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Section 1 Vector Functions: Curves and Trajectories

Vector functions take real numbers and return vectors

\begin{align*} \vec{r}:I\subset \amp\mathbb{R} \to \mathbb{R}^n\\ \amp t \to \vec{r}(t) \subset \mathbb{R}^n \end{align*}

These vectors can be of any arbitrary dimension, but we are often interested in 3 dimensional vector functions:

\begin{equation*} \vec{r}(t) = (f(t), g(t),h(t)) = f(t)\vec{i} + g(t)\vec{j} + h(t)\vec{k}, \end{equation*}

In this expression 1  \(f(t)\text{,}\) \(g(t)\) and \(h(t)\) are real functions, \(f,g,h:\mathbb{R}\subset I \to \mathbb{R}\text{,}\) called the component functions of \(\vec{r}(t)\text{.}\)

\(\vec{i}=(1,0,0)\text{,}\) \(\vec{j}=(0,1,0)\) and \(\vec{k}=(0,0,1)\) are the canonical vectors in \(\mathbb{R}^3\text{.}\)
Remark 1.1.
Vector functions are used to represent trajectories in space or curves.

The vector function \(\vec{r}(t)=(5t+1,\ln(t^3),\sqrt{3-t})\) has three component functions:

\begin{align*} \amp f(t)=5t+1,\\ \amp g(t)=\ln(t^3),\\ \amp h(t)=\sqrt{3-t}. \end{align*}

The domain of \(\vec{r}\text{,}\) if not explicitly given, is the largest set in which all the component functions are defined, in this example it is \([0,3)\text{.}\)

Remark 1.3.
All the component funcions of \(\vec{r}(t)\) have to be defined for every point of its domain \(D\text{.}\) The domain, or part of it, may have be given explicitly when coding (when plotting, for example).

Some of the most important concepts in Vector Calculus use the concept of limit, which we define componentwise 2 :

It is possible to define this limit in terms of \(\varepsilon\) and \(\delta\text{,}\) but both definitions can be proven equivalent.
Definition 1.4.

For a vector function \(\vec{r}:I\to \mathbb{R}^3\text{,}\) \(\vec{r}(t)=(f(t),g(t),h(t))\text{,}\) we define the limit of \(\vec{r}(t)\) when \(t \) tends to \(a \in \overline{I}\) 3  as:

\begin{equation*} \lim_{t \to a} \vec{r}(t) = \left(\lim_{t \to a} f(t), \lim_{t \to a} g(t), \lim_{t \to a} h(t) \right), \end{equation*}

if the real functions have limits at \(a\text{.}\)

The limit could, in principle, be calculated in points of the boundary do the domain of \(\vec{r}\text{,}\) which is denoted by \(\overline{I}\text{.}\)

If \(\vec{r}(t)=(1+t^3)\vec{i} + te^{-t}\vec{j}+\frac{\sin(t)}{t}\vec{k}\) then

\begin{align*} \lim_{t \to 0} \vec{r}(t) =\amp \lim_{t \to 0}(1+t^3)\vec{i} + \lim_{t \to 0}te^{-t}\vec{j}+\lim_{t \to 0}\frac{\sin(t)}{t}\vec{k}\\ =\amp \vec{i} +\vec{k}. \end{align*}
Remark 1.6.
Instead of defining everything from scratch for vector functions, we prefer to build upon the knowledge of single variable calculus. This might be an adequate viewpoint for who is interested in applications.
Definition 1.7.

The vector function \(\vec{r}:I\to \mathbb{R}^3\text{,}\) is continuous at \(a \in I\) if, and only if:

\begin{equation*} \lim_{t \to a} \vec{r}(t) = \vec{r}(a). \end{equation*}

(\(\Longrightarrow\)) If \(\vec{r}\) is continuous at \(a\text{,}\) we have that

\begin{align*} \left(\lim_{t \to a} f(t), \lim_{t \to a} g(t), \lim_{t \to a} h(t) \right) =\amp \lim_{t \to a} \vec{r}(t) = \vec{r}(a)\\ =\amp \left(f(a), g(a), h(a) \right) \end{align*}

and \(f(t)\text{,}\) \(g(t)\) and \(h(t)\) are continuous at \(a\text{.}\)

(\(\Longleftarrow\)) If \(f(t)\text{,}\) \(g(t)\) and \(h(t)\) are continuous at \(a\text{,}\) we have that

\begin{align*} \lim_{t \to a} \vec{r}(t) =\amp \left(\lim_{t \to a} f(t), \lim_{t \to a} g(t), \lim_{t \to a} h(t) \right)\\ =\amp \left(f(a), g(a), h(a) \right) = \vec{r}(a) \end{align*}

and \(\vec{r}(t)\) is continuous at \(a\text{.}\)

Remark 1.9.
We will focus on continuous vector functions to represent curves and trajectories. We do this to avoid calling a bunch of erratic points a curve or a trajectory.

Let \(I \subset \mathbb{R}\) be an interval. If \(\vec{r}:I \to \mathbb{R}^n\) (usually with \(n= 2\) or \(3\)) is continuous for all \(t \in I\) and is not constant, then the set

\begin{equation*} C = \{(x,y,z) \in \mathbb{R}^3 | (x,y,z) = r(t), \mbox{ for some } t \in I\} \end{equation*}

is called a space curve.

\(r(t) = (f(t),g(t), h(t))\) is a parametrization of \(C\text{;}\) \(t\) is the parameter and \(x=f(t)\text{,}\) \(y=g(t)\) e \(z=h(t)\) are the parametric equations of \(C\text{.}\)

Python code for plotting a plane curve given by a vector function \(\vec{r}:I \to \mathbb{R}^2\text{.}\)

Find the (max) domain of \(\vec{r}(t)=(\sqrt{16-t}, \sqrt{t-2})\text{.}\) Is this vector function continuous? If it is, use the code above to plot the curve parametrized by \(\vec{r}(t)\text{.}\)
Solution.
\(x(t) = \sqrt{16-t}\) is defined (in the reals) for \(t \leq 16\text{.}\) \(y(t) = \sqrt{t-2}\) is defined (in the reals) for \(t \geq 2\text{.}\) So the domain is \([2,16]\text{.}\)

Python code to crate an animation of a point moving accordingly to some vector function \(\vec{r}:[a,b] \to \mathbb{R}^2\text{.}\)

Let \(P=(-1,0)\) and \(Q=(0,1)\text{.}\) Use the code above to animate a point moving accordingly to these functions and describe the differences in their movement:

  1. \(\vec{r}(t) = P + (Q-P)t\text{,}\) with \(t \in [0,4]\text{.}\)

  2. \(\vec{r}(t) = P + (Q-P)t^2\text{,}\) with \(t \in [0,2]\text{.}\)

  3. \(\vec{r}(t) = P + (Q-P)4\sin(t)\text{,}\) with \(t \in [0,\pi/2]\text{.}\)

Solution.
It is necessary to change the domain \([a,b]\) in the code, the axis limits from \([-1.4,1.4,-1.4,1.4] \) (\([x_0,x_1,y_0,y_1] \) ) to \([-1.4,3.4,-0.4,4.4] \text{.}\) It might be good to also change the number of points n (i took 1000). All the vector functions parametrize the same line segment (from (-1,0) to (3,4)), but with different speeds.
Remark 1.12.
Each parametrized curve has infinitely many parametrizations, which correspond to the different velocities it can be traveled over.

Python code for plotting a space curve given by a vector function \(\vec{r}:I \to \mathbb{R}^3\text{,}\) \(\vec{r}(t)=(x(t),y(t),z(t))\text{.}\)

Change the code above to plot the space curves given by the functions below:

  1. \(\vec{r}(t) = (\sin (t), \cos (t), \cos (t))\text{,}\) with \(t \in [0,2\pi]\text{.}\)

  2. \(\vec{r}(t) = (\sin (t^3), \cos (t^3), \cos (t^3))\text{,}\) with \(t \in [0,1.845]\text{.}\)

Solution.

  1. Change # Parameter t = np.linspace(0,6.283,100), # Dataset x = np.sin(t); y = np.cos(t); z = np.cos(t)

  2. Change # Parameter t = np.linspace(0,1.845,100), # Dataset x = np.sin(t^3); y = np.cos(t^3); z = np.cos(t^3)

In both cases we obtain the same curve:
Remark 1.14.
If \(\vec{r}:[a,b] \to \mathbb{R}^n\) is a parametrization of a space curve, we can replace the variable \(t\) by an expression \(f(t)\text{,}\) for some function \(f:[c,d]\to[a,b]\text{,}\) with \(f([c,d])=[a,b]\text{,}\) to obtain another parametrization \(\vec{r_p}:[c,d] \to \mathbb{R}^n\) of the same curve (but with different velocities).
Find three different parametrizations for the curve \(\vec{r}(t) = (\sin (t), \cos (t), t^2/4)\text{,}\) with \(t \in [0,2\pi]\text{.}\) Test if your vector functions produce the same plot with the codes above (they should).
Solution.
We could choose any parametrization we want, for example: \(\vec{r}(t) = (\sin (2t), \cos (2t), (2t)^2/4)\text{,}\) with \(t \in [0,\pi]\) or \(\vec{r}(t) = (\sin (\sqrt{t}), \cos (\sqrt{t}), t/4)\text{,}\) with \(t \in [0,4\pi^2]\) or \(\vec{r}(t) = (\sin (4t), \cos (4t), (4t)^2/4)\text{,}\) with \(t \in [0,\pi/2]\text{.}\) So we obtain: