Arc Length

We can deduce that the length of a curve with parametric equations ${\begin{cases}x=f(t)\\y=g(t)\end{cases}}$, $a\leq t\leq b$ should be:

$L=\int _{a}^{b}{\sqrt {{\biggl (}{\frac {dx}{dt}}{\biggr )}^{2}+{\biggl (}{\frac {dy}{dt}}{\biggr )}^{2}}}dt$

Since vector functions are fundamentally parametric equations with directions, we can utilize the formula above into the length of a space curve.

Arc length of a space curve

If the curve has the vector equation $\mathbf {r} (t)=\langle f(t),g(t),h(t)\rangle ,a\leq t\leq b$, or, equivalently, the parametric equations $x=f(t),y=g(t),z=h(t)$, where $f',g',h'$ are continuous, then the length of the curve from $t=a$ to $t=b$ is:

$L=\int _{a}^{b}{\sqrt {[f'(t)]^{2}+[g'(t)]^{2}+[h'(t)]^{2}}}dt=\int _{a}^{b}{\sqrt {{\biggl (}{\frac {dx}{dt}}{\biggr )}^{2}+{\biggl (}{\frac {dy}{dt}}{\biggr )}^{2}+{\biggl (}{\frac {dx}{dz}}{\biggr )}^{2}}}dt$}}

For those who prefer simplicity, the formula can be rewritten into:

$L=\int _{a}^{b}|\mathbf {r} '(t)|dt\quad$ or $\quad {\frac {dL}{dt}}=|\mathbf {r} '(t)|$

Example Problems

1. Find the circumference of the circle given by the parametric equations $x(t)=R\cos(t),y(t)=R\sin(t)$ , with $t\in [0,2\pi ]$.

${\begin{aligned}s&=\int \limits _{0}^{2\pi }{\sqrt {\left({\tfrac {d}{dt}}{\big (}R\cos(t){\big )}\right)^{2}+\left({\tfrac {d}{dt}}{\big (}R\sin(t){\big )}\right)^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {{\big (}-R\sin(t){\big )}^{2}+{\big (}R\cos(t){\big )}^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {R^{2}{\big (}\sin ^{2}(t)+\cos ^{2}(t){\big )}}}dt\\&=\int \limits _{0}^{2\pi }Rdt\\&=R\cdot t{\Big |}_{0}^{2\pi }\\&=\mathbf {2\pi R} \end{aligned}}$

2. Find the length of the curve $y={\frac {e^{x}+e^{-x}}{2}}$ from $x=0$ to $x=1$.

${\begin{aligned}s&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {d}{dx}}\left({\frac {e^{x}+e^{-x}}{2}}\right)\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {e^{x}-e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+{\frac {e^{2x}-2+e^{-2x}}{4}}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\frac {e^{2x}+2+e^{-2x}}{4}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\left({\frac {e^{x}+e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\frac {e^{x}+e^{-x}}{2}}dx\\&={\frac {e^{x}-e^{-x}}{2}}{\bigg |}_{0}^{1}\\&=\mathbf {\frac {e-{\frac {1}{e}}}{2}} \end{aligned}}$

Resources

• Arc Length and Curvature, OpenStax