# Arc Length for Parametric Equations

## Theorem

Let $x = \map f t$ and $y = \map g t$ be real functions of a parameter $t$.

Let these equations describe a curve $\CC$ that is continuous for all $t \in \closedint a b$ and continuously differentiable for all $t \in \openint a b$.

Suppose that the graph of the curve does not intersect itself for any $t \in \openint a b$.

Then the arc length of $\CC$ between $a$ and $b$ is given by:

$\ds s = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \rd t$

for $\dfrac {\d x} {\d t} \ne 0$.

## Proof

 $\ds s$ $=$ $\ds \int_a^b \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$ Definition of Arc Length $\ds$ $=$ $\ds \int_a^b \sqrt {\paren {\frac {\frac {\d x} {\d t} } {\frac {\d x} {\d t} } }^2 + \paren {\frac {\frac {\d y}{\d t} } {\frac {\d x} {\d t} } }^2} \rd x$ because $\paren {\dfrac {\frac {\d x} {\d t} } {\frac {\d x}{\d t} } }^2 = 1$, and from corollary to chain rule $\ds$ $=$ $\ds \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \paren {\frac 1 {\frac {\d x} {\d t} } } \rd x$ factoring $\dfrac {\d x} {\d t}$ out of the radicand. No absolute value is needed as length cannot be negative. $\ds$ $=$ $\ds \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \frac {\d t} {\d x} \rd x$ Derivative of Inverse Function $\ds$ $=$ $\ds \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \rd t$ Integration by Substitution

$\blacksquare$