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| ==Resources== | | ==Resources== |
| + | * [https://en.wikibooks.org/wiki/Calculus/Arc_length Arc Length], WikiBooks: Calculus |
| * [https://openstax.org/books/calculus-volume-3/pages/3-3-arc-length-and-curvature Arc Length and Curvature], OpenStax | | * [https://openstax.org/books/calculus-volume-3/pages/3-3-arc-length-and-curvature Arc Length and Curvature], OpenStax |
Revision as of 14:01, 1 October 2021
We can deduce that the length of a curve with parametric equations
,
should be:
![{\displaystyle L=\int _{a}^{b}{\sqrt {{\biggl (}{\frac {dx}{dt}}{\biggr )}^{2}+{\biggl (}{\frac {dy}{dt}}{\biggr )}^{2}}}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76976463e914bfb3e9649f9bf055cc4302ecc79e)
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
, or, equivalently, the parametric equations
, where
are continuous, then the length of the curve from
to
is:
}}
For those who prefer simplicity, the formula can be rewritten into:
or ![{\displaystyle \quad {\frac {dL}{dt}}=|\mathbf {r} '(t)|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7af8a9ad556637028c9716854b306112ad1b3037)
Example Problems
1. Find the circumference of the circle given by the parametric equations
, with
.
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb61ab3a36536f1b8ef4b6e557dda44b7a2349d7)
2. Find the length of the curve
from
to
.
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c37fa7e5bde769230d12416141f2e0b9080ed7f)
Resources