### Length of curves

The basic point here is *a formula obtained by using the ideas of
calculus*: the length of the graph of $y=f(x)$ from $x=a$ to $x=b$ is
$$\hbox{ arc length
}=\int_a^b\;\sqrt{1+\left({dy\over dx}\right)^2}\;dx$$ Or, if the
curve is *parametrized* in the form $$x=f(t)\;\;\;\;\;y=g(t)$$
with the parameter $t$ going from $a$ to $b$, then $$\hbox{ arc length
}=\int_a^b\;
\sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt$$

This formula comes from approximating the curve by straight
lines connecting successive points on the curve, using the Pythagorean
Theorem to compute the lengths of these segments in terms of the
change in $x$ and the change in $y$. In one way of writing, which also
provides a good *heuristic* for remembering the formula, if a small
change in $x$ is $dx$ and a small change in $y$ is $dy$, then the
length of the hypotenuse of the right triangle with base $dx$ and
altitude $dy$ is (by the Pythagorean theorem)
$$\hbox{ hypotenuse }=\sqrt{dx^2+dy^2}=
\sqrt{1+\left({dy\over dx}\right)^2}\;dx$$

Unfortunately, by the nature of this formula, most of the
integrals which come up are *difficult* or *impossible* to
‘do’. But if one of these really mattered, we could still estimate it
by *numerical integration*.

#### Exercises

- Find the length of the curve $y=\sqrt{1-x^2}$ from $x=0$ to $x=1$.
- Find the length of the curve $y={ 1 \over 4 }(e^{2x}+e^{-2x})$ from $x=0$ to $x=1$.
- Set up (but do not evaluate) the integral to find the length of the piece of the parabola $y=x^2$ from $x=3$ to $x=4$.

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##### Calculus Refresher

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