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Math Help - 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 [0, \pi]

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

    which leads to evaluating: 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 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 [0, \pi]

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

    which leads to evaluating: 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 cos^{2}x but to no avail. As always, your help would be greatly appreciated.
    If we have the form

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

    In this case

     cos \theta = tan \theta

    So

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

    Of course, then we have

     \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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  3. #3
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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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  4. #4
    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

    cos \theta = tan \theta

    This is the same thing as saying

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

     cos^2 \theta = sin \theta

    which is the same thing as saying

     1 - sin^2 \theta = sin \theta

    Of course this becomes the quadratic equation

     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

     cos \theta will be equal to  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

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

    In this case

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

    So

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

    Of course, then we have

     \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 \theta integral. If cos(x)= tan(\theta) then -sin(x)dx= sec^2(\theta)d\theta. Since sin(x)= \sqrt{1- cos^2(\theta)}, the integral becomes \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, cos(\theta)\ne tan(\theta)! You mean cos(x)= tan(\theta).



    No, because you can't have "dx" in your \theta integral. If cos(x)= tan(\theta) then -sin(x)dx= sec^2(\theta)d\theta. Since sin(x)= \sqrt{1- cos^2(\theta)}, the integral becomes \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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