Derivatives and integrals of vector functions.

Section 2 Derivatives and integrals of vector functions.

Subsection 2.1 Derivatives of vector functions.

Derivatives of vector functions can be interpreted as tangent vectors (if not zero) for curves or as velocity vectors for trajectories.

Definition 2.1.

The derivative, \(\vec{r}'\text{,}\) of a vector function \(\vec{r}:I \to \mathbb{R}^3\text{,}\) at \(t \in I\text{,}\) is defined as:

\begin{equation*} \vec{r}'(t) = \frac{d\vec{r}}{dt}(t) = \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t)-r(t)}{\Delta t}, \end{equation*}

if this limit exists.

If \(\vec{r}(t)\) represents the position of a point, in meters, at the instant \(t\text{,}\) in seconds; then the quotient,

\begin{equation*} \frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}, \end{equation*}

is vector joining the point \(\vec{r}(t)\) to the point \(\vec{r}(t+\Delta t)\) divided by the time it takes to get from one point to the other, \(\Delta t\text{,}\) thus giving an average vector velocity for the trajectory in this interval in meters per second. The graph below shows the curve parametrized by the vector function \(\vec{r}\) and the vectors from \(\vec{r}(t)\) to \(\vec{r}(t+\Delta t)\) re-scaled by \(\frac{1}{\Delta t}\text{,}\) for \(\Delta t\) equal to 1; 0.5 and 0.1. The smaller \(\Delta t\) is, the closer the resultant vector is to the instant variation of position rate (also known as velocity).

If \(\vec{r}(t)\) describes the trajectory of a moving point-like object, we say that \(\vec{r}'(t)\) is the velocity (vector). The size of this vector \(\|\vec{r}'(t)\|\) is the speed of the object at instant \(t\text{.}\)

Theorem 2.2.

If \(f\text{,}\) \(g\) and \(h\) are differentiable functions at \(t\) and \(\vec{r}(t) = (f(t),g(t),h(t))\text{,}\) than \(\vec{r}\) is differentiable¹ at \(t\) and

\begin{equation*} \vec{r}'(t) = (f'(t),g'(t),h'(t)). \end{equation*}

meaning that the derivative exists at this point

Proof.

This can be calculated directly from the definition of derivative:

\begin{align*} \frac{d\vec{r}}{dt}=\amp \lim_{\Delta t\to 0} \frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t} \\ =\amp \lim_{\Delta t\to 0} \frac{1}{\Delta t} \left((f(t+\Delta t), g(t+\Delta t), h(t+\Delta t))-(f(t), g(t), h(t) ) \right) \\ =\amp \lim_{\Delta t\to 0} \frac{1}{\Delta t} \left(f(t+\Delta t)-f(t), g(t+\Delta t)-g(t), h(t+\Delta t)-h(t) \right) \\ =\amp \lim_{\Delta t\to 0} \left(\frac{f(t+\Delta t)-f(t)}{\Delta t}, \frac{g(t+\Delta t)-g(t)}{\Delta t}, \frac{h(t+\Delta t)-h(t)}{\Delta t} \right) \\ =\amp \left( \lim_{\Delta t\to 0}\frac{f(t+\Delta t)-f(t)}{\Delta t}, \lim_{\Delta t\to 0}\frac{g(t+\Delta t)-g(t)}{\Delta t}, \lim_{\Delta t\to 0}\frac{h(t+\Delta t)-h(t)}{\Delta t} \right) \\ =\amp \left(f'(t), g'(t), h'(t) \right) \end{align*}

Example 2.3.

Let \(\vec{r}(t)=\cos 3t\vec{i}+(te^{-t})\vec{j}+(7+3t^2)\vec{k}\) be a vector function. We can calculate its derivative by calculating derivatives to each of the component functions (that are one variable real functions):

\begin{equation*} \vec{r'}(t)=-3\sin 3t\vec{i}+(e^{-t}-te^{-t})\vec{j}+(6t)\vec{k}. \end{equation*}

The code below can be used to calculate derivatives of vector functions.

If \(\vec{r}'(t) \neq 0\text{,}\) we can define the unit tangent vector, that indicates the tangent direction, but with lenght equal to 1:

\begin{equation*} \vec{T}(t)=\frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}=\frac{\vec{r}'(t)}{\sqrt{\langle\vec{r}'(t), \vec{r}'(t)\rangle}} \end{equation*}

The existence of a derivative \(\vec{r}'(t)\) does not guarantee that the curve parametrized by \(\vec{r}(t)\) has a tangent line. If \(\vec{r}'(t) = 0\text{,}\) it is possible that no tangent direction exists, for example "sharp corners" might be hidden. To illustrate this, consider the vector function \(\vec{r}(t)=t^3\vec{i}+t^2\vec{j}\text{,}\) that is differentiable, with \(\vec{r}'(t)=3t^2\vec{i}+2t\vec{j}\text{,}\) but the curve it parametrizes, has a sharp corner at \(\vec{r}(0) = (0, 0)\text{,}\) in which \(\vec{r}'(0) = 0\text{.}\)

Figure 2.4. A plane curve with a sharp corner, parametrized by the differentiable vector function \(\vec{r}(t)=t^3\vec{i}+t^2\vec{j}.\)

Remark 2.5.

A curve \(C\) is said to be smooth if it can be parametrized by a vector function \(\vec{r}:I \to \mathbb{R}^n\text{,}\) which is differentiable and such that \(\vec{r}'(t)\) is continuous and \(\vec{r}'(t) \neq 0\) for all \(t \in I\text{.}\) Smooth curves have tangents at all points!

If \(\vec{r}'(t) \neq 0\text{,}\) we say that \(\vec{r}'(t)\) is a tangent vector to the curve \(C\text{,}\) parametrized by the function \(\vec{r}\text{,}\) at the point \(\vec{r}(t)\text{.}\) The tangent line to \(C\text{,}\) denoted \(\vec{s}(u)\text{,}\) at the point \(P=\vec{r}(t)\text{,}\) is the line that passes through \(P\) and points in the direction of \(\vec{r}'(t)\text{.}\) So, given a fixed value \(t \in I\text{,}\) the tangent line can be parametrized by

\begin{align*} \amp \vec{l}:\mathbb{R} \to \mathbb{R}^3\\ \amp \vec{l}(u) = u\vec{r}'(t)+\vec{r}(t) \end{align*}

The code bellow plots a spacial curve, the tangent vector at a point and the tangent line.

Exercise 2.6.

For every of the vector functions, use the code above to plot a spacial curve, the tangent vector at the given point and the tangent line (at this point):

\(\vec{r}(t) = (t^2-4,\sin (\pi t), 2\cos (\pi t))\text{,}\) with \(t \in [0,2]\text{,}\) \(t_0 = 1/4\text{.}\)
\(\vec{r}(t) = (t,e^t,e^{-t})\text{,}\) with \(t \in [0,1.5]\text{,}\) \(t_0=1\text{.}\)

Solution.

We need change the functions, their domains and the parameter of the tangent point in code above.

Remark 2.7.

The second derivative of a vector function \(\vec{r}(t)\) is the derivative of its derivative, \(\vec{r}''(t)=(\vec{r}')'(t)\text{,}\) and, if \(\vec{r}(t)\) describes the trajectory of a point-like object, than it can be interpreted as the acceleration the particle is subject to.

Remark 2.8.

A smooth curve \(C\) is said to be parametrized by arc length by \(\vec{r}:I \to \mathbb{R}^n\) iff \(\|\vec{r}'(t)\|=1\) for all \(t \in I\text{.}\) For such parametrization, \(\|\vec{r}''(t)\|\) is the curvature and it measures how much the curve is changing direction.²

See XXX for an example of how to parametrize a curve by arc length and calculate its curvature.

Theorem 2.9.

Let \(\vec{u}(t)=(f_u(t), g_u(t), h_u(t))\) and \(\vec{v}(t)=(f_v(t), g_v(t), h_v(t))\) be differentiable vector functions, \(f(t)\) be a differentiable real function and \(c\in \mathbb{R}\) a scalar. Then, if the functions below are defined for \(t\text{,}\) it holds:

\(\frac{d}{dt}[\vec{u}(t)\pm\vec{v}(t)] = \vec{u}'(t)\pm\vec{v}'(t)\text{.}\)
Proof.
\begin{align*} \frac{d}{dt}[\vec{u}(t)\pm\vec{v}(t)]=\amp\frac{d}{dt}[(f_u(t)\pm f_v(t), g_u(t) \pm g_v(t), h_u(t) \pm h_v(t))] \\ =\amp [(f_u'(t)\pm f_v'(t), g_u'(t) \pm g_v'(t), h_u'(t) \pm h_v'(t))] \\ =\amp (f_u'(t), g_u'(t), h_u'(t)) \pm (f_v'(t), g_v'(t), h_v'(t)) = \vec{u}'(t) \pm \vec{v}'(t). \end{align*}
\(\frac{d}{dt}[c\vec{u}(t)] = c\vec{u}'(t)\text{.}\)
Proof.
\begin{align*} \frac{d}{dt}[c\vec{u}(t)]=\amp\frac{d}{dt}(cf_u(t), cg_u(t), ch_u(t)) \\ =\amp (cf_u'(t), cg_u'(t), ch_u'(t)) \\ =\amp c(f_u'(t), g_u'(t), h_u'(t)) = c\vec{u}'(t). \end{align*}
\(\frac{d}{dt}[f(t)\vec{u}(t)] = f'(t)\vec{u}(t) + f(t)\vec{u}'(t)\text{.}\)
Proof.
\begin{align*} \amp\frac{d}{dt}[f(t)\vec{u}(t)] \\ \amp=\frac{d}{dt}[(f(t) f_u(t),f(t) g_u(t), f(t) h_u(t))] \\ \amp= (f'(t) f_u(t)+f(t) f_u'(t),f'(t) g_u(t)+f(t) g_u'(t), f'(t) h_u(t)+f(t) h_u'(t)) \\ \amp= (f'(t) f_u(t),f'(t) g_u(t), f'(t) h_u(t)) + (f(t) f_u'(t), f(t) g_u'(t), f(t) h_u'(t))\\ \amp= f'(t)\vec{u}(t)+f(t)\vec{u}'(t) . \end{align*}
\(\frac{d}{dt}[\vec{u}(t)\cdot\vec{v}(t)] =\vec{u}'(t)\cdot\vec{v}(t)+\vec{u}(t)\cdot\vec{v}'(t)\text{.}\)
Proof.
\begin{align*} \amp\frac{d}{dt}[\vec{u}(t)\cdot\vec{v}(t)]\\ \amp=\frac{d}{dt}(f_u(t)f_v(t)+g_u(t)g_v(t)+ h_u(t)h_v(t)) \\ \amp = (f_u'(t)f_v(t)+f_u(t)f_v'(t)+g_u'(t)g_v(t)+g_u(t)g_v'(t)+ h_u'(t)h_v(t)+h_u(t)h_v'(t)) \\ \amp = (f_u'(t)f_v(t)+g_u'(t)g_v(t)+ h_u'(t)h_v(t)) +(f_u(t)f_v'(t)+g_u(t)g_v'(t)+ h_u(t)h_v'(t)) \\ \amp = \vec{u}'(t)\cdot\vec{v}(t)+\vec{u}(t)\cdot\vec{v}'(t) . \end{align*}
\(\frac{d}{dt}[\vec{u}(t)\times\vec{v}(t)] =\vec{u}'(t)\times\vec{v}(t)+\vec{u}(t)\times\vec{v}'(t)\text{.}\)
Proof.
\begin{align*} \frac{d}{dt}\amp[\vec{u}(t)\times\vec{v}(t)]\\ =\amp \frac{d}{dt}(g_u(t)h_v(t) - h_u(t)g_v(t), f_u(t)h_v(t) - h_u(t)f_v(t), f_u(t)g_v(t) - g_u(t)f_v(t)) \\ =\amp (g_u'(t)h_v(t) - h_u'(t)g_v(t), f_u'(t)h_v(t) - h_u'(t)f_v(t), f_u'(t)g_v(t) - g_u'(t)f_v(t)) \\ \amp +(g_u(t)h_v'(t) - h_u(t)g_v'(t), f_u(t)h_v'(t) - h_u(t)f_v'(t), f_u(t)g_v'(t) - g_u(t)f_v'(t)) \\ =\amp \vec{u}'(t)\times\vec{v}(t)+\vec{u}(t)\times\vec{v}'(t) . \end{align*}
\(\frac{d}{dt}[\vec{u}(f(t))] = f'(t)\vec{u}'(f(t))\text{.}\)
Proof.
\begin{align*} \frac{d}{dt}[\vec{u}(f(t))] \amp = ((f_u(f(t)))',(g_u(f(t)))', (h_u(f(t)))')\\ \amp=(f'(t)f_u'(f(t)),f'(t)g_u'(f(t)), f'(t)h_u'(f(t))) \\ \amp=f'(t)(f_u'(f(t)),g_u'(f(t)), h_u'(f(t))) \\ \amp= f'(t)\vec{u}'(f(t)) . \end{align*}

Exercise 2.10.

What is the curve parmetrized by \(\vec{r}(t) = (t, \sin (t))\text{,}\) \(t \in [0,8]\text{?}\) For each of the functions defined below, calculate \(\vec{r}(f(t))\text{,}\) animate a point moving by \(\vec{r}(f(t))\) with the code and calculate the speed \(\|(\vec{r}(f(t)))'\|\) to explain the differences in the trajectories:

\(f: [0,1.6] \to \mathbb{R}, \,\,\, f(t)=5t\text{;}\)
\(f: [0,2] \to \mathbb{R}, \,\,\, f(t)=t^3\text{;}\)
\(f: [-8,0] \to \mathbb{R}, \,\,\, f(t)=-t\text{.}\)

Solution.

The curve parmetrized by \(\vec{r}(t) = (t, \sin (t))\) is the part of the graph of \(y=\sin(x)\) over the interval \([0,8]\text{,}\) the speed of the movement is \(\|\vec{r}'(t)\| = \sqrt{1+\cos^2(t)}\text{.}\) The compositions \(\vec{r}(f(t))\) parametrize sections of the graph of \(y=\sin(x)\text{.}\) The domains were adjusted for all of them to plot the same section of the curve. Animating it is clear that the firs two parametrizations change the speed the point moves over the curve and the last one reverses the direction of the movement. We find the following composed functions, tangent vector and speeds:

\(\vec{r}(f(t)) = (5t, \sin(5t)), \,\, (\vec{r}(f(t)))' = (5, 5\cos(5t)),\) \(\|(\vec{r}(f(t)))'\| = \sqrt{25+25\cos^2(t)} = 5\sqrt{1+\cos^2(t)}\text{;}\)
\(\vec{r}(f(t)) = (t^3, \sin(t^3)), \,\, (\vec{r}(f(t)))' = (3t^2, 3t^2\cos(t^3)),\) \(\|(\vec{r}(f(t)))'\| = \sqrt{9t^4+9t^4\cos^2(t^3)} = 3t^2\sqrt{1+\cos^2(t^3)}\text{;}\)
\(\vec{r}(f(t)) = (-t, \sin(-t)), \,\, (\vec{r}(f(t)))' = (-1, -\cos(t)),\) \(\|(\vec{r}(f(t)))'\| = \sqrt{1+\cos^2(-t)}\text{.}\)

Subsection 2.2 Integrals of vector functions.

The integrals of the component functions can be used to obtain primitives of vector functions. An important application is related to Newton's second law \(\vec{F} = m \vec{a}\text{.}\) If we know the force that is being applied to an object (together with its mass, starting position and starting velocity), we can obtain its position by applying these integrals successively.

Definition 2.11.

The definite integral of a vector function³ \(\vec{r}:I \to \mathbb{R}^3\) of \(a\) to \(b\text{,}\) with \(a,b \in \overline{I}\text{,}\) \(\vec{r}(t) = f(t) \vec{i}+ g(t) \vec{j} + h(t) \vec{k}\) is defined as:

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

if these (real) integrals exist.

Analogous to \(\vec{r}:I \to \mathbb{R}^2\text{.}\)

Remark 2.12.

In general, the set \(I \subset \mathbb{R}\) has not to be an interval. If there is a finite number of points in which the vector function in not defined or in which there are discontinuities, it is necessary to divide the domain to have these singular points at the extremes of the intervals and take lateral limits.

The Fundamental Theorem of Calculus can be extended to continuous vector functions⁴, so that if the component functions of \(\vec{r}(t) = f(t) \vec{i}+ g(t) \vec{j} + h(t) \vec{k}\) have primitives \(F(t)\text{,}\) \(G(t)\) and \(H(t)\text{:}\)

\begin{align*} \int_a^b \vec{r}(t)dt = \amp \int_a^b f(t) dt \vec{i} + \int_a^b g(t) dt \vec{j} + \int_a^b h(t) dt \vec{k}\\ = \amp (F(b)-F(a)) \vec{i} + (G(b)-G(a))\vec{j} + (H(b)-H(a))\vec{k}\\ = \amp (F(b)\vec{i} + G(b)\vec{j} + H(b)\vec{k}) - (F(a)) \vec{i} + G(a) \vec{j} + H(a)\vec{k})\\ =\amp \vec{R}(b) - \vec{R}(a), \end{align*}

where \(\vec{R}(t)= F(t) \vec{i}+ G(t) \vec{j} + H(t) \vec{k}\) is a primitive of \(\vec{r}(t)\text{,}\) which means that \(\vec{R}'(t)= \vec{r}(t)\text{.}\)

if \(\vec{r}\) is not continuous, we need to divide the interval to have \(\vec{r}(t)\) continuous inside each sub-interval and apply the theorem to each of the sub-intervals.

We write the indefinite integrals as

\begin{equation*} \vec{R}(t) = \int r(t) dt. \end{equation*}

Python can calculate definite integrals (numerical) of vector functions

or indefinite integrals (symbolic, with sympy).

Watch out!!! There are several kinds of integrals in vector calculus, each in a different context, and this is the first of them. To distinguish among them is very important.

Exercise 2.13.

Suppose that a 1Kg rocket is in the origin of space \((0,0,0)\text{,}\) at rest, and starts moving by the action a force (from its motor) that changes in time and is given by the function \(\vec{r}(t) = (t\cos (t),t\sin (t), 1)\text{,}\) with \(t \in [0, \infty)\text{.}\) Use Newton's Second Law (\(\vec{F}=m\vec{a} \)) to determine the position of the rocket at instant \(t\text{,}\) using the codes above.

Solution.

Using Newton's second law, we find the the acceleration is \(\vec{a}(t)=\vec{r}(t)\text{,}\) so that its speed is

\begin{equation*} \vec{v}(t)=\int_0^t \vec{a}(u) du = (t\sin(t) + \cos(t) - 1,-t\cos(t) + \sin(t),t). \end{equation*}

Then we integrate once more to obtain the position:

\begin{equation*} \vec{p}(t)=\int_0^t \vec{v}(u) du = (-t\cos(t) - t + 2\sin(t),-t\sin(t) - 2\cos(t) + 2, t^2/2). \end{equation*}