, with image \(C = \alpha([a,b]) \subset \mathbb{R}^n\text{.}\)

  1. If \(\alpha(a) = \alpha(b)\text{,}\) then we say that the parametric curve is closed, as the image curve has no endpoints (it is a loop).

  2. If \(\alpha: [a,b] \to \mathbb{R}^n\) is injective, we call the set \(\partial C =\{ \alpha(a), \alpha(b) \}\) consisting only in the endpoints of \(C\) the boundary of the curve, which we denote by \(\partial C\text{.}\)

Consider the function \(\alpha: [0,2 \pi] \to \mathbb{R^2}\) given by

\begin{equation*} \alpha(\theta) = (\cos \theta, \sin \theta). \end{equation*}

It is easy to see that the image \(\alpha([a,b]) \subset \mathbb{R}^2\) is the unit circle \(x^2 + y^2 = 1\text{.}\) Let us check that the conditions in Definition 3.2.1 are satisfied:

  1. \(\alpha\) is certainly a \(C^1\)-function on \(\mathbb{R}\text{;}\) in fact, it is a smooth function, so this is a smooth parametric curve.

  2. \(\alpha'(t) = (- \sin \theta, \cos \theta)\) is never zero over the interval \([0,2 \pi]\text{.}\)

  3. The only two values of \(\theta \in [0,2 \pi]\) that have the same image are \(\theta=0\) and \(\theta = 2\pi\text{,}\) the endpoints of the interval.

Because of the last statement above, this is an example of a closed parametric curve.

Subsection 3.2.2 The tangent or velocity vector

The curve \(C\) itself, as a subset of \(\mathbb{R}^n\text{,}\) is just a bunch of points. However, the parametrization \(\alpha :[a,b] \to \mathbb{R}^n\) gives us more information. One can think of it as parametrizing the trajectory of a particle moving along the curve: the particle starts at the point \(\alpha(a) \in \mathbb{R}^n\text{,}\) and moves along the curve until it reaches the endpoint \(\alpha(b) \in \mathbb{R}^n\text{.}\) As such, we can define a velocity vector, giving the velocity of the particle as it moves along the curve following the given parametrization. Geometrically, the velocity vector is nothing but the tangent vector to the curve at a given point.

Definition 3.2.4. The tangent vector to a parametric curve.

Let \(\alpha: [a,b] \to \mathbb{R}^n\) be a parametric curve, with the map \(\alpha(t) = (x_1(t), \ldots, x_n(t))\text{.}\) The tangent vector, or velocity vector \(\mathbf{T}\) is a vector-valued function

\begin{align*} \mathbf{T}: [a,b] \amp \to \mathbb{R}^n\\ t \amp \mapsto \mathbf{T}(t) =\alpha'(t) = (x_1'(t), \ldots, x_n'(t)). \end{align*}

In other words, we simply differentiate the component functions of the vector-valued function \(\alpha\text{.}\) As \(\mathbf{T}\) is a vector-valued function, it assigns a vector to every point on the curve \(C\text{,}\) namely the tangent vector to the curve at that point.

Subsection 3.2.3 Orientation of a parametric curve

If we think of the parametrization as giving the trajectory of a particle moving along the curve, the tangent vector gives the velocity of the particle at every point on the curve. As we assume that \(\alpha'(t)\) is never zero in the definition of parametric curves Definition 3.2.1, we see that the particle travels along the curve without stopping. Moreover, the velocity vector defines a “direction of travel” along the curve \(C\text{.}\) Thus the parametrization of a curve naturally gives \(C\) an orientation, corresponding to the direction of travel. More precisely, the orientation is specified by the tangent vector.

Definition 3.2.5. Orientation of a curve.

The orientation of a curve \(C \subset \mathbb{R}^n\) is given by a choice of direction on the curve, which can be represented geometrically by an arrow on the curve. There are two distinct choices of orientation on a curve: either the arrow is in one direction or it is in the other.

Note that this naturally generalizes the orientation of an interval that we defined in Definition 3.1.3.

Now we see that parametrizing a curve naturally induces a choice of orientation.

The key here is that, according to the definition of parametric curves Definition 3.2.1, the tangent vector never vanishes, and it varies continuously. In other words, the velocity vector is never zero. Thus a particle traveling along the curve cannot turn around, as it would first have to stop as the velocity vector varies continuously, but it cannot stop. So the particle always travel in the same direction along the curve, which defines an orientation on the curve.

Remark 3.2.7.

Another way of thinking about the fact that a parametrization induces an orientation on the curve is that we can think of the parametrization as not only mapping an interval \([a,b]\) to a curve \(C \subset \mathbb{R}^n\text{,}\) but also as mapping the orientation. When we defined parametric curves in Definition 3.2.1, we could think of the domain of \(\alpha\) as being not only an interval but an interval with a choice of orientation; we then always assume that the domain of \(\alpha\) is given the canonical orientation (in the direction of increasing real numbers). Because of property 2 in Definition 3.2.1, the canonical orientation on the interval is then unambiguously mapped to a choice of direction on the image curve \(\alpha([a,b]) = C\text{,}\) which is the induced orientation on the curve.

Let us go back to our parametrization of the unit circle in Example 3.2.3. We have the function \(\alpha: [0,2 \pi] \to \mathbb{R^2}\) given by

\begin{equation*} \alpha(\theta) = (\cos \theta, \sin \theta). \end{equation*}

We know that its image is the unit circle \(x^2 + y^2 = 1\text{.}\) What is the induced orientation on the circle? We calculate the tangent vector:

\begin{equation*} \mathbf{T}(\theta) = (x'(\theta), y'(\theta) )= (- \sin \theta, \cos \theta). \end{equation*}

Now consider \(\theta = 0\text{.}\) The parametrization maps this point to \(\alpha(0) = (1,0)\text{,}\) so this is the point with coordinates \((1,0)\) on the unit circle. As for the tangent vector, we see that \(\mathbf{T}(0) = (0,1)\text{,}\) and hence the tangent vector at the point \((1,0)\) on the unit circle is pointing upwards. This tells us that we are moving along the curve in a counterclockwise direction, which is the induced orientation on the unit circle.

Subsection 3.2.4 Orientation-preserving reparametrizations

Given a curve \(C \subset \mathbb{R}^n\text{,}\) there isn't a unique choice of parametrization; the curve can be parametrized in many different ways. For instance, let \(\alpha: [a,b] \to \mathbb{R}^n\) be a parametric curve, with \(\alpha(t) = (x_1(t), \ldots, x_n(t))\text{.}\) Now suppose that we think of the parameter \(t\) as a function of a new parameter \(u\text{,}\) that is, \(t=t(u)\text{.}\) Our parametrization then becomes \(\alpha(t(u)) = (x_1(t(u)), \ldots, x_n(t(u))\text{.}\) Assuming that \(t(u)\) is chosen appropriately, this may define a new parametrization \(\beta(u) = (X_1(u),\ldots, X_n(u)) = (x_1(t(u)), \ldots, x_n(t(u))\text{.}\) Going from \(t\) to \(u\) is what is called “reparametrizing the curve”.

We can be a little more precise, using the concept of pullbacks introduced in Definition 2.3.3.

First, it is clear that \(\alpha([a,b]) = \phi^*\alpha( [c,d])\text{,}\) as we are just composing the functions \(x\) and \(y\) with \(\phi\text{,}\) and thus both \(\alpha\) and \(\phi^* \alpha\) have the same image curve. But to check that \(\phi^* \alpha\) is a parametrization, we need to make sure that the three conditions in Definition 3.2.1 are satisfied.

Property one is certainly satisfied, since it is assumed that \(\phi\) can be extended to a \(C^1\)-function on an open set containing \([c,d]\text{.}\) Property two is also satisfied, since

\begin{equation*} (\phi^* \alpha)'(u) = \left( \frac{dx_1}{d \phi} \frac{d \phi}{d u}, \ldots, \frac{dx_n}{d \phi} \frac{d \phi}{d u} \right), \end{equation*}

and by assumption \(\frac{d \phi}{d u}\) never vanishes on \([c,d]\text{.}\) As for property three, it follows since \(\phi\) is assumed to be injective.

Remark 3.2.10.

We note that since \(\frac{d \phi}{d u}\) is continuous, and is never zero on \([c,d]\text{,}\) then it is either everywhere positive on \([c,d]\text{,}\) or everywhere negative.

Since a parametric curve induces a choice of orientation on the curve, and there are only two possible choices of orientation, it will be important to distinguish between reparametrizations that preserve the induced orientation, and those that reverse it.

We simply need to compare the tangent vectors. Let \(T_\alpha\) be the tangent vector associated to the parametrization \(\alpha\text{,}\) and \(T_{\phi^* \alpha}\) be the tangent vector associated to \(\phi^* \alpha\text{.}\) We have:

\begin{equation*} T_{\phi^* \alpha}(u) = \frac{d \phi}{d u} \left( x_1'(\phi(u)), \ldots, x_n'(\phi(u)) \right) = \frac{d \phi}{d u} T_{\alpha} (\phi(u) ). \end{equation*}

It is then clear that if \(\frac{d \phi}{d u} > 0\text{,}\) the orientation is preserved, while if \(\frac{d \phi}{d u} \lt 0\) is is reversed.

We already saw in Example 3.2.8 a parametrization of the unit circle that induces a counterclockwise orientation, namely \(\alpha: [0,2 \pi] \to \mathbb{R}^2\) with \(\alpha(\theta) = (\cos \theta, \sin \theta)\text{.}\) Now consider a second parametrization of the unit circle \(\beta: [- \frac{3 \pi}{2}, \frac{\pi}{2}] \to \mathbb{R}^2\) with \(\beta(t) = (\sin t, \cos t)\text{.}\) What orientation is \(\beta\) inducing? The tangent vector reads:

\begin{equation*} \mathbf{T}_\beta(t) = (\cos t, - \sin t). \end{equation*}

Consider \(t = \pi/2\text{.}\) The parametrization maps this point to the point \(\beta(\pi/2) = (1,0)\) on the unit circle. The tangent vector is \(\mathbf{T}_\beta(\pi/2) = (0,-1)\text{,}\) and hence it points downwards. We conclude that \(\beta\) is inducing a clockwise orientation on the circle, which is the opposite of our original parametrization \(\alpha\text{.}\)

Let us now formulate this in the language of reparametrizations as above. Consider the function \(\phi: [- \frac{3 \pi}{2}, \frac{\pi}{2}] \to [0, 2 \pi]\) given by \(\phi(t) = \frac{\pi}{2} - t\text{.}\) This function is bijective, and \(\phi'(t) = -1\) which is of course never zero. The pullback \(\phi^* \alpha: [- \frac{3 \pi}{2}, \frac{\pi}{2}] \to \mathbb{R}^2\) is given by

\begin{equation*} \phi^* \alpha(t) = (\cos(\frac{\pi}{2}-t), \sin(\frac{\pi}{2}-t) ) = (\sin t, \cos t), \end{equation*}

which is our second parametrization \(\beta\text{.}\) Since \(\phi'(t) \lt 0\text{,}\) we expect \(\alpha\) and \(\beta=\phi^*\alpha\) to induce opposite orientation, which is exactly what we observed.

Subsection 3.2.5 Piecewise parametric curves

To end this section, we note that it will sometimes be useful to consider unions of parametric curves as defined in Definition 3.2.1. This is because our definition is fairly restrictive. It would not allow for curves with kinks or corners, for instance, since we impose that \(\alpha'(t)\) is never zero. Also, since we require that \(\alpha\) is injective except possibly at the endpoints, it would not allow for curves with self-intersection. To deal with these cases, all we need to do is consider the union \(C_1 \cup \ldots \cup C_n\) of a finite number of curves with parametrizations, and such that two distinct components can only have one or two points in common (their endpoints). For instance, in the case of a curve with corners, we will treat it as a union of parametric curves, where each curve starts where the previous one ends.

If a piecewise parametric curve is the union of a number of parametric curves, and each parametric curve is smooth, we call the piecewise parametric curve piecewise smooth.

Remark 3.2.13.

We add one more piece of notation. It will be useful to distinguish between curves that have self-intersection and those that do not. We say that a curve that doesn't intersect itself (except possibly at the endpoints) is simple. With our definition of parametric curves Definition 3.2.1, the image of a parametric curve will be always be simple, as it cannot self-intersect.

Non-simple curves can be studied using piecewise parametric curves, as any non-simple curve can be broken into a number of simple curves.

Consider the triangle with vertices \(A=(0,0)\text{,}\) \(B=(0,1)\) and \(C=(1,0)\text{.}\) We cannot parametrize it as a single parametric curve according to Definition 3.2.1, since we cannot find a parametrization \(\alpha\) that has a non-vanishing tangent vector at the vertices of the triangle. Instead, we split it into the union of three parametric curves, corresponding to the three edges of the triangle:

\begin{align*} \alpha_1:\amp [0,1] \to \mathbb{R}^2, \qquad \alpha_1(t) = (0,t),\\ \alpha_2:\amp [0,1] \to \mathbb{R}^2, \qquad \alpha_2(t) = (t,1-t),\\ \alpha_3:\amp [0,1] \to \mathbb{R}^2, \qquad \alpha_3(t) = (1-t,0). \end{align*}

\(\alpha_1\) parametrizes the edge \(AB\text{,}\) \(\alpha_2\) the edge \(BC\text{,}\) and \(\alpha_3\) the edge \(CA\text{.}\) As each parametric curve is smooth, the union defines a piecewise smooth parametric curve.

Exercises 3.2.6 Exercises

1.

Find a parametrization for the straight line between the points \((0,1,1)\) and \((2,3,3)\) in \(\mathbb{R}^3\text{.}\)

Solution.

We are given two points on the line. The vector \(\mathbf{d}\) whose direction is parallel to the line is given by \(\mathbf{d} = (2,3,3) - (0,1,1) = (2,2,2)\text{.}\) So we can write an equation for the line as

\begin{equation*} \mathbf{r}(t) = (0,1,1) + t(2,2,2), \qquad 0 \leq t \leq 1. \end{equation*}

In the language of this section, this gives us a parametrization of the line, in the form of a map:

\begin{align*} \alpha:\amp [0,1] \to \mathbb{R}^3\\ \amp t \mapsto \alpha(t) = (2 t, 1 + 2 t, 1 + 2 t). \end{align*}

2.

Express the upper half of a circle of radius \(3\) and centered at the point \((1,0)\) as a parametric curve. Determine the orientation induced by your parametrization.

Solution.

The circle of radius \(3\) and centered at the point \((1,0)\) has equation

\begin{equation*} (x-1)^2 + y^2 = 9. \end{equation*}

It is easy to find a parametrization for the circle. We can take, for instance,

\begin{equation*} x-1 =3 \cos t, \qquad y=3 \sin t, \qquad 0 \leq t \leq 2 \pi. \end{equation*}

We want only the upper half of the circle though, so we need to restrict to \(y \geq 0\text{.}\) This amounts to restricting the domain of our parametrization to \(0 \leq t \leq \pi\text{.}\) The resulting parametrization can be written as the map \(\alpha: [ 0,\pi] \to \mathbb{R}^2\) with \(\alpha(t) = (3 \cos t + 1, 3 \sin t).\)

What is the induced orientation? The tangent vector to our parametric curve is \(\mathbf{T}(t) = (- 3 \sin t, 3 \cos t)\text{.}\) Pick a point on the circle, say \((4,0)\text{.}\) This corresponds to \(t=0\text{.}\) At this point, the tangent vector is \(\mathbf{T}(0) = (0,3)\text{,}\) which is pointing upwards. This means that our parametrization induces a counterclockwise orientation around the circle.

3.

Consider the parametric curve \(\alpha: [0, 4 \pi] \to \mathbb{R}^3\text{,}\) \(\alpha(t) = (\cos t, \sin t, 4 t)\text{.}\) What is the shape of the image curve \(C = \alpha([0,4 \pi]) \subset \mathbb{R}^3\text{?}\) What is the induced orientation?

Solution.

Let us write \(\alpha(t) = (x(t), y(t), z(t))\text{.}\) We see that \(x^2(t) +y^2(t) = \cos^2 t + \sin^2 t = 1\text{,}\) so all points on the curve \(C\) lie on the cylinder \(x^2+y^2 = 1\) with radius \(1\text{.}\) Moreover, as \(t\) increases, \(z(t)=4 t\) increases linearly. So the curve is an helix on the cylinder with radius \(1\) centred around the \(z\)-axis.

The curve starts at the point \(\alpha(0) = (1,0,0)\) on the cylinder, and ends at the point \(\alpha(4 \pi) = (1,0,16 \pi)\text{.}\) As \(t\) runs from \(0\) to \(4 \pi\text{,}\) we see that the curve goes twice around the cylinder. Moreover, the tangent vector is \(\mathbf{T}(t) = (-\sin t, \cos t, 4)\text{;}\) in particular, its \(z\)-coordinate is always positive, which means that the tangent vector is always pointing upwards. We conclude that the induced orientation is going upwards along the helix.

4.

Suppose that a particle is moving along the parametric curve \(\alpha:[0,\pi] \to \mathbb{R}^3\) with \(\alpha(t) = (\sin(t^2), \cos(t), t)\text{.}\) Find its velocity at \(t = \pi\text{.}\)
Solution.

The velocity vector is \(\mathbf{v}(t) = (2 t \cos(t^2), -\sin(t), 1)\text{.}\) At \(t = \pi\text{,}\) we get \(\mathbf{v}(\pi) = (2 \pi \cos(\pi^2), 0, 1)\text{.}\) This is the velocity of the particle at \(t = \pi\text{,}\) which is of course a vector as the particle is moving in three-dimensional space.

5.

Consider the curve that is the intersection of the cylinder \(x^2 + y^2 = 1\) and the surface \(z = x^2 - y^2\) in \(\mathbb{R}^3\text{.}\) Find a parametrization for the curve. Is it a closed curve?

Solution.

A point on the cylinder \(x^2 + y^2 = 1\) can be parametrized by \((x,y,z) = (\cos t, \sin t, z)\text{,}\) with \(0 \leq t \lt 2 \pi\) and \(z \in \mathbb{R}\text{.}\) But we want the curve to lie on the surface \(z = x^2-y^2\text{,}\) so \(z\) is fixed as \(z = \cos^2 t - \sin^2 t\text{.}\) We thus get a parametric expression for a point on the curve as \((x(t), y(t), z(t)) = (\cos t, \sin t, \cos^2 t- \sin^2 t),\) with \(0 \leq t \lt 2 \pi\text{.}\) To rewrite this as a parametric curve in the language of this section, we include the endpoint \(t = 2 \pi\text{.}\) We get the parametric curve \(\alpha: [0, 2 \pi] \to \mathbb{R}^3\) with \(\alpha(t) = (\cos t, \sin t, \cos^2 t - \sin^2 t)\text{.}\) This is a closed curve, since \(\alpha(0) = (1,0,1)\) and \(\alpha(2 \pi) = (1,0,1)\text{,}\) i.e. the starting and ending points coincide.

6.

Consider the map \(\alpha:[-1,1] \to \mathbb{R}^2\) with \(\alpha(t)= (x(t), y(t))\text{,}\) where

\begin{equation*} x(t)= \begin{cases} t^2 \amp \text{for } 0 \leq t \leq 1 \\ - t^2 \amp \text{for } -1 \leq t \lt 0 \end{cases}, \qquad y(t) = t^2. \end{equation*}

Show that it is not a parametric curve, according to Definition 3.2.1. What does the image \(C=\alpha([-1,1]) \subset \mathbb{R}^2\) look like?

Solution.

At first one may think that this is valid parametric curve. The map \(\alpha\) is injective, so Property 3 is satisfied. Since its derivative is \(\alpha'(t) = (x'(t), y'(t))\) with

\begin{equation*} x'(t) = \begin{cases} 2 t \amp \text{for } 0\leq t \leq 1 \\ - 2 t \amp \text{for } -1 \leq t \lt 0 \end{cases}, \qquad y'(t) = 2 t, \end{equation*}

\(\alpha'(t)\) exists and is continuous for all \(t \in \mathbb{R}\text{,}\) and thus Property 1 is also satisfied (note that \(\alpha\) cannot be extended to a smooth function however, since \(x'(t)\) is not differentiable at \(t=0\text{,}\) but that's ok, it doesn't have to). However, the problem is with Property 2: we see that \(\alpha'(0) = (0,0)\text{,}\) with \(0 \in [-1,1]\text{;}\) thus Property 2 is not satisfied. We conclude that this is not a parametric curve.

The image curve \(C=\alpha([-1,1])\) is the set of points in \(\mathbb{R}^2\) satisfying the equation \(y = |x|\) between \((-1,1)\) and \((1,1)\text{,}\) which has a corner at the origin. This example highlights one of the reasons why Property 2 is there in Definition 3.2.1. If it wasn't there, this means that we could find a parametrization for the curve \(y=|x|\text{;}\) but we do not want the image curve to have kinks or corners. Property 2 ensures that we cannot find a parametrization for the whole curve \(y = |x|\) between \((-1,1)\) and \((1,1)\) at once.

This is not to say however that we cannot deal with this curve. Just like for the triangle, the idea is to consider it as a piecewise parametric curve. I.e., we realize the line segment from \((-1,1)\) to \((0,0)\) as a (smooth) parametric curve, the line segment from \((0,0)\) to \((1,1)\) as another (smooth) parametric curve, and we take the union of the two parametric curves.