Vector Valued Functions

Section 5.2 Vector Valued Functions

Definition 5.2.1.

Let \(A\) be a subset of \(\RR\text{.}\) A vector-valued function is a function \(f: A \rightarrow \RR^n\) for \(n \geq 2\text{.}\) It has a single-variable real input, but outputs a vector in some higher dimensional space. A vector-valued function can be written as a vector of individual functions.

\begin{equation*} f(t) = (f_1(t), f_2(t), \ldots, f_n(t)) \end{equation*}

The functions \(f_i\) are called the component functions of the vector-valued function. In \(\RR^2\text{,}\) the components will often be written \((x(t), y(t))\text{,}\) and similarly in \(\RR^3\text{,}\) \((x(t), y(t), z(t))\text{.}\)

The single-variable component functions allow me to quickly extend many notions from single-variable calculus to vector-valued functions.

Definition 5.2.2.

The limit of a vector valued function is simply the limit of each component.

\begin{equation*} \lim_{t \rightarrow a} f(t) = \left( \lim_{t \rightarrow a} f_1 (t), \lim_{t \rightarrow a} f_2 (t), \ldots, \lim_{t \rightarrow a} f_2 (t) \right) \end{equation*}

Recall that a single variable function \(f(t)\) is continuous at \(a \in \RR\) if

\begin{equation*} \lim_{x \rightarrow a} f(x) = f(a)\text{.} \end{equation*}

Definition 5.2.3.

A vector-valued function is continuous if and only if each component function is continuous.

Definition 5.2.4.

The derivative of a vector valued function \(f: A \rightarrow \RR^n\) is calculated in each component.

\begin{equation*} \frac{df}{dt} = f^\prime(t) = \left( \frac{df_1}{dt}, \frac{df_2}{dt}, \ldots, \frac{df_n}{dt} \right) \end{equation*}

I can state this explicitly, for \(f(x) = (x(t),y(t))\) with values in \(\RR^2\text{.}\)

\begin{equation*} f^\prime(t) = \frac{d}{dt} f(t) = \left( \frac{dx}{dt}, \frac{dy}{dt}\right) \end{equation*}

Likewise, I can give the explicit form for \(f(t) = (x(t), y(t),z(t))\) with values in \(\RR^3\text{.}\)

\begin{equation*} f^\prime(t) = \frac{d}{dt} f(t) = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) \end{equation*}

The result of a derivative of a vector valued function is still a vector. This may seem odd from the viewpoint of first year calculus, where derivatives measured quantifiable geometric properties, such as slopes of tangent lines. Since the answers here are vectors instead of scalars, I will eventually reconsider those interpretations. A major challenge in extending calculus to several variables is the need to re-adjust my interpretation and intuition concerning derivatives (and eventually integrals). The derivative is no longer the slope of a tangent line.