
\section{Arc Length for Curves}

\begin{enumerate}[1.]
\item 
{\it 
Let $C$ be the curve $x=e^{2t}$, $y=2e^t $, $z=t$ for $t \in\mathbb{R}$. 
Find the length of the curve from $t=0$ to $t=1$.}

\medskip


We have $\bfr'(t)=2e^{2t}\ii + 2e^t \jj + \kk$, so that
$$
|\bfr'(t)|= \sqrt{ 4e^{4t}+ 4e^{2t}+1} = 2e^{2t}+1.
$$
The length is $\int_0^1 (2e^{2t}+1)dt = e^{2t}+t\Big|_0^1 = e^2+1-1=e^2$.


\bigskip


\item 
{\it Let $\bfr(t)=e^t\cos(2t) \ii + e^t \sin(2t)\jj$ be a parametrization of a curve $C$ in $\mathbb{R}^2$. 

\medskip\noindent
(a)
Let $s(b)$ be the length of the curve $\bfr(t)$ from $t=-\infty$ to $t=b$. (This is defined as the limit
as $a\to -\infty$ of the length of $\bfr(t)$ from $t=a$ to $t=b$.) Find $s(b)$ as a function of $b$. }

\medskip

We have
\begin{align*}
x'&= e^t \cos (2t) -2e^t  \sin(2t)\\
y'&= e^t \sin(2t) +2e^t\cos(2t)
\\
|{\bf r}'|^2 &= e^{2t} \Big[ \cos^2(2t)-4\sin(2t)\cos(2t)+4 \sin^2(2t)\\
&+ \sin^2(2t)+4\sin(2t)\cos(2t)+4cos^2(2t) \Big]
\\
&= 5e^{2t}
\\
|{\bf r}'|&=\sqrt{5} e^t
\end{align*}
so that
$$
s(b)=  \int_{-\infty}^b \sqrt{5}e^t dt = \sqrt{5} e^b.
$$

\medskip\noindent
{\it (b) Use part (a) to give an arclength parametrization of $C$.}

\medskip


Let $s(t)=\sqrt{5} e^t$ from (a), then $e^t=s/\sqrt{5}$, so that $t=\ln(s/\sqrt{5})$. Therefore
the arclength parametrization of $C$ in terms of $s$ is
$$
\bfr(s) = \frac{s}{\sqrt{5}} \cos\Big(2\ln \frac{s}{\sqrt{5}} \Big)\ii
+ \frac{s}{\sqrt{5}} \sin\Big(2\ln \frac{s}{\sqrt{5}} \Big)\jj.
$$


\bigskip


\item 
{\it 
The paraboloid $z=x^2+y^2+1$ intersects the surface $x^2-2x+y^2=0$ in a curve. Prove that the length of this curve is less than 15.}

\medskip

 Along the intersection curve, we have $z=x^2+y^2+1$ and $x^2+y^2=2x$, so that $z=2x+1$. Now, notice that the equation of the second surface can be rewritten as
 $$
 (x-1)^2+y^2 =1,
 $$
which is a right circular cylinder with the symmetry axis given by $x=1,y=0$. This gives the parametrization
$$
x=1+\cos t,\ y=\sin t,\ z=2(1+\cos t)+1=2\cos t +3,
$$
with $0\leq t\leq 2\pi$. The arc length integral is
$$
L=\int_0^{2\pi} \sqrt{(-\sin t)^2+(\cos t)^2 + (-2\sin t)^2} dt
=\int_0^{2\pi} \sqrt{1+4\sin^2 t} dt.
$$
Since $|\cos t|\leq 1$, we get
$$
L\leq \int_0^{2\pi} \sqrt{5} dt=2\pi\sqrt{5}\approx 14.05 <15.
$$








\end{enumerate}

