\(\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}\)

Vector Functions

Introduction

Throughout the rest of this course, we'll use vector-valued functions extensively. We've actually already seen an example of a vector function, when we described a line in \(\mathbb{R}^3\) with parametric equations \[\langle x_0,y_0,z_0 \rangle +t\langle a,b,c \rangle\]

If we write it like \[\vec{r}(t) = \langle x_0+at,y_0+bt,z_0+ct \rangle,\] it may be clearer that it's a vector valued function. Basically, for each value of \(t\), a vector-valued function describes a vector, and if we draw these vectors from the origin, the tips of the vectors trace out a path in space (a straight line for the function above).


In general, a vector-valued function has the form (in \(\mathbb{R}^3\)) of \[\vec{r}(t) = \langle f(t),g(t),h(t) \rangle\] or \[\vec{r}(t) = \langle x(t),y(t),z(t) \rangle.\]


Example

For example, consider the function \[\vec{r}(t) = \langle t,t^2,2t+1 \rangle\] (note that \(x(t)=t\), \(y(t)=t^2\), and \(z(t)=2t+1\)). The domain of this function is all \(t\) in \(\mathbb{R}\), but the range is not quite so simple (we'd need to see a graph to get an idea of it).

If we pick a few representative values for \(t\), we can start to build a table of vectors:

\[\begin{array}{c | c} t & \langle x,y,z \rangle\\ \hline 0 & \langle 0,0,1 \rangle\\ 1 & \langle 1,1,3 \rangle\\ 2 & \langle 2,4,5 \rangle\\ \vdots & \vdots \end{array}\]

We can graph this space curve with Matlab using the following command:

        >> ezplot3('t','t^2','2*t+1',[-6*pi,6*pi])
        

Another Example

Graph \(\vec{r}(t)=\langle \cos t,\sin t,t \ln t \rangle\):

        >> ezplot3('cos(t)','sin(t)','t*log(t)',[-6*pi,6*pi])
        

Another Example

Graph \(\vec{r}(t)=\langle 4 \cos t,t^2+2t,4 \sin t \rangle\):

        >> ezplot3('4*cos(t)','t^2+2*t','4*sin(t)',[-6*pi,6*pi])
        

Limits With Vector-Valued Functions

If we have a vector function, the limit of this function as \(t\) approaches some value is just what you would expect (and hope): it is equal to the limit of each component function.

Find \(\displaystyle\lim_{t \to 0^+} \langle \cos t, \sin t, t\ \ln t \rangle\).

Solution

Find the limit of each component function:

\[\begin{align} \lim_{t \to 0^+} \cos t &= 1\\ \lim_{t \to 0^+} \sin t &= 0\\ \lim_{t \to 0^+} t\ \ln t &= 0 \cdot (-\infty) \textrm{ (indeterminate: use L'Hopital's Rule)}\\ \end{align}\]

Using L'Hopital's Rule on the third limit:

\[\lim_{t \to 0^+} t\ \ln t = \lim_{t \to 0^+} \dfrac{\ln t}{1/t} = \textrm{(L)} \lim_{t \to 0^+} \dfrac{1/t}{-1/t^2} = \lim_{t \to 0^+} -t = 0\]

Therefore, the limit is

\[\ans{\lim_{t \to 0^+} \langle \cos t, \sin t, t\ \ln t \rangle = \langle 1,0,0 \rangle}\]

Derivatives of Vector-Valued Functions

Just like with limits, derivatives of vector functions work as you would hope: we can differentiate each component function individually.

\[\begin{align} \vec{r}\ '(t) &= \lim_{\Delta t \to 0} \dfrac{\Delta r}{\Delta t} = \lim_{\Delta t \to 0} \dfrac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}\\ &= \lim_{\Delta t \to 0} \left(\dfrac{x(t+\Delta t)-x(t)}{\Delta t} \hat{i} + \dfrac{y(t+\Delta t)-y(t)}{\Delta t} \hat{j} + \dfrac{z(t+\Delta t)-z(t)}{\Delta t} \hat{k}\right)\\ &= \left(\lim_{\Delta t \to 0} \dfrac{x(t+\Delta t)-x(t)}{\Delta t}\right) \hat{i} + \left(\lim_{\Delta t \to 0} \dfrac{y(t+\Delta t)-y(t)}{\Delta t}\right) \hat{j} + \left(\lim_{\Delta t \to 0} \dfrac{z(t+\Delta t)-z(t)}{\Delta t}\right) \hat{k}\\ &= x'(t) \hat{i} + y'(t) \hat{j} + z'(t) \hat{k} \end{align}\]

Summary

If \(\vec{r}(t) = \langle x(t),y(t),z(t) \rangle\), then \(\vec{r}\ '(t) = \langle x'(t),y'(t),z'(t) \rangle\).

If \(\vec{r}\ ' \neq \vec{0}\) (the zero vector), then \(\vec{r}\ '(t)\) is a tangent vector to the space curve.

Find the first and second derivatives of \(\vec{r}(t) = \langle t,t^2,2t+1 \rangle\).

Solution

\[\ans{\begin{align} \vec{r}\ '(t) &= \langle 1,2t,2 \rangle\\ \vec{r}\ ''(t) &= \langle 0,2,0 \rangle \end{align}}\]

Using L'Hopital's Rule on the third limit:

\[\lim_{t \to 0^+} t\ \ln t = \lim_{t \to 0^+} \dfrac{\ln t}{1/t} = \textrm{(L)} \lim_{t \to 0^+} \dfrac{1/t}{-1/t^2} = \lim_{t \to 0^+} -t = 0\]

Therefore, the limit is

\[\ans{\lim_{t \to 0^+} \langle \cos t, \sin t, t\ \ln t \rangle = \langle 1,0,0 \rangle}\]

Unit Tangent Vector

This vector will come back later; to find the unit tangent vector, simply divide the tangent vector by its magnitude.

The unit tangent vector to \(\vec{r} (t)\) is \[\vec{T}(t) = \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|}\]

Find the unit tangent vector for \(\vec{r}(t) = \langle t^2,4t,4 \ln t \rangle,\ \ \ \ t > 0\).

Solution

First, find \(\vec{r}\ '(t)\):

\[\vec{r}\ '(t) = \left\langle 2t,4,\dfrac{4}{t} \right\rangle\]

Then divide this by its magnitude to find the unit tangent vector:

\[\begin{align} \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} &= \dfrac{\langle 2t,4,4/t \rangle}{\sqrt{4t^2+16+16/t^2}}\\ &= \dfrac{\langle 2t,4,4/t \rangle}{\sqrt{(2t+4/t)^2}}\\ &\implies \ans{\vec{T}(t) = \dfrac{1}{2t+\dfrac{4}{t}} \left\langle 2t,4,\dfrac{4}{t} \right\rangle} \end{align}\]

Try it yourself:

(click on the problem to show/hide the answer)

Find the unit tangent vector for \(\vec{r} (t) = \langle 10,3\cos t, 3\sin t \rangle\).
\(\vec{T}(t) = \langle 0,-\sin t, \cos t \rangle\)

Let \(\vec{r}(t) = \langle t^5,t^4,t^3 \rangle\).

  1. Find the unit tangent vector to this curve at \(t=1\).
  2. Find parametric equations for the tangent line to \(\vec{r}(t)\) at \(t=1\).

Solution

  1. The unit tangent vector to this curve at \(t=1\): \[\begin{align} \vec{r}\ '(t) &= \langle 5t^4,4t^3,3t^2 \rangle\\ \vec{r}\ '(1) &= \langle 5,4,3 \rangle\\ \dfrac{\vec{r}\ '(1)}{|\vec{r}\ '(1)|} &= \dfrac{\langle 5,4,3 \rangle}{\sqrt{25+16+9}} = \dfrac{1}{5\sqrt{2}} \langle 5,4,3 \rangle\\ &= \ans{\left\langle \dfrac{\sqrt{2}}{2}, \dfrac{2\sqrt{2}}{5}, \dfrac{3\sqrt{2}}{10} \right\rangle} \end{align}\]
  2. Parametric equations for the tangent line to \(\vec{r}(t)\) at \(t=1\): \[\begin{align} \vec{r}\ '(1) &= \langle 5,4,3 \rangle = \langle a,b,c \rangle\\ \langle x_0,y_0,z_0 \rangle &= \langle x(1),y(1),z(1) \rangle = \langle 1,1,1 \rangle \end{align}\]

    Putting the starting point and the direction together:

    \[\ans{\begin{align} x &= 1+5t\\ y &= 1+4t\\ z &= 1+3t \end{align}}\]

Try it yourself:

(click on the problem to show/hide the answer)

Find parametric equations for the tangent line to \(\vec{r}(t) = \langle 10,3\cos t,3\sin t \rangle\) at \(t=0\).
\begin{align} x &= 10\\ y &= 3\\ z &= 3t \end{align}

Find the angle between the following two curves at the point where \(t=1\) and \(s=2\).

\[\begin{align} \vec{r}_1(t) &= \langle t,1-t,3+t^2 \rangle\\ \vec{r}_2(s) &= \langle 3-s,s-2,s^2 \rangle \end{align}\]

Solution

Find the angle between their tangent vectors:

\[\begin{align} \vec{r}_1\ '(t) &= \langle 1,-1,2t \rangle \longrightarrow \vec{r}_1\ '(1) = \langle 1,-1,2 \rangle\\ \vec{r}_2\ '(s) &= \langle -1,1,2s \rangle \longrightarrow \vec{r}_2\ '(2) = \langle -1,1,4 \rangle \end{align}\]

Take the dot product to find the angle between them:

\[\begin{align} \langle 1,-1,2 \rangle \cdot \langle -1,1,4 \rangle &= -1-1+8 = 6\\ \cos \theta = \dfrac{6}{\sqrt{1+1+4}\ \sqrt{1+1+16}} &\longrightarrow \ans{\theta = 54.74^{\circ}} \end{align}\]

Most of the derivative rules are what you would expect (the derivative of a sum/difference, constant multiple, etc.), but here are a few extensions of basic derivative rules, the Chain Rule and Product Rule:

  • Chain Rule: \[\dfrac{d}{dt} \left(\vec{u}(f(t))\right) = \vec{u}\ '\left(f(t)\right) f'(t)\]
  • Dot Product Rule: \[\dfrac{d}{dt} \left(\vec{u}(t) \cdot \vec{v}(t)\right) = \vec{u}\ '(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}\ '(t)\]
  • Cross Product Rule: \[\dfrac{d}{dt} \left(\vec{u}(t) \times \vec{v}(t)\right) = \vec{u}\ '(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}\ '(t)\]

Integrals of Vector-Valued Functions

By now it shouldn't come as a surprise that to integrate a vector-valued function, we can integrate the component functions individually. If \(\vec{r}(t) = \langle x(t),y(t),z(t) \rangle\), then \[\int \vec{r}(t)\ dt = \left(\int x(t)\ dt\right) \hat{i} + \left(\int y(t)\ dt\right) \hat{j} + \left(\int z(t)\ dt\right) \hat{k} + \vec{C}\] where \(\vec{C}\) is a constant vector \(\langle c_1,c_2,c_3 \rangle\). For a definite integral,

\[\int_a^b \vec{r}(t)\ dt = \left(\int_a^b x(t)\ dt\right) \hat{i} + \left(\int_a^b y(t)\ dt\right) \hat{j} + \left(\int_a^b z(t)\ dt\right) \hat{k}\]

If \(\vec{r}(t) = 2t\ \hat{i} + \sin t\ \hat{j} + t^2\ \hat{k}\), find \(\displaystyle\int \vec{r}(t)\ dt\).

Solution

\[\ans{\begin{align} \int \vec{r}(t)\ dt &= (t^2+c_1) \hat{i} + (-\cos t+c_2) \hat{j} + \left(\dfrac{1}{3}t^3+c_3\right) \hat{k}\\ &= \left\langle t^2,-\cos t,\dfrac{1}{3}t^3 \right\rangle + \langle c_1,c_2,c_3 \rangle \end{align}}\]

If \(\vec{r}(t) = \langle 2t+1,t^2,t-3 \rangle\), find \(\displaystyle\int_1^2 \vec{r}(t)\ dt\).

Solution

\[\begin{align} \int_1^2 \vec{r}(t)\ dt &= \left(t^2+t \bigg|_1^2\right) \hat{i} + \left(\dfrac{1}{3}t^3 \bigg|_1^2\right) \hat{j} + \left(\dfrac{1}{2}t^2-3t \bigg|_1^2\right) \hat{k}\\ &= \ans{4\hat{i} + \dfrac{7}{3}\hat{j} - \dfrac{3}{2}\hat{k}} \end{align}\]

Try it yourself:

(click on a problem to show/hide its answer)

  1. If \(\vec{r}(t) = \left\langle \dfrac{t}{\sqrt{t^2+2}},e^{-3t},\sin (4t) + 1 \right\rangle\), find \(\displaystyle\int \vec{r}(t)\ dt\).
  2. \(\displaystyle\int \vec{r}(t)\ dt = \left\langle \sqrt{t^2+2},-\dfrac{1}{3}e^{-3t},t-\dfrac{1}{4}\cos 4t \right\rangle + \vec{C}\)

  3. If \(\vec{r}(t) = \left\langle 1,3\cos \dfrac{t}{2},4t \right\rangle\), find \(\displaystyle\int_0^\pi \vec{r}(t)\ dt\).
  4. \(\displaystyle\int_0^\pi \vec{r}(t)\ dt = \left\langle \pi,6,2\pi^2 \right\rangle\)