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Math Help - Application of integral calculus. Find a length

  1. #1
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    Application of integral calculus. Find a length

    Help me please to solve these problems on the application of integral calculus to problems of geometry:
    1. Find a length of the curve y = ln(cos(x)) +2, 0 =< x =< pi/6.
    2. Find a length of the polar curve r = sqrt(2)*exp{fi}, 0 =< fi =< 2pi.
    3. Find a length of the parametric curve x = 3(t - sin(t)), y = 3(1 - cos(t)), 0 =< t =< 2pi.
    4. Find a square which limited the polar curve r = a*cos(3*fi), 0 =< fi =< 2pi.

    For me these problems have been the most difficult some of them, I began to deal with, but my answers were not correct.
    Tell me please, what program would be easier to do the drawings for these tasks.

    Thank advance!
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Help me please to solve these problems on the application of integral calculus to problems of geometry:

    1. Find a length of the curve y = ln(cos(x)) +2, 0 =< x =< pi/6.

    Thank advance!
    So 1.{\text{ }}y = \ln \cos x + 2,{\text{ }}0 \leqslant x \leqslant \frac{\pi }{6},{\text{ }}L = ?

    L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} }  = \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {\sqrt {1 + {{\left( {\frac{d}{{dx}}\left( {\ln \cos x + 2} \right)} \right)}^2}} }.

    {\left( {\frac{d}{{dx}}\left( {\ln \cos x + 2} \right)} \right)^2} = {\left( { - \frac{{\sin x}}{{\cos x}}} \right)^2} = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.

    \sqrt {1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}  = \sqrt {\frac{1}{{{{\cos }^2}x}}}  = \frac{1}{{\cos x}}{\text{ }}\left( {0 \leqslant x \leqslant \frac{\pi }{6}} \right).

    L = \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{dx}}{{\cos x}}}  = \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{\cos x}}{{{{\cos }^2}x}}dx}  =  - \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{\cos x}}{{{{\sin }^2}x - 1}}dx} =

    =  - \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{d\left( {\sin x} \right)}}{{{{\sin }^2}x - 1}}}  = \left. { - \frac{1}{2}\ln \left| {\frac{{\sin x - 1}}{{\sin x + 1}}} \right|} \right|_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} =

    =  - \frac{1}{2}\left( {\ln \left| {\frac{{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2} - 1}}{{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2} + 1}}} \right| - \ln \left| { - 1} \right|} \right) =  - \frac{1}{2}\ln \frac{1}{3} = \frac{{\ln 3}}{2}.

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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Help me please to solve these problems on the application of integral calculus to problems of geometry:

    2. Find a length of the polar curve r = sqrt(2)*exp{fi}, 0 =< fi =< 2pi.

    Thank advance!
    r = \sqrt 2 {e^\varphi },{\text{ }}0 \leqslant \varphi  \leqslant 2\pi ,{\text{ }}L = ?

    L = \int\limits_\alpha ^\beta  {\sqrt {{r^2} + {{r'}^2}} d\varphi }  = \int\limits_0^{2\pi } {\sqrt {{{\left( {\sqrt 2 {e^\varphi }} \right)}^2} + {{\left( {\sqrt 2 {e^\varphi }} \right)}^\prime }^2} d\varphi }  =

    = \int\limits_0^{2\pi } {\sqrt {2{e^{2\varphi }} + 2{e^{2\varphi }}} d\varphi }  = 2\int\limits_0^{2\pi } {{e^\varphi }d\varphi }  = \left. {2{e^\varphi }} \right|_0^{2\pi } = 2{e^{2\pi }} - 2.

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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Help me please to solve these problems on the application of integral calculus to problems of geometry:

    4. Find a square which limited the polar curve r = a*cos(3*fi), 0 =< fi =< 2pi.
    Thank advance!
    So we have r = a\cos 3\varphi ,{\text{ }}0 \leqslant \varphi  \leqslant 2\pi ,{\text{ }}S - ?

    Solution

    \frac{S}{6} = \frac{1}{2}\int\limits_\alpha ^\beta  {{r^2}d\varphi }  = \frac{{{a^2}}}{2}\int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {{{\cos }^2}3\varphi d\varphi }  =

    = \frac{{{a^2}}}{4}\int\limits_0^{{\pi  \mathord{\left/<br />
 {\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {\left( {1 + \cos 6\varphi } \right)d\varphi }  = \left. {\frac{{{a^2}}}{4}\left( {\varphi  + \frac{1}{6}\sin 6\varphi } \right)} \right|_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}}<br />
= \frac{{{a^2}}}{4} \cdot \frac{\pi }{6} = \frac{{\pi {a^2}}}{{24}}.

    So \frac{S}{6} = \frac{{\pi {a^2}}}{{24}} \Leftrightarrow S = \frac{{\pi {a^2}}}{4}.

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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Help me please to solve these problems on the application of integral calculus to problems of geometry:
    3. Find a length of the parametric curve x = 3(t - sin(t)), y = 3(1 - cos(t)), 0 =< t =< 2pi.
    Thank advance!
    x = 3\left( {t - \sin t} \right),{\text{ }}y = 3\left( {1 - \cos t} \right),{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}L = ?

    L = \int\limits_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt}  = \int\limits_0^{2\pi } {\sqrt {{{\left( {\frac{d}{{dt}}3\left( {t - \sin t} \right)} \right)}^2} + {{\left( {\frac{d}{{dt}}3\left( {1 - \cos t} \right)} \right)}^2}} dt} .

    {\left( {\frac{d}{{dt}}3\left( {t - \sin t} \right)} \right)^2} = 9{\left( {1 - \cos t} \right)^2} and {\left( {\frac{d}{{dt}}3\left( {1 - \cos t} \right)} \right)^2} = 9{\sin ^2}t.

    \sqrt {9{{\left( {1 - \cos t} \right)}^2} + 9{{\sin }^2}t}  = 3\sqrt {1 - 2\cos t + {{\cos }^2}t + {{\sin }^2}t}  = 3\sqrt {2 - 2\cos t} =

    = 3\sqrt {2 - 2\left( {1 - 2{{\sin }^2}\frac{t}{2}} \right)}  = 3\sqrt {4{{\sin }^2}\frac{t}{2}}  = 6\left| {\sin \frac{t}{2}} \right| = 6\sin \frac{t}{2}{\text{ }}\left( {0 \leqslant t \leqslant 2\pi } \right).

    L = 6\int\limits_0^{2\pi } {\sin \frac{t}{2}dt}  = 12\int\limits_0^{2\pi } {\sin \frac{t}{2}d\left( {\frac{t}{2}} \right)}  = \left. { - 12\cos \frac{t}{2}} \right|_0^{2\pi } =  - 12\left( { - 1 - 1} \right) = 24.

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