Explanation of the steps on calculating the length of the catenary












3












$begingroup$


Can someone please explain to me what identities have been used in the calculation of the below example and why the limits of integration change from $0$ to $a$? I have tried $sinh^2{ t} = [-1 + cosh (2 t)]/2$ and end up with nothing like the form below.



Example:



The catenary, is the shape of a wire hanging under its own weight and is given by



$$
x(t) = t{bf{e}}_1 + cosh{t} {bf{e}}_2,$$
with $I = [−a,a]$.



The total length of this catenary is



begin{align}
L &=int_ {a}^{−a} {sqrt{1 + sinh^2{ t}}} quad dt \
&=2int_0^a cosh{t} quad dt \
&=2sinh{a}.
end{align}






integration trigonometry proof-explanation







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$endgroup$

















    3












    $begingroup$


    Can someone please explain to me what identities have been used in the calculation of the below example and why the limits of integration change from $0$ to $a$? I have tried $sinh^2{ t} = [-1 + cosh (2 t)]/2$ and end up with nothing like the form below.



    Example:



    The catenary, is the shape of a wire hanging under its own weight and is given by



    $$
    x(t) = t{bf{e}}_1 + cosh{t} {bf{e}}_2,$$
    with $I = [−a,a]$.



    The total length of this catenary is



    begin{align}
    L &=int_ {a}^{−a} {sqrt{1 + sinh^2{ t}}} quad dt \
    &=2int_0^a cosh{t} quad dt \
    &=2sinh{a}.
    end{align}






    integration trigonometry proof-explanation







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Can someone please explain to me what identities have been used in the calculation of the below example and why the limits of integration change from $0$ to $a$? I have tried $sinh^2{ t} = [-1 + cosh (2 t)]/2$ and end up with nothing like the form below.



      Example:



      The catenary, is the shape of a wire hanging under its own weight and is given by



      $$
      x(t) = t{bf{e}}_1 + cosh{t} {bf{e}}_2,$$
      with $I = [−a,a]$.



      The total length of this catenary is



      begin{align}
      L &=int_ {a}^{−a} {sqrt{1 + sinh^2{ t}}} quad dt \
      &=2int_0^a cosh{t} quad dt \
      &=2sinh{a}.
      end{align}






      integration trigonometry proof-explanation







      share|cite|improve this question











      $endgroup$




      Can someone please explain to me what identities have been used in the calculation of the below example and why the limits of integration change from $0$ to $a$? I have tried $sinh^2{ t} = [-1 + cosh (2 t)]/2$ and end up with nothing like the form below.



      Example:



      The catenary, is the shape of a wire hanging under its own weight and is given by



      $$
      x(t) = t{bf{e}}_1 + cosh{t} {bf{e}}_2,$$
      with $I = [−a,a]$.



      The total length of this catenary is



      begin{align}
      L &=int_ {a}^{−a} {sqrt{1 + sinh^2{ t}}} quad dt \
      &=2int_0^a cosh{t} quad dt \
      &=2sinh{a}.
      end{align}






      integration trigonometry proof-explanation




      integration trigonometry proof-explanation






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      share|cite|improve this question













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      edited 2 days ago









      saulspatz

      14.2k21329




      14.2k21329










      asked 2 days ago









      JamesJames

      715




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          2 Answers
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          $begingroup$

          Just as we have the identity $sin^2t+cos^2t=1$, so do we have the identity
          $$cosh^2t-sinh^2t=1$$
          Thus $sqrt{1+sinh^2t}=cosh t$ and
          $$L=int_{-a}^asqrt{1+sinh^2t},dt=int_{-a}^acosh t,dt$$
          $cosh$ is an even function, and for even functions the integral from $-a$ to $a$ is twice the integral from 0 to $a$.
          $$int_{-a}^acosh t,dt=2int_0^acosh t,dt$$
          The antiderivative of $cosh$ is $sinh$, like how the antiderivative of $cos$ is $sin$:
          $$2int_0^acosh t,dt=2[sinh t]_0^a=2sinh a$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Excellent. Thank you very much :)
            $endgroup$
            – James
            2 days ago



















          2












          $begingroup$

          The following identities/facts have been used:




          • For any function $f:[-a,a] to mathbb{R} $ which is both even and integrable, $int_{-a}^a f(x) dx = 2 int_{0}^af(x) dx $. This explains the change in the limits of integration.

          • For any real number $t$, $1+ sinh ^2(t) = cosh ^2 (t)$.

          • For any non-negative real number $x$, $|x| = sqrt{x^2} = x$, used in conjunction with the fact that $cosh ^2 (t) > 0$, for all $t in mathbb{R}$.


          $textit{Side note:}$ The last point is important to be aware of. We cannot always make the simplification $sqrt{x^2} = x$, as it fails when $x < 0$.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Just as we have the identity $sin^2t+cos^2t=1$, so do we have the identity
            $$cosh^2t-sinh^2t=1$$
            Thus $sqrt{1+sinh^2t}=cosh t$ and
            $$L=int_{-a}^asqrt{1+sinh^2t},dt=int_{-a}^acosh t,dt$$
            $cosh$ is an even function, and for even functions the integral from $-a$ to $a$ is twice the integral from 0 to $a$.
            $$int_{-a}^acosh t,dt=2int_0^acosh t,dt$$
            The antiderivative of $cosh$ is $sinh$, like how the antiderivative of $cos$ is $sin$:
            $$2int_0^acosh t,dt=2[sinh t]_0^a=2sinh a$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Excellent. Thank you very much :)
              $endgroup$
              – James
              2 days ago
















            4












            $begingroup$

            Just as we have the identity $sin^2t+cos^2t=1$, so do we have the identity
            $$cosh^2t-sinh^2t=1$$
            Thus $sqrt{1+sinh^2t}=cosh t$ and
            $$L=int_{-a}^asqrt{1+sinh^2t},dt=int_{-a}^acosh t,dt$$
            $cosh$ is an even function, and for even functions the integral from $-a$ to $a$ is twice the integral from 0 to $a$.
            $$int_{-a}^acosh t,dt=2int_0^acosh t,dt$$
            The antiderivative of $cosh$ is $sinh$, like how the antiderivative of $cos$ is $sin$:
            $$2int_0^acosh t,dt=2[sinh t]_0^a=2sinh a$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Excellent. Thank you very much :)
              $endgroup$
              – James
              2 days ago














            4












            4








            4





            $begingroup$

            Just as we have the identity $sin^2t+cos^2t=1$, so do we have the identity
            $$cosh^2t-sinh^2t=1$$
            Thus $sqrt{1+sinh^2t}=cosh t$ and
            $$L=int_{-a}^asqrt{1+sinh^2t},dt=int_{-a}^acosh t,dt$$
            $cosh$ is an even function, and for even functions the integral from $-a$ to $a$ is twice the integral from 0 to $a$.
            $$int_{-a}^acosh t,dt=2int_0^acosh t,dt$$
            The antiderivative of $cosh$ is $sinh$, like how the antiderivative of $cos$ is $sin$:
            $$2int_0^acosh t,dt=2[sinh t]_0^a=2sinh a$$






            share|cite|improve this answer









            $endgroup$



            Just as we have the identity $sin^2t+cos^2t=1$, so do we have the identity
            $$cosh^2t-sinh^2t=1$$
            Thus $sqrt{1+sinh^2t}=cosh t$ and
            $$L=int_{-a}^asqrt{1+sinh^2t},dt=int_{-a}^acosh t,dt$$
            $cosh$ is an even function, and for even functions the integral from $-a$ to $a$ is twice the integral from 0 to $a$.
            $$int_{-a}^acosh t,dt=2int_0^acosh t,dt$$
            The antiderivative of $cosh$ is $sinh$, like how the antiderivative of $cos$ is $sin$:
            $$2int_0^acosh t,dt=2[sinh t]_0^a=2sinh a$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Parcly TaxelParcly Taxel

            41.6k137299




            41.6k137299












            • $begingroup$
              Excellent. Thank you very much :)
              $endgroup$
              – James
              2 days ago


















            • $begingroup$
              Excellent. Thank you very much :)
              $endgroup$
              – James
              2 days ago
















            $begingroup$
            Excellent. Thank you very much :)
            $endgroup$
            – James
            2 days ago




            $begingroup$
            Excellent. Thank you very much :)
            $endgroup$
            – James
            2 days ago











            2












            $begingroup$

            The following identities/facts have been used:




            • For any function $f:[-a,a] to mathbb{R} $ which is both even and integrable, $int_{-a}^a f(x) dx = 2 int_{0}^af(x) dx $. This explains the change in the limits of integration.

            • For any real number $t$, $1+ sinh ^2(t) = cosh ^2 (t)$.

            • For any non-negative real number $x$, $|x| = sqrt{x^2} = x$, used in conjunction with the fact that $cosh ^2 (t) > 0$, for all $t in mathbb{R}$.


            $textit{Side note:}$ The last point is important to be aware of. We cannot always make the simplification $sqrt{x^2} = x$, as it fails when $x < 0$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              The following identities/facts have been used:




              • For any function $f:[-a,a] to mathbb{R} $ which is both even and integrable, $int_{-a}^a f(x) dx = 2 int_{0}^af(x) dx $. This explains the change in the limits of integration.

              • For any real number $t$, $1+ sinh ^2(t) = cosh ^2 (t)$.

              • For any non-negative real number $x$, $|x| = sqrt{x^2} = x$, used in conjunction with the fact that $cosh ^2 (t) > 0$, for all $t in mathbb{R}$.


              $textit{Side note:}$ The last point is important to be aware of. We cannot always make the simplification $sqrt{x^2} = x$, as it fails when $x < 0$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                The following identities/facts have been used:




                • For any function $f:[-a,a] to mathbb{R} $ which is both even and integrable, $int_{-a}^a f(x) dx = 2 int_{0}^af(x) dx $. This explains the change in the limits of integration.

                • For any real number $t$, $1+ sinh ^2(t) = cosh ^2 (t)$.

                • For any non-negative real number $x$, $|x| = sqrt{x^2} = x$, used in conjunction with the fact that $cosh ^2 (t) > 0$, for all $t in mathbb{R}$.


                $textit{Side note:}$ The last point is important to be aware of. We cannot always make the simplification $sqrt{x^2} = x$, as it fails when $x < 0$.






                share|cite|improve this answer











                $endgroup$



                The following identities/facts have been used:




                • For any function $f:[-a,a] to mathbb{R} $ which is both even and integrable, $int_{-a}^a f(x) dx = 2 int_{0}^af(x) dx $. This explains the change in the limits of integration.

                • For any real number $t$, $1+ sinh ^2(t) = cosh ^2 (t)$.

                • For any non-negative real number $x$, $|x| = sqrt{x^2} = x$, used in conjunction with the fact that $cosh ^2 (t) > 0$, for all $t in mathbb{R}$.


                $textit{Side note:}$ The last point is important to be aware of. We cannot always make the simplification $sqrt{x^2} = x$, as it fails when $x < 0$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered 2 days ago









                E-muE-mu

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