When working with objects moving through space, it makes sense to consider them moving over a certain amount of time, \(t.\) Time could be drawn as another dimension on a graph, but most of the time this is unnecessary since time always carries on in the same way (assuming your not dealing with anything traveling near the speed of light.) For this reason, it is often useful to define the position on the \(x\) and \(y\) axis using time, but not writing time as a third axis. This is something that does not work so well with Cartesian equations but is much simpler using vector-valued functions, hence making them incredibly useful in Physics, Machine Learning, and many other subjects.
Vector-Valued Function Definition
Before you get into the details of vector-valued functions, it is important to understand vectors fully.
Vectors
A vector is a mathematical object that has both direction and magnitude.
Fig. 1. A vector can be thought of like an arrow, pointing from one place to another.
Vectors can be written in two different ways,
Column Vector Form: \( \begin{bmatrix} x \\ y \end{bmatrix}, \)
Component Form: \( x \vec{i} + y \vec{j}. \)
These two vectors are equivalent. Numerically, vectors can be added and subtracted by adding or subtracting the individual components. Similarly, they can be multiplied by scalar quantities by multiplying the individual components. In component form, this looks just like collecting like terms and expanding brackets.
Graphically, adding vectors is done by stacking them tip to tip, and subtracting by stacking them tip to tip, but pointing the second vector in the opposite direction. Multiplying numbers by a scalar \(\lambda\) is the same as stacking \(\lambda\) of the same vectors, tip to tip, and if \(\lambda\) is negative, the product will be pointing in the opposite direction.
Finally, given a vector \( v = x \vec{i} + y \vec{j},\) the magnitude \(|\vec{v}|\) and direction angle \(\theta\) of a vector can be calculated using the following formulas:
\[ \begin{align} | \vec{v} | & = \sqrt{ x^2 + y^2 }, \\ \theta & = \tan^{-1}\left({\frac{x}{y}}\right) \end{align} \]
For more information on all of this, see Vectors.
What are Vector-Valued Functions?
Vector-valued functions are just like real-valued functions, but output a vector instead of a scalar.
A vector-valued function is a function that takes a scalar value as input, and gives a vector as output. A vector-valued function of one variable looks like this,
\[ \vec{r}(t) = \begin{bmatrix} f(t) \\ g(t) \end{bmatrix} = f(t) \vec{i} + g(t) \vec{j}. \]
Here, \( f(t)\) and \(g(t)\) are parametric equations.
Given this definition, you can deduce the domain and range of a vector-valued function.
The domain of a vector-valued function is a subset of \(\mathbb{R},\)
The range of an \(n\)-dimensional vector-valued function is a subset of \(\mathbb{R}^n.\)
Here you will focus on vectors in 2 dimensions, meaning the range of the functions will be a subset of \(\mathbb{R}^2.\) It is important to note that it is a subset of \(\mathbb{R}^2\) and not the whole of \(\mathbb{R}^2,\) since you will encounter many vector-valued functions that cannot output to every point in \(\mathbb{R}^2.\)
Examples of Vector-Valued Functions
There are many different types of vector-valued functions, but here you will look at some of the simplest.
Straight Lines
The vector-valued formula for a straight line is
\[ \vec{r}(t) = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} + t \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}. \]
Here, \(\vec{a} = a_1 \vec{i} + a_2 \vec{b} \) is the position vector of a point \(a\) on the line, and \( \vec{b} = b_1 \vec{i} + b_2 \vec{j} \) is a vector that is parallel to the line.
Fig. 2. A line is defined by a vector-valued function using a point on the line, \(a,\) and a vector parallel to the line, \(\vec{b}.\)
Circles and Ellipses
The vector-valued equation for a circle with radius \(a\) is
\[ \vec{r}(t) = \begin{bmatrix} a \cos{t} \\ a \sin{t} \end{bmatrix} \]
Fig. 3. The vector-valued function for a circle can be made using the sine and cosine functions.
An ellipse can be defined similarly, but using \(a\) as the intercept on the \(x\)-axis and \(b\) as the intercept on the \(y\)-axis.
\[ \vec{r}(t) = \begin{bmatrix} a \cos{t} \\ b \sin{t} \end{bmatrix} \]
Fig. 4. The vector-valued function for an ellipse can be defined similarly to that of a circle, but taking into account the different axis intercepts.
Spirals
There are many ways to define spirals in mathematics, but an easy way is to define them similarly to spirals and circles, but with a \(t\) term in front of the trigonometric functions.
\[ \vec{r}(t) = \begin{bmatrix} a t \cos{t} \\ b t \sin{t} \end{bmatrix} \]
Fig. 5, The graph of an spiral, where \(a = b = \frac{1}{2}. \)
Graphing Vector-Valued Functions
When you first learnt to graph Cartesian equations such as \(y = f(x),\) you likely started by drawing a table of values for \(x,\) and then filling in the corresponding values of \(y.\) You could then plot these points and join them up, to create an estimation of the curve. You can do the exact same thing to graph vector-valued functions, but instead starting with the variable \(t\) and using these values of \(t\) to calculate the corresponding values of \(x\) and \(y.\) Let's look at an example of this.
Sketch the graph of \( \vec{r} = t^2 \vec{i} + t \vec{j}, \) for values of \(-4 < t < 4. \)
Solution
First, create a table with three columns, titled \(t, x, y.\) You can fill in the \(t\) column with the integers from \(-4\) to \(4.\)
\(t\) | \(x\) | \(y\) |
-4 | | |
-3 | | |
-2 | | |
-1 | | |
0 | | |
1 | | |
2 | | |
3 | | |
4 | | |
From here, you can start filling in the values. Remember that \(x\) will be the coefficient of the \(\vec{i}\) term, and \(y\) will be the coefficient of the \(\vec{j}\) term. First, let's fill in the \(x\) column by squaring all of the values in the \(t\) column.
\(t\) | \(x\) | \(y\) |
-4 | 16 | |
-3 | 9 | |
-2 | 4 | |
-1 | 1 | |
0 | 0 | |
1 | 1 | |
2 | 4 | |
3 | 9 | |
4 | 16 | |
Next, fill in the \(y\) column. This will be exactly the same as the values on the \(t\) column.
\(t\) | \(x\) | \(y\) |
-4 | 16 | -4 |
-3 | 9 | -3 |
-2 | 4 | -2 |
-1 | 1 | -1 |
0 | 0 | 0 |
1 | 1 | 1 |
2 | 4 | 2 |
3 | 9 | 3 |
4 | 16 | 4 |
Next, plot the \((x,y)\) pairs on a graph.
Fig. 6. The shape of these dots seems to resemble a parabola.
Based on the shape of the plotted points and the fact that the function has a \(t^2\) term in it, it appears to be a parabola. You can draw a curve between these points to get the following curve:
Fig. 7. The finished curve is the parabola \(x = y^2.\)
To see more examples, see Graphing Vector-Valued Functions.
Vector-Valued Functions Formula
The most important formula for vector-valued functions is the formula for arc length, or the length of a curve between two points.
The length of the curve between the points \(t=a\) and \(t=b.\)
The length \(L\) of a curve \(\vec{r}(t) = f(t) \vec{i} + g(t) \vec{j} \) between two point \(a\) and \(b\) is
\[ L = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2} \, \mathrm{d}t. \]
This measures the whole length of the curve as if you had laid a piece of string on the curve and then cut it off and measured it. Let's look at some examples using this formula.
Find the arc length of
\[ \vec{r} = \begin{bmatrix} \sin{(3t)} \\ \cos{(3t)} \end{bmatrix} \]
for \(-4 < t < 2.\)
Solution
Here, \(f(t) = \sin{(3t)}\) and \(g(t) = \cos{(3t)}.\) The formula requires the derivatives of these functions, so you must differentiate them both.
\[ \begin{align} f'(t) & = 3 \cos{(3t)} \\ g'(t) & = 3 \sin{(3t)}. \end{align} \]
From here, you can substitute these into the formula for the arc length.
\[ \begin{align} L & = \int_{-4}^{2} \sqrt{(3 \cos{(3t)})^2 + (3 \sin{(3t)})^2} \, \mathrm{d}t \\ & = \int_{-4}^2 \sqrt{ 9 \cos^2{(3t)} + 9 \sin^2{(3t)} } \, \mathrm{d}t \\ & = \int_{-4}^{2} \sqrt{9 (\cos^2{(3t)} + \sin^2{(3t)})} \, \mathrm{d}t. \end{align} \]
From here, you can use the formula \(\sin^2{x} + \cos^2{x} = 1. \)
\[ \begin{align} L & = \int_{-4}^{2} \sqrt{9 \cdot 1} \, \mathrm{d}t \\ & = \int_{-4}^{2} 3 \, \mathrm{d}t \\ & = [3t]_{-4}^{2} \\ & = 3\cdot 2 - 3 \cdot (-4) \\ & = 18. \end{align} \]
Hence, the arc length is 18 unit length.
Derivatives of Vector-Valued Function
The derivative of vector-valued functions can be found by differentiating each component of the vector-valued function. The derivative of \( \vec{r}(t) = f(t) \vec{i} + g(t) \vec{j} \) is:
\[ \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}(t) = \frac{\mathrm{d}f}{\mathrm{d}t}(t) \vec{i} + \frac{\mathrm{d}g}{\mathrm{d}\mathrm{d}}(t) \vec{j}, \]
assuming that the derivatives of \(f(t)\) and \(g(t)\) with respect to \(t\) exist. This makes sense logically, as it is just like using the addition rule when differentiating any other function. The derivative of a vector-valued function at a point will point in the direction of travel of the function, at a tangent to the curve.
If the vector valued function, call it \(\vec{s}(t),\) represents position on the \(xy\) plane at time \(t,\) then the derivative of this function will be the velocity vector \(\vec{v}(t).\) The magnitude of the velocity vector at time \(t\) is the speed of travel at time \(t.\) Similarly, the differential of the velocity vector will be the acceleration vector, \( \vec{a}(t). \) Let's take a look at differentiating some vector-valued functions.
A particle's position in space is given by the vector-valued function
\[ \vec{s}(t) = \begin{bmatrix} 3t^2 \\ e^t \end{bmatrix}. \]
Find the vector-valued functions for the velocity and acceleration of the particle.
Solution
If you differentiate the position function, you will get the velocity function. This will be,
\[ \vec{v}(t) = \vec{s}'(t) = \begin{bmatrix} 6t \\ e^t \end{bmatrix}. \]
Next, you can differentiate this again to find the acceleration function.
\[ \vec{a}(t) = \vec{v}'(t) = \begin{bmatrix} 6 \\ e^t \end{bmatrix}. \]
To learn more about differentiating vector-valued functions, see Calculus of Vector-Valued Functions.
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