
\section{Curves and Vector-Valued Functions}

\begin{enumerate}[1.]
\item
{\it 
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 finction ${\bf r}:\rr\to\rr^2$ which parametrizes $C$ and such that ${\bf r}(t)$ is differentiable for
all $t$.}
\medskip

We are looking for a differentiable parametrization of the absolute value curve. We claim that $\bfr(t) = (t^{3}, |t^{3}|)$ works. 
This parametrization is clearly a bijection from $\rr$ to $C$, the component $x(t)=t^3$ is differentiable for all $t$, and $y(t)=|t^3|$  is differentiable for all $t\neq 0$. 
It remains to check that $y(t)$ is differentiable at $t = 0$:
\begin{align*}
\lim_{x \rightarrow 0^{+}} |t^3|/t & = \lim_{x \rightarrow 0^{+}} t^3/t = \lim_{x \rightarrow 0^{+} } t^2 = 0,
\\
\lim_{x \rightarrow 0^{-}} |t^3|/t & = \lim_{ x \rightarrow 0^{-}} -t^3/t = \lim_{x \rightarrow 0^{+} } -t^2
=0.
\end{align*}
Since the one-sided derivatives exist and are equal, $y(t)$ is differentiable at $t=0$.

\bigskip

\item
{\it 
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.}

\medskip

The answer is no. Let $\bfr=\cos t\,\ii + \sin t\, \jj + t\, \kk$ and $[a,b]=[0,2\pi]$, so that
$\bfr(b)-\bfr(a) = \ii +2\pi\kk -\ii =2\pi\kk$. But $\bfr'(t) = -\sin t\,\ii + \cos t\,\jj +\kk$ is never parallel to the
$z$ axis.


\bigskip




\item {\it 
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
We need to find $Q= 
\left(\begin{matrix}
c_{11} & c_{12} \\ c_{21} & c_{22}
            \end{matrix}\right)
$ so that $Q$ is orthogonal and the above equations hold.
The first equation 
$$
L
\left(\begin{matrix}
0\\0
            \end{matrix}\right)
= {\bf a}+Q{\bf 0}=  \left(\begin{matrix}
a_1\\ a_2
            \end{matrix}\right)
$$
is true for any $Q$.  Let ${\bf v}=v_1\ii+v_2\jj$, where
$v_1=b_1-a_1$ and $v_2=b_2-a_2$. Then the second equation is
$$
\left(\begin{matrix}
b_1\\ b_2
            \end{matrix}\right)
            =L
\left(\begin{matrix}
1\\0
            \end{matrix}\right)
= \left(\begin{matrix}
a_1\\a_2
            \end{matrix}\right) + \left(\begin{matrix}
c_{11} & c_{12} \\ c_{21} & c_{22}
            \end{matrix}\right)\left(\begin{matrix}
1\\0
            \end{matrix}\right)
  = \left(\begin{matrix}
a_1\\a_2
            \end{matrix}\right)  + \left(\begin{matrix}
c_{11}\\c_{21}
            \end{matrix}\right)     
$$
so that $c_{11}=b_1-a_1=v_1$ and $c_{21}=b_2-a_2=v_2$. 

For $Q$ to be orthogonal, the second column must be orthogonal to the first one, therefore a multiple of 
$ \left(\begin{matrix}
-v_2\\v_1
            \end{matrix}\right)$. We also have the condition $\det(Q)=1$:
$$            
1=\det(Q)= \left|\begin{matrix}
v_1 & -\lambda v_2 \\ v_2 & \lambda v_1
            \end{matrix}\right| = \lambda(v_1^2+v_2^2).
            $$
But $v_1^2+v_2^2=|AB|^2=1$ by assumption, so that $\lambda=1$ and $Q=\left(\begin{matrix}
v_1 & - v_2 \\ v_2 & v_1
            \end{matrix}\right)
            $.

\medskip


{\it (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
We compute
\begin{align*}
L(\bfr(t))
& = \left(\begin{matrix}
a_1\\a_2
            \end{matrix}\right) + \left(\begin{matrix}
v_1 & - v_2 \\ v_2 & v_1
            \end{matrix}\right)\left(\begin{matrix}
t\\ g(t) 
            \end{matrix}\right)
            \\[2mm]
& =\left(\begin{matrix}
a_1\\a_2
            \end{matrix}\right) +
 \left(\begin{matrix}
v_1t - v_2 g(t) \\ v_2 t + v_1 g(t)
            \end{matrix}\right) 
\\[2mm]
&=
\left(\begin{matrix}
a_1+tv_1  - v_2 g(t) \\ a_2+tv_2 + v_1 g(t)
            \end{matrix}\right) 
= {\bf w}(t) + g(t){\bf n},
\end{align*}
where
${\bf w}(t)= {\bf a} + t{\bf v}$ for $t\in[0,1]$ is a parametrization of the line segment from $A$ to $B$, and ${\bf n}=-v_2\ii + v_1\jj$ is orthogonal to ${\bf v}$. We also have $|{\bf n}|=1$, using again that $|AB|=1$.

\bigskip

{\it (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 $\mathcal{C}_1$.
}



\medskip

Let $A=(3,4)$ and $B=\left(\frac{7}{2}, 4+\frac{\sqrt{3}}{2}\right)$, and note that
$${\bf v}=\overrightarrow{AB}= \frac{1}{2}\ii + \frac{\sqrt{3}}{2}\jj$$
is a unit vector. 
We use (b) with ${\bf n}=-v_2\ii+v_1\jj=- \frac{\sqrt{3}}{2}\ii +  \frac{1}{2}\jj$, and with the
line segment $\overline{AB}$ parametrized by
\begin{align*}
{\bf w}(t)&= {\bf a} + t{\bf v}= (3\ii+4\jj)+t\left(\frac{1}{2}\ii +\frac{\sqrt{3}}{2}\jj\right).
\end{align*}
This gets us the parametrization
\begin{align*}
L(\bfr(t)) & = {\bf w}(t) + \sin(t){\bf n}= (3\ii+4\jj)+t\left(\frac{1}{2}\ii +\frac{\sqrt{3}}{2}\jj\right)
+\sin(t) \left(-\frac{\sqrt{3}}{2}\ii + \frac{1}{2} \jj\right)\\
&= \left( 3 + \frac{t}{2}-\frac{\sqrt{3}\sin(t)}{2}\right)\ii
+\left( 4 + \frac{\sqrt{3}t}{2}+\frac{\sin(t)}{2}\right) \jj
\end{align*}















\end{enumerate}

