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\centerline{MATH 136 -- Calculus 2}
\centerline{Arc Length}
\centerline{March 13, 2020}
\vskip 10pt
\noindent
{\it Background}
We have now seen why the arc length of a graph $y = f(x)$ for $x = a$
to $x = b$ can be computed by the integral 
$$\text{Arc Length} = \int_a^b \sqrt{1 + (f'(x))^2}\ dx$$
It is actually relatively rare that these integrals can be computed
exactly using the FTC.  Many of them require numerical approximations
because the complicated integrand tends to have no elementary 
antiderivative.  The following cases are a few of the ones
that \emph{can be} evaluated by the FTC, either because the $\sqrt{1 + (f'(x))^2}$
is rather simple, or because the $1 + (f'(x))^2$ is set up to be a perfect 
square(!)
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\noindent
{\it Questions}
\vskip 10pt
\begin{enumerate}
\item Find the arc length of the curve $y = 4 + x^{3/2}$ for $x = 0$ to $x = 4$.
\item[] {\it Solution:}  With $f(x) = 4 + x^{3/2}$, $f'(x) = \frac{3x^{1/2}}{2}$ and the 
arc length integral is 
$$L = \int_0^4 \sqrt{1 + \left(\frac{3 x^{1/2}}{2}\right)^2}\ dx = \int_0^4 \sqrt{1 + \frac{9x}{4}}\ dx$$
We can evaluate this by letting $u = 1 + \frac{9x}{4}$ and $dx = \frac{4}{9}\ du$.
$$L =  \int_{u = 1}^{u = 10} \frac{4}{9} u^{1/2}\ du = \left. \frac{8}{27} u^{3/2}\right|_1^{10} = 
\frac{8}{27}(10\sqrt{10} - 1).$$
\item Find the arc length of the curve $y = \frac{x^{3/2}}{3} - x^{1/2}$ for $x = 2$ to $x = 8$.
\item[] {\it Solution:}  Here for $f(x) = \frac{x^{3/2}}{3} - x^{1/2}$, we have 
$f'(x) = \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2}$.  When we compute the quantity inside
the square root in the arc length integrand a sort of algebraic miracle happens:
$$1 + (\frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2})^2 = 1 + \frac{x}{4} - \frac{1}{2} + \frac{1}{4x}
= \frac{x}{4} + \frac{1}{2} + \frac{1}{4x}$$
is also the perfect square $\left(\frac{x^{1/2}}{2} + \frac{x^{-1/2}}{2}\right)^2$.  So 
we must compute 
\begin{align*}
\int_2^8 \frac{x^{1/2}}{2} + \frac{x^{-1/2}}{2}\ dx 
&= \left. \frac{x^{3/2}}{3} + x^{1/2}\right|_2^8\\
&= \frac{11\sqrt{8}}{3} - \frac{5\sqrt{2}}{3}\\
&= \frac{17\sqrt{2}}{3}
\end{align*}
(using $\sqrt{8} = 2\sqrt{2}$ to simplify).  
\item The most important example of this type from a practical point of 
view is probably the following.  Chains or wires hanging under their own weight suspended
from their endpoints have the shape of a curve called the \emph{catenary}.  Up 
to scaling, a catenary can always be given by an function of the form 
$$f(x) = \frac{e^x + e^{-x}}{2}.$$ 
This function is also called the \emph{hyperbolic cosine} and there are several sections
in our textbook dealing with these and related functions.  
Find the arc length of the catenary curve above for
$x = -1$ to $x = 1$.  
\item[] {\it Solution:}  This one is similar to the previous one in that 
when we compute $1 + (f'(x))^2$ for $f(x) = \frac{e^x + e^{-x}}{2}$, the result
is also a perfect square:
$$1 + \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x}}{4} + \frac{1}{2} + \frac{e^{-2x}}{4} 
= \left(\frac{e^x + e^{-x}}{2}\right)^2.$$
Hence the arc length integral is 
\begin{align*}
\int_{-1}^{1} \sqrt{\left(\frac{e^x + e^{-x}}{2}\right)^2}\ dx &= \int_{-1}^1 \frac{e^x + e^{-x}}{2}\ dx\\
&= \left. \frac{e^x - e^{-x}}{2}\right|_{-1}^1\\
&= \frac{e - e^{-1}}{2} - \frac{e^{-1} - e}{2}\\
&= e - e^{-1}.
\end{align*} 
\end{enumerate}
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