
\section{Curves and Vector-Valued Functions}\label{sec:lecture11}

\subsection{Continuity and Derivatives for Vector-Valued Functions}

In this chapter, we will work with \textbf{vector-valued functions} of a single variable.  These are functions $\mathbf{r} : I \rightarrow \mathbb{R}^n$ (for now, we will have $n = 2$ or $n = 3$), where $I \subset \mathbb{R}$ is an interval.  We can always write the function $\mathbf{r} = \mathbf{r} (t)$ in terms of its component functions:
$$
\mathbf{r} (t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}
$$
where $x,y,z : I \rightarrow \mathbb{R}$ are scalar-valued functions.  


\begin{definition}
The vector-valued function $\mathbf{r}: I \rightarrow \mathbb{R}^n$ is called \textbf{continuous} if and only if all component functions $x(t), y(t)$ and $z(t)$ are continuous on the interval $I$.
\end{definition}


A useful interpretation is to think of $\mathbf{r}$ as describing the motion of an object in space, so that the coordinates of that object at time $t$ are $(x(t), y(t), z(t))$. The assumption that $\bfr$ is continuous guarantees that the object does not make any instantaneous jumps from one location to another.           

\begin{definition}
A \textbf{curve} is the set $\{\mathbf{r}(t) : t \in I \} \subset \mathbb{R}^n$ of values attained by a continuous vector-valued function.  We refer to $\mathbf{r}$ as a \textbf{parametrization} of the curve.
\end{definition}

With our previous interpretation of vector valued functions,  we interpret the curve $\{\mathbf{r}(t) : t \in I \}$ as the path travelled by a particle or object over time, but without any information on where the particle was at any given time $t \in I$. Note that the same curve may have many different parametrizations, corresponding to the object travelling the same path but at different speed and possibly in different directions.


\begin{definition}
The vector-valued function $\mathbf{r}: I \rightarrow \mathbb{R}^n$ is \textbf{differentiable} on an open interval $I$ if and only if all component functions $x(t), y(t)$ and $z(t)$ are differentiable on $I$.  In this case, we write
$$
\frac{d \mathbf{r}}{dt} = \mathbf{r}' (t) : = x' (t) \mathbf{i} + y' (t) \mathbf{j} + z' (t) \mathbf{k}.
$$
Similarly, we will say that $\mathbf{r}$ is $C^k$ on an interval $I$ if and only if all of its component functions are $C^k$ on that interval.
\end{definition}

Following our intepretation, the function $\mathbf{r}'(t) = \mathbf{v}(t)$ is the (instantaneous) velocity of our moving object at time $t$, defined as the limit of the \textbf{average velocity} on the interval $(t, t + \Delta t)$:
$$
\frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t} \rightarrow \mathbf{v}(t) \textrm{ as } \Delta t \rightarrow 0.
$$

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .35\linewidth]{figures/11fig1.pdf}
\caption{The particle at times $t$ and $t + \Delta t$,  as well as the instantaneous velocity at time $t$.  The green arrow indicates the direction of motion of the particle along its associated curve.}
\end{minipage}
\end{figure}


The function $\mathbf{v}$ is, itself, vector-valued, so that it has both \textit{magnitude} and \textit{direction}. 
The magnitude of instantaneous velocity, given by $v(t) : = |\mathbf{v}(t)|$, is the \textbf{speed}  at time $t$.  Further, the \textbf{acceleration} at time $t$ is given by the formula $\mathbf{a}(t) = \mathbf{v}' (t) = \mathbf{r}'' (t)$.


\subsection{Some Examples}
The following are examples of curves in $\mathbb{R}^2$ and $\mathbb{R}^3$, with some associated parametrizations.

\begin{example}\label{ex:line}
Let $P_0 : = (x_0,y_0, z_0)$ be some point in $\mathbb{R}^3$ and let $\mathbf{v}$ be some (constant) vector.  Then, a straight line through $P_0$ in the direction $\mathbf{v}$ can be parametrized as
$$
\mathbf{r}(t) : = \vec{OP_0} + t \mathbf{v} = \mathbf{r}_0 + t \mathbf{v}, \quad t \in \mathbb{R},
$$
where $\mathbf{r}_0 = \vec{OP_0}$. With this parametrization, we have $\mathbf{v} (t) = \mathbf{r}' (t) = \mathbf{v}$, so that the velocity of $\mathbf{r}$ is constant at all times $t \in \mathbb{R}$.

However, the same line can also be parametrized as:
\begin{equation}\label{eq:ex1a}
\mathbf{r}_2(t) = \mathbf{r}_0 + 2t \mathbf{v}
\end{equation}
\begin{equation}\label{eq:ex1b}
\mathbf{r}_3(t) = \mathbf{r}_0 + t^3 \mathbf{v}
\end{equation}
Both \eqref{eq:ex1a} and \eqref{eq:ex1b} describe the same curve. However, in \eqref{eq:ex1a}, the velocity is now $2 \mathbf{v}$ (so double that of our original $\mathbf{r}$); whereas, in \eqref{eq:ex1b}, the velocity is $3 t^2 \mathbf{v}$, and so is variable in time $t \in \mathbb{R}$.
\end{example}


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .25\linewidth]{figures/11fig2.pdf}
\caption{Two different velocities along the same line, discussed in Example \ref{ex:line}.}
\end{minipage}
\end{figure}



\begin{example}\label{ex:circle}
The function $\mathbf{r}(t) = \cos (t) \mathbf{i} + \sin (t) \mathbf{j}$ gives a parametrization of the unit circle $x^2 + y^2 = 1$ in the $xy$-plane. The velocity and acceleration are 
$$
\mathbf{v}(t) : = - \sin (t) \mathbf{i} + \cos (t) \mathbf{j}
$$ 
$$\mathbf{a}(t) = - \cos (t) \mathbf{i} - \sin (t) \mathbf{j}
$$
\end{example}



\begin{example}\label{ex:helix}
The equation
$$
\mathbf{r}(t) = \cos (t) \mathbf{i} + \sin (t) \mathbf{j} + t \mathbf{k},
$$
describes a helix curve. This curve wraps around the $z$-axis, and is contained in the surface of the cylinder $x^2 + y^2 = 1$. The velocity, speed and acceleration are given by
$$
\mathbf{v}(t) = - \sin (t) \mathbf{i} + \cos (t) \mathbf{j} + \mathbf{k}
$$
$$
v(t) : = |\mathbf{v}(t)| = \sqrt{\sin^2 (t) + \cos^2 (t) + 1^2} = \sqrt{2}
$$
$$
\mathbf{a}(t) : = - \cos (t) \mathbf{i} - \sin (t) \mathbf{j}
$$
\end{example}



\begin{example}\label{ex:deformedhelix}
The function $\mathbf{r}(t) = \cos (t) \mathbf{i} + \sin (t) \mathbf{j} + t^3 \mathbf{k}$ parametrizes a ``deformed'' helix, where the loops (corresponding to full revolutions about the $z$-axis) are compressed in the $z$-direction for small $|z|$ and elongated for large values of $|z|$.
\end{example}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .30\linewidth]{figures/11fig3.png}
\caption{The helical curve described in Example \ref{ex:helix}.}
\end{minipage}
\end{figure}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .30\linewidth]{figures/11fig5.png}
\caption{The deformed helix in Example \ref{ex:deformedhelix}.}
\end{minipage}
\end{figure}
\pagebreak



We have the following differentiation rules for vector-valued functions.  These follow easily from the single-variable differentiation rules applied to the component functions.

\begin{proposition}
Suppose that $\mathbf{u}, \mathbf{v} : I \rightarrow \mathbb{R}^3$ are vector-valued functions and $\lambda : I \rightarrow \mathbb{R}$ is scalar-valued, all of whom are differentiable.  Then,
\begin{eqnarray*}
(\mathbf{u} + \mathbf{v})' & =  & \mathbf{u}' + \mathbf{v}' \\
(\lambda \mathbf{u})' & = & \lambda' \mathbf{u} + \lambda \mathbf{u}' \\
(\mathbf{u} \cdot \mathbf{v})' &=& \mathbf{u}' \cdot \mathbf{v} + \mathbf{u}\cdot \mathbf{v}' \\
(\mathbf{u} \times \mathbf{v})' &=& \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'
\end{eqnarray*}
Further, if $s : J \rightarrow I$,  we have the following \textbf{vector-valued Chain Rule}
$$
\frac{d}{dt} \big[ \mathbf{u}(s(t)) \big] = s' (t) \mathbf{u}' (s(t)).
$$
We often use this identity when \textbf{reparametrizing curves} (more on this to come).
\end{proposition}


\section*{Worked Problems}

In Questions 1 and 2, we parametrize curves at the intersection of two surfaces. In each case, it helps that one of those surfaces projects to a simple curve (circle or line) in the $xy$-plane, and so does the 3-dimensional curve we need to parametrize. Therefore we can parametrize $x(t)$ and $y(t)$ first (we already know how to do it for both the line and the circle), and then we use the equation of the other surface to find $z(t)$. In general, if no such simplifications are available, parametrizing an intersection curve can be significantly more difficult.

Question 4 is important: we will use this equivalence many times in this class.



\begin{enumerate}[1.]
\item{\it Parametrize the curve of intersection of the cylinder $x^2 + y^2 = 4$ and the plane $z = 3x+5y$.}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .25\linewidth]{figures/11fig6.png}
\caption{The curve described in Worked Problem 1.}
\end{minipage}
\end{figure}

\medskip
\noindent
Let $x(t) = 2 \cos t$ and $y(t) = 2 \sin t$, so that 
$$
x^2 + y^2 = 4 (\cos^2 t + \sin^2 t ) = 4.
$$
Then,  for any $t,z\in \mathbb{R}$, the point $(x(t), y(t), z)$ lies on the cylinder $x^2 + y^2 = 4$.  For the point to also lie in the plane $z=3x+5y$, we let $$z(t) = x(t) + y(t) = 6\cos t + 10 \sin t.$$ Our parametrization is then
$$
\mathbf{r} (t) : = 2 \cos t \mathbf{i} + 2 \sin t \mathbf{j} +  (6\cos t + 10 \sin t ) \mathbf{k},
$$
and we choose $t \in [0, 2 \pi]$, which corresponds to one full revolution of the curve. 

\bigskip

\item{\it Parametrize the curve of intersection of the paraboloid $z = x^2 + y^2$ and the plane $x = 2y$.}

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .45\linewidth]{figures/11fig7.png}
\caption{The curve described in Worked Problem 2.}
\end{minipage}
\end{figure}

\medskip
\noindent
Set $y(t) = t$. Then from the equation of the plane, $x(t) = 2 y (t) = 2 t$, and from the equation of the paraboloid, $z(t) = x^2(t) + y^2(t) = 5 t^2$.  Our parametrization is then
$$
\mathbf{r}(t) = 2 t \mathbf{i} + t \mathbf{j} + 5 t^2 \mathbf{k}, \quad t \in \mathbb{R}.
$$

\bigskip

\item{\it Find a formula for $\frac{d}{dt} |\mathbf{r} (t) |$.}

\medskip
\noindent

By the Chain Rule for scalar functions, we have
\begin{eqnarray*}
\frac{d}{dt} \vert \mathbf{r} (t) \vert & = & \frac{d}{dt} \sqrt{\mathbf{r} \cdot \mathbf{r}} = \frac{1}{2 \sqrt{\mathbf{r} \cdot \mathbf{r}}} \cdot  \frac{d}{dt} (\mathbf{r} \cdot \mathbf{r}) \\
\quad & = & \frac{1}{2 \sqrt{\mathbf{r} \cdot \mathbf{r}}} \big( \mathbf{r}' \cdot \mathbf{r} + \mathbf{r} \cdot \mathbf{r}' \big) \\
\quad & = & \frac{2 \mathbf{r} \cdot \mathbf{r}'}{2 \sqrt{\mathbf{r} \cdot \mathbf{r}}} \\
\quad & = & \frac{\mathbf{r} \cdot \mathbf{r}'}{\vert \mathbf{r}\vert}
\end{eqnarray*}

\bigskip

\item{\it Suppose $\mathbf{r}:I\to \RR^n$ is differentiable, where $I$ is an interval. Prove that $|\mathbf{r}|$ is constant if and only if $\mathbf{v} \cdot \mathbf{r} = 0$ for all $t\in I$. 

\medskip

(Geometrically, the condition that $|\mathbf{r}| = a$ for some constant $a\geq 0$ means that the trajectory of $\bfr(t)$ must lie on the sphere of radius $a$ centered at the origin. The condition that  $\mathbf{v} \cdot \mathbf{r} = 0$ means that 
$\mathbf{r} \perp \mathbf{v}$. Here and below, we use the convention that the zero vector is perpendicular to any vector.)
}

\medskip
\noindent
For any $a\in[0,\infty)$, we have
the chain of equivalences
$$
\vert \mathbf{r} \vert = a\  \Leftrightarrow\  \vert \mathbf{r} \vert^2 = a^2\  \Leftrightarrow\  \mathbf{r} \cdot \mathbf{r} = a^2 
$$
So, $|\bfr|$ is constant if and only if $\bfr\cdot\bfr$ is constant, if and only if $\frac{d}{dt} \big( \mathbf{r} \cdot \mathbf{r} \big)= 0$.
However,
$$
\frac{d}{dt} \big( \mathbf{r} \cdot \mathbf{r} \big) = \mathbf{r}' \cdot \mathbf{r} + \mathbf{r} \cdot \mathbf{r}' = 2 \mathbf{r}' \cdot \mathbf{r} = 2 \mathbf{v} \cdot \mathbf{r}
$$
Combining the above, we get that $|\bfr|$ is constant if and only if 
 $\mathbf{v} \cdot \mathbf{r} = 0$ for all $t\in I$.  
 
\end{enumerate}

\begin{remark}
(a) By the same argument, $\vert \mathbf{v} \vert$ is constant if and only if $\mathbf{a} \cdot \mathbf{v} = 0$ for all $t$. 

\smallskip\noindent
(b) If instead of $\mathbf{r} \cdot \mathbf{v} = 0$
we have $\mathbf{r} \cdot \mathbf{v} > 0$ for all $t$, then
$$
\frac{d}{dt} \big(\vert \mathbf{r} \vert^2 \big) = 2 \mathbf{r} \cdot \mathbf{v} > 0
$$
and so $\vert \mathbf{r} \vert$ is increasing in $t$. Similarly, if $\mathbf{r} \cdot \mathbf{v} < 0$, then $\vert \mathbf{r} \vert$ is decreasing in $t$.
\end{remark}



\section*{Practice Problems}




\begin{enumerate}[1.]
\item
{
Let $C$ be the curve in the $xy$-plane that consists of the half-line $y=-x$, $x\leq 0$, and the half-line $y=x$,
$x\geq 0$. Give an example of a vector-valued function ${\bf r}:\rr\to\rr^2$ which parametrizes $C$ and such that ${\bf r}(t)$ is differentiable for
all $t$.}

\bigskip

\item
{
Suppose that ${\bf r}:[a,b]\to\rr^3$ is a parametrized curve such that ${\bf r}$ and ${\bf r}'$ are continuous on
$[a,b]$ and ${\bf r}'(t)\neq 0$. Does there have to exist a $t_0\in(a,b)$ such that $\bfr (b)-\bfr(a)=(b-a) \bfr'(t_0)$?
If yes, prove it. If no, give a counterexample.}


\bigskip

\item 
Let $A=(a_1,a_2)$ and $B=(b_1,b_2)$ be two points in $\rr^2$ such that $|AB|=1$.
Let also ${\bf a}=\overrightarrow{OA}=a_1\ii+a_2\jj$.


\medskip

(a) Find an orthogonal $2\times 2$ matrix $Q$ such that
\begin{itemize}
\item $\det(Q)=1$,
\item the mapping $L:\rr^2\to\rr^2$, defined by
$$
L
\left(\begin{matrix}
x\\ y
            \end{matrix}\right)
= {\bf a} + Q \left(\begin{matrix}
x\\ y
            \end{matrix}\right)
$$
satisfies 
$
L
\left(\begin{matrix}
0\\0
            \end{matrix}\right)
= \left(\begin{matrix}
a_1\\ a_2
            \end{matrix}\right)
$ and
$
L
\left(\begin{matrix}
1\\0
            \end{matrix}\right)
= \left(\begin{matrix}
b_1\\ b_2
            \end{matrix}\right).
$
 
 \end{itemize}


\medskip



(b) 
Let $\bfr:[0,1]\to \rr^2$ be given by $\bfr(t)=t\ii+g(t)\jj$, where
$g:[0,1]\to\rr$ is continuous on $[0,1]$ and satisfies $g(0)=g(1)=0$. This parametrizes a curve $\mathcal{C}$.

Let $L$ be as in (a). Then $L(\bfr(t))$ parametrizes the curve obtained from $\mathcal{C}$ by translating and rotating it so that the initial point is moved to $A$ and the terminal point is moved to $B$. (This is a fact from linear algebra that you do not have to prove.)

Evaluate $L(\bfr(t))$ in terms of components. Prove that we may choose a parametrization ${\bf w}:[0,1]\to\rr^2$ of the line segment $\overline{AB}$, and a unit vector ${\bf n}$ perpendicular to $\overrightarrow{AB}$, so that 
$$
L(\bfr(t)) = {\bf w}(t) + g(t){\bf n}.
$$


\medskip

 (c) 
Let $\mathcal{C}$ be the curve in $\rr^2$ parametrized by ${\bf r}(t)=t\ii+\sin(\pi t)\jj$,
$t\in[0,1]$. Let $\mathcal{C}_1$ be the curve in $\rr^2$ obtained from $\mathcal{C}$ by translating and rotating it in the plane so that the initial point is moved to $(3,4)$ and the terminal point is moved to $\left(\frac{7}{2}, 4+\frac{\sqrt{3}}{2}\right)$. 
Parametrize $\mathcal{C}_1$.










\end{enumerate}




