Difference between revisions of "Arc Length"

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We can deduce that the length of a curve with parametric equations <math>\begin{cases} x=f(t) \\ y=g(t) \end{cases} </math>, <math>a\le t\le b </math> should be:<blockquote><math>L=\int_a^b\sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2}dt </math></blockquote>Since vector functions are fundamentally parametric equations with directions, we can utilize the formula above into the length of a space curve.
 
We can deduce that the length of a curve with parametric equations <math>\begin{cases} x=f(t) \\ y=g(t) \end{cases} </math>, <math>a\le t\le b </math> should be:<blockquote><math>L=\int_a^b\sqrt{\biggl(\frac{dx}{dt}\biggr)^2+\biggl(\frac{dy}{dt}\biggr)^2}dt </math></blockquote>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===
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==Arc length of a space curve==
  
 
If the curve has the vector equation <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle,a\le t\le b</math>, or, equivalently, the parametric equations <math>x=f(t),y=g(t),z=h(t)</math>, where <math>f',g',h'</math> are continuous, then the length of the curve from <math>t=a</math> to <math>t=b</math> is:
 
If the curve has the vector equation <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle,a\le t\le b</math>, or, equivalently, the parametric equations <math>x=f(t),y=g(t),z=h(t)</math>, where <math>f',g',h'</math> are continuous, then the length of the curve from <math>t=a</math> to <math>t=b</math> is:
 
:<math>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</math>}}
 
:<math>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</math>}}
 
For those who prefer simplicity, the formula can be rewritten into:<blockquote><math>L=\int_a^b|\mathbf{r}'(t)|dt\quad </math> or <math>\quad\frac{dL}{dt}=|\mathbf{r}'(t)|</math></blockquote>
 
For those who prefer simplicity, the formula can be rewritten into:<blockquote><math>L=\int_a^b|\mathbf{r}'(t)|dt\quad </math> or <math>\quad\frac{dL}{dt}=|\mathbf{r}'(t)|</math></blockquote>
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===Example Problems===
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1. Find the circumference of the circle given by the parametric equations <math>x(t)=R\cos(t),y(t)=R\sin(t)</math> , with <math>t\in[0,2\pi]</math>.
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:<math>\begin{align}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\\
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&=\int\limits_0^{2\pi}\sqrt{\big(-R\sin(t)\big)^2+\big(R\cos(t)\big)^2}dt\\
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&=\int\limits_0^{2\pi}\sqrt{R^2\big(\sin^2(t)+\cos^2(t)\big)}dt\\
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&=\int\limits_0^{2\pi}Rdt\\
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&=R\cdot t\Big|_0^{2\pi}\\
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&=\mathbf{2\pi R}\end{align}</math>
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2. Find the length of the curve <math>y=\frac{e^x+e^{-x}}{2}</math> from <math>x=0</math> to <math>x=1</math>.
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:<math>\begin{align}s&=\int\limits_0^1\sqrt{1+\left(\frac{d}{dx}\left(\frac{e^{x}+e^{-x}}{2}\right)\right)^2}dx\\
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&=\int\limits_0^1\sqrt{1+\left(\frac{e^{x}-e^{-x}}{2}\right)^2}dx\\
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&=\int\limits_0^1\sqrt{1+\frac{e^{2x}-2+e^{-2x}}{4}}dx\\
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&=\int\limits_0^1\sqrt{\frac{e^{2x}+2+e^{-2x}}{4}}dx\\
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&=\int\limits_0^1\sqrt{\left(\frac{e^{x}+e^{-x}}{2}\right)^2}dx\\
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&=\int\limits_0^1\frac{e^{x}+e^{-x}}{2}dx\\
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&=\frac{e^{x}-e^{-x}}{2}\bigg|_0^1\\
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&=\mathbf{\frac{e-\frac1e}{2}}\end{align}</math>
  
 
==Resources==
 
==Resources==
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* [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
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==Licensing==
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Content obtained and/or adapted from:
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* [https://en.wikibooks.org/wiki/Calculus/Arc_length Arc Length, WikiBooks: Calculus] under a CC BY-SA license

Latest revision as of 10:15, 2 November 2021

We can deduce that the length of a curve with parametric equations , should be:

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

Example Problems

1. Find the circumference of the circle given by the parametric equations , with .

2. Find the length of the curve from to .

Resources

Licensing

Content obtained and/or adapted from: