Calculus of Vector-Valued Functions

Section 3.2 Calculus of Vector-Valued Functions

References.

OpenStax Calculus Volume 3, Section 3.2
¹
openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions
.
Calculus, Early Transcendentals by Stewart, Section 13.2.

Introduction.

Here, finally we see some calculus. Much of it is simply extending to 3D what was seen in Calculus 2 for curves in the plane; for example, see OpenStax Calculus Volume 3, Section 1.2

openstax.org/books/calculus-volume-3/pages/1-2-calculus-of-parametric-curves

and the notes for Section 1.2, Calculus of Parametric Curves.

Topics.

Subsection 3.2.1 Derivatives of Vector-Valued Functions

We can build derivatives of vector functions from derivative of components, but the definition can also be done from first principles, with difference quotients:

Definition 3.2.1. Derivative of \(\vec{r}\).

The derivative \(\vec{r}'\) of a vector function of variable \(t\) is given by

\begin{equation} \frac{d\vec{r}}{dt} = \vec{r}' = \lim_{h \to 0}\frac{\vec{r}(t+h)-\vec{r}(t)}{h}.\tag{3.2.1} \end{equation}

It can be checked that for \(\vec{r}(t) = \vector{f(t),g(t),h(t)}\text{,}\) the derivative (if it exists) is the vector of derivatives of the components:

\begin{equation} \frac{d\vec{r}}{dt} = \frac{d}{dt}\vector{f(t),g(t),h(t)} = \vector{f'(t),g'(t),h'(t)}.\tag{3.2.2} \end{equation}

Thus the derivative exists if and only if all of the component derivatives exists.

Subsection 3.2.2 Properties of the Derivative of Vector-Valued Functions

The familiar differentiation rules for sums, products and compositions have natural counterparts for vector functions:

Theorem 3.2.2.

For \(\vec{u}\) and \(\vec{v}\) differentiable vector functions, \(f\) a differentiable scalar (real-valued) function, and \(c\) a scalar constant,

\(\displaystyle \ds\frac{d}{dt}[\vec{u}(t)+\vec{v}(t)] = \vec{u}'(t)+\vec{v}'(t)\)
\(\displaystyle \ds\frac{d}{dt}[c\vec{u}(t)] = c\vec{u}'(t)\)
\(\displaystyle \ds\frac{d}{dt}[f(t)\vec{u}(t)] = f'(t)\vec{u}(t) + f(t)\vec{u}'(t)\)
\(\displaystyle \ds\frac{d}{dt}[\vec{u}(t) \cdot \vec{v}(t)] = \vec{u}'(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}'(t)\)
\(\displaystyle \ds\frac{d}{dt}[\vec{u}(t) \times \vec{v}(t)] = \vec{u}'(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}'(t)\)
\(\displaystyle \ds\frac{d}{dt}[\vec{u}(f(t))] = \vec{u}'(f(t))f'(t)\)

Note: as always with the cross product, the order matters in Item v.

Subsection 3.2.3 Tangent Vectors and the Principle Unit Tangent Vector

For any value of \(t=a\text{,}\) where the derivative vector \(\vec{r}'(a)\) exists and is non-zero, it is tangent to the space curve \(C\) at point \(P\) with position vector \(\vec{r}(a)\text{,}\) and so is called a tangent vector to \(C\) at \(P\text{.}\) The line through \(P\) with this tangent vector is the tangent line to \(C\) at \(P\) with equation

\begin{equation*} \vec{L}(t) = \vec{r}(a) + t \, \vec{r}'(a) \end{equation*}

It will often be useful to consider the principle unit tangent vector

\begin{equation*} \T(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}. \end{equation*}

and then the tangent line at the point \(P\) can be written as

\begin{equation*} \vec{L}(s) = \vec{r}(a) + s \, \T(a) \end{equation*}

Here the parameter \(s\) is used because it corresponds to arc-length along this line, as will be discussed in Section 3.3.

The existence of a tangent direction given by \(\vec{r}'(t)\) and thus of this unit tangent vector is what guarantees that the curve has no "corners", as with the graph of a differentiable function, so this important "niceness" condition has a name:

Definition 3.2.3.

A space curve is smooth if it is given by \(\vec{r}(t)\) on interval \(I\) with both \(\vec{r}\) and \(\vec{r}'\) continuous, and with \(\vec{r}' \neq \vec{0}\) except possibly at the endpoints of \(I\text{.}\) This is equivalent to the existence of the unit tangent vector \(\T(t) = \vec{r}'(t)/\| \vec{r}'(t) \|\text{.}\)

If the derivative is zero at a finite number of points, the curve is called piecewise smooth.

Example 3.2.4.

Consider the plane curve \(\vec{r}(t) = \langle x, y \rangle = \langle t^3, t^2 \rangle\text{.}\) It is differentiable with derivative \(\vec{r}'(t) = \langle 3 t^2, 2 t \rangle\text{,}\) but for \(t=0\) this derivative is zero.

One way to see what happens there is to eliminate the parameter \(t\text{:}\) \(t = x^{1/3}\) so \(y = x^{2/3}\) and \(dy/dx = \frac{2}{3} x^{-1/3}\) so the curve is not differentable at the origin (it has a cusp there).

Thus, this curve is only piecewise smooth.

Subsection 3.2.4 Integrals of Vector-Valued Functions

Like derivatives, definite integrals of vector functions can be built from first principles with Riemann sums, and one gets the predictable result in terms of integrals of components:

For \(\vec{r}(t)=\vector{f(t),g(t),h(t)}\text{,}\)

\begin{equation*} \int_a^b \vec{r}(t) \, dt = \vector{\int_a^b f(t) \, dt,\int_a^b g(t) \, dt,\int_a^b h(t) \, dt} \end{equation*}

Indefinite integrals work likewise: the one new detail to note is that there is a constant of integration for each component:

\begin{equation*} \int \vec{r}(t) \, dt = \vector{\int f(t) \, dt + C_1, \int g(t) \, dt + C_2, \int h(t) \, dt + C_3} \end{equation*}

Equivalently the constant of integration is a vector \(\vec{C} = \vector{C_1, C_2, C_3}\text{.}\)

Study Guide.

Study Section 3.2 of Calculus Volume 3

⁷

openstax.org/books/calculus-volume-3/pages/3-2-calculus-of-vector-valued-functions

; in particular

All the Definitions, Theorems, Examples and Checkpoints.
One or several exercises from each of the following ranges: 41–50, 51–54, 55–58, 59–61, 62, 63, 63, 64.

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