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Thread: Arc Length

  1. #1
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    Arc Length

    I could use some help determining the arc length determined by the function:

    f(x) = sin x over the interval $\displaystyle [0, \pi]$

    I know that I need to use the formula $\displaystyle s = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$

    which leads to evaluating: $\displaystyle s = \int_{0}^{\pi} \sqrt{1 + cos^{2}x} dx$,

    which looks like it should be easy to do, but I'm just missing it. I've tried Pythagorean Indentity and power reduction formulas to replace $\displaystyle cos^{2}x$ but to no avail. As always, your help would be greatly appreciated.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by kaiser0792 View Post
    I could use some help determining the arc length determined by the function:

    f(x) = sin x over the interval $\displaystyle [0, \pi]$

    I know that I need to use the formula $\displaystyle s = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$

    which leads to evaluating: $\displaystyle s = \int_{0}^{\pi} \sqrt{1 + cos^{2}x} dx$,

    which looks like it should be easy to do, but I'm just missing it. I've tried Pythagorean Indentity and power reduction formulas to replace $\displaystyle cos^{2}x$ but to no avail. As always, your help would be greatly appreciated.
    If we have the form

    $\displaystyle \sqrt{a^2 + x^2} $ we let $\displaystyle x=atan \theta $

    In this case

    $\displaystyle cos \theta = tan \theta $

    So

    $\displaystyle \int \sqrt { 1 + cos^2 \theta } dx = \int \sqrt { 1 + tan^2 \theta } dx $

    Of course, then we have

    $\displaystyle \int \sqrt { 1 + tan^2 \theta } dx = \int \sqrt { sec^2 x } dx = \int sec x dx $

    Change your bounds, and evaluate the integral. No problem!
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    Thanks Allan.

    So you're essentially using trig substitution? I considered it, but it threw me using tan x in terms of cos x. I appreciate your help.
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    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by kaiser0792 View Post
    Thanks Allan.

    So you're essentially using trig substitution? I considered it, but it threw me using tan x in terms of cos x. I appreciate your help.
    Absolutely, but if you recall if we let

    $\displaystyle cos \theta = tan \theta $

    This is the same thing as saying

    $\displaystyle cos \theta = \frac { sin \theta }{ cos \theta } $

    $\displaystyle cos^2 \theta = sin \theta $

    which is the same thing as saying

    $\displaystyle 1 - sin^2 \theta = sin \theta $

    Of course this becomes the quadratic equation

    $\displaystyle 0 = sin^2 \theta + sin \theta - 1 $

    This has a value, and is therefore no problemmmmo!

    In other words, what i'm trying to say is that

    $\displaystyle cos \theta $ will be equal to $\displaystyle tan \theta $ for some value of theta. We need to find that out!
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    Excellent points and explanation. I am very grateful.
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  6. #6
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    Quote Originally Posted by AllanCuz View Post
    If we have the form

    $\displaystyle \sqrt{a^2 + x^2} $ we let $\displaystyle x=atan \theta $

    In this case

    $\displaystyle cos \theta = tan \theta $
    No, $\displaystyle cos(\theta)\ne tan(\theta)$! You mean $\displaystyle cos(x)= tan(\theta)$.

    So

    $\displaystyle \int \sqrt { 1 + cos^2 \theta } dx = \int \sqrt { 1 + tan^2 \theta } dx $

    Of course, then we have

    $\displaystyle \int \sqrt { 1 + tan^2 \theta } dx = \int \sqrt { sec^2 x } dx = \int sec x dx $

    Change your bounds, and evaluate the integral. No problem!
    No, because you can't have "dx" in your $\displaystyle \theta$ integral. If $\displaystyle cos(x)= tan(\theta)$ then $\displaystyle -sin(x)dx= sec^2(\theta)d\theta$. Since $\displaystyle sin(x)= \sqrt{1- cos^2(\theta)}$, the integral becomes $\displaystyle \int \frac{sec^4(\theta)}{\sqrt{1- tan^2(\theta)}} d\theta$ and I don't think you are any better off.
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    No, $\displaystyle cos(\theta)\ne tan(\theta)$! You mean $\displaystyle cos(x)= tan(\theta)$.



    No, because you can't have "dx" in your $\displaystyle \theta$ integral. If $\displaystyle cos(x)= tan(\theta)$ then $\displaystyle -sin(x)dx= sec^2(\theta)d\theta$. Since $\displaystyle sin(x)= \sqrt{1- cos^2(\theta)}$, the integral becomes $\displaystyle \int \frac{sec^4(\theta)}{\sqrt{1- tan^2(\theta)}} d\theta$ and I don't think you are any better off.
    Yeah...that's right, i was drunk last night lol definately my bad on that one
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