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Thread: Another Arc Length Problem

  1. #1
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    Another Arc Length Problem

    I've attached the problem I'm working on, I keep getting negative answers. I know I'm left with the square root of 5 next to the integral (if I'm not mistaken that is), which would end up being sqrt[5] times x evaluated from -3 to 1 and I keep getting negative answers. Help a brotha out?
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  2. #2
    Senior Member Twig's Avatar
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    hi

    hi

    When you have a curve $\displaystyle y =f(x) $

    $\displaystyle \int_{\gamma} \, ds = \int_{a}^{b} \sqrt{1+f'(x)^{2}} \, dx $

    So in this case, we get:

    $\displaystyle L = \int_{-3}^{1} \sqrt{1+2^{2}}\, dx = 4\sqrt{5} $
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  3. #3
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    Quote Originally Posted by Twig View Post
    hi

    When you have a curve $\displaystyle y =f(x) $

    $\displaystyle \int_{\gamma} \, ds = \int_{a}^{b} \sqrt{1+f'(x)^{2}} \, dx $

    So in this case, we get:

    $\displaystyle L = \int_{-3}^{1} \sqrt{1+2^{2}}\, dx = 4\sqrt{5} $
    How did you get that though? Because you'd have sqrt[5] minus the sqrt[5] times -3?
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    I've attached another similar problem, here I don't know how the heck to find the integral of 144x^1/4, it ends up being crazy, what would I do in this situation?
    Attached Thumbnails Attached Thumbnails Another Arc Length Problem-picture-2.png  
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  5. #5
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    Quote Originally Posted by fattydq View Post
    I've attached another similar problem, here I don't know how the heck to find the integral of 144x^1/4, it ends up being crazy, what would I do in this situation?
    $\displaystyle y = 5 + 8x^{\frac{3}
    {2}} \hfill \\$

    $\displaystyle y' = 12x^{\frac{1}
    {2}} \hfill \\$

    $\displaystyle 1 + \left( {y'} \right)^2 = 1 + 144x \hfill \\$

    $\displaystyle {\text{Arc Length}} \hfill \\$

    $\displaystyle \int\limits_0^1 {\sqrt {1 + 144x} } {\text{ }}dx \hfill \\$

    $\displaystyle {\text{Substitue, }}1 + 144x = t^2 \hfill \\$

    $\displaystyle \Rightarrow 144dx = 2tdt \Rightarrow dx = \frac{{tdt}}
    {{72}} \hfill \\$

    $\displaystyle {\text{As }}x \to 0,{\text{ }}t \to 1{\text{ and as }}x \to 1,{\text{ }}t \to \sqrt{145} \hfill \\$

    $\displaystyle = \frac{1}
    {{72}}\int\limits_1^{\sqrt{145}} {t^2 dt} = \left. {\frac{1}
    {{72}}\left( {\frac{{t^3 }}
    {3}} \right)} \right|_1^{\sqrt{145}} = ....... \hfill \\ $
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  6. #6
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    Quote Originally Posted by fattydq View Post
    I've attached another similar problem, here I don't know how the heck to find the integral of 144x^1/4, it ends up being crazy, what would I do in this situation?
    $\displaystyle x = y + y^3 \hfill \\$

    $\displaystyle \frac{{dx}}
    {{dy}} = 1 + 3y^2 \hfill \\$

    $\displaystyle 1 + \left( {\frac{{dx}}
    {{dy}}} \right)^2 = 1 + \left( {1 + 3y^2 } \right)^2 = 2 + 6y^2 + y^4 \hfill \\$

    $\displaystyle {\text{Arc Length}} \hfill \\$

    $\displaystyle \int\limits_2^5 {\left( {2 + 6y^2 + y^4 } \right)} {\text{ }}dy \hfill \\$
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  7. #7
    Senior Member Twig's Avatar
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    Quote Originally Posted by fattydq View Post
    How did you get that though? Because you'd have sqrt[5] minus the sqrt[5] times -3?


    $\displaystyle \int_{-3}^{1} \sqrt{5}\, dx = \sqrt{5}\, \int_{-3}^{1} \, dx = \sqrt{5}(1-(-3)) = \sqrt{5}(4) = 4\sqrt{5} $
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