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Math Help - Length of a curve

  1. #1
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    Length of a curve

    Can someone help with these two questions.

    Determine the length of the given curve

    y=x^2 from 0 to \frac {\sqrt 3} {2}

    y=\ln x from 1 to \sqrt 3
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  2. #2
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    Quote Originally Posted by chengbin View Post
    Can someone help with these two questions.

    Determine the length of the given curve

    y=x^2 from 0 to \frac {\sqrt 3} {2}

    y=\ln x from 1 to \sqrt 3
    Hi

    The elementary displacement along a curve is

    dl = \sqrt{dx^2 + dy^2}

    If y = f(x) then dy = f'(x) \:dx

    Therefore dl = \sqrt{1 + f'^2(x)}\:dx

    The length of the curve is L = \int_{x_0}^{x_1}\sqrt{1 + f'^2(x)}\:dx
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  3. #3
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    Thanks, I know that, but I just can't get the answer in my solution book. So I was hoping someone will write a step by step instruction of solving it.
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  4. #4
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    If you apply the formula for the first one

    L = \int_{0}^{\frac{\sqrt{3}}{2}}\sqrt{1 + 4x^2}\:dx

    which can be solved by parts
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  5. #5
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    And I don't know how to integrate \int \frac {dx} {cos^3x}
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by chengbin View Post
    And I don't know how to integrate \int \frac {dx} {cos^3x}
    try to do so

    \int {\frac{{dx}}{{{{\cos }^3}x}}}  = \int {\frac{{\cos x}}<br />
{{{{\cos }^4}x}}dx}  = \int {\frac{{\cos x}}{{{{\left( {1 - {{\sin }^2}x} \right)}^2}}}dx}  = \left\{ \begin{gathered}\sin x = t, \hfill \\\cos xdx = dt \hfill \\ \end{gathered}  \right\} =

    = \int {\frac{{dt}}{{{{\left( {1 - {t^2}} \right)}^2}}}}  = \int {\frac{{dt}}{{{{\left( {{t^2} - 1} \right)}^2}}}}  = \int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2}{{\left( {t - 1} \right)}^2}}}}.

    \frac{1}{{{{\left( {t + 1} \right)}^2}{{\left( {t - 1} \right)}^2}}} = \frac{A}{{{{\left( {t + 1} \right)}^2}}} + \frac{B}{{t + 1}} + \frac{C}{{{{\left( {t - 1} \right)}^2}}} + \frac{D}{{t - 1}} \Leftrightarrow

    \Leftrightarrow 1 = A{\left( {t - 1} \right)^2} + B\left( {t + 1} \right){\left( {t - 1} \right)^2} + C{\left( {t + 1} \right)^2} + D\left( {t - 1} \right){\left( {t + 1} \right)^2} \Rightarrow

    \Rightarrow \left. {\begin{array}{*{20}{c}}{t =  - 1} \hfill  \\{t = 1} \hfill  \\{t = 2} \hfill  \\{t = 0} \hfill  \\\end{array} } \right|{\text{ }}\left\{ \begin{gathered}4A = 1, \hfill \\4C = 1, \hfill \\A + 3B + 9C + 9D = 1, \hfill \\A + B + C - D = 1; \hfill \\ \end{gathered} \right. {\text{ }} \Leftrightarrow \left\{ \begin{gathered}A = {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}, \hfill \\B = {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}, \hfill \\D = - {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}, \hfill \\C = {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}. \hfill \\ \end{gathered}  \right.
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  7. #7
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    = \int {\frac{{dt}}{{{{\left( {{t^2} - 1} \right)}^2}}}} = \int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2}{{\left( {t - 1} \right)}^2}}}}.

    (t^2-1)^2=(t-1)^2(t+1)^2

    ???
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  8. #8
    Senior Member DeMath's Avatar
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    \begin{gathered}\boxed{{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \hfill \\{\left( {{t^2} - 1} \right)^2} = {\left( {\left( {t + 1} \right)\left( {t - 1} \right)} \right)^2} = {\left( {t + 1} \right)^2}{\left( {t - 1} \right)^2} \hfill \\ \end{gathered}

    Do you understand it??
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  9. #9
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    OMG, that's the dumbest mistake I've done in a long time.

    Oh I found out how to integrate it using integration by parts.

    I'm not sure how your way of doing it works.

    Anybody help me with the other one? y=lnx
    Last edited by chengbin; March 28th 2009 at 07:04 PM.
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  10. #10
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    For the first one there is a more simpler way by parts

    L = \int\:\sqrt{1 + 4x^2}\:dx

    L = x \:\sqrt{1 + 4x^2} - \int \frac{4x^2}{\sqrt{1 + 4x^2}}\:dx

    L = x \:\sqrt{1 + 4x^2} - \int \frac{1+4x^2-1}{\sqrt{1 + 4x^2}}\:dx

    L = x \:\sqrt{1 + 4x^2} - \int\:\sqrt{1 + 4x^2}\:dx + \int \frac{1}{\sqrt{1 + 4x^2}}\:dx

    L = x \:\sqrt{1 + 4x^2} - L + \frac12\:\int \frac{1}{\sqrt{1 + u^2}}\:du where u = 2x

    2L = x \:\sqrt{1 + 4x^2} + \frac12\:sinh^{-1}u

    L = \frac12\:x \:\sqrt{1 + 4x^2} + \frac14\: sinh^{-1}(2x)
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  11. #11
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    Quote Originally Posted by running-gag View Post
    For the first one there is a more simpler way by parts

    L = \int\:\sqrt{1 + 4x^2}\:dx

    L = x \:\sqrt{1 + 4x^2} - \int \frac{4x^2}{\sqrt{1 + 4x^2}}\:dx

    L = x \:\sqrt{1 + 4x^2} - \int \frac{1+4x^2-1}{\sqrt{1 + 4x^2}}\:dx

    L = x \:\sqrt{1 + 4x^2} - \int\:\sqrt{1 + 4x^2}\:dx + \int \frac{1}{\sqrt{1 + 4x^2}}\:dx

    L = x \:\sqrt{1 + 4x^2} - L + \frac12\:\int \frac{1}{\sqrt{1 + u^2}}\:du where u = 2x

    2L = x \:\sqrt{1 + 4x^2} + \frac12\:sinh^{-1}u

    L = \frac12\:x \:\sqrt{1 + 4x^2} + \frac14\: sinh^{-1}(2x)
    Actually over here in my school, I was advised to

    remember these formulas


    \int{\sqrt{x^2 + a^2}}dx = \frac{x\times \sqrt{x^2 +a^2} }{2} + \frac{a^2(log|x+\sqrt{x^2+a^2}|)}{2} + C <br />

    <br />
\int{\sqrt{x^2 - a^2}}dx = \frac{x\times \sqrt{x^2 -a^2} }{2} - \frac{a^2(log|x+\sqrt{x^2-a^2}|)}{2}+ C<br />

    <br />
 \int{\sqrt{a^2-x^2}}dx = \frac{x\times \sqrt{a^2 -x^2} }{2} - \frac{a^2(sin^{-1}(~\frac{x}{a})~)}{2} + C<br />


    Though we were asked to remember it, I never did it, alwas

    liked the proving stuff, its better , Better to prove it

    A
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  12. #12
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by chengbin View Post
    can anyone help with y=lnx?
    Even if it takes many posts tell us, No spoon feeding now, learn from first

    1) What is f'(x) in this case ?
    2) What was written(Formula) in RG's first post
    3)How to put
    4) how to integrate
    Show these steps and the place you hanged this time!
    Last edited by ADARSH; March 29th 2009 at 09:03 AM.
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  13. #13
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    I got y=x^2 now. It was a pain the in butt to do it. I'll just write my steps here to ask if there is a better way.

    y=x^2 From 0 to \frac {\sqrt3} {2}

    \frac {dy} {dx}=2x

    L= \int^\frac {\sqrt3} {2}_0 \sqrt{1+4x^2}dx

    Let x=\frac {1} {2} tan\theta

    dx=\frac {d\theta} {2cos^\theta}

    When x=0, \theta=0


    When x=\frac {\sqrt3} {2}, \theta=\frac {\pi} {3}

    L=\int^\frac {\pi} {3}_0 \sqrt{tan\theta+1}dx

    L=\int^\frac {\pi} {3}_0 (\frac {1} {cos\theta}) (\frac {1} {2cos^2\theta})

    L=\frac {1} {2} \int^\frac {\pi} {3}_0 \frac {d\theta} {cos^3\theta}

    \frac {d\theta} {cos^3\theta}=\frac {d\theta} {cos\theta cos^2\theta}

    =\bigg [ \frac {tan\theta} {cos\theta} \bigg ]^\frac {\pi} {3}_0 - \int^\frac {\pi} {3} \frac {sin\theta cos\theta} {cos^2\theta} d\theta

    =2\sqrt3 - \int^\frac {\pi} {3}_0 \frac {sin^2\theta} {cos^3\theta}d\theta

    2\sqrt3 - \int^\frac {\pi} {3}_0 \frac {d\theta} {cos^3\theta} + \int^\frac {\pi} {3}_0 \frac {d\theta} {cos\theta}

    2\int^\frac {\pi} {3}_0=2\sqrt3+\int^\frac {\pi} {3}_0 \frac {d\theta} {cos\theta}

    \int \frac {d\theta} {cos\theta}=\int \frac {cos\theta} {cos^2\theta}d\theta

    \int \frac{cos\theta} {1-sin^2\theta} d\theta

    Let u=sin\theta du=cos\theta d\theta

    =-\int \frac {du} {u^2-1}

    \frac {1} {2} \int (\frac {1} {u+1} - \frac {1} {u-1})du

    =1/2(\ln|u+1|-\ln|u-1|

    =\frac {1} {2} \ln \frac {|sinx+1|} {|sinx-1|}

    =2\sqrt3+\bigg [ \frac {1} {2} \ln \frac {|sinx+1|} {|sinx-1|}\bigg ]^\frac {\pi} {3}_0

    =2\sqrt3+\frac {1} {2} \ln \frac {|\sqrt3+2|} {|\sqrt3-2|}

    =2\sqrt3+\ln |\sqrt3+2|

    L=\frac {\sqrt3} {2} + \frac {1} {4} \ln (\sqrt3+2)

    This is one ugly integral.
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  14. #14
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    Quote Originally Posted by chengbin View Post
    [snip]
    Anybody help me with the other one? y=lnx
    Make the substitution u^2 = 1 + x^2.
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