Arc Length

We can deduce that the length of a curve with parametric equations Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} x=f(t) \\ y=g(t) \end{cases} } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\le t\le 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=\frac{e^x+e^{-x}}{2}} from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}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-\frac1e}{2}}\end{align}}

Resources

• Arc Length, WikiBooks: Calculus
• Arc Length and Curvature, OpenStax